Organic Chemistry II
Organic Chemistry II 320N
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This 480 page Bundle was uploaded by Roy Gallegos on Saturday September 27, 2014. The Bundle belongs to 320N at University of Texas at Austin taught by Colapret, John in Fall2013. Since its upload, it has received 643 views. For similar materials see Organic Chemistry II in Chemistry at University of Texas at Austin.
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Date Created: 09/27/14
LAST NAME UTEID Instructions This exam will be closed book No notes books calculators or molecular models will be allowed The exam will be comprised of two distinct parts Part I will consist of a series of multiple choice questions Your answers to these questions must be submitted on a Scantron bubble sheet The answer sheet will be provided for you but you will need your own 2 pencils Only answers marked on the bubble sheet will be graded Answers to Part I questions marked on the exam itself cannot not be graded No regrades will be possible on Part I of the exam Part II consists of questions for which you will need to write out your answers using structures andor words This part of the exam will be hand graded Answers for the Part II questions that are written in pencil will not be eligible for re grades Answers written in pencil with ink overlay will not be eligible for re grades If you use a pen to answer the Part II questions only blue or black ink is acceptable Answers written in red ink cannot be graded 1 You must have your valid UT ID card or other govemmentissued ID with you You will need to show it to the proctors when you turn in your exam 2 Chapter 4 of the University s General Information catalog outlines this university s policies regarding exams as well as other quizzes administered during the semester Specifically students are expected to remain in the exam room until a test is completed These policies will be strictly enforced with no exceptions You may not leave the room for any reason until you are ready to tum in your exam If you wish to leave the room you will need to turn in your exam to the proctors and you will not be allowed to return Please plan accordingly by using the rest room before the exam starts FIRST NAME CH31ON EXAM Question Value Score 15 16 12 15 G501bCDl3A Sec onll 75 Sec onl 75 Raw Total 150 Grade UTID September 2011 CH310NExam I Objective Test Section Identi z the choice that best completes the statement or answers the question There is only one correct answer please carefully bubble your choice on the scantron sheet 3pts ea 75 pts this section 1 Which feature in the 1H NMR spectrum provides information about the number of types of different protons in a compound a number of signals b chemical shift 0 integral d splitting Which combinations of alkyl bromide and carbonyl compound can be used to prepare the following product by addition of the Grignard reagent derived from the alkyl bromide to the carbonyl compound OH W 1 C H32CHBr CH32CHC H2Br CH3CH2CH2BF CH3CH2BF O 0 Z JL Q W I I 1 2 3 4 a only 1 b only 3 c only 1 and 3 d only 2 and 4 What is the major organic product obtained from the following reaction CH2I2 I 2 ZnCu CH U Ck CH3 CH3 1 2 3 Th Et2O CH 2I P09 IgtuaN s 4 What is the splitting of the signal in the 1H NMR spectrum for the methyl protons of lbomo2methylpropane a singlet b nonet c triplet d doublet 5 What is the major organic product obtained from the following reaction OH O 1 HCECLi Ph 1 2 A 2 H33 Ph O Ph H P09 IgtuaN s 6 Which combinations of alkyl bromide and carbonyl compound can be used to prepare the following product by addition of the Grignard reagent derived from the alkyl bromide to the carbonyl compound OH R J ex CH3Er CHs2CHBr CHsCH2Bi39 CH32CH Er O 1 2 3 4 a only 1 and 2 b only 3 and 4 c only 2 and 3 d only 1 2 and 3 Which of the following alcohols can be prepared from a Grignard reagent and ethylene oxide WOH Ph Ph OH 1 2 OH OH PhWlt W a only 1 b only 1 and 2 c only 1 2 and 3 d 1 2 3 and 4 What is the correct assignment of the names of the following ketones Q C O 1 2 3 a 1 acetaldehyde 2 acetophenone 3 benzaldehyde b 1 acetone 2 phenol 3 benzaldehyde c 1 acetone 2 acetophenone 3 benzophenone d 1 formaldehyde 2 benzaldehyde 3 acetophenone What is the hydrogen deficiency index for a compound with a molecular formula of C10H16O2 1 P0 9quot 2 3 4 10 What is the major organic product obtained from the following sequence of reactions ll 12 1Mg0 PCC VBr an Th 2 0 3 H3O o o o vjT kV k 1 2 3 4 a 1 b 2 c 3 d 4 What is the major organic product obtained from the following sequence of reactions B F Et2O a 3methyl2butanol b 3methyl2butene c 3methyllbutanol d 2methylbutane What is the major organic product obtained from the following reaction 0 O CH3OH OCH3 OCH3 H300 OCH3 an Qk H2804 J K OCH3 1 2 3 4 P09 IgtuaN s 13 Which feature in the 1H NMR spectrum provides information about the electronic environment of the protons in a compound a splitting b chemical shift c integral d number of signals 14 What is the major organic product obtained from the following reaction 1 F h3F CH2 O 2go an 1butene 2methyl 1 propanol 2methylpropene 2butene P99quot 15 What is the relationship between the following two structures CL 202 0 J9 A a resonance structures b tautomers c stereoisomers d constitutional isomers but not tautomers 16 What is the major organic product obtained from the following reaction 1 O 2 OH O Jl an Jxxx xx HCN H CN cm H 3jTL X cm cm P09 IgtoaN n 17 18 19 20 21 Which of the following is the best choice of solvent for the formation of phenyllithium by the reaction of bromobenzene with lithium a ethanol b water c tetrahydrofuran d acetic acid Which of the protons in the following molecule appear furthest down eld in the 1H NMR spectrum 0 CH3CH2COCH2CH3 5 if iii iv a i b ii c iii d iv What is the major organic product obtained from the following reaction 1 PhLi 0 E20 at K 2 H3Oquot 3phenylbutanone 3phenyl2butanol propiophenone PhCOCH2CH3 2phenyl2butanol P99quot Which feature in the 1H NMR spectrum provides information about the number of neighboring protons of each proton in the compound a chemical shift b number of signals c integral d multiplicity Which of the protons in the following molecule appear furthest down eld in the 1H NMR spectrum CH3CH2CH2CH2CI I if iii iv o o iii iv 9 9quot 22 Which of the following mechanisms accounts for the acidbase reaction of ethylmagnesum bromide with acetylene 23 24 25 to give acetylide anion AHQEC cH3cH2 MgBr K H cH3cH2 MgBr H cEc H 1 2 T CH3CH2 lMgBr H cEc H cH3cH2 MgBr 3 4 HCECH P09 IAUJNH Which of the following is not true regarding 1H NMR spectroscopy a 8 for a particular proton is independent of the magnetic field strength of the instrument used a quotdownfieldquot peak appears at a lower Value of 8 quotdeshieldingquot leads to peaks at higher Values of 8 d on a 300 MHz instrument a proton that adsorbs irradiation at a frequency 900 Hz higher than the adsorption of TMS appears at 8 3 ppm 09quot Which of the following bonds has the most ionic character a C Cu b C Li c C Zn d C Sn What is the major organic product obtained from the following reaction I E 2butene E 2iodo2butene 1 iodo2methylpropene methylcyclopropane CH32Cu Li an P99quot CH310NExam I Answer Section MULTIPLE CHOICE N 9 5B o 3SE7iESSoggto1cu14gts 9 ANS ANS ANS ANS ANS ANS ANS ANS ANS OOUUU3gtU3gtUU3gt quotH 8 CD 395 er U ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS gtUUUU3gtUUUOOUU3gtOUUUU typo PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS UJUJUUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJUJ ID CH 310 N EXAM 1 UTEID PART 2 Free Style Answer Format 75pts in this section 1 Reactions Provide the reagents for the following reactions Number the steps if more than one is required 3 pts per box OH CH3 OH racemic o it 1 1 C H 2 H3O OH MgBr 1 OH OH CH2CH3 CH3 racemic racemic CH 310 N EXAM 1 2 Mechanism Provide a mechanism for the following transformation Show all important flows of electrons charges and intermediates Where indicated in the structure n a box draw the intermediates 16 pts 0 UTEID KEY OH CH3OH 10 eq H3O cat A 7 OCH3 ltgt V 6 H H T Protonated BenzaldehydeShow both resonance structures Ph Final product amp regeneration of the acid catalyst H3C 0 H3O 39 J Show the loss of a proton Ph H ph C H H cH3 H cH3 J K Show the attack by methanol V W H O T Ph H K J CH 310 N EXAM 1 UTEID KEY 3 Texas twostep Provide both the reagents in square boxes and the product rounded box for these synthesis Note that a second reagent may be required in the square boxes For example the acid step of a Grignard addition 12 pts V V OH o HZQO4 1 PhMgBr39 2 H3O 39 K J 0 39 Ho H2 1 EtMgBr C H30 CH3 2 H3O heat J OH Ph CHZCH3 4 Multistep synthesis Provide the reagents for the following conversions 15 pts BI392 Ph cH3 heat or hv HNCH3 H2 Ni Ph or NaBH CN c CH3 3 H2 1M9 CH2Br CH3 N Ph c CH3 H2 10 2 CH3CHCb 3 H30quot CH3NH2 OH Ph 1 HZCTO4 0 Ph L c CH3 H2 CH 310 N EXAM 1 UTEID 5 Unknown The NMR spectra of an unknown compound are shown below Propose a structure for the unknown in the box and assign the protons to the spectra on the NMR using the abc labeling 8 pts CeHn0z 6 3 2 1 7peaks lit it J 6 s it 35 i 1 i 0 PPM Proposed Structure 132 0 493 114 32 O 229 11 CH 310 N EXAM 1 UTEID N 6 Unknown The NMR spectra of an unknown compound are shown below Propose a structure for the unknown in the box and assign the protons to the spectra on the NMR using the abc labeling 9 pts CsH1o0z PPM W W 734 699 372 701 R 368 734 699 K J END OF EXAM SECTION 12 LASTNAME HRSTNAME UTEID COVERPAGE Instructions This exam will be closed book No notes books calculators or molecular models will be allowed The exam will be comprised of two distinct parts Part I will consist of a series of multiple choice C H 3 0 N questions Your answers to these questions must be X submitted on a Scantron bubble sheet The answer E M 3 sheet will be provided for you but you will need your own 2 pencils Only answers marked on the bubble sheet will be graded Answers to Part I questions marked on the exam itself cannot not be Question Va I ue Score graded No regrades will be possible on Part I of the 1 4 CX3II1 1 2 Part II consists of questions for which you will need 16 G501bCDl3A Answers for the Part II questions that are written in 18 questions only blue or black ink is acceptable pencil will not be eligible for re grades Answers Answers written in red ink cannot be graded Section p to write out your answers using structures andor 12 written in pencil with ink overlay will not be eligible 1 You must have your valid UT ID card or words This part of the exam will be hand graded 18 for re grades If you use a pen to answer the Part II other govemmentissued ID with you You will need to show it to the proctors when Section I 0 you turn in your exam 2 Chapter 4 of the University s General Raw Total 1 Information catalog outlines this G ra d e university s policies regarding exams as well as other quizzes administered during the semester Specifically students are expected to remain in the exam room until a test is completed These policies will be strictly enforced with no exceptions You may not leave the room for any reason until you are ready to tum in your exam If you wish to leave the room you will need to turn in your exam to the proctors and you will not be allowed to return Please plan accordingly by using the rest room before the exam starts CH320NExam Ill Answer Section MULTIPLE CHOICE 1 ANS B PTS 2 ANS C PTS 3 ANS C PTS 4 ANS B PTS 5 ANS D PTS 6 ANS D PTS 7 ANS B PTS 8 ANS C PTS 9 ANS B PTS 10 ANS C PTS 11 ANS C PTS 12 ANS B PTS 13 ANS A PTS 14 ANS B PTS 15 ANS D PTS gtgtgtgtgtgtgtgtgtgtgtgtgtgtgt CH 320 N EXAM 3 UTEID K TM PART 2 Free Style Answer Format 90pts in this section 1 Multistep synthesis Provide the products for the following reactions Be sure to include any stereo or regiochemistry if necessary 14 pts 0 9 0 0 AA P J CH3CH2CH2BI39 gt OCH e 3 ocH3 2 H20 3 LDA 1 HF Draw the correct enolate k 78 C OCH3 O NaOCH3 M 1CH3CO2CH3 OCH3 QL pf OCH3 H20 OCH3 R 1 9 0 Draw the correct enolate 1 NaOCH3CH30H 2 PhCH2Br O O O O O A 1 N H CH PohH ao OCH3 CH2Ph 2 2 quot393 H2Ph CH 320 N EXAM 3 UTEID KEY 2 Diels Alder Reactions Provide the cycloadducts for the following reactions You must include the correct stereochemistry 12 pts Must show cis CO2Et K CO2Et E W co Et K CO2Et C 2 CO2Et Must show cis g 4 CO2Et cN i EtO2C NC cO2Et L Must show trans CN CN Must show trans J CH 320 N EXAM 3 UTEID K i 3 Reactions Provide the product from each reaction Be sure to include any stereo or regiochemistry if necessary 16 pts f W 1M9 NR 2co2 Br 3 HCI C02H kl 1 k J 3r2FeBr3 N02 AICI3 CH3COCAICI3 N02 HN03 f N02 N f N02 COCH3 CIOCCH3 r J O N CH3 AICI3 KOH 2 N02 NZH4 L 4 A H2Ni N02 NH2 CH2CH3 NH2 k CH 320 N EXAM 3 UTEID PEY 4 Texas two step Each of these problems are two step synthesis of a product In the boxes around the arrows provide the reagents for the reaction note that a hydrolysis step eg to finish a Grignard reaction may be required In the rounded rectangle draw the structure of the compound from the first reaction 4 pts per scheme Scheme 1 CO H l CCOCH2CH3IEt3N 2 CHZOH OR 1 39AH CH3CH2COOHl H cat 2 H30 OR 9 9 t I kOl Scheme 2 f 0 N 1 03 NaOH A 2 M923 orNaHSO3 Scheme 3 o O H3CO2C Na0CH3 1 NaOH 3 2 3 2 HCI 10 O cH2oU co2H CH 320 N EXAM 3 UTEID K 5 Mechanism THE Robinson Ring annulation Provide a mechanism for the following transformation Show all important flows of electrons charges and intermediates Use hydroxide and water whenever necessary Draw methylvinyl ketone in the right box amp show how it is connected 18 pts 0 O O NaOHaq A H eOH O O OH T dehydrate O eOH deprotonate proton transfer O W W ltr ow 9 O 11gt makeabond make39a39b0nd IL Hint these are CC bond f arming reactions o H28 Cf 3 1 CH J proton transielx O H2CJ protonate 11 CH 320 N EXAM 3 UTEID K 6 Reactions Provide the product from each reaction Be sure to include any stereo or regiochemistry if necessary 18 pt f N o OH 0 CO2CH3 J OH 1Br HOAc 1 1 2 Kc tBu COZCH3 1 LDA 2 CH3CHO NaBH4 3 H20 0 1 NaOH 1 3 eq CH3MgBr 2 H3O 2 H3O 3 A 1 LiAlH4 2 H3O r N r N OH OH o OH amp CHZOH L k J L J END OF EXAM SECTION 12 NHW Set 1 Answer Section MULTIPLE CHOICE 1 ANS D PTS 2 ANS D PTS 3 ANS A PTS 4 ANS B PTS 5 ANS B PTS 6 ANS D PTS 7 ANS A PTS 8 ANS A PTS 9 ANS B PTS 10 ANS B PTS 11 ANS B PTS 12 ANS A PTS 13 ANS A PTS 14 ANS B PTS 15 ANS D PTS 16 ANS B PTS 17 ANS A PTS 18 ANS C PTS 19 ANS B PTS 20 ANS C PTS 21 ANS D PTS 22 ANS B PTS 23 ANS C PTS 24 ANS A PTS 25 ANS C PTS lgtgtlgtlgtgtlgtgtgtgtgtlgtgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgt HW Set 2 Answer Section MULTIPLE CHOICE quotquotF 3 3 9 quot39quotF 3 O 9 39 quot39gt quot ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS nOUgtgtcuowgtgtnwowgtnnUowgtnuuwgt PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS gtgtlgtlgtgtlgtgtgtgtgtlgtgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgt HW 320N Set 3 Answer Section MULTIPLE CHOICE 1 ANS A PTS 4 2 ANS A PTS 4 3 ANS A PTS 4 4 ANS B PTS 4 5 ANS C PTS 4 6 ANS D PTS 4 7 ANS B PTS 4 8 ANS C PTS 4 9 ANS C PTS 4 10 ANS C PTS 4 11 ANS A PTS 4 12 ANS A PTS 4 13 ANS C PTS 4 14 ANS D PTS 4 15 ANS C PTS 4 16 ANS B PTS 4 17 ANS C PTS 4 18 ANS C PTS 4 19 ANS B PTS 4 20 ANS A PTS 4 21 ANS B PTS 4 22 ANS A PTS 4 23 ANS B PTS 4 24 ANS C PTS 4 25 ANS B PTS 4 320 N HW Set 4 Answer Section MULTIPLE CHOICE 1 ANS C PTS 2 ANS D PTS 3 ANS D PTS 4 ANS A PTS 5 ANS A PTS 6 ANS B PTS 7 ANS D PTS 8 ANS D PTS 9 ANS C PTS 10 ANS A PTS 11 ANS B PTS 12 ANS B PTS 13 ANS D PTS 14 ANS B PTS 15 ANS D PTS 16 ANS C PTS 17 ANS A PTS 18 ANS C PTS 19 ANS C PTS 20 ANS B PTS 21 ANS B PTS 22 ANS B PTS 23 ANS A PTS 24 ANS D PTS 25 ANS D PTS lgtgtlgtlgtgtlgtgtgtgtgtlgtgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgt 320NHW 5 Answer Section MULTIPLE CHOICE quotquotF 3 3 9 quot39quotF 3 O 9 39 quot39gt quot ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS ANS gtgtUgtUgtUgtgt0wOgtO0wgtOUUUgtUOgtgt PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS PTS gtgtlgtlgtgtlgtgtgtgtgtlgtgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgtlgt CH320 N Fall 2013 NHW Set 1 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 Which of the following bonds has the most ionic character a C Si b C Cu c C Zn d C Li 2 Which of the following molecules are deprotonated by ethylmagnesium bromide a phenol PhOH b propyne CH3CsCH c dimethylamine CH32NH d all of these 3 Which of the following mechanisms accounts for the acid base reaction of ethylmagnesum bromide with acetylene to give acetylide anion A H 3EC A 1 cH3cH2 MgBr H cH3cH2 MgBr H cc H 1 2 x 1 cH3cH2 MgBr H cc H cH3cH2 MgBr H cc H 3 4 09939 BOONI 4 What is the major organic product obtained from the following sequence of reactions Li H20 Cl Et2O a 2pentanol b pentane c 1 pentene d E2 pentene Due 9202013 5 What is the major organic product obtained from the following reaction 10D 1WltOH 2 M Mgar 2H3O XL OH 3 4 a 1 b 2 c 3 d 4 6 Which of the following alcohols can be prepared from a Grignard reagent and ethylene oxide OH M Ph OH 2 only 1 and 2 only 1 and 4 only 1 2 and 3 only 2 and 4 990quot Due 9202013 7 8 9 What is the major organic product obtained from the following reaction I N 3 4 a 1 b 2 c 3 d 4 What reactive intermediates are involved in the following reaction gtlt the trichloromethyl anion C3C and dichlorocarbene C2C the trichloromethyl cation Cl3C the cyclohexyl carbocation the cyclic chloronium ion derived from cyclohexene cHcH CH33CO39K an P90quot What is the major organic product obtained from the following reaction CI H CL Cl cig 0 H H cH CI3 Z 7 ac ac ac CHs3CO39K Z H A cH3 Q C C H30 1 CH3 H30 2 H H3C 3 CH3 H30 4 CH3 09939 hOON l Due 9202013 10 What is the IUPAC name of the following compound 11 12 H 3 methy 3 phenylpropanol 3 phenybutanal 3 phenyl 1 butanone 3 phenybutanoic acid P9939 What is the correct assignment of the names of the following ketones O O O H3cJJcH3 I CH3 I I I I I I g 2 pg 3 g a 1 acetone 2 phenol 3 benzaldehyde b 1 acetone 2 acetophenone 3 benzophenone c 1 formaldehyde 2 benzaldehyde 3 acetophenone d 1 acetaldehyde 2 acetophenone 3 benzaldehyde What is the major organic product obtained from the following reaction 1 PhLi o Et2O an JJ 2 H3O 2 phenyl 2 butanol 3 phenyl 2 butanol 3 phenybutanone propiophenone PhCOCH2CH3 P9939 Due 9202013 13 What is the major organic product obtained from the following reaction 14 1 PhLi o Et2O AA 2 H3CI H 1 phenyl 1 butanol 1 pheny 2 butano 2 phenyl 1 butanol butyrophenone PhCOCH2CH2CH3 990quot Which combinations of alkyl bromide and carbonyl compound can be used to prepare the following product by addition of the Grignard reagent derived from the alkyl bromide to the carbonyl compound OH gt C H32CHBr CH32CHC H2Br CH3CH2CH2Br CIHsCH2Br O 0 0 JL WW 1 2 3 4 a only 1 b only 3 c only 1 and 3 d only 2 and 4 Due 9202013 15 Which combinations of alkyl bromide and carbonyl compound can be used to prepare the following product by addition of the Grignard reagent derived from the alkyl bromide to the carbonyl compound OH R J ex CH3Er it IIs2quot iBr CHsCH2Bi39 CH32CH Br 1 2 3 4 a only 1 and 2 b only 3 and 4 c only 2 and 3 d only 1 2 and 3 16 What is the major organic product obtained from the following reaction I O 2 I 1 CN K MON 4JCN 09939 hOOM l Due 9202013 17 18 19 What is the major organic product obtained from the following reaction OH 1 HCECLi Ph O 1 an A 2 Ham Ph D Ph H 09939 hOON l What is the major organic product obtained from the following reaction 1 F h3F CH2 O K 1 butene 2butene 2 methypropene 2 methyl1propanol 2 H30quot an 990quot What is the major organic product obtained from the following reaction I 1 F h3PCHCH3 0 2methyl2heptene 3ethyl2 pentene Z 2 methyl3heptene E 2 methyl3 heptene 990quot Due 9202013 20 22 What is the correct assignment of the names of the following functional groups 4 HO OH EH30 OH CH3O OCH3 5 enol 2 hydrate 3 acetal 4 hemiacetal acetal 2 hydrate 3 enol 4 hemiacetal hydrate 2 hemiacetal 3 acetal 4 enol enol 2 hydrate 3 hemiacetal 4 enol 990quot L L L L What is the major organic product obtained from the following reaction 0 CH3OH OCH3 H3CO Thu amplk H2804 O OCH3 OCH3 2 3 09939 hOOM l What is the major organic product obtained from the following reaction H O4 O O H 3 4 OH OH O O OH a 1 b 2 c 3 d 4 4 OCH3 Q4 Due 9202013 23 What is the correct assignment of the names of the following functional groups NCH3 N NH2 fE 2 1 imine 2 amine 3 hydrazone 4 oxime 1 hydrazone 2 amine 3 imine 4 oxime 1 oxime 2 imine 3 amine 4 hydrazone 1 imine 2 hydrazone 3 oxime 4 amine 990quot 24 What is the major organic product obtained from the following reaction O NNH2 NH2 NH O H2NNH2 NHNH2 1 2 3 4 P9939 hOOM l 25 What is the major organic product obtained from the following reaction ONH2 NH NOH 0 H2NOH T1 ONH2 1 2 3 4 P9939 hOOM l CH 320 N Fall 2013 HW 320N Set 3 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 What is the major organic product obtained from the following reaction I o 1 2 O IlaCH H20 H H an C39 O k Q 09939 BOONI 2 What is the major organic product obtained from the following reaction 0 0 NaOH H20 A L an X Ph H 0 o o 0 Ph Ph Ph Ph 1 2 3 4 99939 BOOMI Due 1162013 3 What is the major organic product obtained from the following reaction 0 xxx xR fJLRR O 0 Ph H Ph t 2 2 O H Ph H 0 A RLK X 3 09939 hOOM l 4 Which of the following compounds can be prepared by a mixed aldol condensation by treatment of a mixture of two carbonyl compounds and NaOH 0 O Ph xRJ4 ylLPh Phxx h HN 1 2 C O R fNx mH x Ph 3 4 99939 BOOMI Due 1162013 5 What is the major organic product obtained from the following reaction 0 O O O O O NaOH H20 an bb A 2 3 I 4 6 What is the major organic product obtained from the following reaction 0 O O 0 O NaOH H20 an 1 2 3 4 7 What is the major organic product obtained from the following reaction 1 LDA 78 0c 1 O 2 O 0 0 2JL H 09939 hOON l 09939 hOON l 09939 hOON l Due 1162013 8 What is the major organic product obtained from the following reaction 0 o H o 1LDA78 C 2 o k 1 2 O TEL 3H3o o 3 4 a 1 b 2 c 3 d 4 Due 1162013 9 What is the major organic product obtained from the following reaction Hint 2 moles of the ester are required C 1NaOCH2CH3 CH3CH2OH OCH CH T 2 3 2H3O C O or O OCH2CH3 1 2 C O O O QH2CH3 QH2QH3 3 4 a 1 b 2 c 3 d 4 Due 1162013 10 What is the major organic product obtained from the following reaction O 1 NaOCH2CH3 CH3CH2OH CH3CH2O TI OCH2CH3 2 H3O o o O o OCH2CH3 OCH2CH3 1 2 o 4 o OCH2CH3 O o O 2CH3 3 99939 BOOMI 11 What is the major organic product obtained from the following reaction 0 O o CH3OH o o H p7 Ph OCH3 Ph OCH3 2 I o Ph 0CH3 OCH 3 Ph 4 09939 hOOM l CD Due 1162013 12 What is the major organic product obtained from the following reaction 0 0 OH 0 NaOCH3 cH3oH 0 Ph OCH3 Ph OCH3 O Jk Ph OCH3 Ph Ph Ph eh 1 2 OCH3 O Ph39lOCH3 Ph OCH3 Ph 4 2 H3O 0 Ph 3 09939 hOOM l 13 What is the major organic product obtained from the following reaction 1 CH3Li 2 H3O 39 09939 hOOM l Due 1162013 14 What is the major organic product obtained from the following reaction 0 O o 0 HO 1 CH3Li an 2 H3O 39 1 2 15 What is the major organic product obtained from the following reaction 09939 hOOM l O o N H T N 1 NI 0 Iquot h 09939 hOOM l Due 1162013 16 What is the major organic product obtained from the following reaction a 2 KM N EJ illxlK Il T 2 09939 BOOMI 17 Which of the following are conjugated dienes CEO 1 2 3 a only 1 b only 2 c only 1 and 2 d 1 2 and 3 Due 1162013 18 Which of the following compounds is the most reactive dienophile CH3 quot392C393quot392 CH3 cOOCH3 CH 3 1 2 3 4 a 1 b 2 c 3 d 4 19 Which of the following is the least reactive dienophile COOCH3 COOCH3 sgt N02 cH3 coocH3 2 3 P9939 hOOM l 20 Which of the following is the least reactive diene in Diels Alder reactions 1 2 3 4 P9939 hOOM l 10 Due 1162013 21 Which of the following is the least reactive diene in Diels Alder reactions lt R K P9939 hOOM l 22 What is the major organic product obtained from the following reaction CO2CH3 CO2CH3 3O2CH3 CO2CH3 co2cH3 1 ii 9 CO2CH3 COZCH3 co2cH3 co2c 3 CO2CH3 3 P9939 hOOM l 11 Due 1162013 23 What is the major organic product of the following sequence of reactions 0 1uAwu 1 2eq NaH x A 2H 3 22eqMe O quot quot3939 quot3939 O o OCH3 OCH3 OCH3 OCH3 1 O 2 llOCH3 MOCH3 OCH OCH J00 3 K 1 3 3 4 P9939 hOOM l 24 What is the major organic product of the following sequence of reactions P9939 BOONI 12 Due 1162013 25 Which of the following are intermediates in the acid catalyzed aldol reaction of propanal to form 2methy 2pentenal enol enolate tetrahedral carbonyl intermediate aldol 39gt quotquot only 1 and 2 only 1 3 and 4 only 2 3 and 4 1 2 3 and 4 990quot 13 CH320 N Fall 2013 320 N HW Set 4 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 CHQCH3 b COCH3 c Br d NH3 What is the major organic product obtained from the following reaction CH3 C39lCZHClg IIIIICH3 l jx EH3 Hl3C ecu melt 09939 BOONI Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 N02 b OH c CH3 d CI Which of the following undergoes the most rapid sulfonation upon treatment with fuming sulfuric acid a benzene b benzoic acid c benzonitrile d nitrobenzene Due 11122013 5 What is the correct assignment of the names of the following substituted benzenes OCH3 HH3 CH3 1 2 3 a 1 anisole 2 aniline 3 toluene b 1 benzaldehyde 2 anisole 3 toluene c 1 anisole 2 xylene 3 toluene d 1 phenol 2 aniline 3 anisole 6 What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic CH2 CH2 CH3 N02 HNO3 N02 H3804 CH3 O2N CH3 CH3 CH3 N02 1 2 N03 3 4 7 Which of the following undergoes the most rapid bromination upon treatment with BraFeBr3 a bromobenzene b benzene c nitrobenzene d anisole CH3 P9939 hOOM l Due 11122013 8 What is the major organic product obtained from the following reaction COOH CDCH COOH COCH COOH HNC3 N02 H28C34 an NO Br Br EH32 Br Br 1 2 3 4 a 1 b 2 c 3 d 4 9 Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH OH N02 N02 1gt3gt2 1gt2gt3 2gt3gt1 3gt2gt1 P90quot Due 11122013 10 What is the major organic product obtained from the following reaction 0 Br COOH COOH COOH 900 Brz Br K FeBr3 K X l I I l l 39 B 1quot W l r CH3 CH3 CH3 CHQBF CH3 1 2 3 4 a 1 b 2 c 3 d 4 11 What is the major organic product obtained from the following reaction 09939 BOONI Due 11122013 12 What is the major organic product obtained from the following reaction CH3 CH3 O CH3 o T T JC cHltcHgt2 cHltcHgt2 2 AICI3 1 2 O CH3 CI CHCH33 T cHltcH3gt2 cHcH3gt2 3 4 09939 BOONI 13 What is the major organic product obtained from the following reaction CCH3 ITr OCH3 QCH3 QCH3 EH2 l FEBF3 an Br CN CN Br CN CN 1 2 3 4 a 1 b 2 c 3 d 4 Due 11122013 14 15 What is the major organic product obtained from the following reaction 09939 hOON l What is the major organic product obtained from the following reaction CH3 CH3 Br CH3 CH3 BF2 Br FeBm an CH3 CH3 Br Br CH3 CH3 1 2 3 4 gr a 1 b 2 c 3 d 4 Due 11122013 16 What is the major organic product obtained from the following reaction 990quot CH3 Cl AICI3 CH3 CH3 CH3 lt 2 1 F3 E hOOlJl 17 What is the major organic product obtained from the following sequence of reactions 990quot iZnHg HIEI pb pb e e K V e ci I2ci I2cH3 I 9 3 at 1 2 3 4 Due 11122013 18 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic 19 20 substitution reactions 3 Br CH3 N02 b Cl OCH3 COCH3 C CH3 NH2 Br d N02 COCH3 COOH Which of the following sets of substituents are all deactivating groups in electrophilic aromatic substitution reactions a CH3 NH2 OH b CH3 Br t C COCH3 N02 Br d Cl OH CHZCH3 What is the major organic product obtained from the following reaction H3 P9939 BOOMI What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic 12 Cl FeCI3 an Cl CCH33 Clt3Hsgts CCH CCH33 CCH33 a 1 b 2 c 3 d 4 Due 11122013 22 What is the major organic product obtained from the following reaction 23 CH3 CH3 PK CH3 803 X l H2SCL1 l l quothxt l quot at l at I I I I I V I K g 80 H 3 SO3H 1 2 3 4 P9939 honoa Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH CH3 1gt2gt3 2gt1gt3 3gt2gt1 1gt3gt2 990quot Due 11122013 24 What is the major organic product obtained from the following reaction Hint Brg in HOAc works like Brg in ABr3 Br2 Br CH3CO2H Br CH3 CH3 CHQBF CH3 CH3 a 1 b 2 c 3 d 4 25 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic substitution reactions a Cl CH3 CN b CN N02 COCH3 c Br OH COCH3 d CI OH CH3 10 CH320 N Fall 2013 320 N HW Set 4 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 CHQCH3 b COCH3 c Br d NH3 What is the major organic product obtained from the following reaction CH3 C39lCZHClg IIIIICH3 l 4 EH3 Hl3C ecu melt 09939 BOONI Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 N02 b OH c CH3 d CI Which of the following undergoes the most rapid sulfonation upon treatment with fuming sulfuric acid a benzene b benzoic acid c benzonitrile d nitrobenzene Due 11122013 5 What is the correct assignment of the names of the following substituted benzenes OCH3 HH3 CH3 1 2 3 a 1 anisole 2 aniline 3 toluene b 1 benzaldehyde 2 anisole 3 toluene c 1 anisole 2 xylene 3 toluene d 1 phenol 2 aniline 3 anisole 6 What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic CH2 CH2 CH3 N02 HNO3 N02 H3804 CH3 O2N CH3 CH3 CH3 N02 1 2 N03 3 4 7 Which of the following undergoes the most rapid bromination upon treatment with BraFeBr3 a bromobenzene b benzene c nitrobenzene d anisole CH3 P9939 hOOM l Due 11122013 8 What is the major organic product obtained from the following reaction COOH CDCH COOH COCH COOH HNC3 N02 H28C34 an NO Br Br EH32 Br Br 1 2 3 4 a 1 b 2 c 3 d 4 9 Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH OH N02 N02 1gt3gt2 1gt2gt3 2gt3gt1 3gt2gt1 P90quot Due 11122013 10 What is the major organic product obtained from the following reaction 0 Br COOH COOH COOH 900 Brz Br K FeBr3 K X l I I l l 39 B 1quot W l r CH3 CH3 CH3 CHQBF CH3 1 2 3 4 a 1 b 2 c 3 d 4 11 What is the major organic product obtained from the following reaction 09939 BOONI Due 11122013 12 What is the major organic product obtained from the following reaction CH3 CH3 O CH3 o T T JC cHltcHgt2 cHltcHgt2 2 AICI3 1 2 O CH3 CI CHCH33 T cHltcH3gt2 cHcH3gt2 3 4 09939 BOONI 13 What is the major organic product obtained from the following reaction CCH3 ITr OCH3 QCH3 QCH3 EH2 l FEBF3 an Br CN CN Br CN CN 1 2 3 4 a 1 b 2 c 3 d 4 Due 11122013 14 15 What is the major organic product obtained from the following reaction 09939 hOON l What is the major organic product obtained from the following reaction CH3 CH3 Br CH3 CH3 BF2 Br FeBm an CH3 CH3 Br Br CH3 CH3 1 2 3 4 gr a 1 b 2 c 3 d 4 Due 11122013 16 What is the major organic product obtained from the following reaction 990quot CH3 Cl AICI3 CH3 CH3 CH3 lt 2 1 F3 E hOOlJl 17 What is the major organic product obtained from the following sequence of reactions 990quot iZnHg HIEI pNo pNo sr sr K V sr ci I2ci I2cH3 I 9 3 at 1 2 3 4 Due 11122013 18 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic 19 20 substitution reactions 3 Br CH3 N02 b Cl OCH3 COCH3 C CH3 NH2 Br d N02 COCH3 COOH Which of the following sets of substituents are all deactivating groups in electrophilic aromatic substitution reactions a CH3 NH2 OH b CH3 Br K C COCH3 N02 Br d Cl OH CHZCH3 What is the major organic product obtained from the following reaction H3 P9939 BOOMI What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic 12 Cl FeCI3 an Cl CCH33 Clt3Hsgts CCH CCH33 CCH33 a 1 b 2 c 3 d 4 Due 11122013 22 What is the major organic product obtained from the following reaction 23 CH3 CH3 P CH3 803 X l H2SCL1 l l quothxt l quot at l at I I I I I V I K g 80 H 3 SO3H 1 2 3 4 P9939 honoa Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH CH3 1gt2gt3 2gt1gt3 3gt2gt1 1gt3gt2 990quot Due 11122013 24 What is the major organic product obtained from the following reaction Hint Brg in HOAc works like Brg in ABr3 Br2 Br CH3CO2H Br CH3 CH3 CHQBF CH3 CH3 a 1 b 2 c 3 d 4 25 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic substitution reactions a Cl CH3 CN b CN N02 COCH3 c Br OH COCH3 d CI OH CH3 10 CH320 N Fall 2013 320 N HW Set 4 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 CHQCH3 b COCH3 c Br d NH3 What is the major organic product obtained from the following reaction CH3 C39lCZHClg IIIIICH3 l EH3 Hl3C ecu melt 09939 BOONI Which of the following substituents is orthopara directing and deactivating in electrophilic aromatic substitution reactions 3 N02 b OH c CH3 d CI Which of the following undergoes the most rapid sulfonation upon treatment with fuming sulfuric acid a benzene b benzoic acid c benzonitrile d nitrobenzene Due 11122013 5 What is the correct assignment of the names of the following substituted benzenes OCH3 HH3 CH3 1 2 3 a 1 anisole 2 aniline 3 toluene b 1 benzaldehyde 2 anisole 3 toluene c 1 anisole 2 xylene 3 toluene d 1 phenol 2 aniline 3 anisole 6 What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic CH2 CH2 CH3 N02 HNO3 N02 H3804 CH3 O2N CH3 CH3 CH3 N02 1 2 N03 3 4 7 Which of the following undergoes the most rapid bromination upon treatment with BraFeBr3 a bromobenzene b benzene c nitrobenzene d anisole CH3 P9939 hOOM l Due 11122013 8 What is the major organic product obtained from the following reaction COOH CDCH COOH COCH COOH HNC3 N02 H28C34 an NO Br Br EH32 Br Br 1 2 3 4 a 1 b 2 c 3 d 4 9 Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH OH N02 N02 1gt3gt2 1gt2gt3 2gt3gt1 3gt2gt1 P90quot Due 11122013 10 What is the major organic product obtained from the following reaction 0 Br COOH COOH COOH 900 Brz Br K FeBr3 K X l I I l l 39 B 1quot W l r CH3 CH3 CH3 CHQBF CH3 1 2 3 4 a 1 b 2 c 3 d 4 11 What is the major organic product obtained from the following reaction 09939 BOONI Due 11122013 12 What is the major organic product obtained from the following reaction CH3 CH3 O CH3 o T T JC cHltcHgt2 cHltcHgt2 2 AICI3 1 2 O CH3 CI CHCH33 T cHltcH3gt2 cHcH3gt2 3 4 09939 BOONI 13 What is the major organic product obtained from the following reaction CCH3 ITr OCH3 QCH3 QCH3 EH2 l FEBF3 an Br CN CN Br CN CN 1 2 3 4 a 1 b 2 c 3 d 4 Due 11122013 14 15 What is the major organic product obtained from the following reaction 09939 hOON l What is the major organic product obtained from the following reaction CH3 CH3 Br CH3 CH3 BF2 Br FeBm an CH3 CH3 Br Br CH3 CH3 1 2 3 4 gr a 1 b 2 c 3 d 4 Due 11122013 16 What is the major organic product obtained from the following reaction 990quot CH3 Cl AICI3 CH3 CH3 CH3 lt 2 1 F3 E hOOlJl 17 What is the major organic product obtained from the following sequence of reactions 990quot iZnHg HIEI p p K V ci I2ci I2cH3 I 9 3 at 1 2 3 4 Due 11122013 18 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic 19 20 substitution reactions 3 Br CH3 N02 b Cl OCH3 COCH3 C CH3 NH2 Br d N02 COCH3 COOH Which of the following sets of substituents are all deactivating groups in electrophilic aromatic substitution reactions a CH3 NH2 OH b CH3 Br C COCH3 N02 Br d Cl OH CHZCH3 What is the major organic product obtained from the following reaction H3 P9939 BOOMI What is the major organic product obtained from the following reaction Hint consider steric factors as well as electronic 12 Cl FeCI3 an Cl CCH33 Clt3Hsgts CCH CCH33 CCH33 a 1 b 2 c 3 d 4 Due 11122013 22 What is the major organic product obtained from the following reaction 23 CH3 CH3 PS CH3 803 X l H2SCL1 l l quothxt l quot at l at I I I I I V I K g 80 H 3 SO3H 1 2 3 4 P9939 honoa Which of the following has the compounds shown in the correct order of decreasing acidity ie more acidic gt less acidic OH OH CH3 1gt2gt3 2gt1gt3 3gt2gt1 1gt3gt2 990quot Due 11122013 24 What is the major organic product obtained from the following reaction Hint Brg in HOAc works like Brg in ABr3 Br2 Br CH3CO2H Br CH3 CH3 CHQBF CH3 CH3 a 1 b 2 c 3 d 4 25 Which of the following sets of substituents are all orthopara directing in electrophilic aromatic substitution reactions a Cl CH3 CN b CN N02 COCH3 c Br OH COCH3 d CI OH CH3 10 CH32O N Fall 2013 320NHW 5 Multiple Choice Identijy the choice that best completes the statement or answers the question There is only one correct response for ea ch question Carefully record your answers on the Scantron sheets provided in class 4 pts each 1 Which carbon in the following carbohydrate is the anomeric carbon 990quot 2 What is the major organic product obtained from the following reaction 2 1 ll39allO2HCl 7 CH N CN W 2 KCNCuCN 339 2 1 ll A llH2 Cll2ll H2 V V I I Ne l quot c H3 L V4J39 cH3 09939 BOONI Due 1252013 3 What is are the major organic products obtained from the following reaction HO HO HO HO CHZOHO CH3OH cH2ocI3 CHZOHO H OOH3 an OH OH OH OH OH OH OH OH OOH3 OH 1 2 3 a only 1 b only 2 c only 2 and 3 d 1 2 and 3 4 Which of the following amino acids has a non poar side chain a histidine b arginine c glutamine d valine 5 What is the approximate value of the pKa of the ocCO2H of amino acids a 2 b 5 c 9 d 12 6 Which of the following is not true sodium soaps are prepared by saponification of triglycerides a by product of the saponification of triglycerides is glycerol soap molecules assemble into micelles grease molecules absorb on to the surface of micelles 990quot 7 Which of the following amino acids is not chiral a leucine b glycine c alanine d proline Due 1252013 8 What is the major organic product obtained from the following reaction 0 O O 1 LiAH4 2 H20 NcH3gt2 NH2 NccH3gt2 WNltcH32 1 2 3 09939 hOOM l 9 Which of the following amino acids is a secondary amine a proline b glutamine c cysteine d aspargine 10 Which of the following has a D configuration CH0 CH0 CH0 HO H H CH H CH HC H HC H HO H H OH HO H H CH 1 CH2OH 2 CH2OH 3 CH2OH a only 1 and 2 b only 1 and 3 c only 2 and 3 d only 1 2 and 3 11 What is the approximate value of the pKa of the ocNH3 of amino acids a 2 b 5 c 9 d 12 12 Which of the following amino acids is not an L isomer a senne b valine c glycine d proline Due 1252013 13 What is the major organic product obtained from the following sequence of reactions O 1LiAH4 JK Br Nacm 2 H20 H2 Ni V 1 an 1 V 2 H HN P9939 hOOM l 14 What is the major organic product obtained from the following reaction 1 2 o O 1 LiAH4 2 H20 H NCH3 T NHCH3 a xv A1x x NHCH3 OH P9939 hOOM l Due 1252013 15 What is the major organic product obtained from the following reaction cno COOH CH2OH CHO COOH H oH H2 H OH H OH 0 H OH Ho H N Ho H Ho H Ho H Ho H H OH H OH H OH H OH H OH H OH H OH H OH H OH H OH cH2oH CH2OH c 2o CH2OH COCH 1 2 3 4 a 1 b 2 c 3 d 4 16 Which of the following is a disaccharide a glucose b fructose c sucrose d mannitol 17 What is the major organic product obtained from the following reaction CO2H CHO CHO CH OH HO oxldase Ho H HO H HO H OH H OH H CH H OH OH H oH H CH H oH CHQOH CO2H CH2OH 1 2 3 a only 1 b only 2 c only 3 d only 1 and 2 Due 1252013 18 What is the major organic product obtained from the following reaction NH2 Br OH N02 NH2 1 lla lO2 HCI 2 H3F O2 an Br Br Br OH 1 2 3 4 09939 hOON l 19 Which of the following amino acids has a polar side chain a isoleucine b valine c phenylalanine d threonine 20 Which carbon in the following carbohydrate is the anomeric carbon i CHC HOH HT OH Hf HO H H OH iv CHZOH a i b ii C iii d iv Due 1252013 21 What is the major organic product obtained from the following reaction CH3 N02 NCH32 OH NH2 Kw H T V7 T T CH3 CH3 CH3 CH3 09939 hOOM l 22 Which of the following amino acids has an aromatic side chain a tyrosine b glutamine c aspargine d valine 23 Which of the following represent a ketose CH2OH Hoc 2 OH 0 O HO O OH OH H OH HO 2 HO H OH CHZOH OH HO OH H OH 1 2 3 CH2OH a only 1 and 2 b only 1 and 3 c only 2 and 3 d 1 2 and 3 24 Which of the following amino acids has a basic side chain a lysine b senne c leucine d tyrosine Due 1252013 25 Which of the following undergoes the most rapid sulfonation upon treatment with fuming sulfuric acid a benzene b benzoic acid c benzonitrile d nitrobenzene 13 Nuclear Magnetic Resonance Spectroscopy 131 Nuclear Spin States Energy states and their orientation spin 39s units for magnetic field 132 Orientation of Nuclear Spins in an Applied Magnetic Fir effect of nuclear spins in magnetic fields 133 Nuclear Magnetic Resonance concept of resonance signals diamagnetic current sheiding deshieding ppm amp TMS 134 An NMR Spectrometer basic parts 135 Equivalent Hydrogens be able to label classify upfield amp downfield 136 Signal Areas integration and its use 137 Chemical Shift electronegativity effects hybridization and diamagnetic effects ring current 138 Signal Splitting and the n 1 Rule application and use 139 The Origins of Signal Splitting theory of splitting and recognition of patterns s d t etc multiple splittings 1313 Interpretation of NMR Spectra Major functional groups splitting patterns multiple splitting 15 Organometallic Compounds 151 Organomagnesium and Organolithium Compounds Rxns with carbonyls amp epoxides Mechanism 152 Lithium Diorganocopper Gilman Reagents Coupling with halides 153 Carbenes and Carbenoids Insertion reactions carbene amp dihalocarbenes SimmonsSmith rxn 16 Aldehydes and Ketones 161 Structure and Bonding Polarity properties bond characteristics 162 Nomenclature trivial names of compounds on slides basic nomenclature 163 Physical Properties solubility trends BP amp or MP trends spectral properties 164 Reactions Reaction theme addition to give tetrahedral intermediate 165 Addition of Carbon Nucleophiles Grignards to give ROH organolithiums acetylide cyanideuse of cyanohydrins 166 The Wittig Reaction use with carbonyls 167 Addition of Oxygen Nucleophiles Hydrates hemi aceta acetal formation mechanisms acid amp base 168 Addition of Nitrogen Nucleophiles imines and enamines formation and hydrolysis mechanism 169 KetoEnol Tautomerism acidity of 0c hydrogens significance mechanismacidic amp basic 1610 Oxidation RHO amp ROHuse of all Cr6 reagents Silver oxideO2 1611 Reduction metal hydride reductions cat H2 Clemmenson Woff Kishnermechanism 1612 Reactions at an ocCarbon racemizationmechanism halogenation acidic amp basic conditionsmechanism Learn to Get More and Do More at Dummiescom Start with FREE Cheat Sheets Ir CV9 It Cheat Sheets Include 7c39 Checklists Charts Common Instructions And Other Good Stuff To access the Cheat Sheet created specifically for this book go to wwwdummiescomcheatsheetorganicchemistryz 1 Get Smart at Dummiescom I ll Dummiescom makes your life easier with 10005 of answers on everything from removing wallpaper if to using the latest version ofWindows Check out our Videos Illustrated Articles Step by Step Instructions Plus each month you can win valuable prizes by entering our Dummiescom sweepstakes Want a weekly dose of Dummies Sign up for Newsletters on Digital Photography Microsoft Windows amp Office Personal Finance amp Investing Health ampWeness Computing iPods amp Cell Phones eBay Internet Food Home amp Garden Find outquotHOW at Dummiescom Sweepstakes not currently available in all countries visit Dummies com for of cial rules Organic Chemistry II FOR DUMMIE6 Organic Chemistry II FOR DUMMIVEH by John T Moore EdD and Richard H Langley PhD WILEY Wiley Publishing Inc Organic Chemistry II For Dummies Published by Wiley Publishing Inc 111 River St Hoboken NJ 070305774 wwwwileycom Copyright 2010 by Wiley Publishing Inc Indianapolis Indiana Published simultaneously in Canada No part of this publication may be reproduced stored in a retrieval system or transmitted in any form or by any means electronic mechanical photocopying recording scanning or otherwise except as permit ted under Sections 107 or 108 of the 1976 United States Copyright Act without either the prior written permission of the Publisher or authorization through payment of the appropriate percopy fee to the Copyright Clearance Center 222 Rosewood Drive Danvers MA 01923 978 7508400 fax 978 6468600 Requests to the Publisher for permission should be addressed to the Permissions Department John Wiley amp Sons Inc 111 River Street Hoboken NJ 07030 201 7486011 fax 201 7486008 or online at ht tp wwwwileycomgopermissions Trademarks Wiley the Wiley Publishing logo For Dummies the Dummies Man logo A Reference for the Rest of Us The Dummies Way Dummies Daily The Fun and Easy Way Dummiescom Making Everything Easier and related trade dress are trademarks or registered trademarks of John Wiley amp Sons Inc and or its af liates in the United States and other countries and may not be used without written permission All other trademarks are the property of their respective owners Wiley Publishing Inc is not associated with any product or vendor mentioned in this book LIMIT OF LIABILITYDISCLAIMER OF WARRANTY THE PUBLISHER AND THE AUTHOR MAKE NO REPRESENTATIONS OR WARRANTIES WITH RESPECT TO THE ACCURACY OR COMPLETENESS OF THE CONTENTS OF THIS WORK AND SPECIFICALLY DISCLAIM ALL WARRANTIES INCLUDING WITH OUT LIMITATION WARRANTIES OF FITNESS FOR A PARTICULAR PURPOSE NO WARRANTY MAY BE CREATED OR EXTENDED BY SALES OR PROMOTIONAL MATERIALS THE ADVICE AND STRATEGIES CONTAINED HEREIN MAY NOT BE SUITABLE FOR EVERY SITUATION THIS WORK IS SOLD WITH THE UNDERSTANDING THAT THE PUBLISHER IS NOT ENGAGED IN RENDERING LEGAL ACCOUNTING OR OTHER PROFESSIONAL SERVICES IF PROFESSIONAL ASSISTANCE IS REQUIRED THE SERVICES OF A COMPETENT PROFESSIONAL PERSON SHOULD BE SOUGHT NEITHER THE PUBLISHER NOR THE AUTHOR SHALL BE LIABLE FOR DAMAGES ARISING HEREFROM THE FACT THAT AN ORGANIZA TION OR WEBSITE IS REFERRED TO IN THIS WORK AS A CITATION ANDOR A POTENTIAL SOURCE OF FURTHER INFORMATION DOES NOT MEAN THAT THE AUTHOR OR THE PUBLISHER ENDORSES THE INFORMATION THE ORGANIZATION OR WEBSITE MAY PROVIDE OR RECOMMENDATIONS IT MAY MAKE FURTHER READERS SHOULD BE AWARE THAT INTERNET WEBSITES LISTED IN THIS WORK MAY HAVE CHANGED OR DISAPPEARED BETWEEN WHEN THIS WORK WAS WRITTEN AND WHEN IT IS READ For general information on our other products and services please contact our Customer Care Department within the US at 8777622974 outside the US at 3175723993 or fax 3175724002 For technical support please visit www wiley com techsupport Wiley also publishes its books in a variety of electronic formats Some content that appears in print may not be available in electronic books Library of Congress Control Number 2010926849 ISBN 9780470178157 Manufactured in the United States of America 10 9 8 7 6 5 4 3 2 1 WILEY About the Authors John T Moore EdD grew up in the foothills of western North Carolina He attended the University of North Carolina Asheville where he received his bachelor s degree in chemistry He earned his master s degree in chemistry from Furman University in Greenville South Carolina After a stint in the United States Army he decided to try his hand at teaching In 1971 he joined the chemistry faculty of Stephen F Austin State University in Nacogdoches Texas where he still teaches chemistry In 1985 he went back to school part time and in 1991 received his doctorate in education from Texas AampM University For the past several years he has been the coeditor along with one of his former students of the Chemistry for Kids feature of The Journal of Chemical Education In 2003 his rst book Chemistry For Dummies was published by Wiley soon to be followed by Chemistry Made Simple Broadway and Chemistry Essentials For Dummies Wiley John enjoys cooking and making custom knife handles from exotic woods Richard H Langley PhD grew up in southwestern Ohio He attended Miami University in Oxford Ohio where he received bachelor s degrees in chemistry and in mineralogy and a master s degree in chemistry His next stop was the University of Nebraska in Lincoln Nebraska where he received his doctorate in chemistry Afterwards he took a postdoctoral position at Arizona State University in Tempe Arizona followed by a visiting assistant professor position at the University of Wisconsin River Falls In 1982 he moved to Stephen F Austin State University For the past several years he and John have been graders for the freeresponse portion of the AP Chemistry Exam He and John have collaborated on several writing projects including 5 Steps to a Five AP Chemistry and Chemistry for the Utterly Confused both published by McGrawHill Rich enjoys jewelry making and science ction Dedication John I dedicate this book to my wife Robin sons Matthew and Jason my wonderful daughterinlaw Sara and the two most wonderful grandkids in the world Zane and Sadie I love you guys Rich I dedicate this book to my mother Authors Acknowledgments We would not have had the opportunity to write this book without the encouragement of our agent Grace Freedson We would also like to thank Chrissy Guthrie for her support and assistance in the early portion of this project and to Sarah Faulkner who helped us complete it We would also like to thank our copy editor Caitie Copple and our technical editors Susan Klein and Joe Burnell Many thanks to our colleagues Russell Franks and Jim Garrett who helped with suggestions and ideas Rich would also like to acknowledge Danica Dizon for her suggestions ideas and inspiration Thanks to all of the people at Wiley publishing who help bring this project from concept to publication Publisher s Acknowledgments We re proud of this book please send us your comments at http dummies Custhelp com For other comments please contact our Customer Care Department within the US at 8777622974 outside the US at 3175723993 or fax 3175724002 Some of the people who helped bring this book to market include the following Acquisitions Editorial and Composition Services Media Development Project Coordinator Patrick Redmond Project Editors Sarah Faulkner L dG h N39kk39G 1 Christina Guthrie ayoutan rap ms 1 1 atey P fdL Alb S 39RS39h Senior Acquisitions Editor roo rea ers aura ert oss1ty m1t Lindsay Sandman Lefevere Indexer Sham Shock Copy Editor Caitlin Copple Special Help Jennifer Tebbe Assistant Editor Erin Calligan Mooney Senior Editorial Assistant David Lutton Technical Editors Susan J Klein PhD Joe C Burnell PhD Editorial Manager Christine Meloy Beck Editorial Assistants Jennette ElNaggar Rachelle S Amick Cover Photos Haywiremedia I Dreamstimecom iStock Cartoons Rich Tennant www the5thwave com Publishing and Editorial for Consumer Dummies Diane Graves Steele Vice President and Publisher Consumer Dummies Kristin FergusonWagstaffe Product Development Director Consumer Dummies Ensley Eikenburg Associate Publisher Travel Kelly Regan Editorial Director Travel Publishing for Technology Dummies Andy Cummings Vice President and Publisher Dummies TechnologyGeneral User Composition Services Debbie Stailey Director of Composition Services Contents at a Glance 7 Part I Brushing Up on Important Organic Chemistry I Concepts 7 Chapter 1 Organic Chemistry 11 Here We Go Again 9 Chapter 2 Remembering How We Do It Mechanisms 17 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping 31 Chapter 4 Conjugated Unsaturated Systems 53 Chapter 5 Seeing Molecules Spectroscopy Revisited 67 Part II Discovering Aromatic And Not So Aromatic 79 Chapter 6 Introducing Aromatics 81 Chapter 7 Aromatic Substitution Part 1 Attack of the Electrophiles 93 Chapter 8 Aromatic Substitution Part 11 Attack of the Nucleophiles and Other Reactions 111 Part III Carbonyls Good Alcohols Gone Bad z 727 Chapter 9 Comprehending Carbonyls 123 Chapter 10 Aldehydes and Ketones 137 Chapter 11 Enols and Enolates 161 Chapter 12 Carboxylic Acids and Their Derivatives 187 Part IV Advanced Topics Every Student39s Nightmare t 2 79 Chapter 13 Amines and Friends 221 Chapter 14 Metals Muscling In Organometallics 249 Chapter 15 More Reactions of Carbonyl Compounds 261 Chapter 16 Living Large Biomolecules 281 Part V Pulling It All Together 309 Chapter 17 Overview of Synthesis Strategies 311 Chapter 18 Roadmaps and Predicting Products 327 Part 1 The Part of 0 33 7 Chapter 19 Ten Sure re Ways to Fail Organic Chemistry II 339 Chapter 20 More than Ten Ways to Increase Your Score on an Organic Chemistry Exam 343 Appendix Named Reactions 347 349 Table of Contents 0 7 About This Book 1 Conventions Used in This Book 2 What You re Not to Read 2 Foolish Assumptions 2 How This Book Is Organized 3 Part I Brushing Up on Important Organic Chemistry I Concepts 3 Part II Discovering Aromatic And Not So Aromatic Compounds 3 Part III Carbonyls Good Alcohols Gone Bad 3 Part IV Advanced Topics Every Student s Nightmare 4 Part V Pulling It All Together 4 Part VI The Part of Tens 4 Icons Used in This Book 4 Where to Go from Here 5 Part I Brushing Up an Important Organic Chemistry I Concepts 7 Chapter 1 Organic Chemistry II Here We Go Again 9 Recapping Organic Chemistry I 10 Intermolecular forces 10 Functional groups 11 Reactions 11 Spectroscopy 11 Isomerism and optical activity 12 Looking Ahead to Organic Chemistry II 14 Chapter 2 Remembering How We Do It Mechanisms 17 Duck Here Come the Arrows 17 Coming Around to Curved Arrows 19 Getting Ready for Some Basic Moves 20 Bond A lone pair 21 Bond A bond 21 Lone pair A bond 22 Combining the Basic Moves 22 Intermediates 24 Keys to substitution and elimination mechanisms 25 Revisiting FreeRadical Mechanisms 27 Organic Chemistry II For Dummies Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping 31 Getting Acquainted with Alcohols 31 Structure and nomenclature of alcohols 32 Physical properties of alcohols 33 Making moonshine Synthesis of alcohols 34 What will they do besides burn Reactions of alcohols 40 Introducing Ether Not the Ether Bunny 46 Structure and nomenclature of ethers 46 Sleepy time Physical properties of ethers 46 Synthesis of ethers 47 Reactions of ethers 49 Summarizing the Spectra of Alcohols and Ethers 51 Chapter 4 Conjugated Unsaturated Systems 53 When You Don t Have Enough Unsaturated Systems 53 Conjugated systems 53 The allylic radical 54 Butadiene 55 Delocalization and Resonance 56 Resonance rules 56 Stability of conjugated unsaturated systems 57 Reactions of Conjugated Unsaturated Systems 57 Put in the second string Substitution reactions 57 Electrophilic addition 59 More than a tree DielsAlder reactions 62 Passing an Exam with DielsAdler Questions 65 Indentifying the product 65 Identifying the reactants 66 Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited 67 Chemical Fingerprints Infrared Spectroscopy 68 Double bonds 68 Triple bonds 69 0H and NH stretches 69 CH stretches 69 Suntans and Beyond Ultraviolet and Visible Spectroscopy 70 Not Weight Watchers Mass Watchers Mass Spectroscopy 72 The molecular ion 72 Fragmentation 73 No Glowing Here NMR Spectroscopy 73 Proton 74 Carbon13 77 Table of Contents Part II Discovering Aromatic And Not So Aromatic p 79 Chapter 6 Introducing Aromatics 81 Benzene Where It All Starts 81 Figuring out benzene s structure 81 Understanding benzene s resonance 83 The stability of benzene 84 Physical properties of benzene 85 Organic math H1 ickel s Rule 85 Other aromatics 87 Smelly Relatives The Aromatic Family 87 Nomenclature of the aromatic family 87 Derivatives of benzene 88 Branches of aromatic groups 89 Black Sheep of the Family Heterocyclic Aromatic Compounds 89 Aromatic nitrogen compounds 90 Aromatic oxygen and sulfur compounds 90 Spectroscopy of Aromatic Compounds 90 IR 91 UVvis 91 NMR 91 Mass spec 92 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles 93 Basics of Electrophilic Substitution Reactions 94 Reactions of Benzene 95 Halogenation of benzene 95 Nitration of benzene 96 Sulfonation of benzene 97 FriedelCrafts Reactions 99 Alkylation 99 Acylation 100 Why Do an Alkylation 101 Changing Things Modifying the Reactivity of an Aromatic 101 Lights camera action Directing 102 Turning it on turning it off Activating and deactivating 107 Steric hindrance 109 Limitations of Electrophilic Substitution Reactions 109 X Organic Chemistry II For Dummies Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles and Other Reactions 111 Coming Back to Nucleophilic Substitution Reactions 111 Mastering the Mechanisms of Nucleophilic Substitution Reactions 112 Losing and Gaining Mechanisms of EliminationAddition Reactions 114 Benzyne 114 The eliminationaddition mechanism 114 Synthetic Strategies for Making Aromatic Compounds 115 Brie y Exploring Other Reactions 117 Part III Carbonyls Good Alcohols Gone Bad 727 Chapter 9 comprehending Carbonyls 123 Carbonyl Basics 123 Considering compounds containing the carbonyl group 124 Getting to know the acidic carbonyl 127 Polarity of Carbonyls 128 Resonance in Carbonyls 129 Reactivity of the Carbonyl Group 130 Spectroscopy of Carbonyls 130 Infrared spectroscopy 130 UltravioletVisible electronic spectroscopy 131 Nuclear magnetic resonance NMR spectroscopy 132 Mass spectroscopy 134 Chapter 10 Aldehydes and Ketones 137 Meeting Alcohol s Relatives Structure and Nomenclature 137 De ning Physical Properties of Aldehydes and Ketones 139 Creating Aldehydes and Ketones with Synthesis Reactions 140 Oxidation reactions 140 Reduction reactions 142 Other reactions 143 Taking Them a Step Further Reactions of Aldehydes and Ketones 146 Nucleophilic attack of aldehydes and ketones 147 Oxidation of aldehydes and ketones 156 The BaeyerVilliger reaction 158 Checking Out Spectroscopy Specs 160 Chapter11 Enols and Enolates 161 Getting to Know Enols and Enolates 161 Enough already Structure of enols and enolates 162 I thought I saw a tautomer 163 Studying the Synthesis of Enols and Enolates 164 Table of Contents Thinking Through Reactions of Enols and Enolates 166 Haloform reactions 166 Aldol reactions and condensations 168 Addition reactions to unsaturated aldehydes and ketones 173 Other enolaterelated reactions 178 Miscellaneous reactions 180 Chapter 12 Carboxylic Acids and Their Derivatives 187 Seeing the Structure and Nomenclature of Carboxylic Acids and Derivatives 188 Structure 188 Nomenclature 188 Checking Out Some Physical Properties of Carboxylic Acids and Derivatives 193 Carboxylic acids 193 Esters 193 Amides 194 Considering the Acidity of Carboxylic Acids 194 Determining How Carboxylic Acids and Derivatives Are Synthesized 196 Synthesizing carboxylic acids 196 Developing acyl halides with halogen 199 Removing water to form acid anhydrides 200 Uniting acids and alcohols to make esters 202 Bringing acids and bases together to create amides 206 Exploring Reactions 208 Generous carboxylic acids 209 Simple acyl halide and anhydride reactions 210 Hydrolysis of esters 210 Amide reactions ester s cousins 211 Other reactions of carboxylic acids and derivatives 213 Taking a Look at Spectroscopy and Chemical Tests 217 Identifying compounds with spectral data 218 Using chemical tests 218 Part IV Advanced Topics Every Student39s Nightmare P J 2 79 Chapter 13 Amines and Friends 221 Breaking Down the Structure and Nomenclature of Nitrogen Compounds 221 Primary amines 222 Secondary and tertiary amines 223 Quaternary amines quaternary ammonium salts 223 Heterocyclics 224 XV X i Organic Chemistry II For Dummies Sizing Up the Physical Properties 225 Understanding the Basicity of Nitrogen Compounds 226 Synthesizing Nitrogen Compounds 227 Nucleophilic substitution reactions 227 Reduction preparations 229 Seeing How Nitrogen Compounds React 233 Reactions with nitrous acid 233 Replacement reactions 235 Coupling reactions of diazonium salts 239 Reactions with sulfonyl chlorides 239 Exploring elimination reactions 241 Mastering Multistep Synthesis 244 Identifying Nitrogen Compounds with Analysis and Spectroscopy 246 Chapter 14 Metals Muscling In Organometallics 249 Grignard Reagents Grin and Bear It 249 Preparation of Grignard reagents 250 Reactions of Grignard reagents 250 Organolithium Reagents 256 Formation of Other Organometallics 257 Putting It Together 258 Chapter 15 More Reactions of Carbonyl Compounds 261 Checking Out the Claisen Condensation and Its Variations 262 Doing the twostep Claisen condensation 262 Circling around Dieckmann condensation 264 Doubling Up Crossed Claisen condensation 265 Other carbanions 266 Exploring Acetoacetic Ester Synthesis 267 De ning Malonic Ester Synthesis 269 Working with Other Active Hydrogen Atoms 273 Reacting with Knoevenagel Condensation 273 Looking at Mannich Reactions 275 Creating Enamines Stork Enamine Synthesis 277 Putting It All Together with Barbiturates 279 Chapter 16 Living Large Biomolecules 281 Delving into Carbohydrate Complexities 282 Introducing carbohydrates 282 Examining the many reactions of monosaccharides 286 Synthesizing and degrading monosaccharides 291 Meeting the Daldose family 293 Checking out a few disaccharides 295 Looking at some polysaccharides 296 Discovering nitrogencontaining sugars 298 Lipids Storing Energy Now So You Can Study Longer Later 299 Pondering the properties of fats 299 Soaping up with saponi cation 300 Table of Contents Bulking Up on Amino Acids and Proteins 301 Introducing amino acids 302 Perusing the physical properties of amino acids 302 Studying the synthesis of amino acids 304 Part V Pulling It All Together 309 Chapter 17 Overview of Synthesis Strategies 311 Working with OneStep Synthesis 312 Tackling Multistep Synthesis 312 Practicing Retrosynthetic and Synthetic Analysis 313 Example 1 314 Example 2 319 Example 3 321 Example 4 322 Example 5 324 Chapter 18 Roadmaps and Predicting Products 327 Preparing with Roadmap Basics 327 Practicing Roadmap Problems 328 Problem one 328 Solution one 328 Problem two 330 Solution two 330 Problem three 332 Solution three 332 Predicting Products 335 Part 1 The Part of pZ 337 Chapter 19 Ten Surefire Ways to Fail Organic Chemistry II 339 Simply Read and Memorize Concepts 339 Don t Bother Working the Homework Problems and Exercises 340 Don t Buy a Model Kit 340 Don t Worry About Falling Behind 340 Don t Bother Learning Reactions 341 If Your Textbook Confuses You Don t Bother with Additional Resources 341 Don t Bother Reading the Chapter before Attending Class 341 Attend Class Only When You Feel Like It 342 Don t Bother Taking Notes Just Listen When You Aren t Sleeping or Texting 342 Don t Bother Asking Questions 342 Organic Chemistry II For Dummies Chapter 20 More than Ten Ways to Increase Your Score on an Organic Chemistry Exam 343 Don t Cram the Night before a Test 343 Try Doing the Problem Sets and Practice Tests Twice 344 Study the Mistakes You Made on Previous Exams 344 Know Precisely Where Why and How the Electrons Are Moving 344 Relax and Get Enough Sleep before the Exam 345 Think Before You Write 345 Include Formal Charges in Your Structures When Appropriate 345 Check That You HaVen t Lost Any Carbon Atoms 346 Include EZ RS cistrans Pre xes in Naming Organic Structures 346 Think of Spectroscopy Especially NMR As a Puzzle 346 Make Sure That Each Carbon Atom Has Four Bonds 346 Appendix Named Reactions 347 G 349 Introduction Welcome to Organic Chemistry II For Dummies We re certainly happy you decided to delve further into the fascinating world of organic chemistry It s a complex area of chemistry but understanding organic chemistry isn t really that difficult It simply takes hard work attention to detail some imagination and the desire to know Organic chemistry like any area of chemistry is not a spectator sport You need to interact with the material try different study techniques and ask yourself why things happen the way they do Organic Chemistry II is a more intricate course than the typical freshman introductory chemistry course and you may find that it s also more involved than Organic 1 You may actually need to use those things you learned and study habits you developed in Organic 1 to be successful in Organic 11 But if you work hard you can get through your Organic 11 course More importantly you may grow to appreciate the myriad chemical reactions that take place in the diverse world of organic chemistry About This Book quot Organic Chemistry II For Dummies is an overview of the material covered in the second half of a typical collegelevel organic chemistry course We have made every attempt to keep the material as current as possible but the field of chemistry is changing ever so quickly as new reactions are developed and the fields of biochemistry and biotechnology inspire new avenues of research The basics however stay the same and they are where we concentrate our attention As you flip through this book you see a lot of chemical structures and reactions Much of organic chemistry involves knowing the structures of the molecules involved in organic reactions If you re in an Organic Chemistry 11 course you made it through the first semester of organic chemistry so you recognize many of the structures or at least the functional groups from your previous semester s study If you bought this book just to gain general knowledge about a fascinating subject try not to get bogged down in the details Skim the chapters If you find a topic that interests you stop and dive in Have fun learning something new If you re taking an organic chemistry course you can use this rather inexpensive book to supplement that very expensive organic textbook 2 Organic Chemistry II For Dummies Comentions Used in This Book We have organized this book in a logical progression of topics your second semester organic chemistry course may progress similarly In addition we set up the following conventions to make navigating this book easier 1 Italics introduce new terms that you need to know 1 Bold text highlights keywords within a bulleted list 1 We make extensive use of illustrations of structures and reactions While reading try to follow along in the associated figures whether they be structures or reactions What You39re Not to Read You don t have a whole lot of money invested in this book so don t feel obliged to read what you don t need Concentrate on the topics in which you need help Feel free to skip over any text in a gray shaded box which we refer to as sidebars Although interesting they aren t required reading Foolish Assumptions We assume and we all know about the perils of assumptions that you are one of the following 1 A student taking a collegelevel organic chemistry course 1 A student reviewing organic chemistry for some type of standardized exam the MCAT for example 1 An individual who just wants to know something about organic chemistry If you fall into a different category you re special and we hope you enjoy this book anyway Introduction 3 How This Book Is Organized The topics in this book are divided into six parts Use the following descriptions and the table of contents to map out your strategy of study Part I Brushing Up on Important Organic Chemistry I Concepts Part I is really a rapid review of many of the concepts found in an Organic Chemistry I course It s designed to review the topics that you need in Organic II We set the stage by giving you an overview of Organic Chemistry II and then review mechanisms Next we cover alcohols and ethers their properties synthesis and reactions followed by an overview of conjugated unsaturated systems We end this review section with a discussion of spectroscopy including IR UVvisible mass spec and of course NMR A whirlwind tour of Organic 1 Part1 Discovering Aromatic And Not So Aromatic Compounds In Part II we concentrate on aromatic systems starting with the basics of structure and properties of benzene and then moving on to related aromatic compounds We even throw in a section of spectroscopy of aromatic compounds Chapters 7 and 8 finish up this part by going into detail about substitution reactions of aromatic compounds You find out all you ever wanted to know and maybe more about electrophilic and nucleophilic substitutions along with a little about elimination reactions Part III Carhony Is Good Alcohols Gone Bad In Part III we cover that broad category of organic compounds called the carbonyls First we give you an overview of carbonyl basics including structure reactivity and spectroscopy Then we go into more detail on aldehydes and ketones enols and enolates and carboxylic acids and their derivatives 4 Organic Chemistry II For Dummies Part IV Advanced Topics Every Student39s Niqh tmare In Part IV we start by taking a closer look at nitrogen compounds and their structure reactivity and reactions Then we move on to organometallic compounds where we meet the infamous Grignard reaction We then finish up this part by addressing some moreinvolved reactions of the carbonyls and biomolecules You pick up some good hints for synthesis and roadmaps here Part V Pulling It All Together In Part V we show you how to pull all the previous information together and use it to develop strategies for designing synthesis reactions We talk about both onestep and multistep synthesis as well as retrosynthetic analysis Then we tackle the dreaded organic roadmaps We all wish we had an organic chemistry GPS here Part VI The Part of Tens In this final part of the book we discuss ten surefire ways to flunk your organic chemistry class so you know what to avoid along with ten ways to increase your grade on those organic chemistry exams Icons Used in This Book NB39R 4 If you have ever read other For Dummies books such as the wonderful Chemistry For Dummies or Biochemistry For Dummies written by yours truly and published by Wiley you recognize the icons used in this book The following four icons can guide you to certain kinds of information This icon is a flag for those really important things that you shouldn t forget as you go deeper into the world of organic chemistry We use this icon to alert you to a tip on the easiest or quickest way to learn a concept Between the two of us we have almost 70 years of teaching experience We ve learned a few tricks along the way and we don t mind sharing Introduction 5 The warning icon points to a procedure or potential outcome that can be dangerous We call it our Don tTryThisAtHome icon We try to avoid getting too technical throughout this book believe it or not but every now and then we can t help but throw something in that is a little more indepth than you might need You won t hurt your education by skipping it Where to Go from Here The answer to this question really depends of your prior knowledge and goals As with all For Dummies books this one attempts to make all the chapters independent so that you can dive right into the material that s causing you trouble without having to read other chapters first If you feel comfortable with the topics covered in Organic Chemistry I feel free to skip Part I If you want a general overview of organic chemistry skim the remainder of the book Take a deeper plunge into a chapter when you find a topic that interests you or one in which you really need help And for all of you no matter who you are or why you re reading this book we hope you have fun reading it and that it helps you to understand and appreciate organic chemistry 6 Organic Chemistry II For Dummies E Brushing Up on Important Organic Chemistry I Concepts The 5th Wave By Rich Tennant JASON WAS ALWAYS LATE FOR CLASSES HELD IN THE ESCHER BUILDING In this part Part I is a review of some general chemistry and Organic Chemistry 1 topics you need a firm grounding in before moving on to Organic Chemistry 11 Different books and different instructors break Organic 1 and Organic 11 material at different places We use the most common break but some Part 1 material may in fact be new to you Even if you covered these concepts last semester some of them have a high vapor pressure and may have escaped between semesters We begin by bringing you up to speed on mechanisms and reminding you how to push electrons around with those curved arrows We jog your memory with a discussion of substitution and elimination reactions and their mecha nisms in addition to free radical reactions Next you review the structure nomenclature synthesis and reac tions of alcohols and ethers and then you get to tackle conjugated unsaturated systems Finally we remind you of spectroscopic techniques from the IR fingerprints to NMR shifts The review in this part moves at a pretty fast pace but we re sure you can keep up Chapter 1 Organic Chemistry II Here We Go Again In This Chapter Reviewing the material you learned in Organic 1 Previewing what you find out in Organic 11 f you re looking at this chapter it s probably because you re getting ready to take the second half of organic chemistry are in the midst of Organic 11 or you re trying to figure out what Organic 11 covers in time to change your major from premed to art history In any respect you probably successfully completed Organic Chemistry 1 Many of the study techniques and coping mechanisms you learned that helped you do well in Organic 1 are helpful in Organic 11 The two primary things to remember are 1 Never get behind 1 Carbon has four bonds In this book we use larger more complex molecules than you may have encountered in Organic 1 We chose to do this because firstly that s the nature of Organic 11 larger and more complex molecules Secondly many of you will be taking biochemistry at some point and to succeed in that subject you need to become comfortable with large involved molecules If you do take biochemistry be sure to check out Biochemistry For Dummies by John T Moore and Richard H Langley Wiley We understand the authors are really great guys To get you started this chapter does a quick review of the topics commonly found in Organic 1 and then gives an overview of what we cover in Organic 11 70 Part I Brushing Up on Important Organic Chemistry I Concepts Recapping Organic Chemistry In Organic I you learned that organic chemistry is the study of carbon compounds Until the mid1800s people believed that all carbon compounds were the result of biological processes requiring a living organism This was called the vital force theory The synthesis or formation of urea from inorganic materials showed that other paths to the production of carbon compounds are possible Many millions of organic compounds exist because carbon atoms form stable bonds to other carbon atoms The process of one type of atom bonding to identical atoms is catenation Many elements can catenate but carbon is the most effective with apparently no limit to how many carbon atoms can link together These linkages may be in chains branched chains or rings providing a vast combination of compounds Carbon is also capable of forming stable bonds to a number of other elements including the biochemically important elements hydrogen nitrogen oxygen and sulfur The latter three elements form the foundation of many of the functional groups you studied in Organic 1 Intermolecular forces You also learned about intermolecular forces in Organic 1 Intermolecular forces forces between chemical species are extremely important in explaining the interaction between molecules Intermolecular forces that you saw in Organic I and see again in Organic II include dipoledipole interac tions London hydrogen bonding and sometimes ionic interactions Dipoledipole forces exist between polar regions of different molecules The presence of a dipole means that the molecule has a partially positive 8 end and a partially negative 8 end Opposite partial charges attract each other whereas like partial charges repel Hydrogen bonding as the name implies involves hydrogen This hydrogen atom must be bonded to either an oxygen atom or a nitrogen atom In non biological situations hydrogen bonding also occurs when a hydrogen atom bonds to a fluorine atom Hydrogen bonding is significantly stronger than a normal dipoledipole force and is stronger than London dispersion forces the forces between nonpolar molecules due to the fluctuations of the electron clouds of atoms or molecules The hydrogen bonded to either a nitrogen or oxygen atom is strongly attracted to a different nitrogen or oxygen atom Hydrogen bonding may be either intramolecular or intermolecular Chapter 1 Organic Chemistry II Here We Go Again In organic reactions ionic interactions may serve as intermolecular or intramolecular forces In some cases these may involve metal cations such as Na or anions such as Cl Cations may include an ammonium ion from an amino group such as RNH3 The anion may be from a carboxylic acid such as RCOO The oppositely charged ions attract each other very strongly Functional groups Carbon is an extremely versatile element because it can form many different compounds Most of the compounds have one or more functional groups which contain atoms other than carbon and hydrogen andor double or triple bonds and define the reactivity of the organic molecule In Organic I you probably started with the hydrocarbons compounds of carbon and hydrogen including the alkenes and alkynes that contained double and single bonds respectively Then you probably touched on some of the more common functional groups such as alcohols and maybe even some aromatic compounds Reactions You encountered a lot of reactions in Organic 1 Every time you encountered a different functional group you had a slew of reactions to learn Reactions that told how the functional group could be formed common reactions that the functional group underwent reactions reactions and more reactions Two of the most important ones you learned were substitution and elimination reactions SN1 SN2 E1 and E2 We hope you learned them well because you ll be seeing them again quite often Spectroscopy In Organic I you probably learned a lot about the different types of spectroscopy and how they re used in structure determinations You discovered how mass spectroscopy can give you an idea about molar mass and what fragments may be present in the molecule You found out that infrared spectroscopy can be used to identify functional groups and you learned to look at the fingerprint region Then finally you progressed to nuclear magnetic resonance NMR spectroscopy one of the main tools of organic chemists which can be used to interpret chemical shifts and splitting patterns to give you more clues about structure 77 72 Figure 11 Cis and trans isomers Part I Brushing Up on Important Organic Chemistry I Concepts Isomerism and optical activity During Organic 1 you were exposed to the concepts associated with isomerism and optical activity You need to be familiar with these concepts in Organic 11 so we take a few minutes here for a brief review Isomers are compounds with the same molecular formula but different structural formulas Some organic and biochemical compounds may exist in different isomeric forms and these different isomers have different properties The two most common types of isomers in organic systems are cistrans isomers and isomerism due to the presence of a chiral carbon Cistrans isomers The presence of carboncarbon double bonds leads to the possibility of isomers Double bonds are rather restrictive and limit molecular movement Groups on the same side of the double bond tend to remain in that position cis while groups on opposite sides tend to remain across the bond from each other trans You can see an example of each in Figure 11 However if the two groups attached to either of the carbon atoms of the double bond are the same cistrans isomers are not possible Cis isomers are the normal form of fatty acids but processing tends to convert some of the cis isomers to the trans isomers H Cl H C CC Cl Cl H Cl Cis isomer Trans isomer Cistrans isomers are also possible in cyclic systems The cis form has similar groups on the same side of the ring while the trans form has similar groups above and below the ring Chiral compounds A carbon atom with four different groups attached is chiral A chiral carbon rotates planepolarized light light whose waves are all in the same plane and has an enantiomer nonsuperimposable mirror image Rotation which may be either to the right dextrorotatory or to the left levorotatory leads to one optical isomer being d and the other being 1 Specific rotation represented 73 Chapter 1 Organic Chemistry II Here We Go Again Figure 12 Representa tions of a molecule with two chiral centers by oc39 where or observed rotation T temperature and D sodium D line is a measure of the ability of a compound to rotate light The specific rotation comes from the observed rotation oc divided by the product of the concen tration of the solution and the length of the container Other than optical activity the physical properties of enantiomers are the same A racemic mixture is a 5050 mixture of the enantiomers A meso compound is a compound with chiral centers and a plane of symmetry The plane of symmetry leads to the optical rotation of one chiral carbon cancelling the optical rotation of another Diastereomers are St I39 OiSOII1 I39S that 8I39 I1 t I18I1tiOII1 I39S RS notation is a means of designating the geometry around the chiral center This method requires the groups attached to the chiral center to be priori tized in order of decreasing atomic weight To assign the center place the lowest priority group the group with the lowest atomic weight on the far side and count the remaining groups as 1 2 and 3 Counting to the right is R and counting to the left is S Any similarity between d and I and R and S is coincidental Some important organic compounds have more than one chiral center Multiple chiral centers indicate the presence of multiple stereoisomers The maximum number of stereoisomers is 2 where n is the number of non identical chiral centers Figure 12 shows the four stereoisomers present in a molecule with two chiral centers Nonsuperimposable mirror images are enantiomers while the other species in the figure are diastereomers Unlike enantiomers diastereomers have different physical properties CH2OH CH2OH CH2OH CH2OH H c o HO c H H c OH HO c H HO 3 H H C3 OH H C3 OH HO C 3 H C C C C H o o H H o o H VEnantiomers EnantiomersV Diastereomers 74 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 13 The Fischer projection formulas Emil Fischer developed a method of drawing a compound to illustrate which stereoisomer is present Drawings of this type called Fischer projection formulas are very useful in biochemistry In a projection formula a chiral carbon is placed in the center of a pattern The vertical lines bonds point away from the viewer and the horizontal lines point towards the viewer Fischer used the D designation if the most important group was to the right of the carbon and the L designation if the most important group was to the left of the carbon See Figure 13 CH0 CH0 H C OH HO C H CHZOH CH2OH Dglyceraldehyde Lglyceraldehyde The use of D and L is gradually being replaced by the R and S system of designating isomers which is particularly useful when more than one chiral carbon atom is present Looking Ahead to Organic Chemistry One of the keys to Organic II is mechanisms the specific way in which a reaction proceeds Recall from Organic I that this involves pushing around electrons showing where they re going with curved arrows We give you a good review of these concepts in Chapter 2 along with some basic reaction moves In Chapter 3 we go into some depth about alcohols and ethers Like Organic 1 when we encounter a new functional group we examine the structure nomenclature properties synthesis and reactions In some courses and textbooks alcohols are covered in the first semester but for those readers who haven t gotten to them yet we include them in this book If you re already comfortable with that material please feel free to skip that chapter and go on to another Chapter 1 Organic Chemistry II Here We Go Again p Z Conjugated unsaturated systems are an important part of organic chemistry so in Chapter 4 we spend a little time talking about those systems setting the stage for our discussion of aromatic compounds that you can find in Chapter 6 To bring you up to speed on spectroscopy we cover the basics in Chapter 5 We give you the executive summary on infrared IR ultravioletvisible UVvis mass spectrometry mass spec and nuclear magnetic resonance NMR In addition many of the chapters in this book have a spectroscopy section at the end where we simply cover the essentials concerning the specific compounds that you study in that chapter Aromatic compounds and their reactions are a big part of any Organic 11 course We introduce you to the aromatic family including the heterocyclic branch in Chapter 6 You may want to brush up on the concept of resonance beforehand Then in Chapters 7 and 8 you find out more than you ever wanted to know about aromatic substitution reactions starring electrophiles and nucleophiles Another important part of Organic 11 is carbonyl chemistry We look at the basics of the carbonyls in Chapter 9 It s like a family reunion where I John one of your authors grew up in North Carolina everybody is related You meet aldehydes ketones carboxylic acids acyl chlorides esters amides and on and on It s a quick peek because later we go back and examine many of these in detail For example in Chapter 10 you study aldehydes and ketones along with some of the amines while in Chapter 11 we introduce you to other carbonyl compounds enols and enolates along with nitroalkanes and nitriles Carboxylic acids and their derivatives are also an important part of Organic 11 We spend quite a few pages looking at the structure nomenclature synthesis reactions and spectroscopy of carboxylic acids While on this topic in Chapter 12 we use a lot of acidbase chemistry most of which you were exposed to in your introductory chemistry course For a quick review look over a copy of Chemistry For Dummies or Chemistry Essentials For Dummies both written by John T Moore and published by Wiley Carbon compounds that also contain nitrogen such as the amines play a significant part of any Organic 11 course You encounter more acidbase chemistry with the amines along with some more reactions We hit this topic in Chapter 13 and give you some tips for multistep synthesis You probably haven t considered the fact that some organic compounds may contain a metal so we give you an opportunity to become familiar with the organometallics in Chapter 14 In this chapter you meet the Grignard reaction It s a very important organic reaction that you may have the opportunity to run in organic lab 76 Part I Brushing Up on Important Organic Chemistry I Concepts You just can t get away from those carbonyls so you get another taste of these reactions many of them named reactions in Chapter 15 You may be able to avoid biomolecules if your course doesn t cover them but if it does Chapter 16 is there for you Finally what s a good organic course without multistep and retrosynthesis along with roadmaps We hope that our tips can ease your pain at this point Roadmaps are the bane of most organic chemistry students but just hang in there There is life after organic chemistry and you may just find in the end that you actually enjoyed organic And for those of you who missed the chemical calculations there s always quantitative analysis and physical chemistry Chapter 2 Remembering How We Do It Mechanisms In This Chapter Analyzing arrows Breaking down basic moves Contemplating combining basic moves Mastering freeradical mechanisms Mechanisms are the key to organic chemistry Understanding the mechanism allows organic chemists to control the reaction and to avoid unwanted side reactions Understanding the mechanism many times allows chemists to increase the yield of product In this chapter you review the basics of mechanisms and their conventions and look at some of the more common ways that electrons shift during a reaction You also see how these individual steps can fit together in the overall reaction mechanism and apply some of these techniques in free radical mechanisms Duck Here Come the Arrows Many types of arrows are used in organic chemistry and each of them conveys information about the particular reaction These arrows include the resonance arrow equilibrium arrow reaction arrow doubleheaded arrow and single headed arrow The resonance arrow a single line with arrow heads at both ends see Figure 21 separates different resonance structures The actual structure is a weighted average of all resonance forms More resonance forms usually indicate a more stable structure One or more of the resonance structures may be useful in predicting what will happen during a reaction a mechanism 78 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 21 The resonance arrow Figure 22 The equilib rium arrows Figure 23 The reaction arrow The equilibrium arrow which has two lines pointing in opposite directions see Figure 22 separates materials that are in equilibrium Materials on each side of the arrow are present Unlike a resonance arrow the materials actually exist and aren t a hybrid If one of the arrows is longer than the other is it indicates that one side of the equilibrium predominates over the other A reaction arrow a single arrow pointing in one direction see Figure 23 simply separates the reactants from the products The use of this arrow usu ally indicates that the reaction proceeds in only one direction unlike the equilibrium arrow Double and singleheaded curved arrows indicate the movement of electrons Doubleheaded curved arrows shown in Figure 24a show the movement of two electrons whereas singleheaded curved arrows Figure 24b indicate the movement of one electron The electrons always move in the direction indicated by the arrow The head point of the arrow is where the electron is going and the tail is the electron s source 79 Chapter 2 Remembering How We Do It Mechanisms Figure 24 Curved arrows indicating the move ment of electrons Coming Around to Curted Arrows NBER V 4 quot Mechanisms like resonance structures utilize curved arrows Resonance structures are ways of illustrating the various resonance forms that contribute to the resonance hybrid If you need more review refer to Organic Chemistry I For Dummies Many of the same rules apply to both however there are some important differences 1 In resonance the electrons don t actually move whereas in mechanisms there is an actual movement of electrons 1 In resonance you should never ever break a single bond however many mechanisms involve the breaking of a single bond Nonetheless you should never ever exceed an octet of electrons for any atom in the second period A mechanism provides a means toward understanding why a reaction occurred When you understand why a reaction occurred you re much closer to understanding organic chemistry Reactions involve the breaking and the forming of bonds The mechanism shows how the electrons move flow to break old bonds and to form new bonds Curved arrows indicate the flow of the electrons from the nucleophile electron donor to the electrophile electron acceptor To be successful in organic chemistry you must know the mechanism for the reaction you re studying 20 Part I Brushing Up on Important Organic Chemistry I Concepts Mechanisms in this book are in general advanced examples of mechanisms appearing in Organic Chemistry I For Dummies A college organic chemistry course presents very few completely new mechanisms Perfecting a few mechanisms goes a long way toward understanding all reaction mechanisms and therefore all organic reactions Although many students feel that memorization is important understanding the mechanism is what s neces sary to comprehend organic chemistry If you simply memorize mechanisms you ll become hopelessly confused by even minor changes however if you understand a mechanism thoroughly you can accommodate any changes Keep two things in mind when drawing curved arrows The tail of the arrow needs to be in the right place and the head of the arrow needs to be in the right place Simple right Don t forget that electrons occupy orbitals Other than radicals the electrons in the orbitals are either bonding pairs or lone pairs This means that the tail of the curved arrow must be at a lone pair or a bonding pair A radical may have the tail of the curved half arrow originating at the unpaired electron The head of the curved arrow indicates where a lone pair is going or where a bond will form Ge ttinq Ready for Some Basic M01es x The tail of a curved arrow has two possible positions and the head of a curved arrow has two possible positions This means that in theory four combinations are possible These combinations are 1 Bond A lone pair 1 Bond A bond 1 Lone pair A bond 1 Lone pair A lone pair The last combination doesn t work at least not in a single step because it tends to force an atom to exceed an octet This leaves only three important types The basic idea behind these reactions is the same An electronrich atom with a lone pair a nucleophile donates that lone pair to an electronpoor atom an electrophile Figure 25 Bond to Ione pair movement Figure 26 Bond tobond movement Figure 27 Two electron pair movements 27 Chapter 2 Remembering How We Do It Mechanisms Bond one pair An example of the bond to lone pair combination is shown in Figure 25 In this example the tail of the curved arrow begins at the bonding pair The head of the curved arrow is at the chlorine atom where it forms a lone pair The overall charge doesn t change The original compound was neutral 0 charge the products are 1 1 0 CHQCHQCH3 CHZCHZCH3 N cH3 c gI gt CH3C 99 CHQCH3 CHZCH3 Bond bond An example of a bondtobond step is shown in Figure 26 The tail of the curved arrow begins at one of the bonding pairs of the double bond the TCbOI1d while the head points to where the new TCbOI1d will form lt9V gt A more common example of this process involves two arrows and the shifting of two electron pairs An example of this process is shown in Figure 27 The tail of the curved arrow again begins at one of the bonding pairs of the TCbOI1d while the head points to where the new bond will form This movement forces the bonding pair between the hydrogen and oxygen to move to the oxygen atom to create a lone pair on the oxygen atom H 39lt5 H H 22 Figure 28 Lone pair to bond movement Part I Brushing Up on Important Organic Chemistry I Concepts When more than one curved arrow is present they should all point in the same general direction and never toward each other or away from each other However curved singleheaded arrows do not necessarily follow this rule Lone pair bond An example of the lone pair to bond step is shown in Figure 28 In this step the tail of the curved arrow begins at the lone pair The head of the curved arrow is going to form the CN bond Notice that there s conservation of the positive charge In any mechanism the overall charge must remain the same G KM NH3 G9 Combining the Basic Moves quot Figure 29 Conversion of tbutyl alcohol to tbutyl chloride A common error in a mechanism is to attempt to do too much in a single mechanism step and a sure sign that you re trying to do too much is having arrows pointing in opposite directions in the step You can have arrows pointing in one direction in a step and in the opposite direction in the next step but resist the temptation to combine these two steps The best way to see how these steps work together is with an example Begin by examining the conversion of tbutyl alcohol to tbutyl chloride This process has a 96 percent yield This is a good thing The overall reaction is shown in Figure 29 CH3 CH3 HCI CH3C OH T CH3C C H20 conc CH3 CH3 Chapter 2 Remembering How We Do It Mechanisms quot Figure 210 Step 1 Lone pair to bond movement Figure 211 Step 2 Bond to Ione pair movement 23 Step 1 This reaction like most reactions involving an acid begins with protonation In this case a lone pair from the oxygen forms a bond to the hydrogen from the hydrochloric acid See Figure 210 The movement of the oxygen lone pair to the hydrogen pushes the bonding pair from the HCl bond onto the chlorine This is a lone pair to bond transfer which induces a bond to lone pair transfer Protonation is almost always the first step in mechanisms involving an acid CH3 CH3 I r H 039 CH3C QH CH3C OH or CH3 CH3 H Step 2 The presence of a positive charge on the oxygen atom is unstable because the oxygen has such a high electronegativity The bonding pair from the C0 bond moves to the positive charge on the oxygen to become a lone pair See Figure 211 In this case this is the ratecontrolling step which is why this is an example of an SN1 mechanism The water molecule formed is a good leaving group which facilitates this reaction The OH group is not a good leaving group This is a bond to lone pair transfer CH3 CH3 l quot93 Slow CH3C OH Rate determining CH3C H20 CH3 H CH3 Step 3 Forming a carbocation is difficult however tertiary carbocations such as this one can form as intermediates or species that exist for a short time during the reaction See Figure 212 The positive charge on the carbon makes this a strong electrophile that seeks a lone pair In the final step of this mechanism the carbocation accepts a lone pair from the chloride ion generated in the first step The transfer is lone pair to bond 24 Figure 212 Step 3 Lone pair to bond movement Figure 213 Nucleophilic attack of a double bond Part I Brushing Up on Important Organic Chemistry I Concepts CH3 CH3 l e CH3 C 2939 CH3 CH3 Each step includes a conservation of charge Conservation of charge is an important part of all mechanisms Intermediates In the preceding mechanism the carbocation was an intermediate a species that exists for a short time during the reaction The form of the intermediate is often essential to understanding the mechanism The curved arrows help you in drawing the intermediate Because you can use curved arrows in only three ways bond to lone pair bond to bond and lone pair to bond you have limited options for drawing intermediates In the next example a nucleophile attacks a double bond See Figure 213 In this case the nucleophile is the hydroxide ion The process begins with the hydroxide ion attacking the carbon atom at one end of the carboncarbon bond This is a lone pair to bond step Next a pair from the TCbOI1d shifts to form another TCbOI1d on the other side of the carbon atom This is a bondto bond transfer Finally a bond to lone pair transfer takes place You need to be Very careful to keep the formal charges correct It may help to remember that charge will be conserved Don t forget The nucleophile is at the tail of the arrow and the electrophile is at the head of the arrow 25 Chapter 2 Remembering How We Do It Mechanisms Some materials may behave as either a base or a nucleophile The hydroxide ion is an example When the nucleophile attacks and removes a hydrogen ion it is behaving as a base When the nucleophile is attacking at some other point than a hydrogen atom it is acting like a nucleophile For example both the methoxide ion CH3O and the tbutoxide ion CH33O are strong bases but only the methoxide ion is a strong nucleophile The tbutoxide ion is too big and bulky to attack efficiently The effect of the bulky nature of the tbutoxide ion on its reactivity is an example of steric hindrance which was discussed in your Organic 1 course and naturally in Organic 1 For Dummies A molecule with a lone pair of electrons to donate can behave as a nucleophile The strength of the nucleophile the nucleophilicity is often related to basicity A strong nucleophile is usually a strong base and vice versa But nucleophilicity and basicity aren t the same Basicity refers to the ability of a molecule to accept bond with an H The base strength is shown by its equilibrium constant On the other hand nucleophilicity refers to the ability of a lone pair of electrons to attack a carbon on an electrophile When working with nucleophiles keep a few additional points in mind 1 Nucleophiles that are negatively charged are stronger nucleophiles than neutral ones 1 Generally nucleophilicity increases as you go down the periodic table 1 Nucleophilicity is decreased by steric hindrance Keys to substitution and elimination mechanisms Four types of mechanisms are inherent to Organic Chemistry I These are substitution reaction mechanisms SN1 and SN2 and elimination reaction mechanisms E1 and E2 The principles of these four types apply to Organic Chemistry II and no review would be complete without a few reminders about these processes The SN refers to a nucleophilic substitution process where some nucleophile attacks an electrophile and substitutes for some part of the electrophile The E refers to an elimination process where the nucleophile attacks an electrophile and causes the elimination of something The 1 and 2 refer to the order of the reaction A 1 first order means only one molecule determines the rate of the reaction whereas a 2 second order means that a combination of two molecules determines the rate of the reaction In many cases two or more of these mechanisms are competing and more than one product may result 26 Part I Brushing Up on Important Organic Chemistry I Concepts Increasing the strength of the nucleophile increases the likelihood of a substitution occurring instead of elimination Increasing substitution on the electrophile tends to increase the likelihood of a firstorder process over a secondorder process This means a tertiary 3 carbon is more reactive than a secondary 2 carbon atom which in turn is more reactive than a primary 1 carbon atom All nucleophilic substitution reactions require a good leaving group Ions like OH RO alkoxide and NH2 are terrible leaving groups and don t normally form The more likely leaving groups in these cases are H20 ROH and NH3 respectively The following four lists summarize the main features of each of these mechanisms Remember the following when working with an SN1 mechanism Me methyl V Reactivity increases in the order Me lt 1 ltlt 2 lt 3 V A racemic mixture results if the attack is on a stereogenic center V It s a twostep process V It requires a good nucleophile V It relies on a carbocation intermediate V The rate depends entirely on the concentration of the electrophile V The intermediate carbocation can undergo rearrangement V Polar solvents especially those with hydrogen bonding promote this type of reaction Good SN1 substrates make stable carbocation intermediates Also solvents that can supply an H protic solvents will stabilize carbocations in SN1 reactions The following are the main features of an SN2 mechanism V Reactivity increases in the order 3 ltlt 2 lt 1 lt Me V Inversion of configuration results if attack is at the stereogenic center V It s a onestep process V It requires a strong nucleophile V The rate depends on the concentration of both the nucleophile and the electrophile V Backside attack occurs V Rearrangement is not possible V Polar aprotic solvents promote this type of reaction 27 Chapter 2 Remembering How We Do It Mechanisms SN2 reactions lead to an inversion of stereochemistry Nucleophilicity is decreased by protic solvents in SN2 reactions The presence of a polar aprotic solvent is a clue that the mechanism is SN2 The main features of an E1 mechanism are as follows V Reactivity increases in the order Me lt 1 ltlt 2 lt 3 V The major product is the most substituted alkene V It s a twostep process V It requires a weak base V The rate depends entirely on the concentration of the electrophile V It uses a carbocation intermediate V Polar hydrogen bonding solvents promote this type of reaction V lt s promoted by high temperatures The following list describes the main features of an E2 mechanism V Reactivity increases in the order Me lt 1 lt 2 lt 3 V The major product is the most substituted alkene V It s a onestep process V The mechanism requires a strong base V The rate depends on the concentration of both the base and the electrophile V The intermediate is periplanar V lt s promoted by high temperatures Revisiting Free RadicaI Mechanisms Freeradical mechanisms obviously involve free radicals A free radical is a species with an unpaired electron In these mechanisms singleheaded curved arrows are the norm In Organic Chemistry 1 these free radicals first appear when examining the chlorination of an alkane such as methane The process begins with an initiation step as shown in Figure 214 All initiation steps increase the number of free radicals The initiation step is homolytic bond cleavage where each of the chlorine atoms receives one of the two electrons originally present in the bond and two chlorine free radicals form The chlorine free radicals like all free radicals are very reactive 28 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 214 Initiation step of a freeradical mechanism Figure 215 Freeradical attack of an alkane 39f cI cI C3 C3 A chlorine free radical attacks an alkane molecule like methane to form hydrogen chloride and a methyl radical see Figure 215 CIH I H I o I I c3 I i The methyl radical once formed is also a very reactive species which attacks other species to continue the reaction through a series of propagation steps All propagation steps maintain the number of free radicals Finally the reaction of a free radical with another free radical gives rise to a termination step All termination steps decrease the number of free radicals In a freeradical mechanism the reaction of a free radical with a molecule results in a free radical and a free radical reacts with a free radical to produce a molecule that isn t a free radical There are two important considerations concerning freeradical mechanisms One of these factors is the identity of the halogen and the other is the stability of the alkyl free radical The more substituted the carbon atom is the more stable the free radical The stability of the alkyl free radicals increase in the following order Me lt 1 lt 2 lt 3 The relative stabilities rarely lead to rearrangement of the free radical The formation of a morestable free radical increases the selectivity of the reaction For this reason the replacement of a particular hydrogen atom by a halogen isn t simply a matter of probability In propane replacement of one of the hydrogen atoms on the central carbon should occur onefourth 28 of the time You may want to draw this reaction to see why this is true However chlorination shows a distribution where replacement occurs at the second carbon about threefourths of the time and for bromination the replacement is almost exclusively on the central carbon atom Table 21 indicates the relative selectivity of chlorine and bromine Chapter 2 Remembering How We Do It Mechanisms Table 21 Selectivity of Chlorine and Bromine in FreeRadical Halogenation of Alkanes 1 2 3 RCH3 RZCHZ R3 H Chlorine CIZ l 39 53 Bromine Brz 1 820 1640 Living with mechanisms Mechanisms are very important in the under standing of organic reactions Many mecha nisms are presented in any organic chemistry course and beginning students can get into trouble in a number of places If you keep the following items in mind while studying and working mechanisms life will be easier 1 You probably won39t use all the reactants in each step of the mechanism 1 Be careful not to mistake a multistep syn thesis problem with a mechanism Both involve a number of steps but curved arrows only appear in mechanisms 1 The solvent isn39t a reactant It may promote a particular type of mechanism however that doesn39t make it a reactant 1 Materials may be added to prevent a buildup of undesired products For example a base may be present to trap released acid These compounds aren39t part of the mechanism 1 Draw out all the atoms in the vicinity of the reaction center especially if there are charges or lone pairs Don39t try to do too much in one step In many cases you should draw the pos sible resonance structures especially for intermediates Not everything will have a resonance structure 1 Keep your goal in mind It39s easy to go off on atangent 1 Don39t attempt to overanalyze the process Pick the best reaction from the ones you studied to get from pointAto point B 1 Ions such as sodium Naf potassium Kf and lithium Lif are usually spectator ions and therefore aren39t part of the reaction mechanism 1 Before going to the next step make sure the step you justfinished is reasonable For exam ple are the electrons moving in the same direction Are the intermediates reason ably stable Do you have like charges close together Is there conservation of charge 1 Acidic conditions may yield cationic or neu tral products Basic conditions may yield anionic or neutral products 1 Watch out for those pesky fivebonded carbon atoms 1 When you finish the mechanism go over each step and check your assumptions Make sure none of your intermediates are unstable 30 Part I Brushing Up on Important Organic Chemistry I Concepts Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping In This Chapter Checking out alcohols How they look and what they re called Taking a look at the synthesis and reactions of alcohols Tackling the basics of ethers Finding out what ethers do Clarifying the spectra of alcohols and ethers Tvo types of organic compounds contain singlebonded oxygen atoms the alcohols and the ethers In the alcohols an oxygen atom is between a carbon atom and a hydrogen atom whereas in an ether an oxygen atom is between two carbon atoms The generic representation of an alcohol is ROH and the generic representation of an ether is ROR39 If the carbon atom attached to the OH group is part of an aromatic ring the compound is a phenol which unlike the alcohols is an organic acid In this chapter you investigate the properties synthesis and reactions of alcohols and ethers So drink up and let s go Getting Acquainted with Alcohols The alcohols are an important group of organic compounds Even though an OH group is present these are not basic compounds but are neutral to weakly acidic materials The OH hydroxyl group is the source of hydrogen bonding which increases the melting and boiling points of the alcohols relative to those of comparably sized alkanes Hydrogen bonding also makes alcohols more soluble in water than less polar materials 32 Figure 31 Primary secondary and tertiary alcohols Part I Brushing Up on Important Organic Chemistry I Concepts Structure anol nomenclature of alcohols In this section we take a look at how you can classify alcohols and the nomen clature of alcohols no just calling them bourbon gin and Scotch won t work Classifying alcohols There are three general categories of alcohols 1 Primary 1 1 Secondary 2 1 Tertiary 3 The categories depend upon the number of carbon atom attached to the alcohol carbon atom Figure 31 illustrates the three types of alcohols Note the number of carbon atoms attached to the boldfaced carbon atoms H H CH3 CH3CH2CH2C OH CH3CH2C CH3 CH3C CH3 H OH OH 1 2 3 To distinguish between the alcohols in the figure simply count the carbon atoms attached to the carbon atom in boldface Primary alcohols have one carbon atom attached to the central carbon secondary alcohols have two and tertiary alcohols have three Naming alcohols The nomenclature of the alcohols is an extension of the rules for the naming of other organic compounds The general changes in the rules for alkanes are 1 The parent chain contains the OH 1 The carbon with the OH gets the smaller number 1 Drop the final e and add 01 The common names of the alcohols consist of the name of the alkyl group and the word alcohol For example CH3OH is methyl alcohol and CH3CH2OH is ethyl alcohol Some examples of naming alcohols are shown in Figure 32 Figure 32 Examples of naming alcohols Figure 33 Two common alkoxides 33 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping Br 4 2 1 CH3CH CH3 CH3CH CH CH3 3 OH OH 2propanol 3bromo2butanol 5 4 3 2 1 0 CH2CH CH2CH CH3 OH 4penten2ol Cycloheptanol Under extreme conditions alcohols may behave as acids and lose an H to leave an anion with the general formula RO These are alkoxides Alkoxides are important in organic synthesis because they are very strong bases and may be good nucleophiles Figure 33 illustrates two common alkoxides CH3 6 CH33 K CH3 Potassium tertbutoxide CH35 N 6 Sodium methoxide Physical properties of alcohols The important physical properties of organic compounds include melting and boiling points and solubility and density Whenever you compare physi cal properties of different compounds stick to compounds with similar molecular weights Melting and boiling points The presence of hydrogen bonding causes alcohols to have significantly higher melting and boiling points than alkanes Figure 34 illustrates the for mation of hydrogen bonds between alcohol molecules Part I Brushing Up on Important Organic Chemistry I Concepts Figure 34 Hydrogen bonding between alcohol molecules Figure 35 Protonation of an alcohol by su u c acid Impurities increase the boiling point and reduce the freezing points of materials Solubility and olensity Alcohols with three or fewer carbons atoms are miscible in water in all proportions The solubility of alcohols in water decreases with increasing number of carbon atoms so that alcohols with more than six carbon atoms are nearly insoluble As the number of carbon atoms increases the solubility in nonpolar solvents increases Alcohols like most organic compounds containing oxygen or nitrogen are soluble in concentrated sulfuric acid because the acid protonates the oxygen atom as illustrated in Figure 35 H 03 H Hso Alcohols tend to be denser than hydrocarbons with comparable carbon content Makiny moonshine Synthesis of alcohols There are a number of methods for synthesizing alcohols Some of the meth ods are suitable only for the preparation of small quantities of alcohol while other methods are industrially important for the synthesis of thousands of gallons of alcohol Figure 36 Catalytic hydration of an alkene to produce an alcohol Figure 37 Mechanism of the Markovnikov addition of waterto an alkene to yield an alcohol 35 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping Hydration of alkenes The hydration of alkenes is one important method of synthesizing alcohols Industrially sulfuric acid is used as a catalyst while smallscale preparations often utilize toxic mercury compounds Definitely not the kind of stuff you want to drink Catalytic hydration of alkenes The general reaction for the catalytic hydration of an alkene to produce an alcohol is shown in Figure 36 and the mechanism is in Figure 37 This pro cess is an example of a Markovnikov addition as seen in Organic Chemistry I H OSO3H H gt H2304 H20 h H CH3 CHCH2 CH33CH3 H 5 H 2 H H CH H Cl 4 CH3 C3CH3 CH3 C3CH3 H H Oxymercurationdemercuration reactions with alkenes Oxymercurationdemercuration is a useful laboratory method for the syn thesis of small quantities of alcohol Like the catalytic hydration reaction this process is an example of Markovnikov addition It s a useful procedure because it tends to result in high yields and rearrangements rarely occur 36 Part I Brushing Up on Important Organic Chemistry I Concepts The general reaction is shown in Figure 38 and you can see a specific exam ple in Figure 39 I I Figure38 THI O 0Xymer Q HgOAc2aq T curation OH HQOAC demer curation of an alkene to yield an O alcohol l O Figure 39 Synthesis H of 2pro H 1 HgOAc2aqTHF panobv I CH3 C CH3 oxymercura CH3 CCH2 2 NaBH4 tiondemer OH curation The numbering of the steps shown in Figure 39 is essential The numbers indi cate that Step 1 must take place before and be separate from Step 2 Hydroborationoxidation reactions with alkenes Hydroborationoxidation is a useful method when the desired product is the antiMarkovnikov alcohol The boranecontaining reactant is normally dibo rane BZH6 or the borane tetrahydrofuran complex BH3THF No matter what hydroboration agent is used it s usually simplified to BH3 in the mecha nism BH3 is a useful reactant because it s a good Lewis acid An example illustrating the general mechanism is shown in Figure 310 The attack in Figure 310 is on the terminal carbon for two reasons 1 Steric more important the outer carbon is more accessible 1 Electronic the 2 carbocation is more stable Figure 310 Synthesis of 1pro panol from propene by hydro boration oxidation Figure 311 Reaction of methycy clopentene by hydro boration oxidation 37 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping H 6 I CH3CH gH2 gt CH3 CIIH2 Forming T 3 H L B H H H l H H T CH3 T 3H2 Breaking H 3 H H H202OH ROH CH3 CH2CH2 OH The transition state is fourcentered which means there is syn addition both groups add to the same face of the double bond Anti addition is where the two groups add to opposite faces of a double bond When doing a reaction you can usually call it quits at this point In reality the process continues until the replacement of all BH with BR to give BR3 as long as the alkene is small A question that may appear on some Organic Chemistry exams is What will be the product of the reaction sequence given in Figure 311 H CH3 OH H H 1 BH3THF m OR 2 H202OH H H CH3 OH CH3 trans cis 38 Figure 312 Production of 12pro panediol from propene Part I Brushing Up on Important Organic Chemistry I Concepts The BH3 undergoes syn addition so the trans isomer will form However the cis isomer can be produced by heating the mixture to 160 degrees Celsius Heating to 160 degrees initiates a series of eliminations and readditions of the boron until the boron ends up on the least sterically hindered carbon Diols from reactions with alkenes The reaction of either cold dilute aqueous potassium permanganate KMnO4 or osmium tetroxide OsO4 to an alkene will form through syn addition a cisglycol Using osmium tetroxide requires a second step using hydrogen peroxide H202 to regenerate the expensive reagent Figure 312 gives an example of this process The reaction of the alkene with a peroxyacid can lead to the transglycol anti addition KMnO4aq OH OH CH3CHCH2 Cold dil CH3CH CH2 Preparation of alcohols by the reoluction of carhonyls The reduction of carboxylic acids or esters requires very powerful reduc ing agents such as lithium aluminum hydride LiAlH4 or sodium Na metal Aldehydes and ketones are easier to reduce so they can use sodium borohy dride NaBH4 Examples of these reductions are shown in Figure 313 Grianarol reagents and production of alcohols A Grignard reagent is an organometallic compound containing magnesium Grignard reagents are very useful in organic synthesis In most cases an organolithium compound can be substituted for a Grignard reagent Figure 313 Examples of the prepa ration of alcohols bythe reduction of carbonyls Figure 314 Mechanism forthe Grignard reac on Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping CH2OH A56 OH O H 1 LiAH4Et2O C OH 2 HH20 1 NaBH4EtOH o 2 HH20 o CH3CH24C mquot39aEtOH CH3CH2gt4CH2 0H OEt The synthesis often begins with the preparation of the Grignard reagent by reacting an alkyl halide with magnesium metal in dry ether Then the Grignard reagent is used to attack a carbonyl group to form a magnesium salt The mechanism is shown in Figure 314 The addition of water causes hydrolysis of the salt to produce an alcohol some examples of which are shown in Figure 315 R R6Mg6X C6 I I CO 5 GRIgx 39 40 Part I Brushing Up on Important Organic Chemistry I Concepts R6Mg6X l 2 H drol sis C H CH3 CH2OH H H Formaldehyde A 1 alcohol Any other aldehyde 0 ll 0 2 Hydrolysis c A R CH CH3 R H A 2 alcohol Any ketone O OH Figure 315 l Synthesis of C 2 Hydrmysis RCCH3 alcohols by the Grignard R R R reacnon T A 3 alcohol What will they do besides hum Reactions of alcohols Alcohols represent a Very Versatile class of compounds for organic synthesis so a Variety of synthetic pathways utilize alcohols Some examples appear in Figure 316 The conversion of alcohols to ethers appears later in this chapter Alkenes H20 O RX ROH 1 5 Aldehydes ketones T Carboxylic acids F 3 16 Alkyl IQUFB 1 halides Summary RCO2H of the reactions of alcohols R0 T Nucleophile Esters Figure 317 Acid catalyzed dehydration of 2pro panol to propene Figure 318 Mechanism of the acid catalyzed dehydration of 2pro panol to propene 47 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping Dehydration of alcohols H20 Alcohols can be dehydrated by a number of procedures One method is to use a dehydrating agent such as phosphorus oxychloride POCI3 and another method is acidcatalyzed dehydration using sulfuric acid IIZSO4 or phosphoric acid H3PO An example of acidcatalyzed dehydration is in Figure 317 and the mechanism is shown in Figure 318 CH3CH CH3 H2804 A CH3CH CH2 Q T CH33CH3 CH3CHCH2 IOH quot Fast H T T Slow CH3C CH3 T CH38 C H H20 Cl Rate determining l 09 H H quot H Tertiary alcohols dehydrate most readily primary alcohols least readily and unsurprisingly secondary alcohols are intermediate This relates to the rela tive stability of the intermediate carbocation The temperature and concen tration of the acid depends upon the type of alcohol A primary alcohol such as ethanol requires concentrated acid and a Very high temperature 180 degrees Celsius while a tertiary alcohol such as tbutyl alcohol requires 20 percent sulfuric acid at 85 to 90 degrees Celsius The process follows an E1 mechanism and produces the thermodynamically more stable product 42 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 319 Oxidation of a primary alcohol to an aldehyde andthento an carbox ylic acid Figure 320 Example of the oxida tion of a primary alcohol to an aldehyde Oxidation of alcohols Both primary and secondary alcohols can be oxidized but tertiary alcohols won t undergo simple oxidation Oxidation of a primary alcohol gives an alde hyde however preventing further oxidation of the aldehyde to a carboxylic acid is difficult Secondary alcohols oxidize to a ketone without the problem of additional oxidation occurring The symbol 0 often appears in organic reactions to indicate oxidation and the symbol H indicates reduction Most organic compounds including alcohols undergo oxidation in the form of a combustion reaction The general reaction for the oxidation of a primary alcohol is in Figure 319 Oxidizing agents that aren t strong enough to oxidize the aldehyde formed in the first step include PCC pyridinium chlorochro mate Sarett s reagent CrO32py and acidic aqueous potassium dichromate K2Cr2O7 The reaction with potassium dichromate requires the distillation of the aldehyde from the mixture before further oxidation can occur Two exam ples are in Figure 320 RCH2OH j RCHO RCOOH H H H C C C O 2 Cr03392PY CH CH c OH CH2C392 H 25 Figure 321 Oxidation of a primary alcohol to a carboxyhc acid Figure 322 Oxidation of a secondary alcohol to a ketone Figure 323 Acid catalyzed reaction of an alcohol with a car boxylic acid to form an ester 43 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping Stronger oxidizing agents such as hot basic potassium permanganate or Jones reagent acidic CrO3 oxidize a primary alcohol to a carboxylic acid as illustrated in Figure 321 The characteristic dark purple color of the perman ganate ion changes to brown indicating the formation of MnO2 CH3 0 CH3CH C CH3 1 KMnO4OH A A CH3CH CH2OH 2 H OH The oxidation of a secondary alcohol to a ketone can also occur through the use of a Variety of oxidizing agents Figure 322 illustrates the oxidation of a secondary alcohol using Jones reagent CH3 OH CF03H2SO4Clll CH3 0 Acetone CH3CH CH CH3 1520o CH3CH C CH3 Conversion of alcohols to esters In the presence of an acid catalyst alcohols react with carboxylic acids to form esters The general reaction is in Figure 323 and you can find a discus sion of the reaction in Chapter 12 R39OH RjcO H R10O OH OR39 44 Part I Brushing Up on Important Organic Chemistry I Concepts Reaction of alcohols as acids In general alcohols are neutral compounds however in the presence of extremely strong bases or an active metal it s possible to force the loss of a hydrogen ion to leave an alkoxide RO ion The general reaction is shown in Figure 324 The amide ion NH2 is a strong enough base to force the loss as is the hydride ion H however the hydroxide ion OH isn t strong enough Poor little guy Active metals include sodium and potassium Specific examples are in Figure 325 The alkoxide ions are useful in organic synthesis because they re good nucleophiles Figure 324 5 St b 6 FZL if d a cc H9 ion from an active metal Akoxide ion alcohol Figure 325 CH3 CH3 Examples of the forma K Hoc cH3 gt 69 9 c cH3 12 H2 tion of an alkoxide CH3 CH3 ion from an alcohol NaH EtOH NaQEt H2 Conversion of alcohols to alkyl halides Sometimes the alcohol group can be replaced with a halogen A number of reagents cause this conversion including the hydrohalic acids llCl HBr and HI sodium bromide with sulfuric acid NaBrHZSO4 hydrochloric acid with zinc chloride llClZnCl2 phosphorus tribromide PBr3 in base or reflux ing with thionyl chloride SOCl2 The acidic reactions utilizing the general mechanism appear in Figure 326 Methanol and most primary alcohols follow an SN2 mechanism while most other alcohols react by S 1 The SN1 reactions lBEIz may involve rearrangement such as a hydride shift N Don t forget that SN2 mechanisms involve an inversion of configuration Figure 326 General steps forthe conversion of an alco hol to an alkyl halide Figure 327 Examples of the forma tion of an alkyl halide from an alcohol 45 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping R Ht FiH2 RH2 3 W H20 R X gt RX Figure 327 illustrates two examples using HCIZnCl2 In the second example a rearrangement takes place OH CI HCI 3 ZnC2 CH3 CH3 HCI CH3C CH CH3 Zncl CH3C CH2CH3 2 H OH Cl A common analytic test used to classify alcohols is the Lucas test This test is for alcohols with six or fewer carbon atoms and therefore soluble in reagent mixture They react with HCIZnCl2 to form an alkyl chloride which is insol uble in the reaction mixture causing the solution to turn cloudy The time required for the reaction to occur indicates the type of alcohol Table 31 summarizes the time required for each type of alcohol Table 31 Lucas Test for Classifying Alcohols Type of Alcohol Time Required to Form an Alkyl Chloride Tertiary 3 Immediately Secondary 2 Five minutes Primary l 24 hours while heating 46 Part I Brushing Up on Important Organic Chemistry I Concepts Introoluciny Ether Not the Ether Bunny Figure 328 Examples of naming ethers Ethers are compounds closely related to the alcohols You probably think of anesthetics when you think of ethers but these compounds are a lot more Versatile than that In this section we look at the structures properties and reactions of ethers Structure and nomenclature of ethers The general formula of an ether is ROR where R and R may be any alkyl or aryl group You can name an ether simply by identifying the R groups and adding the word ether or by naming one R group and the oxygen atom as an alkyoxy group attached to the other R group Examples of both methods are in Figure 328 A cyclic ether containing two carbon atoms and an oxygen atom is an epoxide O Diphenyl ether or phenyl ether 0 CH2CH3 3ethoxycyclohexene Sleepy time Physical properties of ethers In this section we examine some of the physical properties of ethers Meltiny and boiliny points The ethers are only slightly polar therefore their melting and boiling points are only slightly above those of the corresponding alkanes The correspond ing alcohols have significantly higher melting and boiling points Solubility Ethers are much less soluble in water than the alcohols are The solubility is normally less than 8 grams g per 100 milliliters mL The addition of Various Figure 329 Preparation of symmet ric ethers from an alcohol Figure 330 Williamson synthesis of an ether from an alkoxide ion and an alkyl halide Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping 47 inorganic salts to the solution further decreases the solubility of the ether This process is called salting out Ethers are soluble in concentrated sulfuric acid Ethers are good for extracting organic compounds from aqueous solutions because the density of ethers is less than water causing the ether layer to float on top of the water layer Synthesis of ethers Symmetric ethers R R can be prepared by the acidcatalyzed dehydration of primary alcohols However this reaction competes with the acidcatalyzed dehydration of the alcohol to form an alkene Lower temperatures favor ether formation over alkene formation Secondary and tertiary alcohols favor alkene formation The general reaction is shown in Figure 329 H 2ROH 2804 R O R A ROH HR ROH2 H The Williamson ether synthesis is a general method for producing both sym metric and asymmetric R t R ethers This is an SN2 process following the general procedure in Figure 330 The process involves the reaction of an alk oxide ion with an alkyl halide RO39Na RX RoR39 OI OF ArO39Na 39039 The synthesis of cyclic ethers especially epoxides provides important reagents for organic synthesis Industrially ethylene oxide is the most important ether because it is used in the synthesis of many other organic compounds This com pound forms by the catalytic oxidation of ethylene as seen in Figure 331 48 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 331 Production 0 of ethylene CH2CH2 oxide by the 225 C 2 CH2 catalytic oxidation of ethylene Ethwene Ethylene oxide Oxirane An epoxide can also be formed by peroxidation of an alkene This reaction typically employs MCPBA metachloroperoxybenzoic acid now you see why we abbreviated it however CF3COO2H also works Figure 332 illus trates one example and Figure 333 presents a generic mechanism 00 OOH Figure 332 Forming an epoxide by CH3 MCPBA 0 CH3 peroxida C tion of an CH C a39kequote39 CH3 CH3 Ether or CHCI3 CH3 CH3 ll T R iC R j C Figure 333 0 Mechanism 0 forthe form 0 T ing of an v H O H epoxide by peroxida tion of an alkene NG Figure 334 Cleavage of an ether by a hydrohalic acid Figure 335 Mechanism forthe cleavage of an ether by a hydrohalic acid 49 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping Reactions of ethers Ethers are relatively unreactive compounds however they will react under certain circumstances In this section we describe situations where reactions occur In addition to reacting under certain conditions ethers will slowly air oxidize to produce explosive peroxides Acidic cleavage of ethers Hydrohalic acids can cause cleavage of the ether Hydrofluoric acid HF doesn t work as well as the other acids in the group IC1 HBr or HI Secondary and tertiary ethers undergo SN1 reaction while methyl and pri mary ethers undergo SN2 reaction A generic example is in Figure 334 and the mechanism is in Figure 335 ROR HI ROHR HI RH2O A T 39o CH2 CH3 HX39 o 5 OH XCHZCH3 50 Part I Brushing Up on Important Organic Chemistry I Concepts Sulfuric acid and ethers Cold concentrated sulfuric acid will react with ethers to give a soluble prod uct An example of this process is in Figure 336 Figure 336 H Reaction of H 30 ethers with E 0 E F EtO E H804 cold con ef at 3d Diethyloxonium hydrogen sulfate sulfuric acid Reactions of epoxides Epoxides are more reactive than other ethers This is due to ring strain inher ent in any threeatom ring system The most useful reactions are acidic cleav age and nucleophilic cleavage Acidic cleavage is an SN2 mechanism with a pseudocarbocation ion The reaction produces the trans product anti addition Figure 337 shows a typical reaction and Figure 338 illustrates the mechanism Figure 337 H Acid cleav CH3OH age of an epoxide 39 0 T H CH3 H T L Q q i 3 CH CH B CH2CH2 CH239CU 2 2 H20 H OH Figure 338 Mechanisin CH2CH2 for the acid cleavage of an epoxide OH T Anti Figure 339 Nucleophilic cleavage of an epoxide Figure 340 Examples of nucleophilic cleavage of an epoxide 57 Chapter 3 Alcohols and Ethers Not Just for Drinking and Sleeping A generic nucleophilic cleavage appears in Figure 339 and Figure 340 shows two reactions The threemembered ring in the intermediate in Figure 340 is an example of a pseudocarbocation 39039 gtC Clt quotquot 39 3 E H c C Nu NU CH CH3 3 1 N 6CH CH C CH C CH CH3 C CH3 3 3 3 3 3 2H OH H CH3 CH3 CH3 C CH3 633 6 H20H CH3 C CH3 oH O I OH H Summarizing the Spectra of Alcohols and Ethers Tables 32 and 33 summarize the infrared and protonNMR nuclear magnetic resonance spectroscopic properties of alcohols and ethers In the proton NMR the oxygen atom is deshielding Phenols and alcohols rapidly exchange protons so their NMR spectra are solvent dependant The alcohol and ether groups don t have any characteristics absorptions in UVVis spectra 52 Part I Brushing Up on Important Organic Chemistry I Concepts Table 32 Infrared Data Functional Group Absorption Location CH stretch 3000 0H stretch 3600 3200 cm C0 stretch 1150 3 1050 1 cm Strong for ethers ArC0 stretch C0 stretch 1275 1200 cm Table 33 Proton NMR Data Functional Group Chemical Shift 8 In parts per million CHO 334 Ar0H 68 ROH 15 Chapter 4 Conjugated Unsaturated Systems In This Chapter Understanding unsaturated systems Tackling delocalization and resonance Deciphering reactions of conjugated unsaturated systems I he presence of a double or triple bond generally makes a molecule more reactive However if the system is conjugated some stability is restored In this chapter you look at conjugated unsaturated systems especially their reactions And to help prepare you for an exam we also go through some example problems Have fun When you Don39t Have Enough Unsa tum ted Systems An unsaturated system has one or more double or triple bonds You prob ably learned quite a bit about systems containing one double or one triple carboncarbon bond in Organic Chemistry I Now you re looking at systems containing more than one multiple bond Molecules containing more than one multiple carboncarbon bond can be classified into one of two categories those with isolated multiple bonds and those with conjugated multiple bonds Isolated multiple bonds are simply an extension of the behavior of systems containing only one double or triple bond However systems with conjugated multiple bonds behave differently In conjugated systems resonance plays a significant role Conjugated systems A conjugated system is a species with alternating double and single bonds andor a porbital next to a TCbOI1d The porbital may contain zero one or two electrons Some examples of conjugated systems are shown in Figure 41 54 Part I Brushing Up on Important Organic Chemistry I Concepts These systems have a high degree of stability and undergo some unique and unusual reactions CH2CH CHCH2 Ej To H T porbital with 2 electrons Figure 41 some porbital with O electrons pOrbita with 1 eiectron examples of 69 o conjugated H H2C CHCH2 2C C C 2 systems The allylic radical Although propene has only one double bond it can become part of a conju gated system in the form of a radical or an ionic species The radical formed from propene is the allylic radical and it can be formed through the reaction of propene with another free radical such as a chlorine atom Resonance stabilizes the radical as illustrated in Figure 42 and the resonance hybrid is shown in Figure 43 In Figure 42 the resonating electrons in porbitals are shown while the nonresonating electrons in 6bonds are part of the lines indicating the bonds Figure 42 H H H H Resonance P C CLO of the allylic H radical H l H l H H H Figure 43 shows that a partial radical character is on the first and third carbon atoms Both of the carboncarbon bonds are equal and intermedi ate in length between a single and a double bond The allylic radical is more NBEIz stable than a tertiary carbon radical The stability of radicals is in the series allylic gt 3 gt 2 gt 1 gt methyl Figure 43 Allylic radical resonance hybnd Figure 44 Structure of 13butadi ene Figure 45 The equi librium between cis and trans 13butadi ene 55 Chapter 4 Conjugated Unsaturated Systems Butadiene A pair of carboncarbon double bonds separated by a carboncarbon single bond is the basis for a conjugated system The simplest conjugated system is 13butadiene so we use it here to illustrate the behavior of all conjugated systems The structure of 13butadiene is shown in Figure 44 In this com pound the single bond is shorter than the single bond in ethane CH3CH3 The sp2 hybridized carbon atoms are more electronegative than the sp3 hybridized carbon atoms They pull the electrons closer to the nuclei making the atoms smaller and causing the carbon atoms to move closer together Molecular orbital theory predicts some doublebond character between the middle carbon atoms CH2CHCHCH2 At room temperature the cis and trans forms of 13butadiene are in equilib rium The equilibrium favors the trans form with the distribution 5 percent cis and 95 percent trans Figure 45 shows the equilibrium and the structures of the two forms L 56 Part I Brushing Up on Important Organic Chemistry I Concepts De localization and Resonance quot Everyone receives an introduction to the basic concepts of resonance in Organic Chemistry 1 Organic Chemistry 11 requires an extension of the basic rules of resonance to other systems In addition to constructing reasonable resonance structures you also need to understand which structures are more stable Resonance ra es Resonance as you saw in Organic 1 occurs in many systems and you need to be able to recognize when it s going to affect the outcome of a reaction In general resonance makes a species more stable by delocalizing the electrons Delocalization among other things reduces electronelectron repulsion Following is an expanded set of rules for drawing resonance structures This list is an expansion of the rules necessary to understand resonance for Organic Chemistry I You may want to bookmark this list because these rules apply throughout the remainder of this book 1 Resonance structures are simply a way of understanding stability they don t really exist 2 When writing the resonance structures you are only permitted to move lonepair electrons or TC l Ct139OI1S think of the TC l Ct139OI1S as a swinging gate Never ever move any atoms 3 All the structures must have reasonable Lewis structures This includes the same overall charge same number of electrons For example carbon should never have five bonds Second period elements can never exceed an octet of electrons Also keep the following rules in mind 0 A negative charge on a more electronegative atom is more stable than a negative charge on a less electronegative atom 0 Structures that place unlike charges close together or like charges apart are more stable Structures doing the opposite are less important less stable However such structures are not neces sarily impossible 0 Structures with all atoms having complete valence shells are very important In most cases this is an octet 4 No change in the number of unpaired electrons should occur Chapter 4 Conjugated Unsaturated Systems 5 7 5 All atoms that are part of the delocalized atsystem must be coplanar or quotW nearly coplanar An sp3 hybridized carbon atom will not be involved in resonance 6 The presence of resonance leads to stabilization which means that the species is more stable than any of the contributing structures 7 Equivalent resonance structures are equally important to the overall structure 8 The more stable the resonance structure the more important it is the more it contributes to the hybrid In general the more stable resonance structure is the one with more bonds Stability of conjugated unsaturated systems You can estimate the stability of a conjugated versus nonconjugated system by comparing the energy changes For example the enthalpy change for the hydrogenation of 1butene is 127 kJmol kilojoules per mole If conjugation doesn t lead to an increase in stability the energy change for the hydrogena tion of 13butadiene would be about twice this value 254 kJmol However the observed value for the hydrogenation of 13butadiene is 239 kJmol The difference between the predicted 2 x 1butene and the observed value indi cates that resonance stabilizes 13butadiene by about 15 kJmol Reactions o f Coniuyated Unsa tura ted Systems Many students make the mistake of reacting conjugated systems as extensions of the reactions they learned for simple alkenes and alkynes While in some cases this is acceptable the unique nature of these systems makes their chem istry different You investigate some of these features in the next few pages Put in the second striny Substitution reactions The reaction of chlorine with propene illustrates one difference caused by conjugation The products of the reaction depend upon the reaction condi tions as illustrated in Figure 46 58 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 46 Two reac tions of propene with chlorine Figure 47 The general mechanism for allylic halogena tion Cl ClgCCI4 CH2CH CH3 LOW T g CH2 CH CH3 CI Addition reaction CI2A gas phase Low concentration CI C 2C CH2 Substitution reaction Allylic chlorination At low temperature propene behaves like another alkene and undergoes a simple addition of a halogen across the double bond to form 12dichlo ropropane These conditions minimize the possibility of forming chlorine atoms chlorine free radicals and the presence of oxygen traps the few that do form However when the conditions promote the formation of chlorine atoms a substitution occurs to produce 3chloropropene The mechanism Allylic halogenation is a substitution reaction involving a freeradical mecha nism The general mechanism is in Figure 47 The final Xcycles back to the beginning shown with the large curved arrow Initiation X2 2 X Requires heat or light Propagation CH2CH CH3 X CH2CH CH2 X2 gt CH2CH CH2 HX gt CH2CH 3H2 X X Termination Any two free radicals combine Understanding the reaction The question is why doesn t the X in Figure 47 add to the double bond to form the following radical shown in Figure 48 The answer is that this would give a secondary free radical which is less stable than the allylic free radical because it doesn t have resonance stabilization Figure 48 A secondary free radical Figure 49 Allylic bro mination Figure 410 The struc ture of NBS 59 Chapter 4 Conjugated Unsaturated Systems CH2393H CH3 X Allylic bromination Another example of allylic halogenation is shown in Figure 49 Br NBSCCI4 Peroxides or hvr The reactant NBS Nbromosuccinimide shown in Figure 410 is a good source of low concentrations of bromine atoms free radicals O N Br Electrontilic addition The addition of a hydrogen halide such as HBr is an important addition reaction for alkenes often seen in Organic Chemistry 1 However conju gated dienes may behave differently An example is the reaction of HBr with 13butadiene as illustrated in Figure 411 In this case no stereochemistry is implied The distribution of the products depends on the reaction conditions shown in Table 41 The information in the last column of the table indicates the process is reversible and an equi librium results upon heating The equilibrium leads to the production of the more stable product 60 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 411 The reaction of HBr with 13butadi ene Figure 412 The pro tonation of 13butadi ene CH2CH CHCH2 HBr 1 mole H Br CH2CH CHCH2 ACCI4 RT CH2CHCH CH2 H Br Table 41 Distribution of Products in the Reaction of HBr with 13Butadiene 40 C 30 30 40 12addition 15 80 15 14addition 85 20 85 Distribution resulting after heating the 80 reaction products to 40 in the presence of HBr The mechanism Examining the mechanism can help you understand the different results The reaction begins with the protonation of one of the carboncarbon double bonds see Figure 412 by the hydrogen ion from the HBr A primary or a secondary carbocation can be formed by this reaction As seen in Organic Chemistry I a secondary carbocation is more stable than a primary carboca tion Also this secondary carbocation is even more stable because it s allylic and resonance stabilized e ioH3 cHcH oH2l A H CH38H CHCH2 Allylic CH2CH CHCH2 A 10 CH2 CH2CHZCHQ In the second step of the mechanism shown in Figure 413 the bromide ion from the HBr attacks the allylic carbocation at one or the other of the par tially positive carbon atoms Attack on the second carbon gives 12addition while attack on the fourth carbon gives 14addition 67 Chapter 4 Coniugated Unsaturated Systems Figure 413 Bromide attack on the allylic carbocation Figure 414 The reaction diagram for 12addi tion and 14addition 6 5 CH3CHCHCH21 Understanding the reaction The bromide has an equal probability of attacking either carbon atom two or carbon atom four so why is the product mixture not 50 percent of each At low temperatures the bromine doesn t move very far after giving up its H so it s near carbon two 12addition At high temperatures the bromine donates an Ht and can move so it s able to form the more stable product a disubstituted CC Figure 414 uses a reaction diagram to illustrate this situation The twostep process requires a diagram with two hills The first step is the same for both products so the second step is the one that makes the difference At low tem peratures fewer molecules have sufficient kinetic energy to get over the higher barrier Therefore the 12addition product lower barrier is likely to form Energy Kinetic product Starting material 1 2addition Thermodynamic product 1 4addition Reaction The 12addition is rate controlled which leads to the formation of the kinetic product fastest formed This formation is especially noticeable at low temperatures because they always favor the kinetic product which is the reaction product with the lower activation energy barrier In addition few molecules have sufficient energy to surmount the barrier in the reverse direc tion to allow the establishment of an equilibrium 62 BE amp N If Figure 415 A general DielsAlder reac on Part I Brushing Up on Important Organic Chemistry I Concepts The 14addition is thermodynamically controlled This reaction forms the thermodynamic product At higher temperatures more molecules have sufficient energy to cross the second barrier in the reverse direction and establish an equilibrium The equilibrium allows the less stable 12addition product to convert to the more stable 14addition product Low temperatures favor the kinetic product and high temperatures favor the thermodynamic product More than a tree DielsA Ider reactions In the DielsAlder reaction a diene such as 13butadiene reacts with a dienophile such as ethylene to form a product with a sixmembered ring This is an important reaction not only to students trying to pass Organic Chemistry but also in organic synthesis Any of a number of dienes react as long as a conjugated system is present Substituents attached to the conjugated system alter the reactivity of the diene The dienophile is typically a substituted alkene however as you see later in this chapter other species may react The substituents around the double bond also alter the reactivity of the dienophile Figure 415 illustrates the general reaction In this figure the arrows are for apparent not actual electron movement as a means of keeping track of the process The final double bond between carbons two and three is between the positions of the double bonds in the original diene 1 2 2 J 9 200 3 K 3 4 4 The basic reaction remains the same when substituents are present as illus trated in Figure 416 In this example the aldehyde is an electronwithdrawing group the electronegative oxygen pulls electron density away from the double bond The polarity arrow illustrates this electron shift This shift of electron density speeds up the reaction a lower temperature is necessary Conditions As long as a diene and a dienophile are present a DielsAlder reaction occurs However the yield of the reaction can be improved by adjusting the reactiv ity of the reactants Chapter 4 Coniugated Unsaturated Systems Figure 416 DielsAlder reaction of a substituted dienophile Figure 417 The reaction of 13cyco pentadiene with ethylene 63 The presence of good electrondonating groups EDG on the diene results in the diene reacting faster Examples of electrondonating groups are alcohols ethers and amines The dienophile has the opposite requirement that is an electronwithdrawing group EWG facilitates the reaction Examples of electron withdrawing groups are carbonyl groups cyano groups and nitro groups The EDG and EWG groups must be directly attached to the diene or dieno phile If another carbon atom is between the group and the diene or dieno phile the group doesn t count In order for a reaction to occur the diene must be in the cis configuration and not the trans configuration refer to Figure 45 Normally both forms are in equilibrium However the diene can be locked into the cis conformation and facilitate the reaction One way to lock the conformation is to use a ring system The compound 13cyclopentadiene contains a diene locked in the cis conformation Figure 417 illustrates the reaction of 13cyclopentadiene with ethylene CH2 15 Stereoctemistru The DielsAlder reaction is a concerted syn addition meaning the addition is on one side with the stereochemistry of the dienophile preserved in the ste reochemistry of the product If the dienophile is cis then the product is also as and if the dienophile is trans the product is also trans See Figure 418 for the attack by a cis dienophile The reaction in Figure 418 is the reaction of isoprene with maleic anhydride cis 64 Figure 418 The reac tion of a cis dienophile Figure 419 The reaction of a trans dienophile Figure 420 Two pos sible cis products Part I Brushing Up on Important Organic Chemistry I Concepts 0 O O gt O x CH3 CH3 0 O The cis product Figure 419 illustrates the attack by a trans dienophile 0 C 0 CI H H O 39 c In some situations such as in Figure 420 two different cis products may form The two products are the endo and the exoproduct The endoproduct is the major or only product of the reaction The process leading to the endoproduct is Alder endoaddition The endoform is more stable than the exoform O i M gt or O o O O H Figure 421 illustrates the difference between the two forms of the product Chapter 4 Coniugated Unsaturated Systems 65 Figure 421 The endo and exo Endo O H forms 0 Exo H O Passing an Exam with Die Is idler Questions Questions concerning DielsAlder reactions commonly appear on organic chemistry exams In order to gain insight into the reaction itself and to increase your score on your organic exams in this section we look at some typical questions that you might face Two general types of DielsAlder questions commonly appear on organic chemistry exams These may be simple standalone questions or part of a larger question One type of question involves identifying the product and the other type of question involves identifying the reactants Indentifyinq the product What is the DielsAlder product from the reaction in Figure 422 T CN Figure 422 What is the product of this reaction H 66 Part I Brushing Up on Important Organic Chemistry I Concepts Figure 423 The prod uct of the reac onin Figure 422 quot Figure 424 The prod uct of a DielsAlder reac on Figure 425 A diene and a dienophile The product of the reaction in Figure 422 is shown in Figure 423 CN If the product has two CC then the double bond next to the electron withdrawing group was originally part of the dienophile Identifying the reactants What reactants are necessary to form the product in Figure 424 0 CH3 CH3 The reactants necessary to form the product in Figure 424 are in Figure 425 o CHlt CH3 o Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited In This Chapter Investigating infrared spectroscopy Understanding ultraviolet and visible spectroscopy Mastering mass spectroscopy Looking at NMR spectroscopy x A variety of instrumental methods supply information about the struc ture of molecules In general the absorption of energy and many times its subsequent release of energy leads to evidence about the presence of some feature of the molecule In infrared IR spectroscopy the absorption of energy in the infrared region of the spectrum gives information about the types of bonds present Ultravioletvisible UVvis spectroscopy provides info about the molecular orbital arrangement in a molecule Mass spectros copy uses energy to ionize or to break up a molecule into ions The masses of these ions give information concerning the size and structure of the original molecule Finally nuclear magnetic resonance NMR spectroscopy utilizes the absorption of energy in the radio wave portion of the spectrum to give information concerning the environments occupied by certain nuclei especially H and 13C In this chapter you find out how these different types of spectroscopy can be used to learn about organic molecules You may want to get yourself a copy of Organic Chemistry I For Dummies by Arthur Winter Wiley It has some detailed sections on the theory behind instrumentation and exactly how it works In principle you can determine the complete structure of an organic mol ecule from just the IR NMR or mass spectrum of a compound However the process of determination can be very tedious Organic chemists use all the data available when attempting to determine the structure of a compound 68 Part I Brushing Up on Important Organic Chemistry I Concepts For example they gather some information from each of the spectra they have available and combine this data to produce a structure consistent with all the available data When looking at a compound such as CIOHZOO they may look at the IR to determine whether the oxygen atom is part of a carbonyl group an alcohol group or some other group Then they may look at the NMR spectrum to get some idea about the carbon backbone Later in this book you can see the specific functional groups and explore other features of the spectra that are characteristic of the group The methods in this chapter apply to the common types of organic com pounds Unusual compounds have their own characteristic behavior The behavior of these unusual compounds normally isn t relevant until advanced courses in organic chemistry Chemical Fingerprints Infrared Spectroscopy quot Light with energy in the infrared region of the electromagnetic spectrum has enough energy to cause a covalent bond to vibrate The vibrations are due to the stretching and bending of the bonds If the vibration causes a change in the dipole moment of the molecule energy will be absorbed Most organic compounds have several absorptions in the 4000 600 cm region of the spectrum All of the absorptions give information about the structure of the molecule but some absorptions are more useful than others are The region below 1500 cm is the fingerprint region of the IR spectrum The fingerprint region gets its name because this region tends to be unique for every com pound no matter how similar it may be to other molecules Don t get too distracted by the mess of closelyspaced peaks in the fingerprint region Instead look primarily in the important places between 1500 and 2800 cm and above 3000 cm The following sections offer a summary of some of the more important IR absorptions that occur in typical organic compounds Double bonds The carbonoxygen double bond and to a lesser degree the carboncarbon double bond are found in many organic compounds The carbonyl carbon oxygen double bond has a very sharp and intense band around 1700 cm Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited In many cases this band is the most prominent feature of the IR spectrum of compounds containing a carbonyl group The peak due to a carboncarbon double bond is characteristic of alkenes lt s normally around 1650 cm Triple bonds Though not as common as double bonds both carboncarbon triple bonds alkynes and carbonnitrogen triple bonds nitriles are important Both occur in the 2600 2100 cm region of the spectrum They are usually very sharp The carbonnitrogen triple bond tends to give a more intense peak than the alkyne peak 0H and NH stretches Compounds containing a hydroxyl group OH have a strong very broad absorption in the 3600 2500 cm region of the spectrum The most common examples are the alcohols and the carboxylic acids The combination of a broad 3600 2500 cm band with a 1700 cm peak often indicates a carbox ylic acid or amide keep reading The NH stretch in amines occurs in the same general region 3500 3100 cm as the hydroxyl group however the intensity tends to be less In addition primary amines usually have two bands The observation of a 3500 3100 cm band with a 1700 cm peak is often indicative of an amide or carboxylic acid If you ve been paying attention you saw that coming If not see the preceding paragraph CH stretches Nearly every organic compound has one or more carbonhydrogen bonds For this reason the CH stretch isn t as useful as you might think Some guidelines are helpful 1 The CH stretch for hydrogen bound to an sp3 hybridized carbon is in the 3000 2850 cm region 1 For an sp2 hybridized carbon the CH stretch is in the 3150 3000 cm region 1 For an sp hybridized carbon the CH stretch is near 3300 cm 70 Part I Brushing Up on Important Organic Chemistry I Concepts See Figure 51 for samples of the IR spectra of Various functional groups 100 CD 0 C 5 2 E 50 D C 9 I 0 Alcohol 4000 39 39 39 39339039039039 39 39 39 392393903900quot 39 39 39139395390039 39 39 39 39i3900039 39 quotquot500 cm 1 Transmittance CC 4000 3000 H Primary amine 43000quot 39 39 39 393390390039 39 39 39 3923939000quot 39 39 39139395390039 39 39 39 39i3900039 39 quotquot53900 cm 1 l Aro atic 43000 quotquot 0390039 39 39 39 3923939000quot 39 39 391393953900quotquot39i3900039 39 39 quot 3953900 cm1 D 8 f E 8 S I 100 Figure 51 Rabs0rp 2 quot 50 tlon spectra 5 C ofvarlous g func onal groups 100 D 0 0 E e z ll 3 C E I Carbonyl 4 00039 39 39 39 39 0390039 39 39 39 3923939000quot 39 39 39139395390039 39 39 39 391393900039 39 39 39 39 3953900 cm 1 Transmittance 0 4000 3000 2000 cm 1 1500 Transmittance Secondary amine R 4000 3000 2000 1500 1000 500 cm 1 100 D O C TS E E 50 D C 9 I HO C O Carboxyllc acid 4000 quotquot 0390039 39 39 39 3923939000quot 39 39 39139395390039 quot 39 391393900039 39 39 quot 3953900 cm1 Suntan and Beyond Ultraviolet and Visible Ultraviolet and Visible spectroscopy UVVis is an analytical technique useful Spectroscopy in the investigation of some organic molecules Absorption of energy in Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited Figure 52 A typical UVvis spectrum this region of the electromagnetic spectrum can excite an electron from the ground state to an excited state usually from the HOMO highest occupied molecular orbital to the LUMO lowest unoccupied molecular orbital This technique is particularly useful for compounds containing multiple bonds The useful spectral range is usually found between 700 and 200 nm The region below 200 nm is the vacuum ultraviolet which requires special instru mentation and for this reason is less important The spectrum is the result of measuring the absorption of light versus wavelength The position of the absorption maximum wavelength peak position kmax is important as is related to the amount of radiation absorbed molar absorptivity 8 UVvis spectra tend to be much simpler than IR spectra Most UVvis spec tra contain only a few peaks in many cases only one or two Organic compounds with no double bonds or only one double bond typically show absorption only in the vacuum ultraviolet portion of the spectrum The main use of this method is for molecules that have conjugated double bonds The more double bonds in a conjugated system the longer the wavelength is where the molecule absorbs light The presence of eight or more conjugated double bonds can shift the absorption maximum into the visible portion of the spectrum which occurs in many organic dyes Absorptions in the UVvis region of the spectrum is evidence for a conjugated TC l CU39OI1 system Compounds containing carbonoxygen double bonds also exhibit absorp tions in the UVvis region In general these absorptions occur at longer wavelengths than absorptions that are due to carboncarbon double bonds Conjugation shifts the absorption maximum to still longer wavelengths In most cases information from the UVvis spectrum of a molecule is useful to corroborate data and information from other sources This method is also useful in the quantitative determination of the concentration of an absorbing molecule Figure 52 shows a typical UVvis spectrum 08 07 06 J 05 04 03 02 01 Absorbance 200 210 220 230 240 250 260 270 280 290 300 310 320 Wavelength nm 77 72 Part I Brushing Up on Important Organic Chemistry I Concepts Not Weight Watchers Mass Watchers Mass Spectroscopy quot In mass spectroscopy an organic molecule is vaporized and is injected into a mass spectrometer where it undergoes bombardment by electrons This assault causes ionization of some of the molecules producing a molecular ion M Other molecules fragment to produce pieces that may or may not be ions The fragments with a positive charge are important Finally some of the molecules or fragments may rearrange andor undergo further fragmen tation to produce other species some of which are positive ions The mass spectrometer then takes the positive ions and sorts them according to their masses The relative intensity number of ions is plotted versus the massto charge ratio me The result is a mass spectrum In the following sections we investigate the properties of the molecular ion M and the various other ions that result The molecular ion The molecular ion represents the molecular mass of the compound Specifically the molecular mass is the sum of the masses of the most abundant isotope of each element present You can derive an approximate formula for a compound based on its molecular mass especially if only some or all of the following ele ments are present C H N O F or I If the compound contains or may contain nitrogen the nitrogen rule is appli cable According to this rule any molecule with an odd number of nitrogen atoms has an odd mass For example in the compound NCH33 the mass is 59 gmole In some cases the M 1 ion is important The primary source of this ion is the presence of carbon13 This isotope is about 11 percent of all the carbon atoms present The relative intensities of the M 1 to the M ions indi cates the number of carbon atoms present in the molecule To determine the number of carbon atoms you have to calculate the relative intensities of the M 1 peak which you do by multiplying the intensity of the M 1 peak by 100 intensity of the M peak Dividing the relative intensity of the M 1 peak by 11 gives the number of carbon atoms present in the formula The presence of a chlorine or bromine atom results in an intense M 2 peak due to the presence of either two chlorine isotopes chlorine35 and chlorine37 or two bromine isotopes bromine79 and bromine81 In the case of chlorine the M 2 peak is about onethird the intensity of the M Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited peak and in the case of bromine the M and M 2 peaks are of approxi mately equal intensity If more than one chlorine or bromine atom is present the pattern is more complicated but it includes a group of peaks separated by two mass units Sulfur has a less obvious M 2 peak because the abundance of sulfur34 is only 44 percent of sulfur32 A little sulfur33 is also present which contrib utes to the M 1 peak Fragmentation In addition to the molecular ion the molecule generates a number of fragments In general the fragments result by breaking the weakest bonds Different types of compounds often have characteristic fragmentation patterns The alkyl portion of organic compounds gives a number of CXHy fragments If it s possible to form a particularly stable carbocation such as a tertiary car bocation like CH33C an especially intense peak results W The loss of 15 mass units from a molecular ion generally indicates the loss of a methyl CH3 group The loss of 29 mass units often indicates the loss of an ethyl CHZCH3 group Cleavage the breaking of a bond next to a heteroatom any atom other than carbon or hydrogen is also relatively common This is particularly impor tant if fragmentation involves the loss of a very stable molecule such as H20 or CO2 since this would indicate the presences of very particular functional groups and thus would give clues to the structure of the molecule No Glowing Here NM R Spectroscopy NMR spectroscopy allows the organic chemist to see the environment surrounding the nuclei in a molecule NMR isn t applicable to all nuclei how ever most elements have one or more isotopes for which NMR is applicable NMR spectroscopy may be used on nuclei that behave as small magnets Organic chemists usually rely on H proton and 13C as the most important isotopes because most organic compounds contain hydrogen and all organic compounds contain carbon Nuclei that behave like small magnets have a magnetic moment are subject to the influence of other magnets In an NMR spectrometer the sample resides in a large external magnetic field This external field forces the nuclei to align themselves either with the field or against the field Nuclei with one alignment 74 Figure 53 The struc ture of ethanol Part I Brushing Up on Important Organic Chemistry I Concepts can absorb energy and switch to the other alignment and vice versa in a pro cess called spin flipping When spin flipping occurs the nucleus is in resonance The energy required to induce this transition is in the radio frequency region of the electromagnetic spectrum In addition to the external magnetic field other magnetic fields influence the nuclei The electrons in the molecule also have their own magnetic field The field due to the electrons tends to oppose the external magnetic field which results in electron shielding The amount of shielding depends on the number of electrons The more electronshielding taking place the lower the energy requirement for resonance resulting in a downfield shift The position of the absorption is referred to as the chemical shift For proton NMR the chemi cal shift is normally in the 015 ppm parts per million region relative to the standard TMS tetramethyl silane SiCH34 For 13C NMR the chemical shift is normally in the 0200 ppm region Chemically equivalent nuclei absorb at the same energy level Consider for example the structure of ethanol see Figure 53 Three distinct types of hydrogen atoms appear in this structure In the proton NMR spectrum of ethanol discussed in the following section Proton the three hydrogen atoms of the CH3 group are chemically equivalent as are the two hydrogen atoms of the CH2 group and both are different from the hydrogen atom attached to the oxygen Therefore the proton NMR spectrum of ethanol begins with three signals Later in this chapter you see that there s more than that to the NMR spectrum of ethanol H H HCCO H H H Neighboring magnetic nuclei also contribute to the magnetic field surrounding a nucleus which gives rise to coupling A bit suggestive I know but that s what it s called Coupling results in the splitting of some of the absorptions Pro ton Proton NMR spectra follow the generalizations expressed in the previous sec tions but this section discusses some additional factors You find that some of these complications don t occur in 13C NMR spectra quot Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited P It s important to know the chemical shift of each chemically equivalent set of hydrogen atoms A proton adjacent to an atom that has a high electronega tivity has a lower electron density than a proton adjacent to an atom with a low electronegativity Therefore a proton adjacent to an oxygen atom for example comes into resonance at a higher frequency than a proton adjacent to for example a carbon atom lower electronegativity Often drawing all the hydrogens on the molecule can help you see the dif ferent chemical environments different interaction due to nearby hydrogen atoms each hydrogen experiences Integration One important additional aspect of proton NMR spectra is the ability to integrate the spectra Integration provides a means to determine how many hydrogen atoms are in each equivalent set and it involves determining the area under each of the peaks in the spectrum The area is proportional to the number of hydrogen atoms contributing to that particular peak For example in the proton NMR of ethanol the integration of the peak due to the alcoholic hydrogen would represent one hydrogen atom The integration of the peak due to the CH2 group would be twice this value twice the area because twice as many hydrogen atoms contribute to the intensity Finally the inte gration of the CH3 absorption would be three times the OH absorption The total of all the integrations must equal the total number of hydrogen atoms in the compound Coupling Coupling is an old BBC comedy that s sure to make you laugh but never mind that In organic chemistry coupling is another feature of proton NMR spectra Coupling is the result of the interaction of the magnetic field of some hydrogen nuclei with other hydrogen nuclei In general hydrogen atoms attached to the same atom chemically equivalent don t couple Hydrogen atoms on adjacent carbon atoms or separated by a double bond do couple Coupling follows the n 1 rule According to this rule a peak splits into n 1 peaks due to neighboring hydrogen atoms where n is the number of equiva lent hydrogen atoms The amount of splitting is expressed by the coupling constant J In the proton NMR of ethanol the hydrogen atom attached to the oxygen doesn t couple and appears as a single peak a singlet The peak due to the hydrogen atoms in the CH3 group couples with the two hydrogen atoms of the adjacent CH2 group The result is the splitting of the CH3 peak into three peaks a triplet which corresponds to n 2 because CH2 has two hydrogen 76 Figure 54 NMR spectrum of ethanol showing the splitting and intensities Part I Brushing Up on Important Organic Chemistry I Concepts atoms At the same time the peak due to the CH2 group splits into four peaks a quartet which corresponds to n 3 A doublet occurs when n 1 The different types doublet triplet and so on exhibit a characteristic ratio of intensities Doublets are equally intense Triplets have a more intense central peak flanked by two equal peaks of lesser intensity A quartet has two equally intense central peaks with two smaller outer peaks that are equal to each other in intensity See Figure 54 for the NMR spectrum of ethanol CH3 CH2 OH Ethanol 3H Z Ethanol has a simple NMR spectrum because the hydrogen atoms attached to the carbon atoms can only couple with the hydrogen atoms on one adjacent carbon atom However in most organic compounds the hydrogen atoms on one carbon couple with hydrogen atoms on two or three adjacent carbon atoms resulting in a multiplet of indistinguishable peaks appearing in the spectrum For example in the compound CH3CH2CHClBr the CH2 hydrogen atoms have four hydrogen atoms on the adjacent carbon atoms In this case the CH3 hydrogen atoms split the CH2 hydrogen atoms into a quartet and the remaining hydrogen atom on the CHClBr splits each of the members of the quartet into a doublet to give a total of eight peaks Figure 55 Carbon13 spectrum of butyric acid 77 Chapter 5 quotSeeingquot Molecules Spectroscopy Revisited Carbon 73 In general 13C NMR spectra are simpler than proton NMR spectra Instead of integration the height of each absorption peak is approximately proportional to the number of carbon atoms contributing to the peak It s only approxi mately proportional because the more hydrogen attached to the carbon atom the greater the peak height Normally no coupling is present in 13C NMR spectra because the abundance of carbon13 is so low that it s unlikely to have two carbon13 atoms close enough together to couple This lack of coupling greatly simplifies the spec tra You can see this simplification in the carbon13 spectra of butyric acid in Figure 55 0 OH Butyric acid I I I I I I I 160 140 120 I I 39 I 39 I 39 I 39 I 39 I 200 180 100 80 60 ppm 78 Part I Brushing Up on Important Organic Chemistry I Concepts Part II Discovering Aromatic And Not So Aromatic Compounds th Wave B Rich Tennant Poo Bwwrrsu HAP nu REPUTATION oF BEING SOMEWHAT UNAPPROACLIABLE In this part n Part II we spend a lot of time and pages on aromatic systems starting with benzene You examine benzene s structure its resonance stabilization and its stability Next you study benzene derivatives and heterocyclic aromatic compounds and then we address the spectroscopy of these aromatic compounds And in Chapters 7 and 8 we introduce you to aromatic substitution by both electrophiles and nucleophiles and you get to see a lot of reactions and a lot of examples In this part you also start working with many more named reactions Chapter Introducing Aromatics In This Chapter Exploring benzene Coping with the aromatic family Getting acquainted with heterocyclic aromatic compounds Shedding light on the spectroscopy of aromatic compounds organic chemistry has two main divisions One division deals with ali phatic fatty compounds the first compounds you encountered in Organic Chemistry I Methane is a typical example of this type of compound The second division includes the aromatic fragrant compounds of which benzene is a typical example Compounds in the two groups differ in a number of ways The two differ chemically in that the aliphatic undergo freeradical substitution reactions and the aromatic undergo ionic substitution reactions In this chapter you examine the basics of both aromatic and heterocyclic aromatic compounds concentrating on benzene and related compounds Benzene Where It All Starts Benzene is the fundamental aromatic compound An understanding of the behavior of many other aromatic compounds is much easier if you first gain an understanding of benzene For this reason you may find it useful to examine a few key characteristics of benzene which we discuss in the following sections Figuring out benzene 395 structure Benzene was first isolated in 1825 from coal tar Later chemists determined that it had the molecular formula C6H6 Further investigation of its chemical behavior showed that benzene was unlike other hydrocarbons in both struc ture and reactivity 82 Figure 61 Some proposed structures ofbenzene Figure 62 Keku 39s proposed struc tures for 12dibromo benzene Part II Discovering Aromatic And Not So Aromatic Compounds Chemists proposed many structures for benzene However the facts didn t support any of the possibilities until Kekul proposed a ring structure in 1865 Some of the proposed structures including Kekul s are in Figure 61 CH3CE CEC CH3 CH2CH CEC CHCH2 CH2 Dewar Ladenberg Kekul benzene prismane The reaction of benzene with bromine in the presence of an iron catalyst eliminated most of the proposed structures This reaction produced only one monobromo product and three distinct dibromo products ortho meta and para Kekul reasoned that orthodibromobenzene 12dibromobenzene existed in two forms that were in rapid equilibrium and couldn t be isolated These two forms are in Figure 62 Br Br Br Br The ring structure with the rapidly moving double bonds explained many of the facts known about benzene at the time However as more information became available and as chemistry advanced it became obvious that more was going on in this system than just rapidly interconverting structures Chemists determined that only one benzene structure existed not an equilib rium between two related structures Chapter 6 Introducing Aromatics Figure 63 The proposed resonance structures ofbenzene Figure 64 A represen tation of the resonance hybrid of benzene 83 Understanding benzene s resonance Recall from Organic 1 that the concept of resonance was developed to describe the electron structure of a molecule having delocalized bonding by writing all the possible Lewis structures of that molecule The term delo calized bonding refers to a situation in which one or more bonding pairs of electrons are spread out over a number of atoms The development of the concept of resonance came after Kekul had proposed the equilibrium structures for benzene The presence of resonance explains why the carbon carbon bonds in benzene are of equal length and strength The original Kekul structures are in reality not equilibrium structures but contributing structures to the resonance hybrid As contributing structures they have no independent existence The only form present is the hybrid The two resonance forms are shown in Figure 63 and the resonance hybrid is shown in Figure 64 Notice the use of the doubleheaded arrow in Figure 63 To review the resonance arrow and others see Chapter 2 The presence of the double bonds in the resonance structures typically implies that benzene should react like an alkene in terms of addition reac tions and similar reactions However as Figure 65 shows benzene does not react like alkenes 84 Figure 65 Examples of reactions where ben zene does notbehave like an alkene Figure 66 Thehydro gena on ofcydo hexane L3cyco hexadmne and benzene Part II Discovering Aromatic And Not So Aromatic Compounds KMnO4aCl No Reaction NR BF2CCI4 No Reaction NR The pielectrons are delocalized over the entire ring structure not localized between two carbons This contributes to the observed stability of benzene The stability of benzene One way to investigate the stability of benzene is to compare the amount of heat produced by the reactions of benzene to similar compounds that are not aromatic For example a simple comparison of the heat of hydrogenation for a series of related compounds allows us to see the difference Figure 66 shows the hydrogenation of cyclohexane 13cyclohexadiene and benzene which make a suitable set because all three yield cyclohexane H2Ni gt 120 kJmol H2Ni 232 kJmol H2Ni gt 208 kJmol The heat of hydrogenation of cyclohexene is about 120 kJmol kilojoules per mole If the reaction of one double bond releases this amount of energy then the reaction of two double bonds 13cyclohexadiene should release about twice this amount of energy The classical three doublebond benzene should Chapter 6 Introducing Aromatics release three times as much energy The energy change for 13cyclohexadiene is 232 kJmol which is close to the predicted value 240 kJmol However the value for benzene 208 kJmol is far from the predicted value of 360 kJ mol The small value for benzene indicates that it is significantly more stable than a triene This difference is a direct measure of the resonance energy Physical properties of benzene Benzene is a typical nonpolar compound that like other nonpolar com pounds has a low solubility in water It has a characteristic odor which most people find unpleasant It was widely used in academic labs as a solvent but that use has been largely discontinued since it was found that benzene may be carcinogenic Table 61 compares the melting and boiling points of benzene to those of cyclohexane which indicates some differences found in aromatic com pounds The values for benzene are higher than the corresponding values for cyclohexane In most other situations when dealing with compounds with similar structures the lower molecularweight compound has the lower melt ing and boiling point however when comparing benzene to cyclohexane you see that the reverse is true This is because of the delocalization of the electron pairs in benzene Table 61 Melting and Boiling Points of Benzene and Cyclohexane Benzene 78 gmoi Cyclohexane 84 gmol Melting Point 6 C 95 C Boiling Point 80 C 69 C Organic math Hiiee s Rule Just what makes a substance such as benzene aromatic We know that ben zene is more stable than expected The increased stability of benzene is due to resonance having a stabilizing influence Following are the requirements for a species to be aromatic 1 The compound must contain a ring system 1 The system must be planar or nearly planar 86 Q lt Part II Discovering Aromatic And Not So Aromatic Compounds gl3El Figure 67 Several aromatic species where n 1 Figure 68 Two additional aromatic systems 1 Each atom in the ring must have an unhybridized porbital perpendicu lar to the plane of the ring This means that an sp3 atom cannot be part of an aromatic system 1 The ring system must have a Hiickel number of TC l Ct139OI1S This last requirement is an important characteristic of all aromatic systems It s known as Hiickel s rule or the 4n 2 rule To apply this rule begin by assigning 4n 2 number of TC l Ct139OI1S in a cyclic system Next solve for n and if n is an integer a whole number the system is aromatic In the case of benzene 4n 2 6 so n 1 One is an integer a Hiickel number so the last requirement to be classified as an aromatic is satisfied Figure 67 contains several aromatic species with n 1 Br39 0 C N Benzene Pyridine Cyclopentadienyl Tropylium ion anion Naphthalene and anthracene see Figure 68 are aromatic systems with n 2 and 3 respectively Naphthalene Anthracene Common examples of systems often mistaken as being aromatic because of their alternating double and single bonds are cyclobutadiene and cyclo octatetraene shown in Figure 69 In the case of cyclobutadiene 4n 2 4 giving 11 05 while for cyclooctatetraene 4n 2 8 so that n 15 In these two compounds 11 is not an integer so these systems are antiaromatic non aromatic Antiaromatic systems nonHiickel systems are less stable than aromatic or normal systems Figure 69 Two anti aromatic syste ms 87 Chapter 6 Introducing Aromatics Cyclobutadiene Cyclooctatetraene Other aromatics A number of other molecules in addition to those shown in Figures 67 and 68 are aromatic The first five possible values of n are 0 1 2 3 and 4 These numbers correspond to 4n 2 values of 2 6 10 14 and 18 respectively Pyridine refer to Figure 67 illustrates the fact that aromatic compounds are not necessarily hydrocarbons However the replacement of the nitrogen in pyridine with oxygen places an sp3 hybridized atom in the ring so the system is no longer aromatic Two species in Figure 67 are not molecules but ions Aromaticity is not restricted to molecules The cyclopentadienyl ion and tropylium ion cyclo heptatrienyl ion are aromatic species where n 1 Smelly Relatives The Aromatic Family Benzene is an important and common aromatic compound However many other aromatic compounds are based on benzene the substituted benzenes In this section we take a look at the properties and reactions of some of these compounds Nomenclature of the aromatic family The nomenclature naming of aromatic compounds begins with numbering the ring positions This numbering is similar to the numbering of cycloalkane systems The key group is at the number one position see Figure 610 Older usage uses ortho 0 in place of 12numbers meta m in place of 13numbers and para p in place of 14numbers 88 Part II Discovering Aromatic And Not So Aromatic Compounds Figure 610 Two examples demonstrat ing different ways of naming Cl Br N02 aromatic c0mp0und3 mdrchlorobenzene Cl Br 35dibromonitrobenzene Derivatives of benzene The nomenclature adopted by the IUPAC International Union of Pure and Applied Chemistry for some additional aromatic systems is shown in Figure 611 The symbolism for Xylene indicates that two methyl groups are present The methyl groups may be at the one and two positions ortho Xylene the one and three positions metaXylene or the one and four posi tions paraXylene Alternate names are 0Xylene mXylene and pXylene In the other cases one group is attached at the number one position All num bering begins at this position seas Cumene Ahlllhe Toluene dc Phenol Xylene Figure 611 Some com mon names adopted by the IUPAC system through common usage COOH so3H ac Benzoic acid Benzenesulfonic acid Chapter 6 Introducing Aromatics Branches of aromatic groups Aromatic groups may be branches on other systems The general name for an aromatic branch is my Two examples of aromatic branches are in Figure 612 Be careful to notice that the third example the benzyl group is not truly aromatic The benzyl group consists of an aromatic ring and an anti aromatic CH2 group For a branch to be truly aromatic the connection must be directly to the ring Additional ways of indicating the presence of a phenyl group include Ph C6H5 and 1 CH3 CH2Ja Figure 612 Some aro matic group names plus the benzyl group j Phenyl ptolyl Benzyl Black Sheep of the Family Heteroeyelie Aromatic Compounds Up to this point this chapter has discussed aromatic systems composed exclusively of rings of carbon atoms But aromatic systems can contain het eroatoms which in this case means any atoms in the ring other than carbon NBEI9 Heteroatoms such as O S N that are doublebonded to other atoms in a ring can t donate lonepair electrons to the pi system because their p electrons are already involved in the double bond Singlebonded heteroatoms can donate a single lonepair to the pi system but not two because one lonepair must be in an unhybridized p orbital orthogonal at 90 degrees to the sp2 ring plane 90 Part II Discovering Aromatic And Not So Aromatic Compounds Aromatic nitrogen compounds In addition to pyridine refer to Figure 67 other aromatic nitrogen systems exist and show up on organic chemistry tests Two examples of aromatic nitrogen systems are shown in Figure 613 NH2 Figure613 N N Two lt nitrogen N N containing I H N aromatic H systems T Pyrrole Adenine Aromatic oxygen and sulfur compounds Two additional examples of aromatic compounds containing heteroatoms are shown in Figure 614 In both compounds the heteroatom has two lone pairs However only one of the pairs is in a porbital perpendicular to the plane of the ring The other electron pair is in the plane of the ring Figure 614 f A sufur and an Q 8 oxygen containing aromatic compound Furah Thiophene Spectroscopy of Aromatic Compounds Spectroscopy provides many clues to the identity of a compound Aromatic compounds because of the delocalization of the electrons have unique fea tures in their spectra In fact spectral evidence can indicate what atoms or functional groups are attached to the aromatic ring or whether the ring itself contains an atom other than carbon Chapter 6 Introducing Aromatics 9 7 IR The CH bends are characteristic of the substitution around the aromatic ring Aromatic compounds have a characteristic CH peak near 3030 cm In addition the infrared spectra can contain the following features V Up to four ring stretches exist in the 1450 1600 cm region and two stronger peaks are in the 1500 1600 cm region V In the infrared spectra of monosubstituted benzenes usually two Very strong peaks appear one between 690 and 710 cm and one between 730 and 770 cm V Orthodisubstituted benzenes show a strong peak between 680 and 725 cm and a Very strong peak between 750 and 810 cm V Metadisubstituted benzenes show a strong peak between 735 and 770 cm V Paradisubstituted benzenes show a very strong peak between 800 and 860 cm LN1is The conjugated TC l Ct139OI1S of a benzene ring have characteristic absorptions in the ultravioletVisible spectrum that include the following V A moderately intense band appears at 205 nm and a weaker band shows in the 250275 nm region V Conjugation outside the ring leads to additional absorptions NMR The nuclear magnetic resonance spectrum of aromatic compounds com monly contain the following features V The proton NMR of aromatic species have characteristic peaks between 8 60 and 8 95 downfield The downfield shift is due to the aromatic system The aromatic ring has a ring current which gives rise to an induced field V The 13C NMR of aromatic species generally absorb in the 8 100170 region 92 Part II Discovering Aromatic And Not So Aromatic Compounds Mass spec The stability of the aromatic system leads to fragments containing aromatic species The following tips help you interpret the mass spectral data of aro matic compounds 1 The mass spectra of monosubstituted benzenes often contain the C6H5 ion mz 77 1 Other common species are the benzyl cation C6H5CH2 mz 91 and the tropylium ion C7H7 mz 91 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles In This Chapter Exploring electrophilic substitution Checking out what happens in benzene reactions Tackling FriedelCrafts reactions two ways Figuring out how to modify the reactivity of an aromatic Understanding the limitations of electrophilic substitution quot romatic systems are pretty stable they resist reacting Nevertheless a number of reactions involving aromatic systems can be carried out However with the exception of combustion the conditions required by the antiaromatic systems for reactions that you studied in your first semester organic course are different than the conditions for aromatic systems In this chapter we focus on the substitution attack on an aromatic molecule by an electrophile Throughout the chapter we show you how these electro philic substitutions occur first by using benzene as an example Once you become familiar with electrophilic substitution on benzene you re ready to see what happens when substituted aromatic molecules replace the benzene molecule Don t try to memorize the mechanisms in this chapter as separate unrelated entities These mechanisms are closely related and fit together quite nicely Look for the relationships concentrate on understanding how and why the reactions occur as they do and avoid simple memorization 94 Part II Discovering Aromatic And Not So Aromatic Compounds Basics of Electrophilic Substitution Reactions Figure 71 The general mechanism for an elec trophilic substitution reac on One type of reaction that can involve aromatic systems is an electrophilic substitution reaction Like the substitution reactions you learned in your first semester of Organic Chemistry this process involves the substitution of something for a hydrogen atom In this reaction a nucleophile the aromatic system attacks the electrophile The stability of the aromatic system makes it a poor nucleophile though so a very strong electrophile is needed to force the reaction to occur For example the electrophile bromine Br2 is strong enough to attack the double bond in an alkene but it s not strong enough to attack an aromatic system However the Br ion works because it s a much stronger electrophile In general E is the symbol generally used for an electrophile The elec trophile attracts electrons from the Tcsystem of the aromatic ring to form an intermediate Loss of a hydrogen ion from the intermediate to a base completes the reaction The general mechanism is shown in Figure 71 This mechanism is the key to all electrophilic substitution reactions You need to grasp this basic mechanism and be able to recognize it in each of the mecha nisms in this chapter and the next In the mechanism in Figure 71 the hydrogen ion I1 doesn t simply fall off the ring For this loss to occur a base must be present to pull it away In this particular case a wide variety of bases would work Structure III in Figure 71 represents an arenium ion more commonly called a sigma complex Figure 72 shows the energy changes occurring during the reaction E gt Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles y m I Figure 72 Energy changes during an L electrophilic substitution reac on T Progress of reaction Reactions of Benzene In this section we use benzene as a typical aromatic compound to study three basic reactions halogenation sulfonation and nitration In the case of halogenations the electrophile is the X ion X C1 or Br In sulfonation and nitration the electrophiles are S03 and NO2 respectively In each case part of the mechanism involves the generation of the electrophile quotW Don t try to memorize a nitration mechanism as a separate mechanism from halogenations or sulfonation Let your understanding of one mechanism rein force your understanding of other mechanisms Halogenation of benzene In a halogenation reaction a catalyst is necessary to generate the electro phile The most common catalysts are the Lewis acids AIX3 and FeX3 For chlorination X is Cl and for bromination X is Br Adding iodine 1 requires slightly different reaction conditions Figure 73 illustrates the general reac tion for the chlorination of benzene Figure 74 shows a partial mechanism for the reaction Figure 73 C39 The chlo rination of benzene AICI3 C2 gt HCI 96 Figure 74 The partial mechanism forthe chlo rination of benzene NBER Q Part II Discovering Aromatic And Not So Aromatic Compounds The reaction works equally well with AlCl3 AlBr3 FeCl3 FeBr3 and a number of other Lewis bases Some catalysts can also be generated through reactions like 2 FeBr2s Br2l 2 FeBr3s CI CI HCI AICI3 The mechanism begins with the attack on the chlorine molecule by aluminum chloride This step would be the same if ironlII chloride were the catalyst The Clt ion attracts a pair of electrons from the benzene to form an interme diate species The presence of resonance in this intermediate stabilizes it and helps the reaction along These resonance forms and similar forms are important to all the electrophilic substitution mechanisms in this chapter As noted in the earlier section Basics of Electrophilic Substitution Reactions the loss of the hydrogen ion Ht requires the presence of a strong base The chloride ion Cl is a base but it isn t strong enough to accomplish this task However as shown in the mechanism the tetrachloro aluminate ion AlCl4 is a sufficiently strong base This process also regener ates the catalyst so that it s available to continue the process Nitmtion of benzene The generation of the electrophile for a nitration reaction begins with the reaction of nitric acid with sulfuric acid Even though nitric acid is a strong acid it s weaker than sulfuric acid and therefore more baselike so nitric acid acts as a base and donates an OH This process forms the nitronium ion NO2 and water Ht from the sulfuric acid and OH from the nitric acid as shown in Figure 75 The remainder of the reaction is shown in Figure 76 QNBER Figure 75 The forma tion of the nitronium ion Figure 76 The nitration ofbenzene 97 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles The nitrogen atom in all structures in Figure 75 has four bonds This means the nitrogen has a positive formal charge 1 Nitrogen can t have five bonds because no secondperiod element can ever exceed an octet Nitrogen has a zero formal charge when it has three bonds and a lone pair and a negative formal charge 1 when it has two bonds and two lone pairs T E H 6 E N H2S04 e 7 G H79 99 IL J HSOAj H H O NO Hq395c H N02 e P 4 a H The only differences between this mechanism and the halogenation mecha nism see the preceding section for more information on halogenation are the identity of the electrophile and the identity of the base used to remove the hydrogen ion Unlike the base that causes the loss of hydrogen ion in the halogenation reaction the base that removes the hydrogen ion in this mecha nism is the hydrogen sulfate ion HSO4 Nitroaromatic compounds are useful in synthesis because converting the nitro N02 group to an amino NH2 group is relatively easy For example the reaction of nitrobenzene with acidic tinll chloride SnCl2 converts nitrobenzene to aniline an important industrial chemical used in the produc tion of medicines plastics and dyes to name but a few Sulfonation of benzene The generation of the electrophile for the sulfonation generally begins with adding fuming sulfuric acid a mixture of sulfuric acid and sulfur trioxide to benzene The electrophile can then attack the benzene ring in a manner anal ogous to the attack by the nitronium ion The general reaction is shown in Figure 77 and the mechanism is in Figure 78 Notice the similarity between the mechanism in Figures 74 and 78 98 Part II Discovering Aromatic And Not So Aromatic Compounds quot This reaction appears extensively in synthesis problems Keep this reaction in mind when dealing with any synthesis problem involving an aromatic system Sulfonation is easily reversible Simply diluting the fuming sulfuric acid leads to the removal of the SO3H This is an important synthetic technique for pro tecting certain sites from reaction Sulfonation can act as a placeholder while other reactions are performed and then the easy removal of the sulfonic acid group makes the site available for reaction in a later step in a series of reac tion steps Figure 77 H2304 SO3H The SUf0 S03 nation of benzene Fuming SUfUriC aCid 0 Si O Figure 78 The mecha nism forthe su ona on ofbenzene 99 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles Friede Crafts Reactions Figure 79 A typical FriedelCrafts alkylation FriedelCrafts reactions are electrophilic substitution reactions in which the electrophile is a carbocation or an acylium ion The removal of a halide ion from an alkyl halide is the means of generating the carbocation An acylium ion is created by removing a chloride ion from an acid chloride RCOCl Both of these processes require a Lewis acid as a catalyst The most com monly used Lewis acid is aluminum chloride llkylation Figure 79 illustrates a typical FriedelCrafts alkylation Once formed the carbocation is a very strong electrophile A complication that may occur is the rearrangement of the carbocation to a more stable carbocation as seen in SN1 mechanisms of alkyl halides These rearrangements may involve a hydride or other shift Tertiary cations are more stable than either secondary primary or methyl cations Methyl and primary cations are in fact the least stable CH3 CH3C AICI3 HCI Here is the general mechanism of a FriedelCrafts alkylation 1 RC1 AICI3 a R AlCl439 2 R C6H6 a RC6H6 3 RC6H6 AlCl439 a HCI RC6H5 AICI3 Slow In simple mechanisms you encountered in Organic Chemistry 1 methyl and primary carbocations were seriously frowned upon However AICI3 and FeCl3 are such good Lewis acids that even these elusive primary carbocations can form You should still avoid primary carbocations in SN1 mechanisms In reality a true carbocation doesn t form during this reaction The reaction begins with the formation of a complex with the catalyst The complex is of the form AICI3ClR where R has a partial positive charge 8 not a full posi tive charge 1 7 Part II Discovering Aromatic And Not So Aromatic Compounds Figure 710 The gen eration of an acylium ion Figure 711 A Friedel Crafts acylation reac on lcylation A FriedelCrafts acylation is a synthetic method that avoids the problem of rearrangement of the cation Figure 710 illustrates the generation of the elec trophile the acylium ion from an acid chloride The presence of resonance stabilizes the acylium ion and that reduces the possibility of rearrangement O 03 J WAICI3 T AC4 U3 R C R R Figure 711 shows a FriedelCrafts acylation reaction The reaction produces an aryl ketone which is useful in synthesis because it makes it relatively easy to convert the ketone RCOR group to an alkyl R group The procedure involves the catalytic hydrogenation of the aryl ketone and it s particularly useful when the electrophile in a FriedelCrafts alkylation is susceptible to rearrangement Once formed the electrophile behaves like any other electrophile so the mechanism of the attack is the same as that for the previous situation where a nucleophile attacked the electrophile described in the earlier section Basics of Electrophilic Substitution Reactions The attack leads to the for mation of the resonancestabilized sigma complex followed by the loss of a hydrogen ion to a base i i C AICI3 gt C AC439 R c R 9 Acylium ion 0 if R 707 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles The ketone can be reduced with a hot mixture of HCl and zinc amalgam zinc metal dissolved in mercury Why Do an llkylation Figure 712 The oxida tion of an alkylated benzene Alkylated aromatics are useful in organic synthesis If the alkyl group has a hydrogen atom benzylic hydrogen on the carbon adjacent to the ring the alkyl group is susceptible to oxidation A powerful oxidizing agent such as acidic potassium permanganate KMnO4 or acidic dichromate Cr2O72 converts the alkyl group to a carboxylic acid group with the elimination of all carbon atoms except the one connected to the ring Figure 712 illustrates this reaction Any R group with a benzylic hydrogen gives the same product If the ring is disubstituted this can produce a number of isomers Xylenes give phthalic acid isophthalic acid and terephthalic acid CH2CH3 COOH KM nO4aqA or C r2O7239 HA An alkylation using 1chloropropane gives a mixture of products containing both the propyl and the isopropyl group because the primary carbocation propyl rearranges to give the secondary carbocation isopropyl An acyla tion beginning with the acid chloride CH3CH2COCl followed by hydrogena tion yields only the propyl product Chanaina Things Modifying the Reactivity of an Aromatic An electrophilic substitution reaction can take place on a substituted benzene You can replace one of the hydrogen atoms on the substituted benzene with an electrophile A monosubstituted benzene has two ortho hydrogen atoms two metahydrogen atoms and one parahydrogen atom see Figure 713 Based on those ratios substitution of a monosubstituted benzene should be 40 percent ortho 40 percent meta and 20 percent para However you never see this distribution so something else must be hap pening The group already present on the aromatic ring must be influencing further substitution on the aromatic ring 7 Part II Discovering Aromatic And Not So Aromatic Compounds X Figure 713 The ortho ortho ortho meta and para posi tions of a monosub metaf meta stltuted T benzene T para Modifying the reactivity of the aromatic system doesn t invalidate the mecha nisms from earlier in this chapter it just means that you have to go further All substituents influence the resonance in any aromatic system and they may do this in a number of ways V Some substituents withdraw electrons from and others donate electrons to the system V Some substituents may allow extension of the resonance beyond the ring V Substituents may increase the reactivity of the aromatic system activat ing or decrease the reactivity of the aromatic system deactivating V Substituents may direct the electrophile to specific positions around the aromatic ring 0 Some substituents are metadirectors encouraging electrophilic attack upon the meta position The presence of a metadirector means the major or only product of the reaction will be a meta disubstituted benzene 0 Other substituents are orthoparadirectors which facilitate attack at these positions In the next section you see why ortho and paradirectors are a single group Lights camera action Directing A group G attached to a ring makes its presence known in one of the two following ways V An inductive effect due to the pushing or pulling electrons through 6bOI1dS Groups that increase the electron density of the ring are acti vating This helps direct an attacking group to certain positions on the ring Figure 714 How a group G in uences the reactiv ity of an aromatic system NBEl 62 gNBER quotlt2 Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles 7 1 A resonance effect due to the group donating or withdrawing pairs of electrons by creating new TCbOI1dS Groups that withdraw electron den sity from the ring are deactivating which also helps direct an attacking group to certain positions on the ring Figure 714 illustrates activation and deactivation l T In order to understand any chemical process you need to remember that sta bility is the key In the case of aromatic systems resonance is important to stability To find out more about stability and resonance we begin by exam ining the resonance in each of the different sigma complexes that may form listed below and shown in Figure 715 The entering group can attack in one of three relative positions 1 Ortho with the two substituents on adjacent carbons 1 Meta with the two substituents separated by one carbon 1 Para with the two substituents separated by two carbons Sigma complexes always have a positive charge Figure 715 shows that each attack yields three equivalent resonance struc tures This implies that the simple electrophilic attack is not the key to what product will form It must be the identity of the original substituent G All attacks will give three stable resonance structures 7 Part II Discovering Aromatic And Not So Aromatic Compounds Ortho attack IIeta attack l g Q F E E E Para attack Figure 715 The sigma complexes arising from ltgt 4gt ortho meta and para attack T H E H E H E Making a difference with the substituent How can the substituent influence the resonance shown in Figure 715 The answer is that if the substituent can create another resonance structure the sigma complex is further stabilized This additional stabilization leads to a preference for a certain attack If G has a free electron pair next to the ring a positive charge next to it is stabilized by a resonance Two resonance structures in Figure 715 place a positive charge next to G In these two cases an additional resonance form Figure 716 is possible Figure 716 Additional resonance forms aris ing when G has a lone pair of electrons NBER 4quot Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles 705 Q G G H H E E 4 gt H E H E This makes a fourth reasonable resonance structure The more contributing structures present the greater the stabilization of the sigma complex The more stable the hybrid the greater the chances of it being formed Therefore the presence of a substituent capable of donating a pair of electrons favors not only ortho attack but also para attack Such a substituent is an ortho paradirector or in abbreviated form an 0pdirector Refer to Figure 715 paying special attention to the position of the charge If G were an alkyl group it could stabilize an adjacent positive charge by an inductive effect You saw this aspect of alkyl groups in Organic Chemistry I This inductive affect also stabilizes resonance structures with a positive charge on the carbon atom next to G An electrondonating substituent stabilizes an additional resonance form as shown in Figure 716 Thus any electrondonating substituent is an 0pdirector What happens if the substituent is electron withdrawing Figure 717 shows what happens in this case When G attempts to pull electron density from an electronpoor atom the result is a destabilization of the structure for both ortho and para sigma complexes greatly reducing the probability that they will form 706 Part II Discovering Aromatic And Not So Aromatic Compounds Figure 717 The desta bilizing influence of an electron withdrawing substituent Figure 718 Resonance structures of the phe nolate ion NB39R 4quot Figure 719 The resonance hybrid of the phenolate ion G G H E H E Making predictions Many times you can look at the hybrid of the starting material in order to predict where the electrophile will attack For example start by looking at Figure 718 which shows the resonance structures for the phenolate ion The resonance hybrid of the phenolate ion is shown in Figure 719 When drawing resonance structures the overall charge doesn t change from structure to structure Figure 719 illustrates where the electrophile will attack the electronrich region 8 on the ring The two ortho positions are equivalent Otherwise it wouldn t live up to its name If you look at the resonance structures arising when G is a nitro group you see the structures given in Figure 720 The corresponding resonance hybrid is shown in Figure 721 Figure 720 The resonance structures of nitroben zene Figure 721 The resonance hybrid of nitroben zene quot Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles 9 9lt Q39 6 The presence of the partial positive charges 8 at the ortho and para posi e 0 9 399 e 9 0 e 7 N N N f E 5 9 G Q G O9 OQ tions makes an attack at these positions by an electrophile unlikely This leaves the highest relative electron density at the meta position which forces the electrophile to attack meta The lack of a 8 at the metaposition indicates that electrophilic attack is not ideal the ring is said to be deactivated but it will still occur because that s the only option available Any substituent whose atom attached to the benzene contains a lone pair of electrons is orthopara directing but not necessarily a ring activator Substituents without a lone pair on the atom attached to the ring are likely meta directors with the exception of alkyl groups and aromatic rings which turn out to be orthopara directors Turning it on turning it off Activating and deactivating Activation and deactivation deal with how the reactivity of the substituted benzene compares to benzene If the aromatic system is more reactive than benzene the substituent is said to cause activation However if the aromatic system is less reactive than benzene the substituent causes deactivation 707 7 Part II Discovering Aromatic And Not So Aromatic Compounds NBEI Q 4539 Electrophiles seek regions of high electron density or 8 Any substitu ent that gives rise to regions of high electron density facilitates electrophilic attack This is activation Referring back to Figure 719 you see that electron donating groups concentrate a 8 at the ortho and para positions For this reason 0pdirectors are activators The degree of activation reflects how well the 0pdirectors are capable of donating electron density to the aro matic resonance Electrophiles prefer not to attack if no regions of high electron density are available In such cases the ring is deactivated A sufficiently strong electro phile still can with difficulty attack the ring but never in an electrondeficient region 8 As shown in Figure 721 electronwithdrawing groups leave the ortho and para positions with slightly positive charges and the meta position with no charge Activators are 0pdirectors and vice versa while deactivators are mdirectors and vice versa The exception in both cases is halogens Halogens are unusual in that they are 0pdirectors and deactivators This means you should take care any time a halogen is present Halogens are exceptions because they are electronegative electron withdrawing but also have electron pairs next to the ring electron donating Table 71 summarizes what we know about directors and activators and deac tivators When using this table remember two things 1 0pdirectors always beat mdirectors 1 Strong activators always beat weak activators Table 71 Classification of Various Aromatic Substituents OrthoParaDirectors Very strong activators NH2 NHR NR2 OH 0 0R NHC0R 0C0R R C6H5 F Cl Br I Moderate activators Weak activators Mild deactivators MetaDirectors Very strong deactivators NR3 N02 CCI3 CF3 CN S03H C0R COOH COOR CONHZ NH3 Moderate to mild deactivators gNBElgt lt3 quot Chapter 7 Aromatic Substitution Part I Attack of the Electrophiles 7 Pi electronwithdrawing groups direct substitution to the meta position while electrondonating groups direct substitution to the ortho and para position In your first semester of organic chemistry you studied regiochemistry and retrosynthesis The type of director 0p or m is an important aspect of this regiochemistry that you need to consider in any synthesis or retrosynthetic analysis problem Steric hindrance Recall from your first semester of organic chemistry that steric hindrance is the blocking of a reactive site by part of a molecule The presence of a directing group indicates what the major products of an electrophilic substitution reaction are In most cases the reaction also gives a small amount of products not predicted by the directing group Prediction of the major product works well for m directors but what about 0pdirectors Will the ortho or the para product be the major product Two positions are ortho and one is para so the ortho product should be the major product 67 percent and the para product should be the minor product 33 percent Indeed in the absence of steric factors this prediction comes true However groups larger than an ethyl group tend to interfere with attack at the ortho position For example in the case of the halogenation of propylben zene the para product is the major product Many students get into trouble when writing synthetic procedures because they overlook steric effects For example if the task is to synthesize the ortho product starting with propylbenzene you can t rely on the 0pdirecting ability of the propyl group One solution is to place an easily removable substitu ent in the para position leaving only the ortho position available for further attack If the para substituent is a metadirector then both the metadirector and the propyl group favor attack at the same position Limitations of Electrophilic Substitution Reactions In nitration halogenation sulfonation and acylation the reactions are easy to control with temperature The processes deactivate the ring toward fur ther substitution so the reactions inhibit further reaction 7 70 Part II Discovering Aromatic And Not So Aromatic Compounds Figure 722 The result of uncontrolled alkylation of benzene However a FriedelCrafts alkylation can get out of hand The process can con tinue until it replaces all the hydrogen atoms For example the alkylation of benzene can lead to the product pictured in Figure 722 To minimize the pos sibility of multiple alkylations use a large excess of the aromatic compound An amine group limits FriedelCrafts reactions because it reacts with the catalyst so the reaction can t proceed FriedelCrafts alkylation or acylation doesn t take place with groups more deactivating than halogen Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles and Other Reactions In This Chapter Going over the basics and mechanisms of nucleophilic substitution reactions Mastering mechanisms of eliminationaddition reactions Determining synthesis strategies for aromatic systems n this chapter we start by filling you in on nucleophiles before moving on to eliminationaddition reactions If your professor doesn t cover elimina tionaddition reactions do a happy dance and feel free to skip that part of this chapter If you do get to study these reactions we re here to help you out We end the chapter by exploring some important aromatic reactions that don t fit in other categories Coming Back to Nucleophilic Substitution Reactions The basic concepts of nucleophilic substitution reactions appeared in the first semester of organic chemistry These reactions follow SN1 or SN2 mecha nisms In aromatic nucleophilic substitution mechanism we use the desig nation SNAr In SN1 and SN2 mechanisms a nucleophile attacks the organic species and substitutes for a leaving group In aromatic systems the same concepts remain applicable but with some differences that result from the inherent stability of aromatic systems 7 72 Part II Discovering Aromatic And Not So Aromatic Compounds gNBER 6 M353 4quot While aromatic systems often undergo electrophilic substitution reactions they can also undergo nucleophilic substitution In electrophilic substitu tion the aromatic system behaves as a nucleophile whereas in nucleophilic substitution the aromatic system behaves as an electrophile To make an aromatic system change from a nucleophile into an electrophile a strong electronwithdrawing group must be present In addition the aromatic system must have a leaving group similar to the leaving groups seen in Organic Chemistry I Nucleophilic substitution reactions on aromatic systems must have a strong electronwithdrawing group and a good leaving group and the leaving group must be at the ortho or para position relative to the electronwithdrawing group An example of a strong electronwithdrawing group is the nitro group N02 Refer back to Figure 721 in the previous chapter to see the resonance hybrid of nitrobenzene This hybrid has a partial positive charge 8 at the ortho and para positions These partially positive regions are suitable for nucleo philic attack and the meta position is unsuitable So while the nitro group deactivates the ring towards electrophilic attack it activates the ring towards nucleophilic attack Of course no matter how much activation occurs if no leaving group is present no reaction can take place H is never the leaving group Mastering the Mechanisms of Nucleophilic Substitution Reactions NB39R 4quot A traditional SN1 or SN2 mechanism doesn t work for aromatic systems so a new mechanism is necessary This new mechanism is the SNAr or addition elimination mechanism Like an SN2 mechanism an SNAr mechanism begins with the attack of the nucleophile After the initial attack the mechanism is quite different We use the reaction of 3chloronitrobenzene with the hydroxide ion to illus trate in Figure 81 the mechanism for a nucleophilic substitution reaction The different resonance structures represent a Meisenheimer complex The loss of the leaving group reestablishes the stable aromatic system Unlike the positively charged sigma complex the Meisenheimer complex has a negative charge Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles 7 0B I Br Q H 69 G6 66 quot0 66 quot0 66 39b J o 0 oo 39 oo 0 N N N N I3 x F9 4 lt gt gt G G 1 3 QH 393r H p QH 3r QH I Br 9 Figure 81 The mecha nism fora nucleophilic substitution reac on Nucleophilic substitution occurs only when the benzene ring is activated by a strong electronwithdrawing group 7 74 Part II Discovering Aromatic And Not So Aromatic Compounds Losing and Gaining Mechanisms of EliminationAddition Reactions Figure 82 The structure of benzyne An eliminationaddition reaction is another distinct type of reaction mecha nism that occurs in aromatic systems In these mechanisms the elimination involves the loss of an HX molecule While this may seem like a dehydrohalo genation as seen in Organic Chemistry I it really is a different reaction The HX loss leads to the formation of a benzyne intermediate see Figure 82 The mechanism ends with addition to the bond formed by the loss of HX An SNAr mechanism is an additionelimination not an eliminationaddition reaction Benzyne The benzyne molecule refer to Figure 82 is a highly unstable and there fore highly reactive intermediate that forms at high temperatures and high pressures The formation of the carboncarbon triple bond requires two sp hybridized carbon atoms the presence of which makes the structure unsta ble The bond angle about the sp hybridized carbon atoms is 180 degrees which is significantly larger than the 120 degrees of the sp2 present in the ring of benzene A variety of experiments have shown that benzyne is a real molecule However no one yet has been able to isolate this unstable substance The eliminationaddition mechanism The Dow Process utilizes an eliminationaddition reaction to convert chlo robenzene to phenol The proposed mechanism for this reaction is shown in Figure 83 The hightemperature reaction begins with chlorobenzene and aqueous sodium hydroxide Note that this mechanism starts with the hydrox ide attacking as a base beginning dehydrohalogenation to form benzyne The second hydroxide ion attacks as a nucleophile to form a carbanion intermedi ate which behaves as a base in the last step to yield the final product Figure 83 An eimination addition reac on mechanism 775 Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles Experiments with the chlorine attached to a radioactively labeled carbon14 atom produce phenol with 50 percent of the OH attached to the carbon14 atom and 50 percent attached to the adjacent carbon atom This distribution indicates that the second attack by the hydroxide ion has equal probability of attacking either side of the triple bond which is evidence of the existence of the triple bond and therefore of the benzyne molecule The NaOH in the reaction can be replaced with NaNH2 because the amide ion NH2 is similar to the hydroxide ion in that both are strong bases and good nucleophiles the amide ion is a much stronger base The use of sodium amide yields aniline as the product Synthetic Strategies for Making Aromatic Compounds Now you have three basic mechanisms for aromatic rings electrophilic aromatic substitution S Ar and eliminationaddition How do you choose among these The first consideration is what types of other reagents are present If the reagents include an electrophile then the reaction will be electrophilic aromatic substitution The presence of a nucleophile may lead to either SNAr or eliminationaddition If the system meets the three require ments for SNAr then the reaction will follow that mechanism If not it will be an eliminationaddition 7 76 Part II Discovering Aromatic And Not So Aromatic Compounds BE gN I lBEI Q ofquot Figure 84 Two possible synthetic routes quot Nucleophilic substitution reactions on aromatic systems must have a strong electronwithdrawing group and a good leaving group and the leaving group must be at the ortho or para position relative to the electronwithdrawing group Adding the substituents in the correct order is crucial in mastering aromatic synthesis problems Examine the two reaction sequences given in Figure 84 Both sequences involve the same reagents however the order is reversed This shows that the reaction sequence is important You need to plan ahead in any multistep reaction sequence CH3 ortho C o CH3 I C AICI3 AC3 AICI3 II c 0 CH3 C CH3 C CH3Br N R AICI3 39 39 Another thing to consider when designing a reaction is the conditions For example are you promoting the formation of the kinetic or thermodynamic product Figure 85 illustrates this concern Figure 86 illustrates the arrange ment of one of the kinetic products o xylene to the thermodynamic product mxylene Oxidation of an alkylated benzene making an aryl carboxylic acid is a method of converting an orthopara activator into a meta director The reduc tion of a nitro group to make an aryl amine is a way of changing a meta direc tor into an orthopara activator For example on an exam you may be asked to prepare 1bromo3 nitrobenzene from benzene in two steps In this case you must know not only what reactants to use but also the order to use them Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles 7 7 7 C H3 C H3 C H3 CH3 CH3CAICI3 T 00 CH3 Kinetic products CH3CAICI3 ACAICI380 Figure 85 The forma CH3 tion of the kinetic and Thermodynamic product thermo dynamic product T CH3 CH3 CH3 H CH3 CH3 CH3 H H Al I C 3 CH3 H H Methide shift Figure 86 CH3 H The mecha nism forthe conversion of 0xylene HCI AICI3 to mxylene T CH3 Brie fly Exploring Other Reactions Many other reactions involving aromatic systems are possible Many of them are extensions of the reactions you learned in your first semester of organic chemistry For example you may have the hydrogenation reaction of a side 778 Part II Discovering Aromatic And Not So Aromatic Compounds Figure 87 The hydro gena on of the side chain Figure 88 The dehydroha Iogena on of a side chain Figure 89 Markovnikov and anti Markovnikov additions chain shown in Figure 87 or the dehydrohalogenation of a halogenated side chain Figure 88 The reaction in Figure 88 gives only one product because this product extends the conjugated system Figure 89 shows how Markovnikov and antiMarkovnikov additions take place 0S CHQCH3 H2Ni gt 23 Atm Br EtOH a CH CH2CH3 HBr HBr CH2 CH CH3 Peroxide Br Figure 810 shows another pair of reactions for the halogenation of an aro matic compound The reaction of the side chain is a freeradical substitution Figure 811 shows the mechanism of this freeradical substitution Chapter 8 Aromatic Substitution Part II Attack of the Nucleophiles 7 79 CH2CH3 CH2CH3 BT2Fe T Br BrghV T CHCH3 Figure 840 or NBSperoxide Two hao BF gena on reactions e Only product Br hv 2 Br 2 3 Benzylic radical CHQCH3 O HBr BI m 4 Figure 811 The free etC radical substitution CHCH3 mechanisms CHCH3 BF forthe Brz j reaction in Br Figure 810 7 Part II Discovering Aromatic And Not So Aromatic Compounds Part III Carbonyls Good Alcohols Gone Bad The 5th Wave By Rich Tennant T what their alcohol absorption rake isquot In this part arbonyls are a Very large category of organic compounds that includes aldehydes ketones enols and enolates carboxylic acids esters amides and a slew of others In the chapters in this part you look at the structure reactivity and spectroscopy of all of these compounds with special attention to aldehydes and ketones enols and enolates and carboxylic acids You study a lot of reactions including different ways of synthesizing these substances as well as the myriad reac tions that these compounds undergo Finally you examine some chemical tests that help identify these compounds and some spectroscopic clues that can also be used for identification Chapter Comprehending Carbonyls In This Chapter Considering carbonyls Pondering polarity Reviewing resonance Clarifying carbonyl reactivity Shedding light on spectroscopy Fe carbonyl group is part of a wide variety of functional groups that are important not only to organic chemistry but also to biochemistry In fact the carbonyl group is the basis of all the classes of organic compounds dis cussed in the next three chapters of this book Figuring out the basics of the carbonyl group sets a good foundation and makes understanding the classes of organic compounds much easier and isn t making it easier the whole point In this chapter you see some classes of organic compounds that contain the carbonyl group We then investigate polarity and resonance as they relate to the carbonyl group and you have the opportunity to examine some reactions involving it Finally you discover the specifics of the different types of spec troscopy associated with the carbonyl group Carbonyl Basics The carbon atom has sp2 hybridization which means the geometry around it is trigonal planar The planar geometry leaves the carbon atom open to attack from either above or below the plane of the triangle A carbonyl group is a very stable group This stability means that although many reactions alter what s attached to the group few reactions actually change the CO and reactions that destroy the carbonyl group require a great deal of energy Conversely reactions that form a carbonyl group are energetically favorable 724 Part III Carbonyls Good Alcohols Gone Bad Figure 91 The car bonyl group Figure 92 Aldehyde and ketone structure Figure 91 shows the carbonyl group Note This name isn t part of the formal nomenclature of organic compounds but just a simple name for a commonly seen group O O Remember two important facts when examining the chemistry of the car bonyl group First the carbonyl group is Very electrophilic and this pro motes attack by a nucleophile Remember the carbon atom is the target of these attacks Second after the attack there s a strong drive for the car bonyl group to reform Considering compounds con toining the carbonyl group Functional groups containing a carbonyl differ in what s attached to the two open bonds on the carbon atom Hydrogen atoms alkyl groups hydroxyl groups and so on can be attached Depending on what s attached the com pound is given a different functional group name and has a unique set of properties The following sections discuss these different functional groups Aldehyde and ketones Aldehydes and ketones are examples of carbonyl compounds They differ in that either one or two alkyl groups are attached to the carbonyl carbon See Figure 92 and check out Chapter 10 for more discussion of aldehydes and ketones if 0 R C H R c R Aldehyde Ketone R or H attached The similarity between an R an alkyl group and an H plain old hydrogen makes aldehydes and ketones similar in reactivities to each other quot Figure 93 The struc ture of a carboxyhc acid M358 49 Figure 94 The carbox ylate ion Figure 95 The acyl group 725 Chapter 9 comprehending Carbonyls You often see aldehydes represented as RCHO Don t confuse this condensed form with an alcohol which is represented as ROH Carboxylic acids Attaching an OH to a carbonyl function yields a carboxylic acid shown in Figure 93 and covered in more detail in Chapter 12 This is a distinct group whose properties are not simply the sum of the properties of a carbonyl group and an alcohol group Don t confuse a carboxylic acid with a ketone or an alcohol Carboxylic acids have entirely different properties and reactivities than either ketones or alco hols In particular the proton l1 on the oxygen in a carboxylic acid is unusu ally acidic hence the name for reasons we talk about later in this chapter in the section Reactivity of the Carbonyl Group The carboxylic acids like all BronstedLowry acids can lose a hydrogen ion The result is a carboxylate ion shown in Figure 94 390 0 icy group Another common combination is the acyl group which is a carbonyl group with one alkyl group R attached Compounds containing the acyl group get a closer examination in upcoming chapters especially Chapter 12 Figure 95 shows an acyl group 0 R10 7 Part III Carbonyls Good Alcohols Gone Bad Icy chlorides The acyl group appears in other compounds such as the acid chlorides which are also known as acyl chlorides see Figure 96 To find out about the synthesis and reactions of these types of compounds check out Chapter 12 Figure 96 0 The acyl chloride R C structure CI Acid anhydride Two carboxylic acid groups may combine to form an acid anhydride with the general structure shown in Figure 97 We discuss this combination in more detail including synthesis and reactions in Chapter 12 O Figure 97 RC The acid 0 anhydride RC structure O Esters A carboxylic acid may react with an alcohol to form an ester see Figure 98 These esters are used quite a bit in the flavor and perfume industry You can see their synthesis and reactions in much more detail in Chapter 12 Figure 98 0 The struc ture of an R C ester 0 R lmides The interaction of a carboxylic acid with ammonia or an amine may form an amide The amide derived from ammonia has an NH2 group attached to the Figure 99 Primary secondary and tertiary amides Figure 910 The or and B hydrogens of a carbonyL Figure 911 Resonance stabilization of an anion due to the loss of a hydrogen ion 727 Chapter 9 comprehending Carbonyls carbonyl group As you can see in Figure 99 primary and secondary amines yield amides with the nitrogen attached to one or two alkyl groups Tertiary amines don t combine to form amides O O O R C R C R C T T T H H Rquot Primary amide Secondary amide Tertiary amide Ge tting to know the acidic carbonyl Carboxylic acids aren t the only acidic carbonyl compounds Acidity can occur in other ways for example hydrogen atoms attached to a carbon adja cent to the carbonyl group the or hydrogen atoms only not the B hydrogens are acidic see Figure 910 203 C I O I Q I O I Q Q I TI Q EILOLIE The acidic nature of the or hydrogen atoms is due to stabilization of the anion formed as a result of the loss of a hydrogen ion See Figure 911 0 II c 9 ll ILOLI H I C I H 7 Part III Carbonyls Good Alcohols Gone Bad Polarity of Carbonyls Figure 912 Polarity of the carbonyl group The electronegativity difference between carbon and oxygen causes the car bonyl group to be polar with one part becoming negatively charged while another part becomes positively charged see Figure 912 This is an induc tion effect Resonance see the next section enhances this polarity Other groups attached to the carbonyl group may also increase the polarity of the group 00 The electronegativity difference makes the oxygen atom 8 and the carbon atom 8 This makes the oxygen atom a nucleophile and the carbon atom an electrophile which sets the stage for nucleophilic attack on the carbon atom examples of which are discussed in Chapter 10 The polarity of the carbonyl group leads to dipoledipole intermolecular forces which in general increase the melting and boiling points of carbonyl compounds above that of comparably sized hydrocarbons The alcohols with their ability to hydrogen bond tend to have higher melting and boiling points Alcohols and carbonyl compounds can hydrogen bond to water mol ecules thereby increasing the watersolubility of these compounds Additional groups attached to the carbonyl for example the OH in the car boxylic acids also contribute to the intermolecular forces The carboxylic acids have higher melting and boiling points than nonpolar molecules of simi lar molecular masses due to the addition of hydrogen bonding to the inter molecular forces present Converting a carboxylic acid to an ester decreases the intermolecular forces due to the loss of hydrogen bonding converting a carboxylic acid to a carboxylate ion increases the strength of the intermo lecular forces because it creates the possibility of iondipole interactions as well as ionic bonding Like the carboxylic acids the primary and secondary amides have a hydro gen bonding contribution to their intermolecular forces Chapter 9 comprehending Carbonyls 7 Resonance in Carbongls Figure 913 Resonance enhancing the polarity of the car bonyl group quot Figure 914 Stabilization of the carbonyl region in an amide by resonance Resonance enhances the polarity of the carbonoxygen bond in the carbonyl group The effect of resonance is shown in Figure 913 Shifting the electron from the double bond to the oxygen gives the oxygen a negative charge and the carbon a positive charge This charge separation is important in the reac tivity and polarity of the carbonyl group Cy OLD Make sure you keep in mind the charge distribution that s present in the righthand structure while you re studying any reactions involving a carbonyl compound In amides the lone electron pair on the nitrogen atom promotes resonance stabilization of the carbonyl region see Figure 914 This stabilization is important not only to amides but also to the secondary structure of proteins C I39M Cy Resonance further increases the nucleophilic nature of the oxygen atom and the electrophilic nature of the carbon atom In many cases the nucleophilic ity of the oxygen leads to its protonation adding an Ht in acidic media The increase in polarity due to resonance contributes to the strength of the intermolecular forces present in the compound Part III Carbonyls Good Alcohols Gone Bad Reactivity of the Carbonyl Group Figure 915 Nucleophilic attack on the carbonyl group Figure 916 Resonance stabilizing the carbox ylate ion A simple nucleophilic attack on a carbonyl group is shown in Figure 915 This process is reviewed in Chapter 2 and is further illustrated in the discus sion of reactions in Chapter 10 0 H Cl C l Nuc The reactivity of the carbonyl group is enhanced by resonance which sta bilizes the carboxylate ion see Figure 916 This increased stability of the carboxylate ion means that it s easier for a hydrogen ion to leave the carbox ylic acid Thus the resonance is one factor in accounting for the acidity of carboxylic acids c 9quot Spectroscopy of Carbonyls Unsurprisingly organic compounds containing carbonyl groups exhibit many of the features of other organic compounds For example they normally have CH and CC stretches in the appropriate region of their infrared spectra In this section we focus on the spectral properties of the carbonyl group and properties induced by the mere presence of the carbonyl group Infrared spectroscopy The CO in all carbonyl groups has a very intense band in the 1700 cm region of its infrared spectrum In many cases this band is the most prominent feature of the spectrum Table 91 shows some typical carbonyl stretches See Chapter 5 for a discussion of IR stretches and peaks Chapter 9 comprehending Carbonyls 7 Figure 917 IR spectra of carbonyl and Carbox ylic acid Table 91 Carbonyl Stretches in Various Functional Groups Carbonyl Stretch cnr Aldehyde 1740 1720 Ketone 1725 1680 Carboxylic acid 1725 1700 Ester 1750 1735 Acid anhydride 1850 1725 two bands Acid halide 1800 Amides 1650 Conjugation of the carbonyl group tends to shift the stretch to a lower fre quency Resonance may occur when a double bond or nitrogen atom is adja cent to the carbonyl group The CH stretch for the aldehyde hydrogen tends to be a weak band in the 2960 2700 cm region of the spectrum Carboxylic acids also have the characteristic broad OH stretch around 3600 3000 cm See Figure 917 0 Tv 39 C Carbonyl 1000 5c0 L O O Transmittance O O 0 4000 3000 2000 cm 1 1500 L O O 0 ll Transmittance O O HO C Carboxylic acid 1500 1000 4000 3000 2000 cmquot Ultravioletvisible electronic spectroscopy In general if no conjugation occurs the UVVisible absorption of carbonyl compounds is below 215 nanometers nm and for this reason isn t Very useful However in a few cases a second absorption may be present You can see some examples of the additional absorption band in Table 92 8 Part III Carbonyls Good Alcohols Gone Bad Table 92 UVVisible Absorptions in Various Functional Groups Compound nanometers nm Aldehyde 290295 Ketone 275285 Acid chloride x235 Nuclear magnetic resonance NM R spectroscopy The two important types of NMR spectroscopy are proton CH NMR and 13C NMR Even though the oxygen can t be directly observed at work in these techniques the presence of the electronegative oxygen atoms influences them both Proton NM R In the proton NMR the presence of the electronegative oxygen tends to shift the position of the chemical shift downfield This can be seen in Table 93 and in the proton NMR spectra of propanal Figure 918 Table 93 Proton NMR Shifts Proton NMR Partspermillion ppm Aldehyde 95 105 Carboxylic acid 1012 3C NMR In carbon13 NMR the presence of the oxygen also influences the chemical shift of the carbon atoms Table 94 shows the chemical shifts of the carbonyl carbon atom for several classes This can also be seen in the NMR spectrum of butyric acid in Figure 919 Figure 918 Proton NMR spectra of propanal and carbon13 spectra of cyclo hexanone II CH3CH2CH JL IL Chapter 9 comprehending Carbonyls C AIL luILu1uIJIuuuII 12 11 10 ppm quotquotquotmquotquotquotm ILuIJLuIuILLIIuI 3 H 99 98 97 ppm 25 24 ppm CH34S I I I IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 9 8 7 6 5 4 3 2 1 0 ppm6 0 I H23 3H2 H2CCCH2 H2 I I I I I I 220 200 180 140 120 100 I I I I 80 60 40 20 0 160 ppm 6 Aldehyde 95105 Carboxylic acid 1012 Table 94 3936 Shifts for the Carbonyl Carbon C NMR Partspermillion ppm Aldehyde 190200 Ketone 200220 Carboxylic acid 177185 Ester 170175 Part III Carbonyls Good Alcohols Gone Bad 0 MOH Butyric acid Figure 919 Carbon13 spectrum of I I butyric acid 200 180 T ppm I I I I 160 140 120 100 80 60 40 20 0 Mass spectroscopy Carbonyl compounds exhibit the normal fragmentation that happens with organic compounds but they also display some other important features Smaller aldehydes usually have a prominent CHO peak me 29 For ketones fragmentation may occur on either side of the carbonyl group which can influence the position of the group The base peak is often CH3CO me 43 for methyl ketones and aliphatic and aromatic acids may have a COOH peak me 45 The loss of CO2 appears in the spectra of many dicarboxylic acids and substi tuted carboxylic acids Aromatic acids often exhibit a prominent OH loss fol lowed by CO loss For example in acid anhydrides a break may occur on either side of the connecting oxygen atom and this is followed by the loss of CO This break on either side is important for anhydrides with different R groups The cleavage of the NR bond is important for amides with Nsubstituted alkyl groups A primary amide usually has a strong peak at me 44 which corre N358 sponds to OCNH2 Don t forget to consider the nitrogen rule when interpreting the mass spectra of amides See Chapter 5 to review the nitrogen rule Chapter 9 comprehending Carbonyls Table 95 and Figure 920 show partial mass spectral data for representative carbonyl compounds You can analyze each of the spectra to try to explain each of the peaks listed Table 95 Mass Spectroscopy me Data for Figure 920 Compound A Compound B Compound 0 me me me 43 100 77 28 44 100 58 90 105 100 58 25 71 15 148 5 72 5 100 5 86 30 115 10 O A C CHCH2CH2CH3 H CH3 0 B C Figure 920 CHCH32 Carbonyl compounds CH3 see Table O l 9395 for mass C 0 spectra data CH T 3 H CH2CH3 Take a look at Figure 921 It has the IR proton NMR and carbon13 NMR of phenylethanal Practice picking out the Various peaks bands and splitting 736 Part III Carbonyls Good Alcohols Gone Bad 100 90 80 o 70 5 T 50 8 co 50 E 40 I 30 20 10 91000 3600 3200 2800 2400 2000 1800 1600 1400 1200 1000 800 600 Wavenumber cm39 039 H a C b a b Cde AL 97 975 970 955 370 355 350 m II I I maI I II6 5Hppm 5 e CH Figure 921 dCba H The IR def proton CH NMR and CH CH carbon13 2 NMR C a C CDCIS b lMS spectra c A L Z7 IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII ethana 220 200 180 160 140 120 100 80 60 40 20 0 T 6Oppm Chapter 10 Aldehydes and Ketones In This Chapter Examining the structure and physical properties of aldehydes and ketones Finding out how aldehydes and ketones are formed Mastering the reactions of aldehydes and ketones Reviewing their spectroscopy n this chapter we focus on two types of compounds containing a carbonyl group aldehydes and ketones see Figure 101 The simplest aldehyde is methanal formaldehyde CHZO and the simplest ketone is propanone acetone C3H6O Figure 101 0 O The general bJ structure R C H R C R39 of alde hydes and Aldehyde Ketone Meeting lleowI s Relatives Structure and Nomenclature The basic nomenclature for aldehydes and ketones follows that of other organic compounds The following steps are the keys 1 Find the longest chain containing the carbonyl group to determine the parent name Replace the final e of the hydrocarbon with an al for aldehydes or an one for ketones 2 Number the longest chain so the carbonyl has the lowest number 0 Part III Carbonyls Good Alcohols Gone Bad 3 Identify and name all substituents attached to the longest chain 4 Place the names of the substituents alphabetically in front of the name of the longest chain 5 For a ketone add a number to indicate the position of the carbonyl group No number is necessary for aldehydes since the carbonyl group is always at carbon number one Some nomenclature examples are found in Figure 102 O10 O10 CH3CH2CH H CH3CH2CH CH2 H CH3 CI 2methylbutanal 3chloropentanal 0 CH3 0 C T CH3CH CH2CH H CH3CHCHCH2C CH3 Figure 102 The struc CHTCH3 4hexene2one We and 2ethyl4methylpentanal nomen clature of some alde hvdes and CH3CH2C CH2C CH3 ketones 24hexanedlone As seen in most categories of organic compounds the simpler members of the family also have common names Some of the common names and sys tematic names are in Figure 103 Chapter 10 Aldehydes and Ketones Figure 103 Common and system atic names of some simpler compounds O H if i V C C H H CH3 H Formaldehyde Acetaldehyde methanal ethanal Benzaldehyde 0 if CH3 CjCH3 C Acetone propanone Benzophenone Defining Physical Properties of Aldehydes and Ketones lBEB 2quot The carbonyl group is a polar group see Chapter 9 to review polarity The polarity of the group leads to stronger intermolecular forces than in nonpolar substances such as the hydrocarbons The dipoledipole forces present are weaker than the hydrogen bonding present in alcohols Therefore the melting and boiling points of carbonyl compounds are higher than the hydrocarbons and lower than the alcohols Aldehydes and ketones have similar melting and boiling points but are above those of ethers When comparing melting points boiling points and solubility you must use compounds of similar masses Aldehydes and ketones with six or fewer carbon atoms are soluble in water The presence of the oxygen atom which may be protonated means they re dissolvable in concentrated sulfuric acid 74 0 Part III Carbonyls Good Alcohols Gone Bad Creating lldehyaies and Ketones with Synthesis Reactions Figure 104 Forming an aldehyde through oxidation of a primary alcohol Figure 105 An example of ketone formation via oxidation of a second ary alcohol Aldehydes and ketones can be synthesized in a number of ways In most cases but not all either an oxidation or a reduction is necessary Oxidation may be a problem in the preparation of an aldehyde because if care isn t taken a carboxylic acid may be formed instead Oxidation reactions The oxidation of a primary alcohol produces an aldehyde while the oxidation of a secondary alcohol produces a ketone Tertiary alcohols don t undergo simple oxidation If you aren t careful the oxidation of a primary alcohol may pass by the aldehyde to form a carboxylic acid To isolate an aldehyde either the oxidizing agent must be weak or you have to separate the aldehyde from the oxidizing agent before further oxidation can occur One way to separate the aldehyde is to distill the aldehyde from the alcohol which has a higher boiling point An example of the oxidation of a primary alcohol is shown in Figure 104 and two examples of the oxidation of a secondary alcohol are shown in Figures 105 and 106 CHOH e B C O392 JJ 2 CHZCI2 Citronellol 82 OH 0 CCH33 CCH33 3tertbutylcyclohexanol 3tertbutylhexanone 91 N 82 C F207 T H20HOACA Figure 106 Another example of oxidation of a second ary alcohol to form a ketone Figure 107 The ozon olysis of an alkene to form a ketone and in this casean aldehyde Chapter 10 Aldehydes and Ketones 74 7 CH3 OH CH3 0 Testosterone 82 o To form ketones or aldehydes a wide Variety of oxidizing agents work including air The most commonuseful oxidants contain chromium or manganese Ozone for ozonolysis is also a useful oxidant to form aldehydes and ketones from alkenes Different organic chemistry instructors emphasize different oxidizing agents Make sure you know the ones your instructor uses You cannot oxidize a tertiary alcohol other than by burning it The oxidation of an alkene with ozone followed by treatment with zinc in the presence of acid gives aldehydes andor ketones The reaction breaks the carboncarbon double bond and changes each of carbon atoms of the CC to a carbonyl group Figure 107 shows an example of the ozonolysis of an alkene O O CH O L C 2 ZnIit CH3 CH3 0 l H H 70 o Part III Carbonyls Good Alcohols Gone Bad On an exam you may be asked to determine the structural formula of the starting alkene given the ozonolysis products A useful technique is to work backward from the products of ozonolysis By cutting off the oxygens and then combining the two pieces you get the starting alkene Reduction reactions The stability of the carbonyl group limits the number of reagents that are sufficiently strong reducing agents to force a reduction The hydride ion H in various forms is a very strong reducing agent in addition it s a potential nucleophile and a strong base However the basic character of the hydride ion limits its usefulness as a nucleophile For this reason forming a complex with the hydride ion and boron or aluminum in order to minimize its basic character is helpful Compounds such as sodium borohydride NaBH4 and lithium aluminum hydride LiAlH4 are examples of good reducing agents containing complexed hydride ion Lithium aluminum hydride LAH is a stronger reducing agent than sodium borohydride By using some reducing agents an aldehyde or a ketone can be reduced back to an alcohol but in this section our emphasis is upon the reduction of a compound to form an aldehyde or a ketone An acid chloride can be reduced to form an aldehyde If an acid chloride isn t available an acid chloride can be formed from a carboxylic acid by refluxing the carboxylic acid with thionyl chloride SOCl2 Many reducing agents work and a commonly used reducing agent is lithium tritertbutoxyaluminum hydride LiAlHOCCH333 at low temperature The actual reducing agent is the hydride ion H This complexed hydride is a weaker reducing agent than either sodium borohydride or LAH Figure 108 shows an example of this reduction reaction Shown parenthetically in this figure is the conversion of a carboxylic acid to an acid chloride by refluxing with thionyl chloride Many reactions are run at 78 degrees Celsius because this temperature can be reached by using dry ice to cool the reaction mixture Chapter 10 Aldehydes and Ketones 7 Figure 108 Reduc on of an acid chloride to form an aldehyde and con version of carboxyhc acid to acid chloride if if CC LiAf7E cgtIE33u3H gt CH SOC2 i COH Other reactions In addition to the methods previously described in this chapter there are numerous other ways to make aldehydes and ketones depending on the starting materials These include using alkynes doing a FriedelCrafts acylation of an acid chloride and an aromatic compound using organic nitriles and the use of carboxylic acid We examine each of these in the following sections Beginning with an alkyne An alkyne can be converted to either an aldehyde or a ketone To form an aldehyde you begin with a terminal alkyne Figures 109 and 1011 show the formation of a ketone and an aldehyde from an alkyne The reaction in Figure 1011 always gives a ketone while the reaction in Figure 109 gives an aldehyde from a terminal alkyne and a ketone from any other alkyne The sia in Figure 1010 is a siamyl group CH32CHCH3CH We discuss enols and tautomerization in Chapter 11 t Part III Carbonyls Good Alcohols Gone Bad Figure 109 The conver sion of an alkyne to an aldehyde or ketone Figure 1010 The siamyl sia group Figure 1011 The conver sion of an alkyne to a ketone CECH CHC B Sia Sia2BH 8 la H202OH H CHCH CH OH Tautomerization 2 CO Enol CH3 CH3CHCHCH3 O CH3CH23CE CH Hg2 1hexyne H20H CH3CH23C39 CH3 2hexanone 78 Utilizing FriedelCrafts acylation A ketone can also be formed with a FriedelCrafts acylation The process requires an acid chloride and an aromatic compound An aldehyde can t be formed by this procedure because the appropriate acid chloride formyl chloride IICOCI is unstable and decomposes to carbon monoxide and hydrogen chloride Figure 1012 illustrates the preparation of acetophenone from benzene and acetyl chloride Creating ketones two ways with organic nitriles Organic nitriles react with either Grignard reagents or organolithium compounds to form ketones Organolithium compounds tend to be more reactive than the Grignard reagents and for this reason are useful when reacting with stubborn nitriles Figure 1013 illustrates the conversion of a nitrile to a ketone with a Grignard reagent Figure 1012 The prepa ration of acetophe none from benzene and acetyl chloride Figure 1013 Using a Grignard reagent to convert a nitrile to a ketone Figure 1014 A partial mechanism forthe conversion of an alkyl hahdeto a nitrile which reacts to form a ketone with a Grignard reagent Chapter 10 Aldehydes and Ketones 7 if C AICI3 CH3 CH3CC Acetophenone 95 O GEN 1 EtMgBrEt2O C 2 H20Ht quot E Propiophenone 89 Figure 1014 shows a partial mechanism for the conversion of a nitrile to a ketone by the reaction with a Grignard reagent 5 5 CH3CH2MgBr J CH2CH3 5 5 P N 1 r K MgBr NaCN DMSO H20Ht CH3CH2Br CHZCH3 CH3CH2quot C so Part III Carbonyls Good Alcohols Gone Bad Forming from carboxylic acid Finally a carboxylic acid can be converted to a ketone with an organolithium reagent The reaction requires two moles of organolithium per mole of carboxylic acid because one mole must react with the acid hydrogen and the second mole attacks the carbonyl group Figure 1015 illustrates the mechanism for this reaction 0 O CH3CH2CH2C 6 CH3CH2CH2C OH 5 m OLi 5 LiCH2CH3 5 CH3CH3 O OLi CH3CH2CH2C LiH2CH3 CHZCH3 6 Figure 1015 H20 The mecha nism for the reaction of OH an organo H20 quot quot cH3oH2cH2co j CHsCH2CH2C OH 2 LiOH compound with a car CH CH boxylic acid CHQCH3 2 3 Taking Them a Step Further Reactions of lldehydes and Ketones When considering the reaction of carbonyl groups remember the polarity of the carbonoxygen bond the hybridization of the carbon atom sp2 and the bond angles of the planar group Figure 1016 summarizes these features Part III Carbonyls Good Alcohols Gone Bad Figure 1018 The R H R R behaviorof C Cb 36 a carbonyl group in R R39 acidic 30 media Oxygencontaining nucleophiles In the presence of acid an alcohol may react with a carbonyl group to produce a hemiacetal or an acetal This is an example of alcohol addition and the process begins with the mechanism shown in Figure 1018 Figure 1019 shows the general structures of hemiacetals and acetals Figure1019 39 l 39 l The general C R C R structures R O R O ofhemiac 39 39 etals and acetals Hemiacetal Acetal In general hemiacetals are too unstable to isolate however they readily form when an aldehyde dissolves in an alcohol Cyclic hemiacetals are more stable if the ring consists of five or six atoms Many carbohydrates form hemiacetals or acetals through an internal reaction of an alcohol with the carbonyl group present in the same molecule Check out Chapter 16 for more on this topic Figure 1020 illustrates the general mechanism for forming a hemiacetal and an acetal Note that all steps in Figure 1020 are reversible W The mechanism in Figure 1020 provides the foundation for many other mech anisms in Organic Chemistry 11 Understanding this mechanism backwards and forwards is very useful to you 750 Part III Carbonyls Good Alcohols Gone Bad Figure 1021 The use of an acetal to protect an aldehyde group from oxidation Figure 1022 A dithiol provides an easier method to reduce a carbonyl group The alcohol in hemiacetalacetal formation can be replaced with a thiol RSH In addition a glycol can be replaced with a dithiol and then you can follow a procedure similar to the one outlined in Figure 1021 This procedure leads to an easy method for reducing a carbonyl Figure 1022 The reaction with a thiol is in the presence of the Lewis acid boron trifluoride BF3 H Raney nickel high yied 00 0 CH3CH3 Chapter 10 Aldehydes and Ketones Figure 1023 The reaction of methyl amine with propanal to form an imine 757 Nitrogencontaining nucleophiles Many different nitrogencontaining nucleophiles can attack a carbonyl group In this section you consider primary amines RNH2 secondary amines RZNH hydroxylamine NHZOH and hydrazine NHZNHZ Primary amines add to aldehydes and ketones to form imines RNCR2 Secondary amines react to form enamines R392CCRNR2 Hydroxylamine reacts to form an oxime R2CNOH Hydrazine reacts to form a hydrazone R2CNNH2 Primary amines R NH2 Figure 1023 shows the reaction of a primary amine with an aldehyde the product of which is an imine With the exception of the last step the mechanism for the formation of an imine is similar to the mechanism for the formation of a hemiacetalacetal In the formation of an acetal the final step involves the positive charge on the protonated hemiacetal attacking a second alcohol to form an acetal However in the corresponding step in the formation of an imine the loss of a hydrogen ion is easier than a second attack CH3NH2 CH CH C N CH gt 3 2 CH3CH2 C Ether 3 H anhydrous Na2SO4 H H20 The CN bond can exist in both E and Z isomers This isn t important for symmetrical ketones but it is important for aldehydes and unsymmetrical ketones The same is also true for the oximes in the later section on hydroxylamines Secondary amines RQNH Figure 1024 shows the reaction of a secondary amine with a ketone to form an enamine With the exception of the last step the mechanism for the formation of an enamine is similar to both the mechanism for the formation of a hemiacetalacetal and the mechanism for the formation of an imine In this case because losing a hydrogen ion from a carbon adjacent to the CN carbon occurs more easily than a second attack an enamine forms The hydrogen loss is on the side that yields the more stable alkene 752 Part III Carbonyls Good Alcohols Gone Bad Figure 1024 The reaction of dimethyl amine with propanone acetone Figure 1025 The forma tion of an oxime by the reaction of hydroxyl amine with propanal CH CH CH CH Q 3 Q 3 CO H N gt C N K C CH3 CH3 2 03 gtCN 0 CH2 CH3 An enamine The more stable alkene is the more substituted alkene Hgdroxylamine NH 20H Figure 1025 shows the reaction of hydroxylamine with a carbonyl to form an oxime Because pure hydroxylamine isn t Very stable in this reaction the source of the hydroxylamine is hydroxylamine hydrochloride NHZOHHCI The mechanism of this reaction is very similar to the mechanism leading to the formation of an imine CH2C H3 0 HZNOH L HONC H H Oxime CH2C H3 Hydmzine NH ZNH 2 The formation of a hydrazone is illustrated in Figure 1026 Hydrazones form when a carbonyl reacts with hydrazine The mechanism for the formation of a hydrazone is also similar to the formation of an imine and an oxime Again the presence of the CN means that both E and Z isomers can form Chapter 10 Aldehydes and Ketones Figure 1026 0 N NH2 The forma I tion of a C C hydrazone CH2CH3 CH2CH3 by the reac tion of a ketone with hydrazine A hydrazone At one time hydrozones were important in the identification of ketones However with advances in spectroscopic methods using NMR data to identify the compound is easier Hydrazones are important intermediates in a WolffKishner reduction a procedure for reducing a carbonyl group An example of a WolffKishner reduction appears in Figure 1027 and the mechanism is in Figure 1028 N NH2 l N NH 0 H C CH CH 2 3 CH2CH3 10020000 A hydrazone H 39 N2 H Figure 1027 CH The Woff Kishner CHZCH3 reduc on of a hydrazone 82 P Part III Carbonyls Good Alcohols Gone Bad H OIH N N N N ll H ll quot c 9 C CHQCH3 CHZCH3 Ahydrazone H NN C39 CH2CH3 HOH N2 I Figure 1028 H The l A H mechanism 3 C of a Woff CHZCH3 T CH2CH3 Kishner reduc on The formation of the carbanion near the end of the mechanism would seem to make the yield low because forming a carbanion is difficult However the loss of the very stable nitrogen molecule N2 promotes the reaction Carboncontaining nucleophiles Ylides pronounced ilids are important compounds containing a negative carbon atom adjacent to a positive heteroatom The two important types of ylides are those that contain phosphorus and those that contain sulfur Figure 1029 The forma tion of a phosphorus ylide Figure 1030 The mechanism of methy enetriphe nyphos phine attacking a carbonyl group 755 Chapter 10 Aldehydes and Ketones The negative carbon atom serves as the nucleophile Phosphorus ylides are important to the Wittig reaction which converts a carbonyl to an alkene The Wittig reaction begins with the preparation of a phosphorus ylide a Wittig reagent Figure 1029 illustrates to formation of a typical phosphorus ylide by an SN2 mechanism in this case methylenetriphenylphosphine The reagent is a hybrid of the two resonance forms illustrated in the figure The mechanism in Figure 1030 shows how the ylide formed in Figure 1029 attacks a carbonyl ltgt3P CH3Br 3 cH3 Br T Ph9 Y39 M th Itquot h I h h 39 phosphme e y ripbreonnyigeosp onium 99 BuLiTHF CH2 T EH2 Acts as a carbanion Methylenetriphenylphosphine O lt 3F EgtH2 C Cg3 CH2 0 Pltlgt3 Triphenylphosphine oxide p9h Part III Carbonyls Good Alcohols Gone Bad Sulfur ylides behave similarly to phosphorus ylides but the final products are different Figure 1031 shows the mechanism for the preparation of a sulfur ylide and the reaction of the sulfur ylide with a carbonyl group Notice that the mechanism for the formation of the sulfur ylide is similar to the formation of a phosphorus ylide However the last step in the sulfur ylide mechanism is an internal SN2 reaction which eliminates the original thioether dimethyl sulfide The reaction of a sulfur ylide with a ketone yields epoxides whereas the product of a phosphorus ylide with a ketone is an alkene Trimethylsulfonium 3 l iodide V CH 39 CH3 CH3 A thioether NaH DMSO Na c3H2 s CH3 CH3 CH3 T CH3 C Figure 1031 The mecha nism forthe formation of a sulfur ylide and attack by the sulfur CH3 o O ylide on a V gt carbonyl x S CH2 9 F CH2 CH3 Oxidation of aldehydes and fee tones Aldehydes easily oxidize to carboxylic acids or to carboxylates In fact preventing the oxidation of an aldehyde is difficult Ketones oxidize with difficulty since a change in the backbone must first take place 757 Chapter 10 Aldehydes and Ketones Figure 1032 The oxida tion of an aldehyde to a carboxylic acid Figure 1033 The reac tion of an aldehyde with Tollen s reagent Aldehydes Aldehydes are easy to oxidize They slowly oxidize in the presence of air thus in the laboratory many old open bottles of aldehydes are acidic In practice the oxidation of an aldehyde may employ several reagents Figure 1032 shows one common reagent mixture the Jones reagent CrO3HZSO4acetone Another common procedure is a twostep reaction where basic potassium permanganate oxidizes the aldehyde to a carboxylate ion and the second step involves the acidification of the product to form the carboxylic acid 0 CrO H so CH3CH24C ketozne 4 CH3CH24C H 0 C 85 O For years Tollen s Reagent AgNH32OH was used in the identification of aldehydes Aldehydes reacted with Tollen s reagent to deposit silver metal on the walls of the reaction vessel forming a mirror An example of a positive Tollen s test is in Figure 1033 CH2 CH2 72 o Ketones Aldehydes are easy to oxidize but ketones are more challenging The two important oxidation reactions of ketones are the oxidation with a strong oxi dant and the iodoform test Part III Carbonyls Good Alcohols Gone Bad Figure 1034 The oxida tion of a ketone with hot basic permanga nate Figure 1035 Theiodo form test Strong oxidants such a hot basic potassium permanganate oxidize a ketone with an alteration of the carbon backbone Figure 1034 illustrates the oxidation of a ketone with permanganate followed by acidification to produce a carboxylic acid The oxidation cleaves the carboncarbon on one side of the carbonyl group COOH 1 KMnO4aqOH A COOH 2 H 79 o For years the iodoform test was a laboratory method for the identification of a methyl ketone a ketone where one of the R groups is a methyl group A positive test produced the compound iodoform lodoform CHI3 is a yellow precipitate with a characteristic odor The oxidation utilizes sodium hypoiodite which is generated in situ by the reaction of iodine with sodium hydroxide Figure 1035 shows an example of the iodoform test N80H 2 o l o H NaO H3 CH3 R CH3 R O The BaeyerVilliger reaction The BaeyerVilliger reaction uses a peroxyacid or peracid RCO3H which is an oxidizing agent and a nucleophile An example of a peracid is mCPBA metachloroperbenzoic acid shown in Figure 1036 The peroxyacid inserts an oxygen atom next to the carbonyl to form a carboxylic acid from an aldehyde or an ester from a ketone Both aldehydes and ketones are susceptible to a BaeyerVilliger reaction see Figure 1037 Chapter 10 Aldehydes and Ketones c Figure 1036 The structure of mCPBA Figure 1037 The mecha nism of a Baeyer Villiger reac on 0 CI 0 0 H O l3 oCo RCO H CH3 3 CH3 RCOOH z z C on R V of C CH3 A phenyl migration H fA C R o During the reaction a group migrates The mechanism is unusual in that the migration is effectively the movement of a carbanion An aldehyde or an unsymmetrical ketone has two possible different groups that could migrate but the group that actually migrates is the one with the higher migratory aptitude The relative ranking of migratory aptitude is H gt phenyl gt 3 gt 2 gt1 gt Me py Part III Carbonyls Good Alcohols Gone Bad Checking Out Spectroscopy Specs In this section we take a quick look at the characteristic spectra of aldehydes and ketones The following list describes the characteristic absortion of aldehydes and ketones in the infrared region V Both aldehydes and ketones exhibit a Very strong characteristic absorption at z1700 cm V Conjugation shifts the band to lower frequency V The hydrogen attached to the carbonyl group of aldehydes gives two bands at 29002820 and 2775 2700 cm V A 2720 cm shoulder indicates a saturated aldehyde The NMR spectra of aldehydes and ketones contain the following characteristics V The proton NMR of the aldehyde hydrogen appears in the region of 8 910 This high shift is due to the inductive effect electronegative oxygen V In 13CNMR the carbonyl carbon of aldehydes appears in the 8 190200 region and for ketones in the 8 200220 region The following bands appear in the UVVis spectra of aldehydes and ketones V A weak band for aldehydes and ketones is between 270300 nanometers nm V In both cases conjugation shifts the band to 300350 nm Aldehydes and ketones commonly exhibit the following characteristics in their mass spectra V Smaller aldehydes usually have a large me 29 peaks corresponding to the CHO ion V Methyl ketones usually have a large me 43 peak corresponding to the CH3CO ion V Ketones usually fragment on either side of the carbonyl group and help in positioning the carbonyl group Chapter 11 Enols and Enolates In This Chapter Exploring enols and enolates Common intermediates Considering synthesis of enols and enolates Looking at how enols and enolates react A ldehydes and ketones are the starting materials for a large number of reactions and these reactions often involve intermediates known as enols and enolates The mechanisms in this chapter are similar to each other and to mechanisms seen in the last chapter Many organic chemistry students have trouble because they treat every mechanism as a member of a large group of independent entities and not different aspects of a small number of related entities So as you go through this chapter examine each of these mechanisms and focus on the fact that many of these mechanisms are different examples of the same mechanism Getting to Know Enols and Enolates Before we look at enols and enolates we need to examine some aspects of carbonyl groups not covered in Chapter 10 A key feature of many carbonyl groups is that a hydrogen atom is attached to the occarbon the carbon atom next to the carbonyl group The Ka acid dissociation constant for the ochydrogen is about the same as for an alcohol 1019 to 1020 or pKa 19 to 20 The acidity of this hydrogen atom is partly because of the electronwithdrawing power of the oxygen and partly because of the resonance stabilization of the conjugate base resulting from the loss of H The resonance stabilization is the more important factor Figure 111 shows the result of hydrogen ion loss and the two resonance structures contributing to the resonance hybrid which is the enolate ion Part III Carbonyls Good Alcohols Gone Bad Figure 111 The acidity of the orhydrogen atom andthe resonance structures of the conjugate base the enolate ion Figure 112 The attack of an enolate ion by a hydrogen ion 0 llL ix Ll I k I H 1 fin 1 Enough already Structure of enols and enolates The enolate ion formed in Figure 111 is a conjugate base that reacts with a hydrogen ion However the hydrogen ion can attack in two possible places Figure 112 shows what forms when a hydrogen ion attacks at each of these two sites Attack at the occarbon yields the keto form the original carbonyl compound Attack at the oxygen atom produces the enol form an alcohol adjacent to a carboncarbon double bond The presence of a negative charge on the enolate ion means that it acts as a nucleophile In general the carbon acts as the nucleophile but in special cases the oxygen may be the nucleophile 5 p H 0 s I o II I I o c c c c Ketoform Enolateion H Enolform 02 Chapter11 Enols and Enolates 7 Figure 113 An ocB dicarbonyl compound gNBE Figure 114 The sta bilization of an ocB dicarbonyl compound through hydrogen bonding and resonance I thought I saw a tautomer The keto and enol forms seen in Figure 112 are readily interconvertable isomers called tautomers The interconversion of these two forms is tautomerization Normally the keto form predominates because its carbon oxygen double bond is more stable than the enol form s carboncarbon double bond The small value for the Ka of the ochydrogen means that only very strong bases can readily remove the hydrogen ion However in ocB dicarbonyl compounds the enol form predominates because an internal hydrogen bond stabilizes the enol form Figure 113 shows an oc3 dicarbonyl compound Because of the influence of the second carbonyl group the ochydrogen becomes more acidic than a simple carbonyl A stable hydrogenbonded and resonancestabilized species can form so acidity is enhanced see Figure 114 0 0 II I ll c c c B ococ39 B j The formation of five and sixmembered rings tends to increase the stability of a species as shown in Figure 114 Hydrogen bond Hyglrogen bond goH c39j ebH 6C to LL H JJ 4 C C Tautomerization doesn t occur without an ochydrogen Most ketones do contain one or more ochydrogen atoms so they undergo tautomerization These ketones exist in equilibrium with the enol form In most cases the ketone form predominates at equilibrium but in a few cases the enol is particularly stable and it predominates 1 Part III Carbonyls Good Alcohols Gone Bad NBER 4quot Ketoenoltautomerization is not resonance The ketone and enol forms are different compounds that are in equilibrium Studying the Synthesis of Enols and Enolates Tautomerization can be induced through the addition of an acid or base See the previous section for details on tautomerism We begin here by investigating the racemization the formation of both enantiomers of the compound shown in Figure 115 and we use this reaction to investigate both the acid and base mechanisms The acidcatalyzed mechanism is shown in Figure 116 and the basecatalyzed mechanism is shown in Figure 117 CH3 0 I 3 C T H T CC H or OH Figure 115 I The race H H mization of l O a ketone Chwal CC through l the action CH3 of acid or base In the acidcatalyzed mechanism Figure 116 the protonation of the intermediate cation may produce either enantiomer because the hydrogen ion may attack the carbon from either side One if by land two if by sea Equal amounts of each enantiomer result and give a racemic mixture In the basecatalyzed racemization Figure 117 the carbon is also vulnerable to attack from either side Attack from one side gives one enantiomer while attack from the other side gives the other enantiomer The attack results in deprotonation Since the probability for attack from either side is equal a racemic mixture again results Chapter11 Enols and Enolates 7 3 39 CH3 6 TH O lt I O uIwCJC H nlllcjc H H H26 Figure 116 CH3 OH The begin l ning of the C mechanism forthe acid catalyzed Enol achiral racemization CH3 O CH3 0 5 I nnICC m CFC H Enolate achiral H Figure 117 CH The begin l 3 OH ning of the CC mechanism forthe base ata39VZed Enol achiral racemization In each of these mechanisms the steps involve the gain or loss of a hydrogen ion The enolate ion shown in Figure 117 is an important nucleophile and is important to many other mechanisms Under acidic conditions the mechanism begins with the gain of a hydrogen ion but under basic conditions the mechanism begins with the loss of a hydrogen ion Z Part III Carbonyls Good Alcohols Gone Bad A mechanism can be under acidic or basic conditions it can t be under acidic and basic conditions That is to say H acid and OH base are never in the same mechanism To produce an enolate the ochydrogen ion must be removed Strong bases such as hydroxide OH and alkoxide OR ions only form a small amount of enolate because these bases aren t sufficiently strong However forming an enolate is easier if you use one of these bases and an ocB dicarbonyl compound Thinking Through Reactions of Enols and Enolates Figure 118 This compound undergoes a haloform reac on Enols and enolates undergo several types of reactions that are important in organic chemistry synthesis In this section we take a look at the major reactions that these compounds undergo Ha Ioform reactions In Chapter 10 you see the iodoform test as a means of identifying methyl ketones Here we reexamine this reaction in light of enols and enolates Haloform reactions involve the interaction of methyl ketones with alkaline halogen X2 where X Cl Br or 1 solution to give haloform CHX3 plus a carboxylate ion The process begins with attack on the ochydrogen of the methyl group followed by attack by a halogen The addition of each halogen atom makes the remaining hydrogen atoms on the same carbon more acidic due to electron withdrawal by the halogen The process is useful in the synthesis of a carboxylate carboxylic acid with one less carbon atom Any compound that oxidizes to a methyl ketone also gives a haloform reaction because halogens are also oxidizing agents For example the compound shown in Figure 118 reacts Chapter11 Enols and Enolates 76 7 Figure 119 illustrates the mechanism for the haloform reaction The mechanism involves a repeated series of base attacks removal of an ochydrogen followed by the reaction with the halogen until all three ochydrogen atoms are replaced Then the base attacks the carbonyl carbon to induce the loss of a carbanion CX3 The highly reactive carbanion quickly attacks and removes the hydrogen from the carboxylic acid group i 010 la O O D f C C Z GL3 39H Cls l 69 C1 9 390 3 o F CI3 Figure 119 OH 7 O The mecha 0X nism of the C haloform reaction CHI3 0 Part III Carbonyls Good Alcohols Gone Bad BE gN I Figure 1110 Mechanisms forthe halogenation of a ketone The haloform reaction is a useful method of preparing a carboxylic acid carboxylate ion with one less carbon It is one of the Very few cases where carbanion loss occurs It s only possible because the three halogen atoms are capable of stabilizing the negative charge A ketone can be halogenated even when it isn t a methyl ketone This process can be either acid or base catalyzed The general mechanism is shown in Figure 1110 X2 I H 39 or OH l Aldo reactions and condensations The reaction of a carbonyl aldehyde or ketone with a base produces an enolate ion a nucleophile This nucleophile attacks any electrophile What happens when you add a base to a carbonyl with no electrophile present It turns out that a reaction still occurs because the carbonyl group itself is an electrophile As the enolate forms it can attack the carbonyl group of another aldehyde or ketone molecule This is an aldol reaction or aldol condensation also called an aldol addition Chapter11 Enols and Enolates 7 Figure 1111 An example of an aldol reac on gNBER 61 61 61 quotca lt2 quot 539 E Ketones are less reactive towards the nucleophile In Organic Chemistry I you saw that alkyl groups are electron donating In ketones the presence of the two alkyl groups attached to the carbonyl do a better job at compensating for the 8 on the carbon atom than do one alkyl group and a hydrogen atom in an aldehyde For this reason aldehydes are more reactive than ketones In the aldol reaction aldehydes or ketones with ochydrogen atoms react in the presence of dilute base to give an aldol a Bhydroxyaldehyde or ketone An aldol contains an aldehyde or ketone and an alcohol group in the 3position Figure 1111 shows the general reaction The aldol formed in this reaction is 2methyl3hydroxypentanal which forms with an 86 percent yield T hat s really good when we were taking organic we would have killed at times for a 5percent yield o o CH3 O OH CH3C H2C cH3cH2c CH3CH2CHCHC H H H OH The product of an aldol reaction of RR39CO is RR39CO2 R is an alkyl group and R may be an H aldehyde or an alkyl ketone Figure 1112 shows the mechanism for the aldol reaction Notice that in the last step the hydroxide ion formed is a weaker base than the alkoxide ion Forming a stronger base from a weaker base is very highly unlikely The aldol formed by the aldol reaction especially if heated can react further The heating causes dehydration loss of H20 and the overall reaction involv ing an aldol reaction followed by dehydration is the aldol condensation The product of an aldol condensation favored by the presence of extended con jugation is an ocBunsaturated aldehyde an enal or ketone The mechanism for dehydration Figure 1113 begins where the mechanism of the aldol reac tion Figure 1112 ends This process works better if extended conjugation results The aldol reaction and condensation are reversible The product of an aldol condensation of RR39CO is RR39CO2 H20 7 Part III Carbonyls Good Alcohols Gone Bad Figure 1112 The mechamsm of the aldol reac on Figure 1113 The mechamsm forthe dehydra on ofthe productof mwamol reac on CH38HC lt gt 39 H20 CDFQ C CH3CH2CHCHC Alkoxlde C H O P CI4 1 Fl CH3 0 CH3CH2CHCHC OH Fl CDF1 OH CH3 0 CH3CH2CH lt3 Clt 39QH CH3cH2CH cG clt H2O H H D H 777 Chapter11 Enols and Enolates Figure 1114 An aldol cyclization reac on leading to the formation of a five membered ring Crossed aldol condensations An aldol reactioncondensation occurs when the enolate ion from an aldehyde or ketone attacks a molecule of the parent compound If however two different carbonyl compounds are present a crossed aldol reaction condensation occurs Crossed aldol condensation reactions may present a problem because multiple products can be formed For example a mixture of carbonyl A and carbonyl B can give two different enolates each of which can then attack either an A or a B molecule Therefore four products are possible A2 A B B A and B2 The formation of multiple products limits the practicality of crossed aldol condensations Following are the two ways to increase the practicality of a crossed aldol condensation 1 One way is to begin with one aldehyde A and slowly add the other aldehyde B This decreases the chances of two of the products B2 and B A from forming leaving A2 and A B 1 The other method is to choose a compound that doesn t have any ochydrogen atoms because a compound with no ochydrogen cannot form an enolate For example if B has no ochydrogen then only two products are possible A2 and A B Examples of aldehydes with no ochydrogen atoms are formaldehyde and benzaldehyde Aldo cyclization A molecule that contains two carbonyl groups may undergo an internal aldol condensation Ideally either two or three carbon atoms should be between the carbonyl groups The carbonyl groups need to be at positions one and four or one and five relative to each other In these cases a five or sixmembered ring forms both of which are stable and help facilitate the reaction A seven membered ring can be formed this way but other sizes don t form easily Figures 1114 and 1115 show examples of aldol cyclization reactions These reactions lead to the formation of a five and sixmembered ring respectively The mechanism for the preparation of jasmone Figure 1116 illustrates the general mechanism for this process NaOH 14diketone 7 0ef Part III Carbonyls Good Alcohols Gone Bad Figure 1115 An ado cyclization reac on leading to the formation of a six membered ring NaOH T EtOH 15diketone NaOH T EtOH Jasmone 0 CH3 CH3 90 Figure 1116 j The mechanism forthe formation of 0 3 H O jasmone V G O I Chapter11 Enols and Enolates 7 P Figure 1117 An example of a Claisen Schmidt reac on forthe formation of an ocB unsaturated ketone Figure 1118 The general structure of an ocB unsaturated aldehyde or ketone Addition reactions to unsaturated aldehyde and fee tones The ocBunsaturated aldehydes or ketones prepared earlier in this chapter are useful as starting materials for a number of reactions In this section you investigate some of these reactions ClaisenSchmidt reaction The ClaisenSchmidt reaction Figure 1117 produces an ocBunsaturated aldehyde or ketone the general structure of which is shown in Figure 1118 The ClaisenSchmidt reaction is a crossed aldol condensation o O o CH l J H R CHCCH A ccH c cH3 3 3 R O ccH c 3 oc The addition reactions to an ocBunsaturated aldehyde or ketone may be a simple addition of a nucleophile or conjugate addition Conjugate addition involves tautomerization see the earlier section I thought I saw a tautomer for details The general structure of the product of simple addition is shown in Figure 1119 while the general structure of the product of conjugate addition is in Figure 1120 The mechanism for conjugate addition is in Figure 1121 7 Part III Carbonyls Good Alcohols Gone Bad Figure 1119 The general structure of the product of simple addition to an ocB unsaturated aldehyde or ketone Figure 1120 The general structure of H O OH the product l of conjugate CCC 4 CCC addition l i l to an ocB Nu unsaturated Keto En0 aldehyde or ketone Figure 1121 The mechanism for conjugate addition to an ocB unsaturated aldehyde or ketone Keto 51039 Chapter11 Enols and Enolates Figure 1122 The resonance hyb d andhs contnbu ng resonance structures resm ng froni nudeophmc a ackina conjugate addmon reac on Figure 1123 The formation of a simple addition product and a conjugate addition product Conjugate addition occurs because there are two sites on the electrophile where a nucleophile can attack The structure of the resonance hybrid and the two resonance structures contributing to the hybrid are shown in Figure 1122 The presence of this resonance is apparent in the infrared spectrum because the carbonyl stretch shifts to a longer wavenumber 1 1 1 1 1 1 if jCCC j Cj Jacjcficj 3905 lJe 5 5 Nucleophilic attack The simple addition and the conjugate addition reactions compete with each other Figure 1123 shows an example of a reaction with both the simple addition and the conjugate addition products In this reaction an increase in the size of the alkyl group on the Grignard reagent leads to an increase in the yield of the conjugate addition product while substitution on the CC leads to an increase in the yield of the simple addition product 0 OH CH3CHCH C H339V399Bquot CH3CHCH C CH3 CH3 CH3 Simple addition product 72 o O CH3 CH CH2 C CH3 CH3 Conjugate addition product 20 o 775 7 X Part III Carbonyls Good Alcohols Gone Bad Michael addition The Michael addition is an enolate ion addition to an ocB unsaturated carbonyl This reaction takes advantage of the increased acidity of a hydrogen atom that s oc to two carbonyl groups This enolate ion is very stable so it s less reactive than normal enolates The morestable enolate leads to a greater control of the reaction so that only one or two products form instead of multiple products from a less stable and therefore more reactive enolate An example of this type of reaction is in Figure 1124 with the mechanism in Figure 1125 0 E1 T l 393 O CH2 CH2 C 1 NaOEtEtOH C O o o oH CH FIgure1124 2H 3 An example CH2 H2C CH CH3 CH3 CO ofaMichae CCH3 0 940 addition Et O O lit 0 ET o o lt9 W o OEt 59 H2CCH CH CH2 CH V 3 Et C CH C CH O 0 3 0 3 C O CH2CH c K Figure 1125 The O 0 CH3 mechanism 0 H for the CH2CH2C Michael C CH CH3 addition CH3 C reaction In Figure 1124 0 Figure 1126 Some Michael donors Figure 1127 Some Michael acceptors 777 Chapter11 Enols and Enolates The addition is to the alkene carbon atom furthest from the carbonyl group This is position four so this is a 14addition or a conjugate addition If the addition had been to the carbonyl carbon this would be a 12addition The 1 in 14addition or 12addition refers to the addition of a hydrogen to the carbonyl oxygen atom to form an enol The Very stable enolate ion is a Michael donor all of which react like the enolate in Figure 1125 Some important Michael donors are shown in Figure 1126 The ocB unsaturated carbonyl is a Michael acceptor Figure 1127 shows some important Michael acceptors which behave like the ocB unsaturated carbonyl in Figure 1125 0 R e 0 av R lt9 0 NH2 0 on HAJK AJK Figure 1128 shows the mechanism for another Michael addition NO2 7 Part III Carbonyls Good Alcohols Gone Bad 9 O Figure 1128 0 H3 The Q o O i O mechanism G for a CH2CH2C H cH2 oH o o A O H EEK EEK 03 H CH2CH C CH3 eo o Michael addition reac on CH3 CH3 Other enoate reated reactions A number of species such as nitroalkanes and nitriles have an acidic ochydrogen atom These compounds can lose a hydrogen ion to produce an anion that is analogous to and reacts like an enolate ion Nitroalkanes The loss of the acidic hydrogen from a nitroalkane produces a resonance stabilized anion This anion a nucleophile is capable of attacking a carbonyl compound Synthetically the advantage of using a nitroalkane is that reducing the nitro group to an amine group is easy Figure 1129 shows the formation and use of the anion from a nitroalkane followed by reduction Figure 1129 The use of a nitroalkane as a substitute for an enolate ion Chapter11 Enols and Enolates Resonance stabilized o R39 c H cc H NO2 R39 R Easily reduced l H2Ni R R cH c NH2 Nitriles The ochydrogen atoms of nitriles are also acidic Figure 1130 shows the formation of the resonancestabilized ion and the reaction of the anion with a carbonyl compound These reactions are important in some synthesis reactions 779 7 Part III Carbonyls Good Alcohols Gone Bad R C3H CE N C Z R R W lt9 9 c3H cE N CH C N Resonance stabilized T 0 Figure 1130 R39C The use of a nitrile as 39 H a substitute R R for an 00 enolate ion CN Miscellaneous reactions A number of other reactions involve directly or indirectly enols or enolates Other additions These reactions mainly involve conjugate additions to carbonyl compounds by nucleophiles such as the cyanide ion CN and primary or secondary amines RNH2 or RZNH Figure 1131 shows the conjugate addition by the cyanide ion and Figure 1132 shows the conjugate addition by a secondary amine Chapter11 Enols and Enolates Figure 1131 The conjugate addition of a cyanide ion Figure 1132 The conjugate addition of a secondary amine CH3 0 CH 0 D 1CN 3 C CH C C CH3 C 2 H CH3 CH3 CH3 CN CH3 CN 88 CH C CH I 3 A 0 I 3 O J CH3CCH C lt gt CH3 CCH C CH CN 3 CN CH3 O O CH3CH C T CHCH C CH3 l CH3 CH3 CH3 92 o e to 5 O W J cH3 CHC j CIDH2gHC CH H N CH3 CH3 N 3 Ul CH3 CH3 CH3 Cannizzaro reaction The Cannizzaro reaction is a redox reaction which requires a concentrated base and a carbonyl group with no ochydrogen atoms Normally the oxidation converts an aldehyde to a carboxylate carboxylic acid while the reduction generates an alcohol Figure 1133 shows an example of a Cannizzaro reaction 787 782 Figure 1133 An example of a Cannizzaro reac on Part III Carbonyls Good Alcohols Gone Bad quot9 Ci rgt C C H H H39 I W OH Klydrde transfer 0 II C H r o H I I CO HC E L PO quotStrong acidquot quotStrong basequot i l T C H C OH A crossed Cannizzaro reaction is similar to a normal Cannizzaro reaction however two different aldehydes are reacting Normally one of the aldehydes is formaldehyde because there are fewer chances for side reactions It also has the advantage of being cheap The reactions in Figures 1134 and 1135 are crossed Cannizzaro reactions using an excess of formaldehyde The excess of formaldehyde increases the probability of the initial attack by the hydroxide being on the formaldehyde instead of the other aldehyde Figure 1135 shows the synthesis of pentaerythritol Acid catalysis In general we use bases to initiate many of the reactions in this chapter However acids can catalyze some reactions Figure 1136 shows the formation of 4methyl3penten2one by acid catalysis Chapter11 Enols and Enolates 7 Figure 1134 A crossed Cannizzaro reac on with excess formalde hyde Figure 1135 Another crossed Cannizzaro reac on with excess formalde hyde Figure 1136 The formation of 4methy3 penten2 one by acid catalysis if CH 39039T39C39H39 L CHZOH C so H H H O O QKX CH2OH H II r 3 CH2OH CCCBDH 5 H n 1 3 c b c CH CH3 or cI2 or Ecng 7 H lt93 at Gt J C CHCCH3 9H lt8H CH 8H2 CH3 l J cHquotcH3 HH 6 CH3 0 CH3 1 cl F c3 6 two 393 393 CH CHlt4 2 39H CH CHl CH3 CH3 P y Part III Carbonyls Good Alcohols Gone Bad Robinson annulation The Robinson annulation begins with a Michael addition followed by an aldol condensation see the earlier section Michael addition for more info on that reaction An example of a Robinson annulation is shown in Figure 1137 Another example the reaction of 2methyl13cyclohexandione with methyl Vinyl ketone is given in Figure 1138 The mechanism of the reaction through the formation of Compound 1 is in Figure 1139 and mechanism for the reaction of Compound 1 to give the final product is in Figure 1140 0 CH3 0 CH3 OH H gt O 01 O 0 CH3 IOH39 Michael addition product CH3 CH3 0 O Figure 1137 A Robinson a 39ati CH3 Robinson annulation product Chapter11 Enols and Enolates 7 O O T CH3 CH3 0 Figure 1138 O OHquot The H 0 MeOH CH2 CH2 CjCH3 Robinson CH2CHC CH3 annulation O 0 Compound 1 applied to O H the reaction of 2methy CH3 H20 13cyco 55 hexandione withmethyl vinyl ketone O O 0 CH3 CH3 L H3bH39 9TKGHFCEHD CH 00 j ltH20 2 3 O O O O 9 CH3 10 CH 0 39 ll 1 CH 593 CH cH2 cH c cH3 2 3 O O F 1139 HOH Igure The 0 beginning CH3 0 stepsinthe mechanism CH2 CH2 C CH3 inFigure 1138 0 Compound 1 p Part III Carbonyls Good Alcohols Gone Bad 0 0 CH3 0 H r CH3 II I ltH CH2CH2C CH2 gt 0 Compound 1 1842 O O 0 CH3 Figure 1140 The final steps of the mechanism beganin Figure 1139 Chapter 12 Carboxylic Acids and Their Derivatives In This Chapter Tackling the names and forms of carboxylic acids and their derivatives Analyzing the physical properties of these compounds Assessing how they are synthesized Looking at their reactions Determining them with spectroscopy and chemical tests Fis chapter continues your examination of compounds containing the carbonyl group in this case carboxylic acids and their derivatives You see many carbonyl compounds in Chapter 9 and in Chapter 10 you see the role of carbonyl groups in aldehydes and ketones Many of the characteristics of the carbonyl group seen in those compounds are part of the chemistry of all carbonyl compounds So remember while reading this chapter that you re looking at an extension of those characteristics and reactions and not a totally different set of compounds And if you skipped Chapter 10 you ll probably find it helpful to go back and read it first Acidbase chemistry is important when studying carboxylic acids because in general they are significantly stronger acids than the ochydrogen atoms seen in Chapter 9 The carboxylic acids are stronger acids because of the inductive effect of the carbonyl oxygen and the oxygen to which the hydrogen is bonded These carboxylic acids are the most important organic acids You find them in citrus fruits citric acid vinegar acetic acid aspirin acetylsalicylic acid and numerous other natural and synthetic compounds as well on numerous organic exams In this chapter you explore the structure synthesis and reactions of these acids and acids like them Part III Carbonyls Good Alcohols Gone Bad Seeing the Structure and Nomenclature of Carbaxylic Acids and Derivatives In Chapter 9 you see the basic structure of each of the carboxylic acids and carboxylic acid derivatives In this chapter we focus on the carboxylic acids and related compounds such as esters acyl chlorides and acid anhydrides and we also include some information on amides see Chapter 13 for an additional examination of amides Before you can get into synthesis and reactions though you need to understand the structure and nomenclature of these compounds Structure You see in Chapter 10 that aldehydes and ketones contain a carbonyl group attached to carbon or hydrogen atoms In the case of carboxylic acids and their derivatives a carbonyl group is attached to an electronegative element such as oxygen chlorine or nitrogen The presence of these elements tends to increase the 8 charge on the carbonyl carbon which makes the carbon atom more susceptible to nucleophilic attack The general formula for a carboxylic acid is RCOOH where R may be hydrogen or any alkyl or aryl group The derivatives vary slightly from that formula 1 In esters the OH is replaced with a OR 1 In acyl chlorides the OH is replaced with a Cl 1 In acid anhydrides two carboxylic acid molecules join with the removal of a water molecule to produce a molecule where an oxygen atom joins two carbonyl groups Nomenclature In Chapter 9 you see how to identify the different types of compounds containing the carbonyl group Now you take a look at the naming nomenclature of these compounds Finding out what carboxylic acids are called When naming carboxylic acids the final e of the hydrocarbon is replaced with either ic acid common name or oic acid IUPAC name The carbonyl carbon on the acid is position one When naming the salts the ic of the acid name is replaced with ate Chapter 12 Carboxylic Acids and Their Derivatives 7 Figure 121 Some typical carboxyhc acids Figure 122 Two carboxyhc acid salts Figure 121 shows the structures of some carboxylic acids and their common names and IUPAC names are given in the following list Figure 122 shows the structures and names of some of carboxylic acid salts V I formic acid or methanoic acid V II acetic acid or ethanoic acid V 111 propionic acid or propanoic acid V IV butyric acid or butanoic acid V V Valeric acid or pentanoic acid V VI 4methylpentanoic acid 0 O O H c CH3C CH3CH2C D OH H OH D OH 0 0 CH3CH2CH2C CH3CH2CH2CH2C V OH V OH O CH3CHCH2CH2C OH CH3 VI ll ll CH3CH216C cO K 0 Na Sodium stearate Potassium benzoate Designating dicarboxylic acids Molecules may contain more than one carboxylic acid group The dicarboxylic acidswhich contain two carboxylic acid groups are Very important in areas such as organic synthesis Many dicarboxylic acids have the general formula HOOCCH2nCOOH Table 121 lists how the names of the dicarboxylic acids relate to the Value of n 7 Part III Carbonyls Good Alcohols Gone Bad Table 121 Some Dicarboxylic Acids H006CH2nCOOH Value of n Dicarboxylic Acid n 0 Oxalic acid n1 Malonic acid n 2 Succinic acid n3 Glutaric acid n 4 Adipic acid A few important unsaturated dicarboxylic acids are shown in Figure 123 The position of the acid groups in a dicarboxylic acid is significant 1 If the two acid groups are ortho the acid is phthalic be sure to pronounce the th The phthalic acids are examples of aromatic dicarboxylic acids 1 If the two acid groups are meta the acid is isophthalic 1 If the two carboxylic acid groups are para the acid is terephthalic ii i i T C C C H Figure 123 HO C OH H0 C Two OH unsaturated H H H C dicarboxylic acids 0 T Maleic acid Fumaric acid Examining the nomenclature of esters As you see later in this chapter in the section Uniting acids and alcohols to make esters esters come from an alcohol and an acid The name of an ester reflects this origin The alcohol name appears first as an alkyl and the acid name comes second with the suffix ate replacing the ic acid part of the acid name Two examples of ester structures and names are in Figure 124 Figure 124 Examples of two esters with their names Figure 125 Examples of two acid anhyd des with their names Chapter 12 Carboxylic Acids and Their Derivatives O Ethyl acetate or CH3 C ethyl ethanoate O CH2 CH3 ii C OCH3 CH2 Dimethyl malonate OCH3 O Naming acid anIydrides Acid anhydrides form by joining two acids together When naming replace the word acid with the word anhydride For example two acetic acid mol ecules join to form acetic anhydride Dicarboxylic acids may react internally to form an acid anhydride See Figure 125 for some examples T C CH3 o Acetic anhydride CH3 0 i C 0 3methoxyphthalic anhydride ii OCH3 O Labeling acyl chlorides When naming an acyl chloride you simply need to replace ic acid with yl chloride See Figure 126 for three examples 797 7 Part III Carbonyls Good Alcohols Gone Bad Figure 126 Examples of three acyl chlorides with their names Figure 127 Examples of amides with their names T CH3C Acetyl chloride CI CH O 3 l lsobutyryl chloride or 0 CH3 H C methylpropanyl chloride Cl 0 ll Benzoyl chloride CI Clarifying amide nomenclature When naming amides replace the ic or 0ic acid with amide Each R group attached to the nitrogen is represented by an N at the beginning of the name See Figure 127 for some examples 0 l Acetamide or CH3C ethanamide NH2 0 CH3CH2CH2CH CH2C Hexanamide NH2 Nmethylpropanamide Chapter 12 Carboxylic Acids and Their Derivatives 7 Checking Out Some Physical Properties of Carbaxylic Acids and Derivatives quot Physical properties of carboxylic acids and derivatives include solubility melting point boiling point and a few other characteristics In this section we examine each class and discuss the most important physical properties In the upcoming section Considering the Acidity of Carboxylic Acids we discuss the most important chemical property of carboxylic acids acidity Carboxy Iic acids Carboxylic acids with six or fewer carbon atoms are soluble in water because of the polarity of the acid functional group and the ability of the acidic hydrogen atom to hydrogen bond Carboxylic acids with more than six carbon atoms react with and dissolve in either aqueous sodium bicarbonate or aqueous sodium hydroxide solution A useful means of distinguishing between larger carboxylic acids and phenols that don t dissolve in water is that the phenols dissolve in aqueous sodium hydroxide but don t dissolve in aqueous sodium bicarbonate Neither sodium hydroxide nor sodium bicarbonate affects the solubility of alcohols In general the carboxylic acids have disagreeable odors high melting points and high boiling points The high melting and boiling points are due to hydrogen bonding In some cases the hydrogen bonding is sufficient to hold two carboxylic acids molecules together as a dimer two molecules held together When this occurs the molecular weight MW appears to be about twice the weight of the acid For example a solution of benzoic acid MW 122 gmol in naphthalene has a molecular weight for the benzoic acid dimer of about 244 gmol Esters In general esters have sweet odors For this reason many are useful in perfumes or as flavorings The boiling points of esters are similar to those of aldehydes and ketones of comparable molar masses which means that the boiling points are lower than comparable alcohols 7 Part III Carbonyls Good Alcohols Gone Bad Amides If one or two hydrogen atoms are attached to the nitrogen atom hydrogen bonding can occur The presence of hydrogen bonding increases the melting and boiling points Considering the Acidity of Carboxylic Acids Figure 128 The induc ve effect Figure 129 Resonance stabiliza tion of the carboxylate ion The carboxylic acids are the most important of the organic acids Notice that many of the mechanisms in this chapter have one or more steps involving H transfer a sure sign of acidity The Ka values acid dissociation constants are normally between 10 4 and 105 indicating that an equilibrium has been established with only a small percentage of the weak acid in its dissociated form As acids they have a sour taste Vinegar is a 4 to 5percent solution of acetic acid Two factors enhance the acid behavior of the carboxylic acids The first factor is an inductive effect see Figure 128 which is a result of the electronwithdrawing power of the two oxygen atoms Note The arrows in the figure indicate the electronwithdrawing power of the two oxygen atoms The other factor is the resonance stabilization of the carboxylate ion see Figure 129 Remember that resonance stabilizes the molecular structure e O O Ill H I FlCOH R ltGgt H 0 8 ll 4 R O RC O e Chapter 12 Carboxylic Acids and Their Derivatives 7 The higher the Ka lower the pKa value the stronger the acid The acidity can be increased by adding electronwithdrawing groups to the R electron donors have the opposite effect For example the acidity of acetic acid increases as chlorine atoms replace hydrogen atoms Acetic acid has Ka 176 x 10 5 chloroacetic acid has Ka 140 x 10 3 dichloroacetic acid has Ka 332 x 102 and trichloroacetic acid has Ka 200 x 10 The distance the electronwithdrawing group is from the carboxylic acid group is also important For example butanoic acid has Ka 15 x 10 5 4chlorobutanoic acid has Ka 3 x 10 5 3chlorobutanioic acid has Ka 89 x 10 5 and 2chlorobutanoic acid has Ka 14 x 103 This shows that the chlorine is more effective the closer it gets to the carboxylic acid group For the aromatic carboxylic acids substituents on the aromatic ring may also influence the acidity of the acid Benzoic acid for example has Ka 43 x 10 5 The placements of various activating groups on the ring decrease the value of the equilibrium constant and deactivating groups increase the value of the equilibrium constant Table 122 illustrates the influence of a number of parasubstituents upon the acidity of benzoic acid Table 122 Comparison of Ka Values of Benzoic Acid to ParaSubstituted Benzoic Acids Benzoic acid Ka 43 gtlt10 5 ParaSubstituted Benzoic Acid Activating Groups K3 Value 0H Ka 28 x 10 5 OCH3 Ka35gtlt10 5 CH3 Ka 43 gtlt10395 ParaSubstituted Benzoic Acid Deactivation Groups K3 Value BrCI Ka 11 gtlt10 4 CHO Ka18gtlt10 CN Ka 28 gtlt10 4 N02 Ka39gtlt10 4 The dicarboxylic acids have two Ka values with Kal gtgt Ka The second Ka value is lower because the loss of the first acidic hydrogen leaves an anion which can backdonate electron density inductive effect The difference between the Ka values decreases as the value of It increases for the series HOOCCH2nCOOH 7 Part III Carbonyls Good Alcohols Gone Bad Determining How Carboxulie Acids and Derivatives lre Synthesized The syntheses of carboxylic acids and their derivatives are important reac tions in organic chemistry In this section you take a look at several ways to make these compounds Synthesizina carboxylic acids A number of methods are used in the synthesis of carboxylic acids Most of these methods involve the oxidation of some organic molecule but other methods can be used In this section we take a look at a few of these methods Oxidation of alkenes The synthesis of carboxylic acids by the oxidation of alkenes is a twostep process In the first step a hot basic potassium permanganate KMnO4 solution oxidizes an alkene and in the second step the oxidized alkene is acidified The process cleaves the carbon backbone at the carboncarbon double bond to produce two smaller carboxylic acid molecules For example oleic acid CH3CH27CHCHCH27COOH yields of mixture of nonanoic acid CH3CH27COOH and nonadioic acid IIOOCCH27COOH Oxidation of aldehydes and primary alcohols The oxidation of either primary alcohols or aldehydes doesn t change the carbon backbone so you end up with a carboxylic acid containing the same number of carbon atoms as the aldehyde or alcohol Alcohols require considerably stronger oxidizing conditions than aldehydes do The oxidation of a secondary alcohol gives a ketone and neither ketones nor tertiary alcohols readily oxidize The oxidants for alcohols include one of the following V Hot acidic potassium dichromate K2Cr2O7 V Chromium trioxide CrO3 in sulfuric acid II2SO4Jones reagent V Hot basic permanganate followed by acidification An example of this type reaction is the conversion of 1decanol CH3CH28CH2OH to decanoic acid CH3CH28COOH Chapter 12 Carboxylic Acids and Their Derivatives 79 7 Figure 1210 The oxida tion of pnitrotol ueneto pnitroben zoic acid The oxidants for aldehydes include one of the following V Any reagent that can oxidize an alcohol V Cold dilute potassium permanganate V A number of silver compounds including AgNH32OH and Ag2O in base followed by acidification V Air over a long period of time An example of this type of reaction is the conversion of hexanal CH3CH24CHO to hexanoic acid CH3CH24COOH Oxidation of alkyl benzenes Strong oxidizing agents are capable of attacking alkyl benzenes if the carbon atom nearest the ring has at least one hydrogen atom attached When this occurs the oxidation removes all of the alkyl group except the carbon atom closest to the ring Oxidizing agents include the following V Hot acidic potassium dichromate solution V Hot 95 degrees Celsius potassium permanganate solution followed by acidification Figure 1210 illustrates the reaction of pnitrotoluene to form pnitrobenzoic acid CH3 00 1 KMnO4H2095 2 H 7 OH NO 2 N02 pnitrotoluene pnitrobenzoic acid 88 o Oxidation of methyl ketones In general ketones don t undergo oxidation however methyl ketones undergo a haloform reaction In a haloform reaction the oxidation converts the methyl group to a haloform molecule usually iodoform CHl3 which leaves the carbon backbone one carbon atom shorter The oxidant in a haloform reaction is sodium hypohalite NaOX which forms by the reaction of sodium hydroxide NaOH with a halogen X2 where X Cl Br or I Figure 1211 illustrates the oxidation of a methyl ketone 798 Part III Carbonyls Good Alcohols Gone Bad Figure 1211 The oxida CH CH C CH CH C tion of a CH methyl 3 ketone NaOX gt o Na CHX3 Hydrolysis of cyanohydrins and other nitrile The basic hydrolysis reaction with water of a nitrile RCN followed by acidification yields a carboxylic acid In general an SN reaction nucleophilic substitution of an alkyl halide is used to generate the nitrile before hydrolysis Figure 1212 illustrates the formation of a carboxylic acid beginning with an alkyl halide O CIDH CH3 Br 1 NaCN 2 OH H20 3 H Figure 1212 Formation of carboxyhc acid from alkyl halide 0 CH CH3 using the I hydrolysis of C a nitrile The product of the reaction in Figure 1212 is fenoprofen an antiarthritic agent Carbonation of Grignard reagents After multiple steps an organic halide can be converted to a carboxylic acid The organic halide converts to a Grignard reagent which reacts with carbon dioxide and then acidification forms the acid Figures 1213 and 1214 illustrate the steps in this process Chapter 12 Carboxylic Acids and Their Derivatives 7 Figure 1213 The use of a Grignard reagent to form car boxylic acid Figure 1214 Another example using a Grignard reagent to form car boxylic acid Br MgBr CH3 CH3 CH3 CH3 M9 T Et2O CH3 CH3 1 CO2Et2O 2 H COOH CH3 CH3 87 CH3 246trimethylbenzoic acid M9 CH3CH2CH2CH2C lit 0 CH3CH2CH2CH2MgC 2 1 CO2Et2O 2 H CH3CH2CH2CH2COOH 73 Pentanoic acid Developing acyl halides with halogen The formation of an acyl halide involves the reaction of a carboxylic acid with a halogen source The common halogen sources are compounds like PX3 PX5 CIOCCOCI oxalyl chloride or SOX2 where X is a halogen The most commonly used acyl halides are the chlorides and the simplistic reaction is RCOOH RCOCI Figure 1215 illustrates the mechanism using thionyl 06M Part III Carbonyls Good Alcohols Gone Bad Figure 1215 Formation of acyl chlo ride by the reaction of carboxyhc acid with thionyl chloride quot chloride SOCI2 as the halogen source One aid in the reaction is the formation of a transition state containing a sixmembered ring This reaction works because the SO2Cl is a better leaving group than Cl quoto CI 39339 CH3Clt J3r3 CH35S OH I l 39 C cH CI 39339 o 39039 o CH3C HCI 802 j CH3AT Cl 0 CI fR H Removing water to form acid anhydrides The name anhydride means without water so that makes it pretty clear that the general idea behind the formation of an acid anhydride is to remove water from a carboxylic acid Both acyl chlorides and acid anhydrides are very effective at removing water In some cases heat can be used to remove water Sodium salt plus acid chloride The reaction of a carboxylic acid with sodium hydroxide NaOH produces the sodium salt of the carboxylic acid The sodium salt then reacts with an acid chloride to form the anhydride Figure 1216 illustrates the final step in this process In this reaction the carboxylate ion behaves as a nucleophile and attacks the carbonyl carbon atom of the acid chloride The reaction of a carboxylic acid with sodium hydroxide also generates water which if not removed reacts with the acid chloride and lowers the yield of the reaction This synthesis can produce either a symmetric anhydride both acids the same or an asymmetric anhydride different acids Chapter 12 Carboxylic Acids and Their Derivatives 0 Z 7 Figure 1216 The reaction of sodium salt of car boxylic acid sodium formate with acid chloride Figure 1217 Acid anhydnde formed by the reaction of acid chloride with carboxyhc acid in the presence of pyridine O quot0 39 CH C 390 3 39 E H C CH C t2O 0 NaC e 3 25 O 0 0 640O N O Acetic formic anhydride lcid plus acid chloride plus pyridine This process that happens when you combine an acid with an acid chloride with pyridine is similar to the reaction of a sodium salt with an acid chloride The pyridine behaves as a base in place of the sodium hydroxide The advantage of this process is that no water forms to react with the acid chloride Figure 1217 illustrates this reaction 0 cH3 clt o CH cO Clt N O 3 CI OH o Q or H lcetic anhydride plus acid The dehydrating properties of an acid anhydride can be used to produce another acid anhydride This is an equilibrium process By heating the mixture the more Volatile acid Vaporizes to shift the equilibrium toward the products Acetic acid from acetic anhydride is useful because it s more Volatile than most other carboxylic acids Figure 1218 illustrates this reaction 202 Part III Carbonyls Good Alcohols Gone Bad Figure 1218 Formation of acid anhy dride by the reaction of carboxyhc acid with ace c anhyd de Figure 1219 The thermal dehydration of a dicarboxylic acid to form a cyclic anhyd de O H O 0 COH CH3C A Clt 2 0 0 CH3 C C O Q 39O 2 CH3Clt OH Cyclic antyolrioles A cyclic anhydride can be formed from a dicarboxylic acid by heating if the anhydride that forms has a five or sixmembered ring If the dicarboxylic acid contains a ring only a cis isomer not the trans isomer reacts Figure 1219 illustrates this type of reaction The black circles in the figure indicate that the cisisomer is reacting to form a cis product o CO2H A C o CO2H 00 Llnitina acids anal alcohols to make esters An ester consists of an alcohol portion and a carboxylic acid portion In the synthesis of an ester these two portions need to be brought together The simplest method is to react an acid with an alcohol in the presence of another alcohol but as you see in the following sections other methods are useful as well Chapter 12 Carboxylic Acids and Their Derivatives 5 Figure 1220 Acid catalyzed formation of an ester from an alcohol and a carboxylic acid Acid plus alcohol This method is called the Fischer esterification It s a condensation reaction where the loss of a water molecule accompanies the joining of the alcohol portion to the acid portion The acid gives up the OH and the alcohol gives up the H to make the water molecule All steps in the mechanism are reversible that is it establishes an equilibrium so removing the ester as soon as it forms is helpful Removal of the ester is normally easy since esters typically have lower boiling points than alcohols and carboxylic acids Figure 1220 illustrates the mechanism for the acidcatalyzed formation of an ester by the reaction of an alcohol with a carboxylic acid KT W C C oH HA o cH3 CH3OH T 39 OH OH H H More positive than normal O CH3 oH H C C EC 1 OH OH H lt gt OH H20 OH OH Part III Carbonyls Good Alcohols Gone Bad Figure 1221 The general method for ester syn thesis from an alcohol and an acid chloride Figure 1222 The reaction of an acid chloride with an alcohol using pyridine to trap the HCI formed lciol chloride plus alcohol This method is easier than the reaction of an alcohol with a carboxylic acid described in the preceding section because acid chlorides are more reactive than acids The reaction forms HCI with either a hydroxide ion or pyridine aiding in the removal of the HCI The general reaction is in Figure 1221 and the reaction using pyridine to scavenge the HCI is in Figure 1222 0 O J CH3OH gt J HCI CH C CH OCH3 0 i R CH3 CH3 CI C J CH3 CH3 N Cl H lciol anhydride plus an alcohol Acid anhydrides are between acid chlorides and carboxylic acids in reactivity so this reaction is more effective than the reaction with a carboxylic acid but less efficient than the reaction with an acid chloride Half of the acid anhydride goes into forming the ester while the other half becomes a carboxylic acid Figure 1223 illustrates this reaction using salicylic acid as the alcohol and acetic anhydride as the acid anhydride to form aspirin an ester 206 Part III Carbonyls Good Alcohols Gone Bad Figure 1225 The forma tion of a methyl ester utilizing diazo methane Figure 1226 The forma tion of an amide by the reaction of an acid chloride with ammonia O10 010 OH 0 Et 0 CH2N2 2 39N2 Bringing acids and bases together to create amides Amides contain an acid portion and an amine portion However unlike the formation of an ester the reaction of a carboxylic acid with an amine is not an efficient method for preparing an amide because as you see in this section the simple reaction of an acid carboxylic acid with a base amine causes interference Fortunately methods similar to many of the other ester synthesis methods are useful in the synthesis of amides From acid chlorides Acid chlorides are Very reactive and they readily react with ammonia primary amines or secondary amines to form an amide Figure 1226 illustrates the reaction of an acid chloride with ammonia Replacing one or two of the hydrogen atoms of ammonia with an organic group will result in an Nsubstituted amide Tertiary amines react with acid chlorides to form a carboxylic acid and an ammonium salt 0 O J T ll HCI ll T F p fir F NH 3I39 H Figure 1227 The prepa ration of phenace n bythe reaction of an amine with an acid anhydnde Figure 1228 The formation of benzamide bythe reaction of ammonia with methyl benzoate Chapter 12 Carboxylic Acids and Their Derivatives h 7 From acid anhydride This process is similar to the formation of an ester by the action of an acid anhydride on an alcohol described in the earlier section Acid anhydride plus an alcohol Half the acid anhydride forms the amide the other half is a leaving group Ammonia primary amines and secondary amines react to produce amides Figure 1227 shows the industrial preparation of phenacetin by the reaction of an amine with an acid anhydride The mechanism for this reaction is similar to the mechanism for the reaction of an acid chloride with an amine refer to Figure 1226 ll NH 2 0 HNCCH CH3C 3 NaOH o T CH 00 39 H20 3 2 CH3C OCHZCH3 O OCHZCH3 From esters Amines also react with esters by a method similar to the reaction of an acid chloride with an amine which was described in the previous section From acid chlorides Figure 1228 illustrates the formation of benzamide by this type of reaction using ammonia and methyl benzoate Again the mechanism is similar to the reaction of an acid chloride with an amine Figure 1226 co2cH3 COZNH2 NH3 CH3OH Ether q Part III Carbonyls Good Alcohols Gone Bad Figure 1229 The reaction of ammonia with a carboxyhc acid to eventually form an amide From carboxylic acids and ammonium salts The reaction of an amine or ammonia with a carboxylic acid first produces an ammonium salt which upon heating loses water and produces an amide This is a low yield process Figure 1229 shows an example of this type of reaction H20 Exploring Reactions Figure 1230 Nucleophilic attack at the acyl carbon atom showing its relationship to an SN2 process Many of the reactions of carboxylic acids and derivatives involve nucleophilic substitution at the acyl carbon atom This is a bimolecular process with the acyl group having a leaving group L The general mechanism is reminiscent of an SN2 mechanism Figure 1230 illustrates the general process and Figure 1231 gives more details on the mechanism 0 0 ll NuCL NuC L F110 Nu V0 R R L The various carboxylic acid derivatives vary in their reactivity stability of the leaving group Acid chlorides for example are more reactive than anhy drides don t leave as easily A summary of the relative reactivities appears in Figure 1232 Chapter 12 Carboxylic Acids and Their Derivatives Figure 1231 6 The general mechanism Nu O p O forthe P N C L G nucleophilic RC gt U gt NuC L3 attack on an l acyl carbon L R H atom Figure 1232 0 O O O O ThereatIve R C gt R C gt R C gt R C gt R C reactivities Ofthe Cap CI 0 OR NH2 OH boxylic acid RC derivatives O NBEI9 The sequence in Figure 1232 not only represents the general reactivity of lt1 carboxylic acid derivatives but also gives information on the ease of synthesis The more reactive a species is the more difficult it is to prepare it and vice versa From this series you can see that synthesizing a less reactive acyl compound from a more reactive acyl compound is always possible Generous carboxylic acids We start this section by giving your brain a rest with some simple material Carboxylic acids are acids Acids donate a hydrogen ion Ht to other species Therefore that s the fundamental reaction of carboxylic acids Rested Good As seen previously these are weak acids although they re stronger than most other organic acids In this chapter we have seen a variety of other reactions such as the formation of an ester that utilize carboxylic acids as one of the reactants Other reactions follow The HellVolhard Zelinsky reaction is a method for forming ochalo acid This is a synthetically useful procedure because the ochalo acids are useful starting materials for other reactions For example the addition of hydroxide ion leads to the replacement of the halogen with an OH group The reaction with ammonia replaces the halogen with NH2 The reaction with cyanide ion CN converts the halide to a nitrile Figure 1233 illustrates this reaction 270 Part III Carbonyls Good Alcohols Gone Bad Figure 1233 H O 3r 0 An example 01 the Hell c 3c 2c 2c 2c 2c c 1 BrP CH3CH2CH2CH2CH2CHC HBr Volhard 2 H20 OH ZemSky OH 2bromoheptanoic acid reaction 90 Simple acyl halide and anhydride reactions Both acyl halides and anhydrides react with water hydrolysis Acyl halides react to form one mole of the carboxylic acid and one mole of the hydrohalic acid HX Anhydrides react to form two moles of carboxylic acid Acyl halides and anhydrides are important reactants for the formation of other carbonyl compounds but you don t need to take up valuable brain space with information about any other acyl halide or anhydride reactions at this time Hydrolysis of esters Esters can undergo hydrolysis using either an acid or a base as a catalyst Hydrolysis always produces an alcohol from the alkyl portion of the ester During acid hydrolysis the acid portion of the ester yields a carboxylic acid During base hydrolysis of an ester which is called saponification the acid portion of the ester yields the carboxylate ion lcid hydrolysis Acid hydrolysis is the reverse of the Fischer esterification seen earlier in the section Acid plus alcohol Figure 1234 illustrates the mechanism Base hydrolysis saponification Saponification base hydrolysis follows a simpler mechanism see Figure 1235 In the reaction one mole of hydroxide generates one mole of alcohol and one mole of carboxylate ion from one mole of ester Based on this stoichiometry the mole relationship as defined by the balanced chemical equation if the number of moles of base is known then the amount of ester is known This stoichiometry is the saponification equivalent used to determine the equivalents of ester Chapter 12 Carboxylic Acids and Their Derivatives 2 7 7 KW 39iO I o I Figure 1234 The mechanism for acid hydrolysis of an ester Figure 1235 The mechanism for base 0 hydrolysis of an ester C BE syn The moles of base moles of ester Amide reactions ester s cousins The reactions of amides have similarities to those of esters Specifically the reactions covered in this section involve the loss or gain of water dehydration or hydrolysis 2 72 Part III Carbonyls Good Alcohols Gone Bad Figure 1236 The mechanism forthe acid hydrolysis of an amide Figure 1237 The mechanism forthe base hydrolysis of an amide lcid or basecatalyzed hydrolysis Acid hydrolysis of an amide yields a carboxylic acid and an ammonium ion The mechanism for acid hydrolysis is shown in Figure 1236 Base hydrolysis of an amide on the other hand yields ammonia and a carboxylate ion You can see this mechanism in Figure 1237 To identify similarities compare these mechanisms to the mechanisms for the hydrolysis of esters refer to Figures 1234 and 1235 to oH oH H l H CuT25 I m3 C RC R ryH2 R NH2 IILH2 H Y W NH4 8H NH3 R c c gH R C o H G C QH R C o gt R c o o R 39 39H2 NH NH2 H 2 02 L H 9 NH3 R C Q Dehydration Amides undergo dehydration Useful dehydrating agents include SOCl2 P 4010 P205 AcO2O and POCI3A The product is a nitrile and in fact dehydration of an amide is one method to produce aryl nitriles Figure 1238 shows the synthesis of 2ethylhexanenitrile from 2ethylhexanamide with a 94percent yield Chapter 12 Carboxylic Acids and Their Derivatives 2 Figure 1238 The synthesis of 2ethyhex anenitrile from 2ethy hexanamide Figure 1239 The decom position of carbonic acid o SOC2 cH3cH2cH2cH2cH c NH2 T CH3CH2CH2CH2CH CN 6 6 80 C CHZCH3 CHZCH3 302 2 HCI Other reactions of carboxylic acids and derivatives In this section we give you a quick look at a few additional and potentially useful reactions of carboxylic acids and compounds derived from carboxylic acids derivatives Carbonic acid derivatives Carbonic acid HZCO3 is a diprotic acid lt s unstable and decomposes to carbon dioxide and water see Figure 1239 00 gt CO2 H20 OH The replacement of both OH groups with chlorine produces carbonyl dichloride also known as phosgene a useful reactant For example phosgene reacts with two moles of alcohol to form a dialkyl carbonate The reaction of phosgene with one mole of alcohol produces an alkyl chloroformate which is a useful intermediate in organic syntheses The reaction of phosgene with four moles of ammonia yields urea and two moles of ammonium chloride NH 4C1 Figure 1240 shows the structures of some of these compounds One useful reaction utilizing alkyl chloroformate is the reaction with an amine in base to form a carbamate urethane Figure 1241 illustrates this reaction 274 Part III Carbonyls Good Alcohols Gone Bad Figure 1240 Some important carbonic acid derivatives Figure 1241 The reaction of an alkyl chlorofor mate with an amine inthe presence of a base Figure 1242 The decom position of carbamic acid O O O O 11 11 ll CICC RO OR H2N NH2 RO C Phosgene Dialkyl carbonate Urea Alkyl chloroformate O 0 ll R39NH2 lcl Ro C OH Ro NHR39 Another useful carbonic acid derivative is carbamic acid Like carbonic acid carbamic acid is unstable see Figure 1242 m CO2 NH3 00 HO NH2 Decarboxylation Decarboxylation is the loss of carbon dioxide which happens easily because of the stability of CO2 Heating Bketo acids to between 100 and 150 degrees Celsius is one example of a decarboxylation reaction The mechanism for the decarboxylation of a Bketo acid is in Figure 1243 Chapter 12 Carboxylic Acids and Their Derivatives 2 d Figure 1243 The mecha nism forthe decarbox ylation of a Bketo acid Figure 1244 The forma tion of the starting material forthe Hunsdiecker reac on Figure 1245 The free radical mechanism of the Hunsdiecker reac on F H 0 C0 OH O l 100 C CO2gt Pa H CCOx C C 0 l Enol Ketone Hunsdiecker reaction The Hunsdiecker reaction is a freeradical reaction for the synthesis of an alkyl halide The starting material comes from the reaction of a silver carboxylate with a solution of a halogen in a solvent such as carbon tetrachloride see Figure 1244 The overall freeradical mechanism is shown in Figure 1245 0 CCI4 v RCOBr R CO239 Ag Big AgBr Initiation Br Propagation O H R a 0 RE R1010 Br 276 Part III Carbonyls Good Alcohols Gone Bad Figure 1246 Additional examples of the Hunsdiecker reac on Figure 1247 A Bhydoxy ester Figure 1248 The Reform atsky reac on Other reagents can be used in the Hunsdiecker reaction as shown in Figure 1246 HgOBr2 CH3CH215CH2COOH T CH3CH215CH2Br CO2 0394 93 COOH I Pb4I2 gt CCI4 C02 The Reformatsky reaction The Reformatsky reaction uses an organozinc intermediate to form Bhydoxy esters see Figure 1247 The general Reformatsky reaction is in Figure 1248 and the mechanism is in Figure 1249 OH i T i C CH3 Br 393 C E CH3 0 Zn OH 0 I I ll c c c l l E CH3CH3 0 Chapter 12 carboxylic Acids and Their Derivatives 2 7 7 T i T i Zn BrC C j Zn C C Et Br Et CH3 0 CH3 0 Co 11 CH3 v r C Figure 1249 0 H 0 The mecha l l nism forthe CCC Reformatsky l l reaction C C Taking a Look at Spectroscopy and Chemical Tests The carbonyl stretch in the 1700 cm region of the infrared spectra of carbonyl compounds is a Very obvious feature of the spectrum for these compounds In this section we look at some other spectral features of carboxylic acids and their derivatives and also at some chemical tests that can help you determine what you re dealing with 2 78 Part III Carbonyls Good Alcohols Gone Bad Identifying compounds with spectral data You can use the unique spectroscopy of carboxylic acids and derivatives described in the following list to help you identify those compounds V Carboxylic acids In addition to the carbonyl stretch the infrared spectra of carboxylic acids also have a broad OH stretch which is often shifted to the 3300 2500 cm region The acid hydrogen is in the 8 1012 region of the proton NMR spectrum V Esters In addition to the carbonyl stretch the infrared spectra contains two CO stretches in the 1300 1050 cm region This is the result of having two different R groups attached to the singly bonded oxygen atom V Acyl chlorides ln acyl chlorides the carbonyl stretch appears in the 1850 1780 cm region V Amides The carbonyl stretch of amides is in the 1690 1630 cm region If one hydrogen atom is attached to the nitrogen atom the amide has an NH stretch If two hydrogen atoms are attached to the nitrogen it has two NH stretches The NH stretches are in the 3500 3300 cm region Other derivatives have similar spectral properties For example acid anhydrides are similar to esters Using chemical tests Carboxylic acids are soluble in either aqueous NaOH or NaHCO3 The other common group of organic acids phenols are weaker than the carboxylic acids Phenols are only soluble in aqueous NaOH Di and trinitrophenols are stronger acids than most other phenols so they are also soluble in aqueous NaHCO3 The neutralization equivalent is a useful means of determining the molecular weight of a carboxylic acid The process begins with a simple neutralization reaction of acid with standard base usually sodium hydroxide The reaction is Acid NaOH Na carboxylate H20 Written in this form you see that the equivalents or milliequivalents of acid are equal to the equivalents of base The equivalent weight of the acid is the grams of acid divided by the equivalents of base The equivalent weight of monoprotic acid is equal to the molecular weight The equivalent weight of diprotic acid is equal to half the molecular weight Part IV Advanced Topics Every Student39s Nightmare The 5th Wave By Rich Tennant Aaron was envied 0r being the Eirst student in his chernistrg class with an i 3peetronneter CHEMIS Pl HTEr4rxnrr F tf5quot In this part Now don t be scared the topics in this part aren t all that advanced First you look at nitrogen compounds especially the amines You study synthesis reactions and then the reactions of these nitrogen compounds We even spend a little time on the synthesis of sulfa drugs Next you shift your focus to organometallics those organic compounds that incorporate metals into their structure Here you study one of the most useful organic reactions the Grignard reaction Then you get a chapter on additional reactions of car bonyl compounds In this chapter we take a look at more condensation reactions as well as some ether synthesis reactions Finally this part finishes up with a chapter on biomole cules If your Organic class doesn t cover this fun and exciting topic feel free to skip this chapter Chapter 13 Amines and Friends In This Chapter Looking at the structure nomenclature and properties Exploring how basicity affects nitrogen compounds Figuring out their synthesis and reactions Assessing multistep synthesis Analyzing and using spectroscopy to determine nitrogen compounds A mines and amides are the most important nitrogencontaining organic compounds Amides are carboxylic acid derivatives which we cover in Chapter 12 In this chapter we focus on amines Amines are nitrogen compounds where the nitrogen atom is attached to one to four organic groups Breaking Down the Structure and Nomenclature of Nitrogen Compounds Amines can be split into two general types primarysecondarytertiary quaternary aliphaticaryl and heterocyclic In aliphaticaryl amines a nitrogen atom is attached to one or more alkylaryl groups In heterocyclic amines the nitrogen atom is part of a ring system Amines may be one of the following V Primary 1 One carbon atom connected directly to the nitrogen atom V Secondary 2 Two carbon atoms connected directly to the nitrogen atom V Tertiary 3 Three carbon atoms connected directly to the nitrogen atom V Quaternary 4 Four carbon atoms connected to the nitrogen atom Part IV Advanced Topics Every Student39s Nightmare Figure 131 Some examples of primary amines Figure 132 Some aromatic amines In quaternary amines which are technically quaternary ammonium salts the nitrogen atom has a positive charge Primary amines The common names of primary amines consist of the name of the alkyl branch followed by the name amine The systematic IUPAC name of primary amines consists of the name of the alkane with the e replaced by the suffix amine Some examples of primary amines appear in Figure 131 If more than one amine group is present you need to use the appropriate prefix For example HZNCHZCHZCHZCHZNHZ is 14butandiamine because the amino groups are attached to the first and fourth carbons CH3CH2NH2 NH2 CH2 NH2 Ethyl amine Ethanamine Cyclopentyl amine Cyclopentanamine Benzyl amme Phenylmethanamine Aniline is a simple aromatic compound composed of an amino group attached to a benzene ring Other aromatic amines are aniline derivatives Some examples of aromatic amines are shown in Figure 132 NH2 CH3 Anmne Toluidine The amine group has a lower priority in numbering the ring positions around the ring than OH and other oxygencontaining groups Some examples showing the lower priority of the amine group are in Figure 133 Chapter 13 Amines and Friends Figure 133 Some primary amines con taining other func onal groups Figure 134 Some secondary and tertiary amines NH2 COOH CH3CH2CHCOOH H2 NH2CH2CH2CCI3 239arninObUtanOiC aCid 4arninO2butanOne NH2 24diaminobenzoic acid Secondary and tertiary amines The common names of secondary and tertiary amines are an extension of the common names of primary amines where the organic groups are named as branches followed by the word amine For example CH3CH23N is triethyl amine The systematic IUPAC names utilize the procedure for primary amines the name of the alkane with the e replaced by the suffix amine plus the names of the remaining organic groups preceded by an N to indicate attachment to the nitrogen atom Some names of secondary and tertiary amines appear in Figure 134 CHsgt2NH N CH2CH2CH3 Dimethyl amine CH3 Nmethylmethanamine Dimethyl propyl amine NNdimethylpropanamine Quaternary amines quaternary ammonium salts Ammonium salts contain a nitrogen atom with four bonds that has a positive charge Fourbonded nitrogen atoms derived from amines are ammonium ions if they re derived from aniline they re anilinium ions If the four bonds are all to carbon atoms the nitrogen atom is quaternary Salts contain a cation named first and an anion named last Typical anions include Cl chloride Br bromide HSO4 hydrogen sulfate or bisulfate and N03 nitrate Figure 135 shows two examples of ammonium ions Part IV Advanced Topics Every Student39s Nightmare Figure 135 Two examples containing ammonium ions Figure 136 Two additional examples of amine nomen clature CH3 6 CH2 NH3 6 Cl CH3 Br CH3 Tetramethylammonium chloride Benzylammonium bromide Two additional examples of amine nomenclature are shown in Figure 136 Note that in the paminobenzoic acid the oxygencontaining group takes precedence in the naming so that the compound is then named as a substituted benzoic acid and not a substituted aniline as is done in the NNdimethylaniline beside it COOH NH2 paminobenzoic acid NNdimethylaniline Heterocyclics Heterocyclics are ring systems containing something other than carbon in the ring The names of some nitrogencontaining heterocyclic compounds with a single nitrogen are listed in Figure 137 Heterocyclic systems may contain more than one nitrogen atom though and some examples are shown in Figure 138 These heterocyclics are important in biological and biochemical systems For a detailed discussion of these topics see Biochemistry For Dummies Chapter 13 Amines and Friends T Figure 137 Some examples of nitrogen containing heterocyclic compounds Figure 138 Some examples of nitrogen containing heterocyclic compounds containing more than one nitrogen atom T Pyridine Pyrimidine I I N H Pyrrole Pyrrolidine Piperidine N lt T N H Pu ne Sizing up the Physical Properties The high electronegativity of the nitrogen atom means that the carbon nitrogen bond of amines is polar This results in an attraction between two polar molecules dipoledipole intermolecular forces which increases the melting and boiling points above those of hydrocarbons However in primary and secondary amines the ability to hydrogen bond overshadows the simple dipoledipole forces present in tertiary amines For this reason primary and secondary amines have much higher melting and boiling points than tertiary amines Hydrogen bonding in amines is not as strong as hydrogen bonding in alcohols therefore the melting and boiling points of amines are lower than those of comparable alcohols See Chapter 3 for a discussion of hydrogen bonding in alcohols R Part IV Advanced Topics Every Student39s Nightmare In general amines with up to about six carbon atoms are soluble in water Primary secondary and tertiary amines react with acids to produce ammonium ions These ammonium ions and quaternary ammonium ions are also soluble in water Amines can be separated from other organic compounds by treating the mixture with an aqueous acid converting the amines to ammonium ions These ammonium ions dissolve in the aqueous acid leaving the other organic materials behind Separating the aqueous layer results in an acid extract containing the ammonium ions Adding base to the acid extract until the solution is basic converts the ammonium ions back to the free amines which then separate from the aqueous layer Many amines have a very distinctive odor like dead fish or worse Understanding the Basicity of Nitrogen Compounds QMBER lt Primary secondary and tertiary amines react with acids to form amine salts This is because of the basic nature of the amines Amines are both BronstedLowry bases they accept hydrogen ions from acids and Lewis bases they furnish an electron pair to Lewis acids As BronstedLowry bases they have Kb values Aliphatic amines have Kb values of approximately 104 and aromatic amines have values near 1010 These values compare to a value of z10 5 for ammonia The increase in the Kb values of aliphatic amines thus making them stronger bases as compared to ammonia is due to the electronreleasing nature of alkyl groups This release of electrons pumps electron density back to the nitrogen atom which stabilizes the positive charge Compared to the aliphatic amines the aromatic amines have lower Kb values This lower value indicates that the product of the protonation of aromatic amines is less stable The decrease in stability is due to a loss in resonance stabilization of the protonated form An amine group attached to an aromatic ring is an activating o pdirector as seen in Chapter 8 The lone pair of electrons on the nitrogen atom makes the amines Lewis bases As Lewis bases they may behave as nucleophiles Because aromatic amines are resonance stabilized they re weaker nucleophiles than alkyl amines Chapter 13 Amines and Friends 7 Synthesizino Nitrogen Compounds Figure 139 The basic reac on for a nucleophilic substitution reaction to produce an amine Amines can be prepared a number of ways These methods include nucleophilic substitution reactions reduction reactions and oxidation reactions Nucleophilic substitution reactions Synthesizing amines with nucleophilic substitution reactions is normally an SN2 process This means that methyl amines react more readily than primary amines and secondary and tertiary amines show Very little reactivity Figure 139 illustrates the basic reaction The resultant amine may react further to give a mixed group of products as shown in the following reaction Using a large excess of ammonia minimizes the chances for multiple alkylations K 82 OH HsN R X Ngt C iisa X T NH2R NH3 CH3CH26CH2Br NH3 CH3CH26CH2NH2 octylamine 45 CH3CH26CH22NH dioctylamine 43 CH3CH26CH23N trioctylamine trace CH3CH26CH24N trace Aromatic halides don t react unless an electronwithdrawing group is attached to the ring For example bromobenzene doesn t react An example showing the reaction when an electronwithdrawing group is present is illustrated in Figure 1310 the nucleophilic substitution attack on pbromonitrobenzene The azide ion is a better nucleophile than amines but it has to be reduced to the amine after nucleophilic substitution Lithium aluminum hydride LiAlH4 in ether followed by treatment with water reduces the azide ion to the amine Figures 1311 and 1312 illustrate two examples of this reaction A Part IV Advanced Topics Every Student39s Nightmare CH3 CH3 T Br N Figure 1310 The nucleophilic CH32NH substitution F attack on obromonitro benzene N02 N02 Azides are explosive releasing nitrogen gas That s why they re used to inflate car air bags CHQCHQBT CH2CH2N NaN3 EtOH NHBI 1 LiAH4ether Figure 1311 The prepa CH2CH2NH2 ration of an amine from 89 an azide Br N3 NHN3 EtOH N8B39 T 1 LiAH4ether Figure 1312 2 H20 Another preparation NH2 of an amine from an 54 azide Chapter 13 Amines and Friends Figure 1313 The prepa ration of potassium phthalimide from phthalimide and potassium hydroxide Figure 1314 Using the Gabriel synthesis to produce an amine The Gabriel synthesis of amines uses potassium phthalimide prepared from the reaction of phthalimide with potassium hydroxide The structure and preparation of potassium phthalimide is shown in Figure 1313 The extensive conjugation resonance makes the ion Very stable An example of the Gabriel synthesis is in Figure 1314 The NZH4 reactant is hydrazine The Gabriel synthesis employs an SN2 mechanism so it works best on primary alkyl halides and less well on secondary alkyl halides It doesn t work on tertiary alkyl halides or aryl halides O KOH lt9 N H T N K O Phthalimide Potassium phthalimide O G 1 N Kt CH2Br 39 DMF cH2NH2 0 g 81 2 N2H4 Reduction preparations We can reduce a number of nitrogen species to an amine In the following sections we take a look at some of the methods that can be used Nitro reductions Organic nitro compounds RN02 can be reduced to amines The R may be either alkyl or aryl Aromatic nitro compounds are easy to prepare and reduce Their preparation utilizes a mixture of nitric acid and sulfuric acid to nitrate the aromatic ring However multiple nitrations may occur potentially causing problems The nitro group can be reduced with a 230 Part IV Advanced Topics Every Student39s Nightmare Figure 1315 The preparation of a nitro compound followed by reduction to an amine Figure 1316 The prepa ration of amphet amine by reduc ve amination number of simple methods including catalytic hydrogenation of the nitro compound or the reaction of the nitro compound with a metal Fe Zn or Sn in the presence of hydrochloric acid followed by the addition of excess base The generic symbol for all of these reductions is H An example of the formation of a nitro compound followed by reduction is shown in Figure 1315 N02 NH2 HNO3H2804 H Reductive amination A number of organic species including amides oximes and nitriles undergo reductive amination a variety of reduction reactions that produce amines In general these processes involve imines RNR or related species Reduction processes include hydrogenation using Raney nickel as the catalyst for nitriles the reaction with sodiumEtOH for oximes and the use of lithium aluminum hydride LiAlH4 for amides or nitriles Figure 1316 illustrates the preparation of amphetamine by reductive amination CH C CH CH Ci 3 C H20 ll H3 N H CHKCHCH3 H2Raney nickel NH2 Amphetamine Figure 1317 In the upper pathway reduc ve amination fails The lower pathway vvorks Figure 1318 The formation of an amine from a ketone via an oxime Figure 1319 The formation of an amine from a nitrile Chapter 13 Amines and Friends 7 In some cases you may run into problems with reductive amination The upper pathway in Figure 1317 illustrates one of the problems a secondary alkyl halide and a weak nucleophile leads to elimination instead of substitution which necessitates the use of the lower pathway OH Br PBrs CH3NH2xs T Cr2O72 HA O NH 1 CIl3NH2 CH3 2 H2Ni Another type of reductive amination is shown in Figure 1318 This reaction illustrates the formation of an amine from a ketone through the formation of an intermediate oxime Figure 1319 shows the conversion of a nitrile to an amine The nitrile can be formed by the action of cyanide ion CN on a halide via an SN2 mechanism N NH2 T O NH2OH RNiH2A or LiAH4 c Part IV Advanced Topics Every Student39s Nightmare Hofmann rearrangement degradation The Hofmann rearrangement is a useful means of converting an amide to an amine The nitrogen of the amide must be primary The reaction results in the loss of one carbon atom Figure 1320 illustrates the generic Hofmann rearrangement reaction and the generic mechanism of the Hofmann rearrangement reaction is shown in Figure 1321 In the reaction an intermediate isocyanate forms The R in Figure 1321 may be alkyl or aryl The first intermediate in the mechanism is resonance stabilized which promotes the reaction it s similar in structure to an enolate In addition the third intermediate is also resonance stabilized Figure 1320 The generic O Hofmann OHquot 2 rearrange R C Br2 RNH2O03 H2OBr ment NH2 reac on o 0 pt R Olt 0Sf Fi oH39igt R C Br NH2 H20 H 3 Figure 1321 An isocyanate O 0 Themecha Q lt RC 4 RC 398 nismforthe r Hofmann rearrange ment reac on C032 H2NR A related reaction is the Curtius rearrangement which replaces the amide with an azide RCON3 The azide can be formed by the reaction of an acyl chloride with sodium azide Chapter 13 Amines and Friends ev Seeing How Nitrogen Compounds React Primary secondary and tertiary amines behave as BronstedLowry bases These amines react like ammonia adding Ht to produce an ammonium ion Amines may also behave as nucleophiles Lewis bases Primary amines are stronger nucleophiles than secondary amines which in turn are stronger nucleophiles than tertiary amines As nucleophiles amines attack acid chlorides to form amides Later in this chapter you see that they re important in the formation of sulfonamides Reactions with nitrous acid Amines react with nitrous acid formed by the reaction NaNO2 Ht HNO2 to give a variety of products The nitrous acid isn t very stable so generat ing it in place from sodium nitrite is necessary Sodium nitrite is a meat preservative and a color enhancer Under acidic conditions nitrous acid dehydrates to produce the nitrosonium ion NO The NO ion is a weak elec trophile that s resonance stabilized See Chapter 7 Figure 1322 illustrates the dehydration of nitrous acid Figure 1322 H The dehy O N H O N H20 69 dration of T N nitrous acid H O H O Tertiary amines don t react directly with acidic sodium nitrite However as seen in Figure 1323 even though the tertiary amine doesn t react its presence activates an aromatic system leading to attack by NO T CH3 CH3 Figure 1323 l The attfexgzlrj N N CH3 N8NO2H CH3 activated m aromatic system ON by Not Activated ring pnitrosoNNdimethylaniline S Part IV Advanced Topics Every Student39s Nightmare As seen in Figure 1324 secondary amines react directly with acidic sodium nitrite to form a nitrosamine These compounds are Very Very toxic Primary amines react under similar conditions to form unstable diazonium salts see Figure 1325 Diazonium salts readily lose the Very stable N2 to form reactive carbocations that are useful in a number of synthetic pathways Figure 1326 shows the resonance stabilization of a diazonium ion Figure 1324 The formation of a H CH3 nitrosamine I I by the N N reaction of a CH3 NaNO2H NQ secondary amine with acidic sodium nitrite Figure 1325 The fa iTiur RNH2 NaNO2H R N X A diazonium sat unstabe salt its decom l N2 position and several possible X outcomes H20 of the H RX carbocation ROH formed by decom Alkene position Chapter 13 Amines and Friends T H lt9 G N N N Figure 1326 Resonance ltgt stabilization of a diazonium ion N2 is a good T leaving group EN Replacement reactions Many kinds of replacement reactions involve nitrogen compounds A good many of these processes described in the following sections utilize diazonium salts Sandmeyer reaction The Sandmeyer reaction utilizes a diazonium salt to produce an aryl halide The process begins by converting an amine to a diazonium salt Decomposition of the diazonium salt in the presence of a copper1 halide places the halide ion into the position originally occupied by the amine The most useful copper1 halides are CuCl and CuBr in addition the copperI pseudohalides such as CuCN also works by placing a nitrile in the position originally occupied by the amine Figure 1327 shows an example of the Sandmeyer reaction 9 NH2 N2 H804 HNO2 T H2804 CH3 CH3 CuBr Figure 1327 Br An example of the Sandmeyer reaction CH3 73 uZ Part IV Advanced Topics Every Student39s Nightmare Replacement by iodide ion This reaction is similar to the Sandmeyer reaction but the halide source is potassium iodide Kl Figure 1328 illustrates this reaction NH2 N2 HNO2 H2804 Nal T I Figure 1328 The prepa ration of an aryl iodide 67 Schiemann reaction The Schiemann reaction is a means of preparing aryl fluorides The process is similar to the Sandmeyer reaction The source of the fluoride is fluoroboric acid HBF4 Figure 1329 illustrates the Schiemann reaction lt9 NH2 N2 Cl HNO2 HCI Cold HBF4 Figure 1329 N2 BF4 The Schiemann N2 BF3 A reac on Figure 1330 Using a dia zonium salt to produce an ether and a phenol Figure 1331 The conver sion of a substituted aromatic amine to a substituted phenoL Chapter 13 Amines and Friends 7 Formation of ethers and phenols Diazonium salts can also be used to form ethers and phenols Reaction of diazonium salt with an alcohol generates an ether while thermal hydrolysis of the diazonium salt yields a phenol Figure 1330 illustrates both formations As seen in Figure 1331 this process also works on substituted aromatic systems 9 NH2 N2 HNO2 T Haq Cold H20A ROH OH O R 9 NH2 N2 HNO2 T Haq Cold H20A N02 N02 OH 86 N02 238 Part IV Advanced Topics Every Student39s Nightmare Figure 1332 The general process for deamination Figure 1333 The forma tion of two different dibromo toluenes Deamination Deamination replaces the amine group with a hydrogen atom This process normally uses hypophosphorous acid H3PO2 The general process for deamination is in Figure 1332 This is a synthetically useful technique that leads to different products than other replacement methods Figure 1333 illustrates the formation of two different dibromotoluenes 9 N2 H H3Po3 B N2 BI Br 6 NH2 NH2 N2 2BT2 08 Fe H C Id CH3 CH3 Br CH3 Br H3PO2 Br Br BT2 3 Fe CH3 CH3 r CH3 Br Br Different products Figure 1334 The formation of an azo compound Y repre sents an activator Chapter 13 Amines and Friends T Coupling reactions of diazonium salts Diazonium salts can attack an aromatic ring that s been activated by certain groups including OH and NR2 The product is an azo compound Figure 1334 illustrates the basic reaction with Y representing the activator Normally para attack occurs because the ortho position is crowded however if the para position is blocked then ortho attack may occur In many cases extended conjugation is present leading to absorption of light in the visible portion of the spectrum Azo dyes are examples of these compounds aEN 0a a Figure 1335 illustrates the mechanism for the formation of phydroxyazoben zene beginning with the reaction of benzenediazonium chloride with an aro matic system activated by an OH The mechanism for the reaction with an amine is similar Figure 1336 illustrates the reaction of a diazonium salt with an amine The product of the reaction in Figure 1336 is pdimethylamino azobenzene Reactions with sulfonyl chlorides Amines can react with sulfonyl chlorides but the product of the reaction depends upon the type of amine 1 Primary amines react to form an Nsubstituted sulfonamide as shown in Figure 1337 1 Secondary amines react to form an NNdisubstituted sulfonamide Figure 1338 1 Tertiary amines react to form salts Figure 1339 24 0 Part IV Advanced Topics Every Student39s Nightmare Figure 1335 N N 9 The OH mechanism for converting benzene diazonium N N chloride to OH phydroxy azobenzene CH3 Figure 1336 NE N3 N CH3 CI The reaction of NN CH3 a diazonium N salt with an CH 3 amine Figure 1337 H O The reaction OH ofaprimary R N CS Ar 3 R N O amine with H S a sulfonyl H 0 chloride 0 AF Figure 1338 The reaction of a secondary amine with a sulfonyl chloride Figure 1339 The reaction of a tertiary amine with a sulfonyl chloride Figure 1340 The general mechanism forthe elimination inthe Hofmann elimination Chapter 13 Amines and Friends 9 7 R 0 R R N CI Ar i R N 0 H p HCI S 0 Ar R39 O 939 o o Nlt CI Ar L R N gtslt Ru 0 Rquot 0 Ar Exploring elimination reactions Elimination reactions involving amines are important synthetic methods They can be used to make a variety of useful organic compounds including alkenes We examine a few of them in this section Hofmann elimination The Hofmann elimination converts an amine into an alkene The process begins by converting an amine to a quaternary ammonium salt that is it has a nitrogen atom with four bonds The general mechanism for the elimination step is in Figure 1340 Figure 1341 illustrates a sample reaction scheme for the Hofmann elimination H ClCH2 A C CH2 H20 NCH33 fl NCH33 24 2 Part IV Advanced Topics Every Student39s Nightmare Fmum134t Asampm reac on scheme forthe Ho nann ehnuna on Figure 1342 Product distribution resulting from the Hofmann elimination Figure 1343 intermedi ate in the Hofmann elimination CH3CH2CH2CH2CH2CH2NH2 i K XS 9 CH3CH2CH2CH2CH2CH2NCH33 Iquot Ag2OH20 69 C H3C HQC HQC HQC HQC H2N 8 H33 O Hquot Agl l sA sA 60 o As you no doubt learned in Organic 1 a more highly substituted double bond is more stable ZaitseV s rule As Figure 1342 shows the reaction does the opposite of ZaitseV s rule That is the leastsubstituted product predominates This happens because the transition state has carbanion character shown in Figure 1343 A 1 carbanion is more stable than a 2 which is more stable than a 23 The process involves antielimination opposite side elimination as shown in Figure 1344 The geometry requires the bulky groups to have the greatest separation lt9 NCH33 OH A 5 T x5 r5 94 6 H OH 55 639 CH3CH2CH CH2 CH3N CH3 CH3 Figure 1344 Anti elimination inthe Hofmann elimination Figure 1345 An example of the Cope elimination Figure 1346 The mechanism of the syn elimination step in the Cope elimination Chapter 13 Amines and Friends Y OHquot H N CH 33 Bulky Cope elimination In the Cope elimination thirty percent hydrogen peroxide H202 is used to produce an amine oxide which upon heating undergoes elimination This is a synelimination process Figure 1345 illustrates the general reaction while Figure 1346 shows the mechanism of the synelimination step 0 0 quot o e 3 CH2N CHs 3 H202 CH2N CH3 Cold H CH3 CH3 An amine oxide A CH2 HO d CH3 sl al QjQ 03 NCH32OH I p CH3 Part IV Advanced Topics Every Student39s Nightmare Mastering Multistep Synthesis lBEI Figure 1347 An example of a multistep synthesis In many cases a desired compound cannot be synthesized directly from readily available materials In these cases a multistep synthesis must be performed Figure 1347 illustrates a multistep synthesis A similar type of problem appears on many Organic Chemistry II exams they re retrosynthetic analysis problems For aromatic amines NR2 is activating 0pdirectors and NR2H is deactivating that is it may interfere with the reactions ll Br 1 Mgether C 2 CO2 OH 3 H gag W C C CH 2 N CH3 HNCH32XS CH3 ii 1 Li0TH4 C N CH3 2 OHH20 l CH3 When attacking a retrosynthetic analysis problem you often know only the formula of the starting material and the desired product in addition the instructor may impose a few other rules The answer to the problem should resemble Figure 1347 Figure 1348 The structure of pamino benzoic acid Figure 1349 The structure of prontosil Chapter 13 Amines and Friends You have many options for attacking multistep synthesis problems In general begin with the desired product and work backwards For example what reactant can produce the final tertiary amine in Figure 1347 After you determine the identity of the reactant you back up one step and determine which reactant can produce the amide given in Figure 1347 After this you back up another step and repeat the procedure until you reach the starting material If you get lost you may need to retrace your steps and redo one or more steps Only try to work from the beginning as a last resort The formation of sulfa drugs is another example of a multistep synthesis The sulfa drugs are bactericides effective against a wide variety of bacteria because they mimic paminobenzoic acid Figure 1348 Many bacteria require paminobenzoic acid which they are unable to synthesize and need to synthesize folic acid Many types of sulfa drugs exist and most of them involve the substitution of one of the hydrogen atoms on the S02NH2 Prontosil Figure 1349 was the first commercially available sulfa drug The metabolism of prontosil produced sulfanilamide NH2 COOH NH2 NH2 NN S02NH2 The procedure for synthesizing sulfanilamide a sulfa drug is a multistep procedure as illustrated in Figure 1350 The first step also works if you substitute an acyl chloride for the acid anhydride The conversion of the amine to an amide converts the strong activator into a medium activator limiting multiple attacks The last step converts the amide back into an amine H Part IV Advanced Topics Every Student39s Nightmare Figure 1350 The multistep synthesis of suHanH amide In the general reactivity scheme acyl chlorides are more reactive than acid anhydrides which are more reactive than carboxylic acids NH2 NHAC NHAC A020 CISO3H xs 3 Q NH3 NH2 NHAC o OH H20 lt s NH b o NH b Sulfanilamide Identifying Nitrogen Compounds with Analysis and Spectroscopy Amines are easily identified because they re readily soluble in dilute acid Sodium fusion converts the amine to the cyanide ion which is detectable by a variety of methods The ready formation and decomposition of diazonium salts discussed in the earlier section Reactions with nitrous acid leads to the identification of primary amines The Hinsberg test see the nearby sidebar is useful in identifying amines The infrared spectra of amines show one or two NH stretches in the 3500 3200 cm region Primary amines usually have two bands while secondary amines usually have one band Obviously since there are no NH bonds tertiary amines have no NH stretch The bands are small and sharp in comparison the corresponding alcohol peaks The hydrogen attached to the nitrogen appears in the 8 15 region of the proton NMR Chapter 13 Amines and Friends 7 Hinsberg test Though ithaslargely been replaced by spectro The second step was to acidify the reaction scopic methods at one time the Hinsberg test mixture The results of the two steps indicated was useful in the characterization of amines the type of amine present Normally this was Thefirst step in the test wasthe reaction of the a good test for up to eight carbon atoms This amine with a benzenesulfonyl chloride in base table lists the results for each type of amine Type of Amine Results of the First Step Results of the Second Step Primary Solution Precipitation Secondary Precipitation No visible change No reaction Tertiary No reaction Solution Part IV Advanced Topics Every Student39s Nightmare Chapter 14 Metals Muscling In Drganometalllcs In This Chapter Coping with Grignard reagents Deciphering organolithium reagents Taking a quick look at other organometallic compounds Testing what you know A metal or semimetal atom can form a covalent bond to carbon in many situations and the resultant compound is an organometallic compound The metaltocarbon bond in these compounds is polar covalent X3C M Metals such as sodium and potassium form ionic organometallic compounds in which the metal forms a cation and the organic portion is a carbanion We look at a couple of types of these organometallic compounds in this chapter particularly those containing magnesium Grignard reagents and lithium organolithium reagents Grignard Reagents Grin and Bear I t NG Grignard reagents organomagnesium compounds are extremely useful in many organic reactions These materials are relatively easy to prepare but they re very sensitive to trace amounts of moisture and air and decompose if either is present Don t use a Grignard reagent when the starting material has an active hydrogen such as carboxylic acids alcohols amines and sulfonic acids because Grignard reagents behave as bases towards them In addition the starting material can t contain other groups that may be attacked by the reagent including other carbonyl groups epoxide groups and nitrile groups F Part IV Advanced Topics Every Student39s Nightmare lBEB 2quot Preparation of Grignard reagents The preparation of a Grignard reagent begins with magnesium metal and dry ether in most cases either diethyl ether or THF tetrahydrofuran The ether cleans the surface of the metal and takes the reagent into solution for reaction If either the ether or the reaction vessel contains moisture the yield is poor The magnesium then reacts with either an alkyl halide or an aryl halide The ease of reactivity decreases in the order R gt RBr gt RC lodides may react too rapidly but chlorides may react too slowly Thus bromides are usually the best The general reaction is RX or ArX Mgether a R5 5MgX or Ar5 5MgX For example the preparation of ethyl magnesium bromide is CH3CH2Br Mg Et2O CH3CH25 5MgBr Pay particular attention to the bond polarity 8398 The carbon atom has a partially negative charge and therefore is a nucleophile and the magnesium atom has a partially positive charge so is an electrophile Reactions of Grignard reagents Grignard reagents behave both as bases and as nucleophiles The basicity leads to the requirement that water be rigorously excluded from the reaction mixture both in the preparation and in the use of the Grignard reagent Basicity Basicity refers to the ability of a Grignard reagent to react with proton donors including weak donors like water The carbanion is the conjugate base of a very weak acid Ka z 10 10 5 This process is R H H R Weak acids WA have strong conjugate bases SB while strong acids SA have weak conjugate bases WB The weaker the acid or base the stronger its conjugate A Grignard reagent behaves as a base towards water an acid An example of this reaction is CH3MgBr H20 e CH4 MgBrOH SB SA WA WB Chapter 14 Metals Muscling In Organometallics 7 Figure 141 The general reaction of a Grignard reagent with a carbonyl As with other acidbase reactions the strong acid and strong base react to produce a weaker acid and a weaker base The strong acid in this case H20 reacts to form the weaker base MgBrOH while the strong base in this case the Grignard reagent CH3MgBr reacts to form the weak acid CH 4 Water a conjugate acid CA and MgBrOH a conjugate base CB constitute a conjugate acidbase pair while the Grignard reagent CB and methane CA constitute another conjugate acidbase pair Another example illustrating the basicity of Grignard reagents is CH3MgBr CH3CECH CH 4 CH3CECMgBr SB SA WA WB This reaction has two conjugate acidbase pairs and the strong acid and base react to produce the weak acid and base The terminal alkyne Ka x 10 25 pKa z 25 is a significantly stronger acid than an alkane The acidity of hydrocarbons increases in the order sp3 lt p2 lt sp NucleopIilicity The nucleophilic reactions of Grignard reagents include reactions that create carboncarbon bonds and the formation of alcohols One way to create a carboncarbon bond is to react a Grignard reagent with a carbonyl compound The result of this reaction is an alcohol derived from an aldehyde Formaldehyde gives a primary alcohol but any other aldehyde gives a secondary alcohol Ketones and esters both react to form tertiary alcohols The general reaction of a Grignard reagent with a carbonyl is shown in Figure 141 The product of the first step in the mechanism in Figure 141 is a salt which undergoes hydrolysis in the next step The hydrolysis an acidbase reaction is shown in Figure 142 6 G 6 0 MgX H H RMQX 8 if 0 if at A 5 Alkoxlde salt 252 Part IV Advanced Topics Every Student39s Nightmare Figure 142 The hydrolysis of the salt formed in Figure 141 Figure 143 The formation of a primary alcohol Figure 144 The synthesis of cyclopentyl methanol Fl R Z H20 vgxo 59 MgX OH SB SA WA WB Only formaldehyde yields a primary alcohol by reaction with a Grignard reagent Figure 143 illustrates the reaction of ethylmagnesium bromide with formaldehyde to form 1propanol Morecomplicated alcohols such as cyclopentylmethanol can be synthesized by this means as shown in Figure 144 6 6 H OH Etmgarf l 4 0 ltgt H ltgt H lc Et Et Alkoxide salt 1prOpanO if C MgBr 1 H H CH20H L f 2 H Any aldehyde other than formaldehyde reacts with a Grignard reagent to produce a secondary alcohol see Figure 145 The general mechanism is the same for any aldehyde see Figure 146 As shown in Figure 147 more than one path may lead to the same alcohol Figure 145 The synthesis of a secondary alcohol bythe reaction of a Grignard reagent with an aldehyde Figure 146 A partial mechanism forthe synthesis of a secondary alcohol CH3 0 CHMgBr H C CH3 CH3 CS3 CH M X 6 g 0 CH3 II C H5 CH2CH3 H C CH2CH3 CH H OH CH CH3 CH3 Chapter 14 Metals Muscling In Organometallics x CH3 OH CH3 9 lt9 0 MgBr H c cH2cH3 254 Part IV Advanced Topics Every Student39s Nightmare Figure 147 Two synthetic paths to the same alcohol Figure 148 The synthesis of a tertiary alcohol through the reaction of a ketone with a Grignard reagent Figure 149 The synthesis of a tertiary alcohol containing a cycohex ane ring O C CH3MgBr OH H 2 H C3CH3 O V H MgBr C H A ketone reacts with a Grignard reagent to produce a tertiary alcohol Examples of this process are shown in Figure 148 and Figure 149 0 OH CH MgBr e 2 H CH C CH 3 CHCCH F 3 3 3 3 CH3 0 HO CH3 CH3MgBr 2 H An alcohol can be formed through the reaction of ethylene oxide with a Grignard reagent As usual nucleophilic attack is carried out by the Grignard reagent This procedure produces a primary alcohol with the addition of two carbon atoms Figure 1410 illustrates the reaction of ethylene oxide with a Grignard reagent Chapter 14 Metals Muscling In Organometallics Figure 1410 The synthesis of 1propanol through the reaction of ethylene oxide with a Grignard reagent Figure 1411 The reac tion of an ester with a Grignard reagent 6 6 6 CH3MgX 2 H CH3CH2CH2OH The reaction of a Grignard reagent with esters is a substitution reaction involving addition and then elimination The process gives a tertiary alcohol Figure 1411 shows an example of the formation of an isotopically labeled tertiary alcohol This reaction involves the attack of the ester by two molecules of the Grignard reagent which adds two identical R groups to the carbonyl carbon 9 lt9 zh ii MgBr 7 C CH2Q 1 6 5 CH2Q l I quotCH3MgX I MCH3 CH3 CH3 OH MCH3 O C 114CH3MgX C 1 4CH3 2 H OH OEt 256 Part IV Advanced Topics Every Student39s Nightmare Organolithium Reagents Figure 1412 The forma tion of an organolith ium reagent Figure 1413 An organo lithium reagent behaving as a base towards a proton deuteron donon In general organolithium compounds behave like Grignard reagents However organolithium compounds tend to be more reactive so they re useful in situations where the reaction with a Grignard reagent is slow or results in a low yield The preparation of an organolithium compound shown in Figure 1412 is similar to the preparation of a Grignard reagent except that lithium replaces the magnesium Two moles of lithium per mole of halide are necessary One lithium yields an organolithium and the other yields a lithium halide mi 6 L39B Dry E20 39 r Organolithium reagents like Grignard reagents are bases that react with proton or deuteron donors Figure 1413 illustrates this reaction In this reaction D20 heavy water is the deuterated form of water in which the hydrogen atoms H are replaced with deuterium atoms D LiOD Like Grignard reagents organolithium compounds react with formaldehyde to produce a primary alcohol Figure 1414 illustrates the reaction of an organolithium reagent with formaldehyde Chapter 14 Metals Muscling In Organometallics M Figure 1414 The prepa ration of a primary alcohol from an organo lithium compound and formal dehyde CH2OH Formation of Other Orqanometallics Grignard reagents react with halides of less electropositive metals to give other organometallic compounds The less electropositive more electronegative metals include Hg Zn Cd Si and the nonmetal P An example of this type of reaction is n RMgX Mxn a MRH n MgX2 Less electronegative metals such as sodium tend to form ionic organometallic compounds These compounds have limited use in synthesis One example of using a different organometallic in a synthesis is CH3 CEC Na CH3 CHO a 2 1 a CH3 CEC CHOH CH3 In general the organometallic compounds presented in this chapter have a metal atom behaving as just another functional group After you consider the polarity difference the compounds behave simply as odd organic compounds However some organometallic compounds such as ferrocene are very different Ferrocene forms from the reaction of ironII ions with cyclopentadienyl ions The cyclopentadienyl ion comes from cyclopentadiene This compound is acidic because the loss of a hydrogen ion shown in Figure 1415 leads to a stable aromatic system Once formed cyclopentadienyl ions readily react with ironII ions to produce the orange stable compound ferrocene FeC5H52 When first synthesized the formation of ferrocene was interesting because of its unusual structure 258 Part IV Advanced Topics Every Student39s Nightmare Figure 1415 The forma tion of the cycopenta dienyl ion Figure 1416 The structure of ferrocene H The structure of ferrocene is shown in Figure 1416 The iron ion is sandwiched between two cyclopentadienyl rings which are freely rotating in solution The cyclopentadienyl rings react like aromatic systems because well they re aromatic The bond is between the iron ion and the TC l Ct139OI1 cloud of the aromatic system Another metal ion may replace the iron in ferrocene Putting It Together Figure 1417 Synthesize this com pound using a Grignard reagent No matter how well you think you understand the material the real test of understanding is your ability to answer questions involving the material For example what if someone asks you What are three methods of preparing the compound in Figure 1417 Via a Grignard reagent OH CjCH3 CH2CH3 Figure 1418 The prepa ration of a Grignard reagent is dependent uponthe identity of Z Chapter 14 Metals Muscling In Organometallics In a question such as this one you need to know not only the basic reactions of Grignard reagents but also the limitations For example if you wish to prepare a Grignard reagent containing an aromatic ring Figure 1418 the substituent Z may be R Ar OR or Cl However Z may not be CO2H OH NH2 SO3H CO2R CN or NO2 because acidic groups generally cause problems by undergoing acidbase reactions with the Grignard reagent Br M9 Et2O 0 Ty Chapter 15 More Reactions of Carbonyl Compounds In This Chapter Shedding light on the Claisen condensation Deciphering acetoacetic and malonic ester synthesis Understanding the Knoevenagel condensation Tackling the Mannich reaction Studying the Stork enamine synthesis Exploring barbiturates n Chapter 11 you see that a hydrogen atom between two carbonyl groups is weakly acidic Ka Values of 1010 to 1014 These compounds are Bdicarbonyl compounds In the same chapter you see that Michael additions can add enolate ions to these compounds In this chapter we examine additional reactions of Bdicarbonyl compounds As you examine the reactions in this chapter keep in mind this basic info The acidity of a hydrogen atom between two carbonyl groups is partly due to the inductive effect weakening of the bond to the hydrogen atom due to the electron density of the bond being pulled towards the nearby electro negative oxygens and the resonance stabilization of the anion formed see Figure 151 The partial negative charge on the carbon atom makes it a nucleophile that is subject to attack by an electrophile N Part IV Advanced Topics Every Student39s Nightmare Figure 151 Resonance stabilization of the anion formed from a Bdicarbonyl compound 02 0 o oH2 C it 30 IO gdi 0 0 quot C H e H o oH c lt gt o oH o lt gt o oH o 05 50 El 6 39 jcjf j H Checking Out the Claisen Condensation and 1 ts Variations The term condensation refers to the joining of two molecules with the split ting out of a smaller molecule The Claisen condensation is used extensively in the synthesis of dicarbonyl compounds In biochemistry it is used to build fatty acids in the body The Dieckmann condensation the crossed Claisen condensation and others with other carbanions are variations of the Claisen condensation In this section we briefly look at these variations Doing the tuo step C Iaisen condensation The Claisen condensation is one method of synthesizing Bdicarbonyl compounds specifically a Bketo ester This reaction begins with an ester and occurs in two steps In the first step a strong base such as sodium ethoxide removes a hydrogen ion from the carbon atom adjacent to the carbonyl group in the ester Resonance stabilizes the anion formed from the ester The anion can then attack a second molecule of the ester which begins a series of mechanistic steps until the anion of the Bdicarbonyl compound forms which in the second reaction step acidification gives the product Chapter 15 More Reactions of Carbonyl Compounds E The general mechanism of the Claisen condensation with ethoxide as the base is shown in Figure 152 Sodium ethoxide is necessary because the starting material is an ethyl ester If the starting material were a methyl ester then the base would be sodium methoxide Choosing a base that matches the type of ester minimizes the formation of other products Note that the final product in Figure 152 is stabilized as illustrated in Figure 153 This is an example of stabilization by ketoenol tautomerization see Chapter 11 to review The driving force is the result of the stability of the carbanion as shown in Figure 151 In this example of condensation two ester molecules join and alcohol a smaller molecule splits out 0 H 1 NaOCH CH 2 c 2H 2 3 CH3CH2OH CH3 ltVOjCH2 CH3 I6CH2CH3 J3 H CH3 OCH2 CH3 O 393 O C CH2 OCH2quotCH3 CH O CH2CH3 H 0 ll 0 C 0 CH3 O CH2CH3 H 9 C CH3C CH O CH2CH3 Qquot 0 39039 Driving force C CH3 CjCH2 OjCH2 CH3 L H Figure 152 CH2CH3 0 The general C mechanism CH3C CH2 O CH2CH3 forthe A Claisen con 96 densation quot T CH2CH3 Part IV Advanced Topics Every Student39s Nightmare Figure 153 Stabilization of the prod uct from Figure 152 quot Figure 154 The Dmckmann conden sa on beghu ng vvnh dhnethyl hexan edioate i ii i i C C C C Enol The Claisen condensation bears some resemblance to the Aldol condensation seen in Chapter 11 The initial step in the mechanisms are Very similar in that in both cases a resonancestabilized ion is formed Circling around Dieckmann condensation Dieckmann condensation is a cyclic Claisen condensation where a molecule attacks itself to form a ring structure Figure 154 shows an example beginning with dimethyl hexanedioate and Figure 155 shows what happens when the carbon chain increases by one carbon atom This process favors the formation of fiVe or sixmembered rings because they re the most stable rings 0 0 o COCH3 ll OCH3 1 Na39OCH3MeOH c C 2 H OCII3 II 82 o Chapter 15 More Reactions of Carbonyl Compounds phM Figure 155 The Dieckmann condensation producing a sixmem bered ring Figure 156 A crossed Claisen con densa on o 0 h COCH3 OCH3 1 Na39OCH3MeOH c C 2 H OCH3 ll Doubling Up Crossed Claisen condensation A crossed Claisen condensation employs two different esters If the esters are A and B the possible products are AA AB BA and BB To minimize the complicated mixture of products one of the reactants must have no orhydrogen atoms If this is ester B then the products would be AA and AB If the concentration of A is Very low then only a small quantity of AA can form Figure 156 illustrates an example of a crossed Claisen condensation where the product is of the AB form 0 0 ll 4 lt3 CH o 3H2 3H2 CH3 CH3 1 0 Et 2 HOAC 0 0 it it C 60 3H2 CH3 266 Part IV Advanced Topics Every Student39s Nightmare Figure 157 The use of sodium ethoxide for a Claisen condensa tion type reac on Figure 158 The use of sodium amide for a Claisen condensa tion type reac on Other carbanions Very strong bases can be used in other reactions that are similar to the Claisen condensation In these reactions carbanions are formed instead of enolates Figures 157 158 and 159 show examples of Various reactions employing three different strong bases sodium ethoxide sodium amide and sodium triphenylmethanide Sodium hydride NaH is a strong base that would also work Any of these Very strong bases could be used in each of the specific reactions 0 W 0 CH3 0 C CH3 CH H 1NaOEtEtOH H 3 C CH 2 H CH H o CH3 3 91 0 o o J3 1 N NH lcl lcl a O 393 2H 2 CH CH3 ll CH2 Cg CH3 CH3 CH3 Write the complete mechanism for each of the reactions in Figures 157 158 and 159 and compare these to the Claisen condensation in order to identify similarities and differences Chapter 15 More Reactions of Carbonyl Compounds R 7 Figure 159 Using sodium tripheny methanide for a Claisen condensa tion type reac on T 3 CH CHCHC 0 Na39CPh3 9 3 3 F C O Verky strong CHC CH2CH3 H C ase 2 H C CH3 i C O c clt LH3 OCH2 CH3 Exploring lcetoacetic Ester Synthesis Acetoacetic ester synthesis is the preparation of substituted acetones and it s an important method for creating a Variety of products It begins with the reaction of acetoacetic ester a dicarbonyl or a similar compound with a strong base to produce a carbanion which then reacts with alkyl halide RX The structure of acetoacetic ester is in Figure 1510 Figure 1511 illustrates an example of an acetoacetic ester synthesis and two possible outcomes Figure 1512 shows the preparation of 2heptanone with a 65 percent yield Via the acetoacetic ester synthesis Figure 1513 presents the preparation of 2benzylcyclohexanone with a 77 percent yield Part IV Advanced Topics Every Student39s Nightmare 0 Figure 151o The CH2C structure of CH acetoacetic 00 3 ester OCH2CH3 O O e CH2C CH C O C CH3 NaOEtEtOH O C CH3 OjCH2 CH3 OjCH2 CH3 RX R O CH C N OEtEtO a 4 O CH3 O R t39t39 epe I Ion C C 0 CH3 T OCH2CH3 Figure 1511 An example R39X of an aceto O acetic ester R synthesis RC c C0 CH3CH2OH and two CH possible 00 3 outcomes T OCH2CH3 Chapter 15 More Reactions of Carbonyl Compounds Figure 1512 The prepa ration of 2heptanone 65 yield Figure 1513 The prepa ration of 2benzycy clohexanone 77 yield CH3 1 N E E H CH2C CH3CH2CH2CH2Br 00 00 CH3 CH2CH2CH2CH2CH3 OCH2 CH3 lt O NaOEtEtOH C O CH2 CH3 OC O T O CH2CH3 O CHQBT V O 0 CH2 CH2 HA C3 O CH2CH3 0 Defining Malonic Ester Synthesis Malonic esters have two ester groups each of which may react as in the acetoacetic ester synthesis due to their similar structure see the preceding section The malonic ester synthesis provides a method for preparing a substituted acetic acid Figure 1514 shows the structure of one type of malonic ester Figure 1515 outlines the basic malonic ester synthesis May repeat in that figure refers to the reaction with a second molecule of RX or R39X 270 Figure 1514 Atypical malonic ester Figure 1515 An outline of the basic malonic ester synthesis Figure 1516 Hydrolysis and decar boxylation Part IV Advanced Topics Every Student39s Nightmare O O CH2CH3 CH2 O CH2CH3 O O O O CH2CH3 gtgtO CH2CH3 CH2 NaOEtEtOH H O CH2CH3 O CH2CH3 O O l RX 0 CH2CH3 Q gtgt O CH2CH3 R R lt R CH o O CH2CH3 0 I o CH239CH3 In most cases the product of the malonic ester synthesis isn t the final product you re looking for Commonly the next step after the reaction in Figure 1515 is hydrolysis and decarboxylation Figure 1516 shows this step CH2 CH3 O Q E O HA 0 9 i O CH239CH3 Figure 1517 Formation of the anion Figure 1518 Formation of compound Afrom the anion in Figure 1517 Chapter 15 More Reactions of Carbonyl Compounds 2 7 7 Synthesis of the anion shown in Figure 1517 is very important since the anion produced is a nucleophile and can undergo further reaction Other strong bases such as sodium hydride NaH can replace the ethoxide ion used in Figure 1517 0 3H2 CH3 O 3H2 CH3 gto O EtO 9 CH2 CH 9 T 9 T O O CH2CH3 CH2 CH3 Once formed the anion has several uses and the following figures illustrate how versatile malonic ester synthesis and its subsequent anion can be Here we are using a typical molecule in a series of reactions Any compound of similar structure could be used Figure 1518 shows the use of the anion to form compound A Hydrolysis and decarboxylation of compound A yields a carboxylic acid Figure 1519 illustrates another synthesis involving compound A A slight change in the procedure presented in Figure 1518 is shown in Figure 1520 0 CH2 CH3 O CH2 CH3 G O O CH CH3CH2CH2CH2Br CH3CH2CH2CH2 CH C mrgt UndA 2 Cl 84 O O O CH2 CH3 CH2 CH3 L H CH3CH24COOH 75 2 h Part IV Advanced Topics Every Student39s Nightmare o C3H2CH3 O C3H2CH3 j 0 CH I E O 3 CH3CH2CH2CH2CH u CH3CH2CH2CH2C CH3 0 Compound A 0 O CH2CH3 CH2CH3 Figure 1519 Another CH example of 3 the use of Compound A CH3CHCH2CHCH COOH in synthesis 2 2 T 74 O CH2 CH3 O CH2 CH3 393 Br 0 G 1 CH BFCH2CH2CH2CH2Br CH2CH2CH2CH2 CH O O 0 I 0 I CH2 CH3 CH2 CH3 EtOEtOH O 0 3H2 CH3 H Br c oEt C l 9 gt 0 4 c oEt gt lt3 T O O CH2 CH3 Figure 1520 Another HA reac on beginning 0 with the anion in C Figure 1517 Chapter 15 More Reactions of Carbonyl Compounds 2 Working with Other Active Hydrogen Atoms Figure 1521 Some electron withdrawing groups Compounds whose structures are similar to Bdicarbonyl compounds also have active hydrogens These compounds have a CH2 sandwiched between two electronwithdrawing groups some examples of which are in Figure 1521 The loss of a hydrogen ion from the sandwiched carbon leaves an anion which can then behave as a nucleophile similar to other nucleophiles seen in this chapter 0 0 it it EN R OR ii c c NO2 X H NR2 O i quot 39loR 39iN Sx II II II Reacting with Knoetenaqel Condensation The condensation of aldehydes and ketones with active hydrogen atoms is called Knoevenagel condensation It is related to an aldol condensation and commonly is used to produce enones a compound with a carboncarbon double bond adjacent to a carbonyl The process requires a weak base an amine A typical example and mechanism are presented in Figure 1522 2 Part IV Advanced Topics Every Student39s Nightmare Fmum152Z Anexampm ofa Knoevenagel conden sa on 3 Et quot 3 JHF1 H13 Et 00 W C 0 Et N M CHC c o 0 1 Et 0 Et H O H I quotc o c o I c o 0 1 Et 0 Et 5 C quotC c o H Cf7 I Chapter 15 More Reactions of Carbonyl Compounds 2 Looking at Mannich Reactions A Mannich reaction is the reaction of formaldehyde with a primary or secondary amine and a compound with an active hydrogen atom The product an amine with a ycarbonyl is called a Mannich base useful in a number of synthesis reactions An example is in Figure 1523 and the mechanism is in Figure 1524 in T in C N C H H CH CH2 CH CH3 CH3 CH3 J HCI O 1 Figure1523 C3 CH2 CH2CH3 The forma CH3 Y CH2 N tion ofa mm Mannich From acetone CH2CH3 base T From formaldehyde 2 a Part IV Advanced Topics Every Student39s Nightmare 9 0 H HO H HCH H CH C1H H N N CH2 CH2 H2 CH2 CH cH 39 39 2 2 CH3 CH3 H3 CH3 CH3 CH3 H 12 H C H 4 39 CH 3H2 CH3 CH3 Iminium ion CH3 CH2 H Figure 1524 The mecha Q nismforthe z formationof C CH2 CH2CH3 a Mannlch CH3 CH2 N base CH2C H3 Chapter 15 More Reactions of Carbonyl Compounds 2 w Creating Enamines Stork Enamine Synthesis Figure 1525 Enamine resonance Figure 1526 Some secondary amines commonly usedin the Stork enamine synthesis Stork enamine synthesis takes advantage of the fact that an aldehyde or ketone reacts with a secondary amine to produce an enamine Enamines are resonance stabilized see Figure 1525 and have multiple applications In the first resonance structure the nitrogen is the nucleophile while in the second resonance structure the carbanion is the nucleophile Some commonly used secondary amines pyrrolidine piperidine and morpholine are shown in Figure 1526 T T ccN R c 9 c g R j 0 N N N H H H Pyrrolidine Piperidine VOFphOine 2 o Part IV Advanced Topics Every Student39s Nightmare Figure 1527 illustrates the formation of an enamine The mechanism is shown in Figure 1528 Figure 1529 illustrates the formation of a 15diketone with a 71 percent yield 0 Figure 1527 H20 N Preparation g j ofan N enamine l H O J H H3 H N H L quotlt H N H p CF H fi R R R R R l 1 a3939s HC C nism of NC 39H H C enamine formation R Fl R R Chapter 15 More Reactions of Carbonyl Compounds 2 0 Figure 1529 The forma tion of a 15diketone 71 yield 0 H K H2CoH CH3 Ki CH3 0 O J J H2CCH2 CH3 H2CCH2 CH3 0 T N H20 gt Putting I t A I Together with Barbitumtes The preparation of barbiturates illustrates many of the synthetic methods covered in this chapter The preparation employs the reaction of urea CONH22 with malonic ester to form barbituric acid The general reaction is presented in Figure 1530 The stable pyrimidine and other resonance forms help drive the reaction By alternating the substituent at carbon number five C5 various pharmacologically active substances can be formed Barbital a sedative and phenobarbital a sleeping aid are shown in Figure 1531 Part IV Advanced Topics Every Student39s Nightmare I T NH2 O C IH6C O CH2 oc2 5cH2 2 EtOH NH2 o c 3H4C I o o Et i OH NC Figure1530 Others lt HO C CH The prepa ration of Nic barbiturates T Apyrimidine Et Et Et Ph 0 c o O c o T T T T Figure1531 NHC NH NHC NH II II examples of O barbiturates T Barbital Phenobarbital Chapter 16 Living Large Biomolecules In This Chapter Exploring carbohydrates from the simple to the complex Observing the reactions synthesis and degradation of monosaccharides Checking out lipids specifically fats Deciphering amino acid structure and how these protein building blocks form A biomolecule is any molecule produced by living cells Some biomolecules include the following all of which we cover in this chapter 1 Carbohydrates Ultimately carbohydrates are the product of photosyn thesis the process in plants that combines carbon dioxide water and energy with chlorophyll and other biomolecules to produce carbohy drates and release oxygen gas The major carbohydrate formed during photosynthesis is glucose Plants and animals sometimes combine simple carbohydrates such as glucose into more complicated carbohy drates such as starch glycogen and cellulose 1 Lipids Lipids are biomolecules that are insoluble in water but soluble in lowpolarity organic solvents such as Et2O and CHCI3 Lipids include fats and oils as well as many other biologically important molecules think waxes and steroids 1 Proteins Proteins are polymers of amino acids Some of the most biologically important proteins are enzymes which act as biological catalysts allowing reactions to occur without the harsh conditions and reagents commonly used in organic chemistry Virtually everything that happens in your body is associated with one or more enzymes Nucleic acids which are polymers of nucleotides are another class of biomolecules but they re beyond the scope of this book for more info about nucleic acids refer to a more advanced text such as Biochemistry For Dummies written by us and published by Wiley P Part IV Advanced Topics Every Student39s Nightmare Deliinq into Carbohydrate Complexities Carbohydrates are either polyhydroxy aldehydes and ketones or substances that form these compounds after hydrolysis The general formula is CxlI20 Normally carbohydrates occur as hemiacetals or acetals hemiketals or ketafs Because carbohydrates are so essential to the existence of living things no organism would have the energy to perform even the basic functions of life without them we spend a great deal of time in this chapter exploring the numerous complexities of carbohydrates First we introduce you to the kinds of carbohydrates that exist Then we explore the reactions synthesis and degradation that affect the simplest carbohydrates monosaccharides Next we get to know some particular monosaccharides the Daldose family before looking at more complex sugars including those containing nitrogen Introducing carbohydrates Monosaccharides which can t be broken down through hydrolysis are the simplest carbohydrates of them all Every other more complex carbohydrate that exists can be broken down into two or more monosaccharides via hydrolysis as you can see from the following 1 Disaccharides break down into two monosaccharides 1 Oligosaccharides break down into anywhere from two to ten monosaccharides 1 Polysaccharides break down into numerous monosaccharides In general the names of carbohydrates end in ose If an aldehyde group is present then the carbohydrate is an aldose If a ketone group is present then the carbohydrate is a ketose However you can also classify carbohydrates in terms of the number of carbon atoms present which leads to names such as triose tetrose pentose and hexose Combinations of these two systems are possible for example glucose pictured in Figure 161 is an aldohexaose Common names for glucose include blood sugar grape sugar and dextrose The sections that follow describe the process of mutarotation in glucose and how glycoside formation can inhibit it Mutarotation Mutarotation is when the straight chain forms of glucose as shown by the Fischer projections are in equilibrium with the cyclic forms of glucose as shown by the Haworth projections For more details on Fischer and Haworth projections see Organic Chemistry For Dummies Figure 162 illustrates the general process occurring during mutarotation notice that two different forms result Chapter 16 Living Large Biomolecules HO H C OH HO C H Figure 161 H C OH The structure of H C OH glucose T CH30H O 1 H c CH2OH l H 2 5CO HV CVOH 39 0 3 T 40 OH K C Ho 4 quotH 2 C Opp C H HF l VOH H H5 oH lt D I CH2OH C3H2OH CH2OH H o H H 3O OH ClO H ClH H I I I pC H EpC IL Figure 162 l l Mutarotation H OH OH in glucose T 0 3 1BEI9 Mutarotation produces two types of cyclic forms called anomers or and B I which differ in their arrangement about the anomeric carbon atom originally the carbonyl carbon atom If the OH on the anomeric carbon atom is down then the structure represents the or anomer if it s up the structure represents the B anomer Due to the equilibrium present one anomer rapidly converts to the other The common anomers have either a five or a sixmembered ring that contains an oxygen atom A sixmembered ring is a derivative of pyran therefore the monosaccharide is a pyranose A fivemembered ring is a derivative of furan Part IV Advanced Topics Every Student39s Nightmare Figure 163 The de va on of pyranose and furanose Figure 164 Different represen tations of glucose which makes the monosaccharide a furanose Figure 163 shows both a pyranose and a furanose The two anomers of glucose refer to Figure 162 are ocDglucopyranose and BDglucopyranose Even though the Haworth projections indicate a flat ring the actual structure of a pyranose ring is the chair conformer see Figure 164 Figure 165 shows one way to represent the equilibrium present during mutarotation The unequal double arrows indicate that the dominant species in each case is the ring form Pyran Furan Pyranose Furanose OCH CH2OH H C oH TH N H0 C H 39 r i H T H C OH OF3 H H C OH H OH CH2OH 02 Figure 165 Representa tion of the muta rotation process gNBE Figure 166 The forma tion of an acetal Figure 167 The forma tion of a glucoside Chapter 16 Living Large Biomolecules orform straight chain 3form T Due to mutarotation a solution of pure or or pure 3 will change to a mixture In the case of glucose the mutarotation gives 36 percent or 64 percent 3 and negligible straight chain The unequal distribution of the two anomers is due to the fact that the OH on the anomeric carbon of the 3 form is equatorial which for a chair conformer is more stable The OH on the anomeric carbon in the or anomer is axial which means this anomer is slightly less stable Glycoside formation The presence of a glycoside which involves the formation of an acetal see Figure 166 or a hemiacetal can block mutarotation Glycosides are different from the original carbohydrates in that they can t undergo mutarotation because the ring is locked a locked ring can t reopen O O CH3 CHsC 2 W30 CH3c H HCI dry H O CH3 An acetal In the case of glucose the acetal is a glucoside see Figure 167 In general the simple process shown in Figure 167 may form either an acetal or a hemiacetal to give a glycoside 2 CH3OH Glucose T Glucoside HCI dry T An acetal of glucose Part IV Advanced Topics Every Student39s Nightmare Following are some additional facts about glycosides to file away in your memory bank 1 Glycosides aren t susceptible to simple oxidation via Fehling s or Tollen s test we explain these tests in greater detail later in this chapter 1 Glycosides don t form osazones see the later Osazone formation section for more on these 1 A glycoside can undergo hydrolysis in an acid but not in a base Examining the many reactions of monosaccharides The presence of alcohol and in some cases an aldehyde group makes monosaccharides susceptible to oxidation whereas the presence of a carbonyl group makes monosaccharides susceptible to reduction Because monosaccharides are the fundamental carbohydrate you need to know what happens in the many reactions in which they re involved The following sections are here to help you out with that Welcome to the nittygritty of monosaccharide oxidation and reduction Oxidation of monosaccharides Oxidizing the aldehyde group present in aldoses is easy oxidizing the carbonyl group in a ketose is far more difficult The susceptibility or lack thereof to simple oxidation is a useful method of distinguishing between aldoses and ketoses The next sections explore the various types of monosaccharide oxidation reactions that can occur Oxidation with a metal ion Carbohydrates such as aldoses that undergo oxidation with metal ions are referred to as reducing sugars Both copperll ions and silver ions are capable of oxidizing aldoses Oxidation by copperll ions is the basis for Fehling s test and Benedict s test whereas oxidation by silver ions is the key to Tollen s test Note These tests work for any sugar with a hemiacetal but they don t work on acetals or ketals The oxidation of an aldose by copperll ions results in the reduction of the copper to copperl oxide which has a formula of Cu2O A positive test has the bright blue copperll solution precipitating brickred copperl oxide The result An aldehyde group becomes a carboxylic acid The oxidation of an aldose or ketose by silver ions results in the reduction of the silver ion to silver metal The silver is normally present as AgNH32OH A positive test has a coating of silver metal precipitating on the sides of the container as a mirror Chapter 16 Living Large Biomolecules 7 Figure 168 Tollen s test Figure 169 The conver sion of an aldose gucoseto an aldonic acid gluconic acid Tollen s test may result in the simple oxidation to a carboxylic acid or it may cause fragmentation of the carbon backbone similar to the oxidation reaction seen with periodic acid see the later related section Figure 168 shows the general reaction that occurs when you conduct Tollen s test H CH2OH CH OH c I oH ll oH I c c oH lt H c oH A9NHs20H o C3 OH H c oH Oxidation with bromine and water The oxidation of an aldose not a ketose with bromine and water results in an aldonic acid An example of this reaction is shown in Figure 169 H H OH HO H BF2H20 T H OH HOH CH2OH Glucose OH HOH HOjH HjOH HjOH CH2OH Gluconic acid Part IV Advanced Topics Every Student39s Nightmare Figure 1610 The oxidation of an aldose glucose by a nitric acid glucaric acid Oxidation with nitric acid Nitric acid oxidizes not only the aldehyde in an aldose but also the alcohol of the highestnumbered carbon atom Figure 1610 shows the oxidation of glucose by nitric acid 0 o H oH quot39 0 Ho H H0 H HNO3 H oH 0 H oH H 0 CHQOH O 0H Glucose Glucaric acid Oxidation with periodic acid Periodic acid is a stronger oxidizing agent than the others mentioned earlier in this chapter It causes oxidative cleavage fragmentation of the carbon backbone Normally a silver ion is present to facilitate the reaction through the precipitation of silver iodate AgIO3 Figure 1611 shows the three general types of oxidative cleavage that occur in carbohydrates and Figure 1612 shows the two additional possibilities that are less reactive Look to Figure 1613 for an example of oxidative cleavage of an aldotriose to produce two formic acid molecules and a formaldehyde molecule Chapter 16 Living Large Biomolecules C OH HIO4 CO j O3 j3OH jC3Q C3 OH HIO4 CO T C3O HOCQ Figure1611 Oxidative 0 HIO4 Hojcio cleavage l e in carbohy CQ HQ CO drates C OH HIO4 H C H e No reaction OH Figure 1612 Less reac C OH tive types Ho4 Ofoxidative CO e No reaction unless heated cleavage l Part IV Advanced Topics Every Student39s Nightmare Figure 1613 Oxidative cleavage of an adotri ose to two formic acid molecules and a form aldehyde molecule Figure 1614 Reducing an aldose to form an alditol o if c C Ho W H if 1 j 4 H O oH H O C CH2OH C HIO4 H0 H 0 CH2OH H C H H Reduction of monosaccIarides Reduction of a ketose yields a secondary alcohol and reduction of an aldose yields a primary alcohol called an alditol A possible reducing agent is hydrogenation in the presence of a catalyst such as platinum another reducing agent is sodium borohydride NaBH4 followed by hydrolysis Figure 1614 illustrates the formation of an alditol H H H oH H OH H OH Ho H 1 NaBH4 Ho H 2 H20 H OH H oH H OH H oH CH2OH CH2OH D I G ucose DGlucitol DSorbitol Osazone formation The carbonyl group and adjacent alcohol oxidizes with excess phenyl hydrazine PhNHNH2 to form an osazone see Figure 1615 Osazone formation is Very important in determining the relationship between Various monosaccharides For example both Dglucose and Dmannose produce the same osazone so they re epimers Epimers differ by only one chiral center which osazone formation destroys Chapter 16 Living Large Biomolecules c 7 Figure 1615 The forma tion of an osazone Figure 1616 The bases for Kiliani Fischer synthesis H o H CN NHPh H oH cN NHPh HOH PhNHNH2xs HOH H oH H oH H oH H oH CH2OH CH2OH Syn tresizing and degrading monosaccmrides To synthesize or degrade a monosaccharide is to convert one monosaccharide into another one Historically these processes were important in understand ing the relationships between Various monosaccharides The next sections give you some insight into specific synthesis and degradation processes KiIiam Fischer synthesis KilianiFischer synthesis is a means of lengthening the carbon backbone of a carbohydrate The process begins with the reaction of hydrogen cyanide HCN with an aldehyde to produce a cyanohydrin Treatment of the cyano hydrin with barium hydroxide followed by acidification yields an aldose with an additional carbon atom as shown in Figure 1616 The formation of the cyanohydrin creates a new chiral center as a racemic mixture 0 CN CHO HCN I 1 BaOH2 CH HOCH HOCH 2 H Figure 1617 shows a specific example of KilianiFischer synthesis 292 Part IV Advanced Topics Every Student39s Nightmare CN CN CHO H oH HO H HO H HCN Q HO H HjOH T H oH H OH H OH H oH H OH CHZOH DArabinose CH2OH CH2OH J Two cyanohydrins 1 BaOH2 2 H T CHO CHO Figure 1617 The con H0H 3940 H version of Darabinose H0jH HOZZH to Dglucose HOH Hj0H and Dmannose HjOH HOH via Kiiani Fischer CH2OH CH2OH synthesis T DGlucose DMannose Ruff degradation Ruff degradation shortens the carbon backbone by one carbon atom The process begins by oxidizing the aldehyde group to a carboxylic acid called a glyconic acid Treatment of the resultant acid with hydrogen peroxide II202 in the presence of ironlll ions Fe3 results in further oxidation converting the carboxylic acid to carbon dioxide and generating an aldose with one less carbon atom Figure 1618 illustrates the Ruff degradation of glucose Chapter 16 Living Large Biomolecules Figure 1618 The Ruff degradation of glucose quot CHO COOH CH0 HOH HOH HQjH HO H Br2H2O Ho H HOH C02 H OH HOH HQH H jOH H jOH C 2Q CH2OH CH2OH DArabinose DGlucose A glyconic acid gluconic acid Meeting the Dadose family Aldose sugars make up a large part of the carbohydrate family but the ones that are really worth knowing are part of the Dfamily The simplest of these Dsugars is the triose glyceraldehyde From there you have 2 tetroses 4 pentoses and 8 hexoses Each of these aldose sugars has an enantiomer Figure 1619 shows the structures of all members of the Daldose family The circle head represents the aldehyde and the lines to the left or right indicate the orientation of the OH on each of the chiral carbon atoms The bottom carbon atom is an achiral CHZOH Note that the mirror image of each aldose in Figure 1619 is the Lenantiomer Figure 1620 indicates the pattern of OH placement for the bottom row of Figure 1619 and Figure 1621 reveals the overall pattern presented in Figure 1619 Note that indicates OH is to the right and indicates it s to the left The following mnemonic gives the bottom row in Figure 1619 from right to left ALL ALTruists GLadly MAke GUmbo In GALlon TAnks P Part IV Advanced Topics Every Student39s Nightmare Glyceraldehyde Threose Erythrose it it E Lyxose Xylose Arabinose Ribose 0 OOOO O00 Figure1619 Z Symbolic representa tionsofthe members ofthe Daldose family T Talose Galactose Idose Gulose Mannose Glucose Altrose Allose ea m as Chapter 16 Living Large Biomolecules Figure 1620 The relative positions of the OH groups in the bottom row of Figure 1619 Figure 1621 The overall pattern in Figure 1619 lBEl T G I G M G A A G T E L X A R T G I G M G A A Checking out a few disaccharides Disaccharides are molecules that break apart into two monosaccharides during hydrolysis Examples include sucrose maltose and cellobiose We cover all three of these disaccharides in the next sections The properties of disaccharides aren t a simple combination of the properties of the two monosaccharides present For example the disaccharide lactose milk sugar is a reducing sugar containing a BDglucose linked to a Dgalactose molecule Sucrose Sucrose pictured in Figure 1622 is a nonreducing sugar containing a glucose joined to a fructose Hydrolysis of sucrose gives a 5050 mixture of the two monosaccharides as invert sugars The name invert sugar derives from the fact that there s an inversion of optical activity upon hydrolysis Sucrose is dextrorotatory as is glucose fructose however is levorotatory The sum of the optical activities of the monosaccharides is levorotatory because fructose more than compensates for the activity of glucose 296 Part IV Advanced Topics Every Student39s Nightmare Figure 1622 The structure of sucrose Figure 1623 The structure of maltose CH2OH H O H CH2OH H O OH H H OH O CH2OH OH OH H I Glucose Fructose Sucrose Maltose Maltose is a disaccharide made of two glucose molecules The linkage involves the oc anomer of the left glucose see Figure 1623 Maltose is a reducing sugar that s sometimes derived from starch or see the later related section The left ring in Figure 1623 is locked and nonreducing however the right ring is reducing and can undergo mutarotation CH2OH CH2OH H H H H o o OH H OH H OH 0 OH Cellobiose The hydrolysis of cellulose a polysaccharide sometimes yields the disaccharide cellobiose When cellobiose a reducing sugar is hydrolyzed two glucose molecules result Unlike maltose which we describe in the preceding section the linkage involves the 3 anomer of the left glucose see Figure 1624 Looking at some polysaccharides Polysaccharides are polymers of monosaccharides usually glucose Examples include starch which is 20 percent amylose and 80 percent amylopectin glycogen animal starch and cellulose We fill you in on all three of these polysaccharides in the following sections Chapter 16 Living Large Biomolecules 7 Figure 1624 The structure of cellobiose CH2OH H o CH2OH OH H H O O OH H OH H OH OH H H OH Starch Starch molecules contain multiple oc anomers of glucose joined by a linkage between the anomeric carbon 1 of one ring and the fourth carbon 4 of the next ring This linkage is therefore known as an 0c 1 A 4 linkage Starch has two components 1 Amylose Amylose is soluble in hot water and nearly insoluble in cold water The chain of glucose units curl into a helical structure with six glucose molecules per loop This form of starch interacts with iodine 12 to produce an intense dark blue color In general amylose has a simple chain structure with each glucose attached to two others The molecular weight of amylose ranges from 150000 to 600000 gmole which indicates that a single molecule of amylose contains anywhere from 800 to 3500 joined glucose molecules During digestion the enzyme Bamylase attacks amylose at the nonreducing end and hydrolyzes the chain into maltose units The enzyme Bamylase is specific for 0c 1 A 4 linkages 1 Amylopectin Amylopectin is similar to amylose except that the glucose chain has branches These branches involve linkages at the CHZOH position 6 which makes them 0c 1 A 6 linkages Amylopectin is watersoluble it also interacts with iodine to form a reddishpurple complex Typically amylopectin is ten times the size of an amylose molecule Digestion requires Bamylase 1 A 4 linkages and a second enzyme to remove the branches as in the 1 A 6 linkages Glycogen Glycogen animal starch is similar to amylopectin but it features more branching and tends to have a higher molecular weight Glycogen occurs in the liver and muscle tissue It interacts with iodine to produce a red color Part IV Advanced Topics Every Student39s Nightmare Figure 1625 A glycosyl amine Figure 1626 An amino sugan Cellulose Cellulose constitutes about 50 percent of wood and about 90 percent of cotton The molecular weight of this polymer is in the 1 to 2 million range Cellulose molecules contain the multiple 3 anomers of glucose joined by a linkage between the anomeric carbon 1 of one ring and the fourth carbon 4 of the next ring This is a B 1 a 4 linkage which is why enzymes specific for or 1 4 linkages and consequently humans can t digest cellulose The chain is typically unbranched and hydrogen bonding between the chains lowers cellulose s solubility in water Treatment of cellulose with a mixture of sodium hydroxide NaOH and carbon disulfide CS2 yields cellulose xanthate The action of acid on cellulose xanthate produces rayon or cellophane Discovering nitrogemcontaininq sugars It may not seem like it but you can replace an alcohol group OH with an amine group NH2 to create a nitrogencontaining sugar Examples of nitrogencontaining sugars include glycosylamines see Figure 1625 and amino sugars see Figure 1626 In glycosylamines the amine group replaces the alcohol group on the anomeric carbon In amino sugars the amino group replaces an alcohol group on a carbon atom that isn t anomeric CH2OH H jo NH2 OHH OH CH2OH I jo OH OH Chapter 16 Living Large Biomolecules l Lipids Storing Energy Now So you Can Study Longer Later Figure 1627 The general structure of fats and oils Lipids are a diverse group of biologically important compounds that are linked only by their similar solubility in nonpolar or slightly polar solvents Only a few lipids make an appearance in your average Organic Chemistry 11 course Fats and fatty acids are highlighted but steroids prostaglandins phospholipids and other lipids generally aren t check out Biochemistry For Dummies written by us and published by Wiley if you want to know more about any of these lipids Pondering the properties of fats A fat is a solid at room temperature triester of glycerin glycerol with fatty acids longchain carboxylic acids An oil is similar to a fat however an oil is a liquid at room temperature The general structure of fats and oils appears in Figure 1627 The melting point of a fat or oil depends on the size of the R groups and the level of unsaturation The smaller the R groups andor the greater the unsaturation the lower the melting point is The types of bonds present also help lower the melting point The long chains of the R groups pack tightly if no double bonds exist but if double bonds are present they create kinks in the chain Chains with kinks don t pack together well thereby lowering the melting point the better the chains pack together the higher the melting point is Naturally occurring examples have fatty acids with an even number of carbon atoms usually 10 to 20 without any branches Double bonds are only present as the cis isomer At least one double bond is present in an unsaturated fat multiple double bonds are present in a polyunsaturated fat t7 Part IV Advanced Topics Every Student39s Nightmare p Srgxk Figure 1628 The general process for producing a soap Figure 1629 The structure of sodium stearate You can hydrogenate some of the double bonds in any polyunsaturated fat Hydrogenation whether complete or partial results in an increase in the melting point a fact that makes it possible to convert a liquid oil to a solid fat Note that partial hydrogenation may convert some of the cis double bonds into trans double bonds thereby producing a trans fat Soapinq up with saponification The saponification the basecatalyzed hydrolysis of an ester of fats has been important since ancient times This process frees the glycerin and releases the fatty acids as carboxylate ions The carboxylate ions along with sodium or potassium ions from the base create a soap refer to Figure 1628 NaOH Fat L Soap OF KOH Sodium stearate shown in Figure 1629 is an example of a type of soap The carboxylate ion in the soap has a hydrophilic or ionic end and a hydrophobic or organic end The presence of the hydrophilic head and hydrophobic tail ends is the key to the behavior of soap The hydrophilic end is water soluble whereas the hydrophobic end is soluble in nonpolar materials such as oil but not the glycerincontaining form O Na 0 C CH216CH3 T T Hydrophilic Hydrophobic Soap works because the nonpolar end dissolves in dirt oil leaving the polar end outside the dirt This combination is known as a micelle see Figure 1630 To the surrounding water molecules the micelle appears as a very large ion These ions are watersoluble and repel each other due to their like charges a behavior that causes them to remain separated Metal ions in hard water Ca2Mg2Fe2 cause a precipitate to form because they react with carboxyl ate ions to form an insoluble material also known as the soap scum hanging around the bathtub or shower Chapter 16 Living Large Biomolecules b 7 Figure 1630 A micelle Detergents are like soaps but they re derivatives of either phosphoric acid or sulfuric acid see Figure 1631 They ve gained popularity because they don t precipitate in hard water S03 Na Sodium sulfonate O S Q Na Fi9quotre163931 Sodium sulfate ester Three gene c detergents O a sulfonate a Sulfate PQ Na ester and a phosphate Na ester T Sodium phosphate ester Bulking up on Amino Acids and Proteins Amino acids are the building blocks of proteins which along with polysac charides and nucleic acids are important biological polymers Much of the behavior of proteins is better described in a biochemistry text However many of the properties of amino acids are as much a part of organic chemis try as they are a part of biochemistry PF Part IV Advanced Topics Every Student39s Nightmare Figure 1632 The general structure of ocamino acids The next sections familiarize you with the basic chemistry of amino acids including their physical properties They also walk you through the many ways in which amino acids can be created synthesized Introducing amino acids Amino acids contain an amino group and a carboxylic acid group The biologically important amino acids are the ocamino acids see Figure 1632 In the ocamino acids the amino group and the carboxylic acid group are attached to the same carbon atom Note that the oc carbon is chiral unless R is a hydrogen atom The naturally occurring ocamino acids are the L enantiomers The difference between ocamino acids is the identity of the R group also attached to the occarbon H I a 0 H2N C C OH R Amino acids link to form amides and create a polyamide In biochemistry the amide gets replaced with a peptide making the proteins polypeptides The sequence of the amino acids dictates the primary structure of the protein Perusina the physical properties of amino acids When proteins undergo hydrolysis you wind up with 22 0c amino acids 20 of which are regular amino acids and 2 of which are derived amino acids Amino acids are amphoteric they possess the characteristics of both acids and bases and can react as either because both acidic and basic groups are present An internal acidbase reaction produces a dipolar ion known as a zwitterion you can see the general structure of one in Figure 1633 Figure 1633 The general structure of a zwitterion Figure 1634 A glycine zwitterion Figure 1635 The action of an acid and a base uponthe zwitterion of alanine R CH3 Chapter 16 Living Large Biomolecules H G l 0 H3N C C 8 R The simplest amino acid is glycine check out a glycine zwitterion in Figure 1634 In glycine the R group is a hydrogen atom The ammonium portion of the zwitterion is acidic and the carboxylate portion is basic The presence of the charges in the zwitterions makes amino acids ionic solids meaning they re nonvolatile The ionic character makes the amino acids soluble in water but not in nonpolar solvents H H3E C3 CO Basic I e A H O Acidic The acidic and basic groups have Ka and Kb values respectively In general the Ka values are approximately 10 9 10 2 pKa 912 and the Kb values are approximately 1039 10 12 pKb 1112 Glycine has a pKa equal to 96 and a pKb equal to 117 Therefore the addition of an acid to a glycine zwitterion leads to protonation of the carboxylate end and the addition of a base leads to deprotonation of the ammonium end see Figure 1635 H H H O I O I 0 HG C C ii H3 C C H C C 3 3 2 H G H l CH3 0H CH3 0 CH3 0 Cation Zwitterion Anion 304 Figure 1636 Determining the isoelectric point pl lBEI Q 2quot Part IV Advanced Topics Every Student39s Nightmare The zwitterion form predominates at the isoelectric point pl Figure 1636 shows where the pl values come from I p p 2 The cation form predominates at a pH lt pl whereas the anion form pre dominates at a pH gt pl In general pKcarb equals pK1 and pKamine equals pK2 However complications may occur because the R group may be acidic or basic The conjugate base of every acid has a pKb that s related to the pKa as pKa pKb 140 or KaKb KW 10 x 10 4 Studying the synthesis of amino acids You can synthesize amino acids in a number of ways although the presence of both an acidic and a basic group may cause some complications Amino acid synthesis serves as the application of previously introduced reactions and most synthetic methods produce a racemic mixture The sections that follow give you a look at the various ways in which amino acid synthesis can occur From ochalogenated acids When the occarbon of a carboxylic acid is halogenated an amino group replaces the halogen in the halogenated acid with excess ammonia Figure 1637 illustrates this reaction From potassium phthalimide When potassium phthalimide is added as part of a sixstep synthesis of an amino acid the resulting amino acid is present as a racemic mixture Figure 1638 illustrates multistep synthesis involving potassium phthalimide Chapter 16 Living Large Biomolecules quot Study reaction sequences such as Figure 1639 to note how the scheme develops and how you can change it to obtain a different product 0 O 1 BrgP CH32CHCH2CH2C T CH32CHCIZCHC 2 H20 OH OH Br T 1 NH3XS Figure 1637 The 0 SVn Eh9SiS CH32CHCH2CHC of an amino 0 v acid from an NH3 ochaogen ated acid R3quot 9UC39n9 45 O O O O C OEt C OEt o I N K Br C H gt N C H C OEt 0 OC OE O O 1 OH T 2 RX Figure 1638 R CHs3CtbCH2C39 The syn 0 OX thesis of an NH2 C OEt amino acid i m9thi0 HCCH2CH2SCH3 NjCCH2CH2SCH3 nine using 2 r potassium COH 3 A C QEt phthalimide 0 Q7 O 306 Part IV Advanced Topics Every Student39s Nightmare quot Figure 1639 The begin ning of the synthesis of an amino acid via an amido malonic ester From amido malonic esters The multistep synthesis shown in Figures 1639 and 1640 leads to the synthesis of an amino acid Note that the diethyl acetamidomalonate shown in Figure 1640 is the product of the reaction in Figure 1639 The example in the figures is a natural amino acid however you can synthesize other not natural amino acids with this procedure Sequences such as those shown in Figures 1640 and 1641 allow you to review many of the different reactions you encountered previously If you don t recall the exact details of one of the reactions go back and review the reaction o C OEt 0 H o H CH3CH23ONO 11 o 2 H2804 o oE 0 Ho Nc 0 l H2Ni o 0 0 O C OEt CH3C C 0Et I C NH cH lt C39 H2N CH l CH3 C OEt C 00 OEt N Diethyl acetamidomalonate Strecker synthesis The Strecker synthesis pictured in Figure 1641 is a relatively simple method for synthesizing an amino acid but it depends on the availability of the appropriate aldehyde Figure 1642 shows a specific example for the synthesis of phenylalanine the resulting amino acid presents itself as a racemic mixture Chapter 16 Living Large Biomolecules BI 7 Figure 1640 The end synthesis of an amino acid via an amido malonic ester Figure 1641 The general process of Strecker synthesis Figure 1642 The synthesis of ipheny alanine via the Strecker synthesis 0 O O 3 OEt Q 3 OEt E e C NH 3H to C NH 3 CH3 CH3 C OEt C Et 0 0 Diethyl acetamidomalonate C BLCH2 OEt O c oa ll H CO2CH3COOH2EtOH CNHC CH2 OEt O O CH3 C OH ll 0 NH2CHCH2 OH O NH2 NH2 0 R C KCN NI4C gt R C CEN HH20 R C C H H H OH O CH C CH CHCN 2 H20 2 H NH2 H O CH2CH C OH NH2 P Part IV Advanced Topics Every Student39s Nightmare Figure 1643 The general process of reduc ve amination Reductive amination Reductive amination shown in Figure 1643 is a biologically important method of synthesizing amino acids The reduction is caused by NADH in the presence of ammonia NADH nicotinamide adenine dinucleotide is a biological reducing agent The mechanism for this reaction and the other biologically important procedure for synthesizing amino acids transamination are more biochemistry than organic chemistry so we don t touch on them here You can check out Biochemistry For Dummies written by us and published by Wiley if you want to know more 0 H I 0 RCCOH gt R C3 C ll NH2 OH Resolution of i amino acids The resolution of i amino acids isn t really a synthetic method but it s certainly useful in the production of a particular amino acid from a racemic mixture In the resolution of i amino acids an enzyme a biological catalyst interacts with only one enantiomer Why you ask Because enzymes are ste reoselective The enzyme leaves one enantiomer unchanged and modifies the other into a different compound which makes it possible to separate the enantiomer from the other compound by a number of techniques After the enantiomer has been separated all that s left is to reverse the process induced by the enzyme Part V Pullinglt All Together The 5th Wave By Rich Tennant 39 3 quotquot 39 i 39 Okag now that the paramedic is here with the de brillator and smelling salts prepare to learn about covalent bondsquot In this part n this part are a couple of chapters designed to help you pull together all those concepts you have been studying In the first chapter we look at some strategies that you can use when designing a synthesis of a particu lar organic compound You know that type of problem Starting with a piece of coal and a glass of water synthe size DNA naming all intermediates Okay maybe not quite that involved but close Then we tackle those dreaded roadmaps walking you through a number of examples and giving you some tips to ease the way Realize that if you re using this book during a course you re almost done when you hit this section Hang in there there s a light at the end of the tunnel and we all hope it s not a train Chapter 17 Overview of Synthesis Strategies In This Chapter Determining strategies for synthesis problems Figuring out a onestep synthesis Considering multistep synthesis for more complicated problems Solving sample problems with retrosynthetic and synthetic analysis NBER oiquot Synthesis is actually the reverse of predicting products In synthesis you have the product of a reaction and you must predict the reaction sequence necessary to form the product While you may have hints as to the identity of the starting material in most cases you need to predict the starting material the reactants and possibly the reaction conditions You encounter two general types of synthesis questions onestep synthesis and multistep synthesis As the name implies a onestep synthesis problem requires one simple answer A multistep synthesis involves more than one reaction and more than one answer may be correct On an organic chemistry exam onestep synthesis questions usually focus on the most recent reactions you have studied whereas multistep synthesis questions usually involve a recent reaction in one step and one or more other reactions from any point in Organic Chemistry I or II in other steps When predicting products you should ask yourself three basic questions and the same questions are also important when considering synthesis problems 1 What kind of reaction 1 What is the regiochemistry of the reaction the chemical environment of the reactive site on this molecule for this particular reaction 1 What is the stereochemistry of the reaction However you need to apply these questions in a different way for synthesis problems than for predicting problems The answers to all three questions are found in the mechanism 3 Part v Pulling It All Together quot If the question starts with a ketone what do you do When predicting products you must consider every reaction you know that begins with a ketone When doing a synthesis problem you must consider every reaction that produces a ketone On an organic chemistry exam if the instructor takes time to draw the specific stereochemistry of a molecule it s a hint that you should carefully consider the stereochemistry of any reactions you use To do any synthesis problem you must know your reactions backwards and forwards This takes time and effort To learn them thoroughly you need to practice over and over During an exam you don t have time to work out every possibility so you must know the reactions Working with One Step Synthesis In a onestep synthesis problem you know that only one reaction is necessary to form a particular product The key is the functional group in that product For example if the functional group is an aldehyde the reaction must be one of the limited number of reactions that you know form an aldehyde Other information may limit your choice of reactions For example the problem may ask you to begin with an alcohol with the same number of carbon atoms or you may have an alkene with a greater number of carbon atoms Next you need to consider regiochemistry Do any of the reactions you know give a product with the correct regiochemistry This may allow you to narrow your choices Finally you must consider the stereochemistry Do any of the reactions you know give you the correct stereochemistry Again this may further limit your choices In some situations more than one reaction can give you the correct product Don t overanalyze the problem in an effort to choose between multiple correct answers just pick one and go with it Tackling Multistep Synthesis In most cases a multistep synthesis is not a simple string of reaction steps You need to look ahead and behind For example you may have a perfectly good reaction that forms a cisalkene but that s the wrong answer because two steps later you need a trans alkene Chapter 17 Overview of Synthesis Strategies 3 If you don t know your reactions you can t possibly solve a multistep synthesis problem Knowing the reactions includes the reactants products conditions regiochemistry and stereochemistry Flashcards are a useful means of learning the reactions On the front side of the card write the reactants and conditions and on the reverse side write the products regiochemistry and stereochemistry Learn the cards in one direction first identifying what s on the back based on what you see on the front and then learn them in reverse knowing what s on the front when you look at the back You must know the reactions backwards and forwards Shuffle the deck often Many times multistep synthesis problems start with an alkane An efficient way of adding a functional group to that alkane is to brominate or chlorinate the alkane through a freeradical reaction A limited number of reactions can add or remove carbon atoms For this reason comparing the carbon backbones of the materials involved in the reaction is useful These reactions require the presence or absence of certain functional groups Practicing Re trosynthe tic and Synthetic Analysis quot Retrosynthetic analysis is a method for tackling synthesis problems especially multistep synthesis problems The application of this technique involves working the problem backwards starting at the final product and ending up with the initial reactants One of the great frustrations you may encounter during an organic chemistry exam is expecting to know the answer to a retrosynthetic analysis question immediately upon first reading the problem Save yourself this frustration and accept the fact that you need to read and reread the problem multiple times When applying retrosynthetic analysis to a multistep synthesis problem you must work backwards If you become lost as a last resort you may want to look at the forward reactions However the forward process often goes off on a tangent or leads to a culdesac After finishing a multistep synthesis problem spend some time working on another problem or task Then come back and check each step in both the forward and the reverse direction You should pay particular attention to both the regiochemistry and the stereochemistry of each step In addition if one of the steps involves a molecule with more than one functional group make sure the reaction only alters the desired functional group 3 Part v Pulling It All Together quot Figure 171 The first ret rosynthetic analysis problem quot To survive multistep synthesis questions on organic chemistry exams you must practice practice and practice The rest of this chapter is dedicated to practicing retrosynthetic and synthetic problems The following examples utilize some of the reactions that you have seen throughout this book Example 7 Figure 171 presents a typical retrosynthetic analysis problem T he presence of a Bdicarbonyl compound indicates that the formation of a carbanion through the loss of a hydrogen ion from the occarbon will probably be important however don t let this distract you from the task at hand The problem asks you to prepare a ketone Your first question should be How can I prepare a ketone One answer to this question is to decarboxylate a Bdicarbonyl compound For the compound in this problem the reaction shown in Figure 172 works Resist the temptation to look ahead on this problem Try to follow the steps and make your own predictions on what to do Your choice may be as good as or better than the one presented here Prepare 3 if CH CH CH CH from C C 3 2 3 W3 CH3 CH2 O39CH2jCH3 0 It may help to redraw the one or more compounds in a retrosynthesis problem so the general shape of the reactants and products are similar Figure 172 One step inthe solution to the problem in Figure 171 CH3 OC T CCH3 A oC CH2CH3 H OH Chapter 17 Overview of Synthesis Strategies 3 7 5 The next step is to determine what reaction produces the reactant in Figure 172 The carboxylic acid can be produced through the hydrolysis of an ester You can now propose the reaction required to produce the particular acid The other compound in Figure 171 is an ethyl ester so this process probably involves an ethyl ester Figure 173 adds the hydrolysis step to the solution to this problem CH O C 3 CCH3 OC CH2CH3 I CH339CH2 1OH39 2 H 39 CH 3 j Figure173 CH3 A 01C Addingthe CH3CH2OH C j hydrolysis CHZCH3 H step OH CH3 CH3 co CH2C H3 3 m Part v Pulling It All Together Next you need to determine how to produce the ester in Figure 173 This is a Bdicarbonyl compound like the starting material This relationship may be important One way to produce this ester is shown in Figure 174 This step requires you to start with a carbanion which can form through the reaction of a strong base with a Bdicarbonyl compound This step is in Figure 175 We use the tbutoxide ion as the base but other bases are acceptable CH3 CH 1 CH 0 C o c 3 Z 3 CH3CH2BT 0 CH3 O C CH2CH3 00 O O CH3CH2 CH3CH2 1OH 2 H 39 Figure174 1 CH3 A 0 CH3 Adding CH3CH2OH C T C th anosteit O C CH2CH3 H CH2CH3 Chapter 17 Overview of Synthesis Strategies 3 7 7 CH OC 3 O i CH339CH2 tOBu CH3 CH 3 O C CH3 0 C CH3CH2BT 3 OiC CHQCH3 Oi Cl 0 CH3CH2 CH3CH2 1OH39 2 H Figure 175 CH3 CH3 Addingthe O O reaction CH3 CH3 with the CH3CH2OH c A gtclt quotb t 39 ne OC CH2CH3 H CH2CH3 3 p Part v Pulling It All Together Now how do you prepare the starting material in Figure 175 The answer is to use a methyl halide such as bromomethane This step appears in Figure 176 CH CH O 3 3 6 CH Br 3 CH CH3 O C O C T CH3 CH2 CH3 CH2 tOBu CH3 CH 1 CH 0 C o c 3 3 CH3CH2BT 0 CH3 O C CH2CH3 QiC ltgt o CH3CH2 CH3CH2 1OH 2 H Fi9ure176 O CH OC 3 Addingthe 3 CH3 reaction CH3CH2OH C C gtC Wlth bI 0m0 CH2CH3 H methane 3 You can link to the starting material from Figure 171 with one more step You need to convert the starting material to the carbanion at the top of Figure 176 The complete answer is in Figure 177 Chapter 17 Overview of Synthesis Strategies 3 7 9 Oi CH3 0 CH3 1 CH3 EtO EtOH e CH3Br CH CH T CHCH3 O O O O O CH3CH2 CH3 CH2 CH3 CH2 tOBu CH CH O C 3 Q 3 CH3CH2BT j lt3 3 OC CHQCH3 Q 1 CH3CH2 CH3CH2 1OH 2 H Figure 177 CH3 CH The sou O O 3 tiontothe CH3 CH3 problem CH3CH2OH C C C presented in OC CH CH j 2 3 H CH2CH3 FIgure171 Example 2 Figure 178 presents a synthetic problem and the answer is given in Figure 179 Try to solve this problem before looking at the answer You may use any inorganic reagents and any organic reagent containing four carbons or less m Part v Pulling It All Together Synthesize in C CH2 CH2 CH3 CH3 CH2 CH2 C from two moles of Figure 178 v Q Q The second 0 l 0 w analysis C C problem 2 CH3 CH2 OCH2 CH3 ii in C C CH3 CH2 pev y H2 y H3 EtO EtOH in ii in l C C C C CH3CH2O CH CH3 CH Br CH3CH2O CH CH3 2 ltgtlti 2 9 CH2Br C CH3CH2O CH CH3 CH3CH2O CH FCH3 C C C C 1gt0H39 II ll 0 O 2 O O ii i T C C C FIgure179 solution CH2 2 C02 CH2 tothe 2 3Vquotth9 HO cH CH3 CH2 Camp3H3 analysis presented in C C C Figure 178 q o q Chapter 17 Overview of Synthesis Strategies 7 Figure 1710 The third analysis problem Figure 1711 Decarb oxylation to produce the desired product shown in Figure 1710 Figure 1712 Preparation of the cyclo propane product Example 3 Another retrosynthetic analysis problem is in Figure 1710 In this problem you may use any inorganic reagents and any organic reagent containing four carbons or less Synthesize O 11 r T CH2 CH3 CH from one mole of C C cH2 CH3 CH2 o CH2 CH2 As seen previously in a Variety of reactions throughout this book the desired product can be formed by decarboxylation Figure 1711 illustrates this decarboxylation step CH3 CH3 OC CH2 oC CH2 2 CCH2 OH You now need to prepare the cyclopropane starting material Figure 1712 shows one way of preparing the cyclopropane starting material CH3 CH3 o C CH CH3 j 2 oC CH2 0 C CH2 L C CH2 1 oH CH CH 4 C CH2 2 O 2 H oC OH O CH3CH2 Part v Pulling It All Together Two steps shown in Figure 1713 are required to prepare the cyclopropane ring Even though not normally required in a retrosynthetic analysis problem Figure 1713 also shows a partial mechanism CH3 CH3 gtBr 9 BrCH2CH2Br CH T CH2 oc o H O O CH3CH2 CH3CH2 C7QEt CH3 CH CH3 oc CH2 3 00 CH T OC CS2 L CCH2 1 OH 2 Figure1713 CHCH2 H C CH2 The forma 0 0 tion ofthe OH cycopro O pane ring CH CH T 3 2 Finally you need to form the carbanion at the beginning of Figure 1713 This step completes this retrosynthetic analysis problem The complete solution is in Figure 1714 Example 4 Figure 1715 presents a fourth problem another retrosynthetic analysis and Figure 1716 shows one possible solution You may use any inorganic reagents and any organic reagent containing four carbons or less Chapter 17 Overview of Synthesis Strategies 3 Figure 1714 The complete solution to the problem presented in Figure H T H cgt Figure 1715 The fourth analysis problem CH3 CH3 CH3 Br Ckl O1 O O 2 CH2 EtO EtOH g3H BrCH2CH2Br C 3H 2 00 00 0 H O 0 O CH3CH2 CH3 CH2 CH3 CH2 G lOEt CH3 CH 3 CH3 o CH2 oc CH2 A C CH 0 C CS2 2 1OH 01C OH O CH3 CH2 Synthesize CH2 CH2 0 P D D C from C OH EtO CH2 OEt e Part v Pulling It All Together OEt OEt oc o C 2 a CH NaHDMF H Q OC OEt OEt CH2Br OH OEt OiC 00 CHCH2 OH gtCH CH2 Figure 1716 OH OEt The solution to the problem A presented CH2quotCH2 in Figure OC 1715 Example 5 Figure 1717 presents one last retrosynthetic analysis problem with its solution in Figure 1718 You may use any inorganic reagents and any organic reagent containing four carbons or less Synthesize Figure 1717 O C The last ret CH2CH C rosynthetlc CH2 OH OH from CH2 analysis 00 problem T CH3 OEt Chapter 17 Overview of Synthesis Strategies 5 OEt OEt O O C NaHDMF 8 CH2 H O O CH2Br CH CH OEt OEt 2 OEt OEt Q C CHCH2 O C CHZCH2 e KtOBu CCH2 CHCH2 OC Q OEt OEt CH3BT OEt OH oclt CH3 CHCH2 oc CH3 CHCH2 C CH2 1 OH C CH2 oc 3 o c OEt QH Figure 1718 A The solu tiontothe H CH3 CHCH2 problem presented C CH2 in Figure 0 1717 Part v Pulling It All Together Chapter 18 Roadmaps and Predicting Products In This Chapter Mastering roadmaps Finding out how to predict products reating roadmaps and predicting the products of a reaction are two challenges found on many Organic Chemistry II exams In Chapter 16 we give you some techniques on how to handle roadmaps In this chapter we apply some of those techniques to show you how to build a roadmap without totally freaking out Then we give you some tips in predicting reaction products Preparing with Roadmap Basics Many Organic Chemistry II exams contain problems known as roadmaps that present you with a collection of facts you use to deduce the identities of a number of compounds Over the years we have seen numerous students throw up their hands when faced with a roadmap problem Indeed many students simply skip the roadmap problems on their exams That s a good way to lose a significant number of points and unnecessarily lower your grade If you know your reactions and the other rules roadmaps aren t as difficult as they seem The secret is to tackle the problem in small pieces first by reading the problem and making a few notes then continuing with the exam Come back to the problem later and make a few more notes Then go to some other part of the exam Continue cycling from the roadmap to other questions on the exam until you have sufficient notes to attempt to solve the roadmap If the roadmap defies solution at this point return to the cycling procedure until you re ready to make another attempt or until that s the only part left on the exam H Part v Pulling It All Together Practicing Roadmap Problems Now you get to take a look at number of roadmap problems Some instructors include spectra on one or more of the compounds in the problem We limit the amount of spectral data so you can focus on the approach to the solution of roadmap problems Look over the question before you look at its solution See if you can develop a plan of attack on your own Take your time Problem one Compounds A and B are both hydrocarbons of molecular weight 98 They re both optically active but have different specific rotations Ozonolysis of A gives formaldehyde CHZO and a ketone ozonolysis of B produces formal dehyde and another aldehyde Treating either A or B with hydrogencatalyst yields C whose molecular weight is 100 C is an optically active compound Give structures for A B and C Solution one On scratch paper make a table consisting of three rows or columns labeled A B and C respectively Each time through the problem try to add some information to this table For example on your first pass you may find A B c cxny cxny c H x y2 The formulas for A and B reflect the fact that these are hydrocarbons while the formula for compound C includes the information that catalytic hydrogenation yields an increase of two mass units corresponding to two hydrogen atoms In addition the hydrogenation reaction indicates that A and B are alkenes On the second pass through you may wish to estimate the values of x and y If the molecular weight is 98 then the maximum value for x is 98 12 g per atom of carbon 8 only whole numbers of atoms are possible A value of 8 doesn t leave much room for hydrogen 2 hydrogen atoms bring the value to 98 If x is 7 then y needs to be 14 while if x is 6 then y needs to be 26 For Chapter 18 Roadmaps and Predicting Products Figure 181 Optically active pos sibilities for seven car bon atoms labeled carbon atoms are chiral an alkane if x is 6 the maximum y is 14 therefore x cannot be 6 or smaller This limits x to 7 which requires y to be 14 This changes the table to A C7H14 B C7H14 C C7H16 The formulas of A and B correspond to CnH2n which means these are simple alkenes with no rings present The formula for C CnH2n2 is that of an alkane also with no rings Since A and B both undergo hydrogenation to form C they must have the same backbone structure which is the same as that of C At this point you may wish to sketch skeleton arrangements for seven carbon atoms There are ten possibilities but only two of the possible arrangements are optically active These two arrangements are shown in Figure 181 The starred carbons are chiral carbons C c o o c c c I o o o o o C I C Formaldehyde and a ketone can only be produced from ozonolysis of A in a limited number of positions which are indicated in Figure 182 The X marks the possible position of the double bond in the structures Two of the three choices in Figure 182 would eliminate the chiral center Thus the double bond in A must be the only one that doesn t involve the chiral carbon atom This gives you not only the structure for A but also the structure for C the hydrogenated version of A Compound B has the same skeleton but has only one place for the double bond if ozonolysis gives formaldehyde and an alde hyde between the last two carbon atoms on the right side of the structure Figure 183 gives the structures of A B and C 330 Part V Pulling It All Together Figure 182 The possible positions X forthe double bond in com pound A Figure 183 The struc tures of A B and C C C C C C C C X IX X C C C C C C I C CH3 CH3 CH3C CH CH3 CH3 CH3 CH CH CH3 CH3 CH3 CH3 CH3 A CH3 CH CH CHCH3 C CH3 B You should now work backwards from your answer to make sure these structures account for the facts in the problem Problem two Unknown compound D which has the formula CIOHIO readily decolorizes a bromine in carbon tetrachloride solution When D is allowed to react with hydrogen and a catalyst E is obtained The M for E has an me Value of 134 When E is heated with a mixture of nitric and sulfuric acid only one product is produced That compound has an M of 179 Heating either D or E in hot chromic acid produces a solid F C8H3O4 whose melting point is in excess of 300 degrees Celsius Give structures for D E and F Solution two Your table begins with three columns D E F C10H10 C H C8H6O4 X Y Chapter 18 Roadmaps and Predicting Products 7 Again the solution involves several passes through the problem gathering data Don t worry if you gather this data in a different sequence than the one shown You just need a collection of facts and the order of acquisition is irrelevant V Compound D has me equal to 130 so the compound at 134 has increased by four hydrogen atoms CIOHI You now know the formulas for all three unknowns You also know that D has either a carboncarbon triple bond or two carboncarbon double bonds The reaction with bromine in carbon tetrachloride confirms this V The degree of unsaturation for compound D is six The conversion of D to E accounts for two of the six The remaining four since they don t react indicate an aromatic ring An insufficient number of carbon atoms are present to account for this degree of unsaturation by rings alone See Organic Chemistry I For Dummies by Arthur Winter Wiley if you re not sure how to determine the degree of unsaturation V The conditions leading to the formation of F indicate a disubstituted aromatic ring In addition the reaction conditions and formula indicate that F is a dicarboxylic acid Disubstitution may be ortho meta or para V The reaction of compound D or E with a mixture of nitric and sulfuric acid nitrates the ring The odd M indicates an odd number of nitrogen atoms and the mass increase 179 134 45 indicates the product is mononitro V The two substituents in D contain four carbon atoms These could be a methyl plus a C3 or two C2 groups V If D is ortho and if one group is methyl and the other group is C3 then four mononitroproducts are given If D is meta then it has four mononitroproducts If D is para it has two mononitroproducts V If D is ortho and if both are C2 then two mononitroproducts are given If D is meta three mononitroproducts are given If D is para then it has one mononitroproduct V The problem states that only one nitration product forms Therefore D and by implication E and F is para substituted In addition both C2 branches must be identical which means each has one degree of unsaturation two alkenes and not one alkyne The branches are CHCH2 Hydrogenation converts the branches to CH2CH3 Heating with chromic acid eliminates the carbon atom furthest from the ring and converts the remaining carbon atoms to carboxylic acid groups COZH Part v Pulling It All Together Figure 184 The structures of D E and I39 The structures of D E and F are found in Figure 184 CH3 CH2 co2H CH2 CO2H CH3 D E F Problem three G and H are isomeric compounds of formula C12H17Br Both give an immediate precipitate upon reaction with alcoholic silver nitrate They both react with NaOEtEtOH to give I a hydrocarbon with an Mt me value of 160 gives five singlets in the NMR the one at 87 being rather broad I decolorizes bromine in carbon tetrachloride and adds one mole of HBr to give G If peroxides are present H is obtained Ozonolysis of 1 produces J and K both of which have a strong band at 1700 cm in the IR K is identical to the product obtained when methylacetylene is allowed to react with mercuric sulfate in aqueous acid J gives two singlets and two doublets in the NMR The low field doublets are in a 23 ratio to the upfield singlets Nitration of J gives only two products both of which have Mt 179 and an M 1 intensity of 1037 One of the two products strongly predominates over the other Give the structures for G through J Solution three As usual begin with a table G H I J K C12H17Br C12H17Br CxHy me 241 me 241 me 160 Chapter 18 Roadmaps and Predicting Products You need to collect the following facts in any order V The signal at 87 in the NMR of indicates the presence of an aromatic ring C6 In addition I has four other nonequivalent hydrogens that are not on a carbon atom that has hydrogen atoms on the adjacent carbon atom J has four NMR signals therefore the hydrogen atoms are in four different environments Worry about the ratios later V The strong band at 1700 cm in the IR of J and K indicates the presence of a carbonyl group V G and H both give an immediate precipitate upon reaction with alcoholic silver nitrate This means that the bromine is probably attached to a tertiary carbon V The reaction with NaOEtEtOH is a dehydrohalogenation which indicates that I is an alkene Since both G and H give 1 they must have similar regiochemistry The me value 160 corresponds to the loss of HBr V The reaction of I with bromine in carbon tetrachloride confirms the prediction that I is an alkene The reaction with HBr means that G is the Markovnikov addition product and the reaction in the presence of peroxides makes H the antiMarkovnikov product V The ozonolysis of along with the IR data indicate that J and K are aldehydes ketones or one of each V Methylacetylene is CH3CECH V The M and M 1 intensity for J indicates that it s a mononitro product Since it has only two products J must be para The predominance of one product over the other means that one of the groups is a better director than the other The previous eight deductions give the foundation for solving the problem The easiest structure to deduce is that of K The reaction of methylacetylene with mercuric sulfate in aqueous acid gives acetone Therefore K is acetone One down four to go Hooray for partial credit The loss of acetone means you have lost C3H6 from I and added an oxygen You now have all the formulas and one structure G H I J K C12H17Br C H Br CIZHI6 CQHIOO C3H6O 12 17 me 241 me 241 me 160 me 134 me 58 334 Part V Pulling It All Together Figure 185 The structures of J and K Figure 186 Finding I Figure 187 The structures of G and H The presence of an aromatic ring in 1 indicates that all five compounds are aromatic J has three carbon atoms that aren t part of the ring and others that are Para substitution indicates the presence of two branches a C1 and a C2 The 23 ratios indicate methyl groups two of them If the C1 is methyl then the C2 must contain the carbonyl which may either be on the carbon next to the ring or on the carbon further away If the carbonyl is next to the ring you have your second methyl group the carbon further from the ring You now have the structure of J See Figure 185 for the structures of J and K C CH3 Rearranging J and K leads us to I see Figure 186 K CH3C CH3 I CH3C CH3 O 0 c CH J V Ozonolysis 3 T C CH3 CH3 CH3 Adding HBr to I gives G and H as seen in Figure 187 G C3H3 3H3 H C3H3 3H3 3 3 CH3 3 C CH3 Br H H Br CH3 CH3 Chapter 18 Roadmaps and Predicting Products d Predicting Products quot Being able to predict the products of a particular reaction is very important It s easy to do when you only have one or two reactions to consider but as you progress through an organic chemistry course the number of reactions increases dramatically Therefore you need to keep some generalities in mind when considering a reaction the type of reaction the regiochemistry of the reaction and the stereochemistry of the reaction In analyzing the type of reaction you should pay close attention to both the starting material and any added reagents In some cases the reaction conditions are also important In general the reactions you studied most recently are important however organic chemistry exams are notorious for including any reaction since the beginning of Organic Chemistry I In all reactions the key is the functional group Each functional group has a limited number of options and you need to be aware of what these are As a last resort when dealing with a reaction that you don t recognize you may want to start a mechanism For example most reactions done in acidic media begin with the protonation of an electronrich site such as a nitrogen oxygen or carboncarbon double bond Information about all three considerations type regiochemistry and stereochemistry appears in the mechanism Don t be distracted by large complicated molecules Focus on the functional group The remainder of the molecule usually goes along for the ride without any changes The regiochemistry is important For example does the reaction follow Markovnikov s rule or is the reaction antiMarkovnikov Or is the substituent a metadirector or orthoparadirector You learned rules such as this for a reason make sure you continually apply them Finally the stereochemistry is important This may be as simple as worrying about the inversion of configuration that occurs during an SN2 mechanism In other cases however it may be important to remember if the reaction is by synaddition or antiaddition Again you should know these rules and always apply them We know that keeping all these rules and facts straight seems like a tremendous task but the secret is practice practice and more practice Eventually reactions and mechanisms will become natural You ll develop a chemical intuition that will save you time and effort In fact you may actually start enjoying the challenge of roadmaps Okay we know we went a little far with the last one but occasionally it really does happen Part v Pulling It All Together Part VI iTequot5 The Paquot 1 Te na t 13 131 wave 5 p run 4 4 gap and a go ua c a o no p n u n o In 040 54 o J p p o u o n u no p p a t up n n p n p U I u o n no a u o I u o lu p a p t lp Q a iv a o J 4 4 a o pa n p p n p p o n u on o n u o a u o a u at o u a o a an v o o a r r up r u g v I r v 4 uu a u a a u a no ctr n u u n I u o n u o o u u u p on a u u o I u o a u I a u n a u an a o n r p p o o r r p n o p u r on p u t r r 4 u a u r 4 n u a u v up a u 4 a u a a u a u a 4 1 Q Iv pa u n n n u o u u p no v o u u a a no av to Dr ugau o Ink n M99 have 9 x mg av E 01 10 ct60 p tgg Iiugei n che39m15 o1quot In this part We know that your major goal in taking Organic 11 is to gain expertise in the specific Organic 11 topics but we also know that you want to maximize your course grade In these two chapters we offer some suggestions that we hope will maximize the results of your study The first chapter is a tongueincheek collection of ten sugges tions to help ensure you don t pass No to be serious these suggestions offer some guidance about general study techniques The second chapter offers some defi nite ideas that will help you increase your individual exam score All you need in addition to these tips is old fashioned hard work Finally as a bonus we include an appendix full of named reactions to help you study Chapter 19 Ten Surefire Ways to Fail Organic Chemistry II In This Chapter Discovering how to avoid the pitfalls of studying organic chemistry Finding out about effective study strategies Understanding how to use every resource available ou probably bought this book to help you succeed in your Organic 11 chemistry course This chapter title caught your eye and you thought I don t want to fail I want to pass Well we decided to tackle the study habits and techniques part of the course in a different fashion than most other lists of advice and these tips are based on what we ve heard our student say throughout our years of teaching So take the advice here in the good natured tongueincheek manner it was intended Simply Read and Memorize Concepts A lot is sometimes said about organic chemistry simply being a memorization course Yes you need to do a lot of memorization but applying the concepts requires much more than simple memorization This is especially true of the second semester of organic chemistry You will not only be applying the new concepts that you are learning but also the concepts from the first semester course It s a lot like learning French or any other foreign language You need to know memorize the vocabulary and rules of grammar but in order to be able to converse in that language you must be able to put all that vocabulary together in the proper way grammar to convey an idea The same is true of organic chemistry g 0 Part VI The Part of Tens Don t Bother Working the Homework Problems and Exercises The key to passing organic chemistry is work work work Your brain needs repetition to form those neural pathways so you must work and rework the homework exercises and exercises And don t be sloppy if you leave out formal charges or ionic charges or don t check the number of bonds for each carbon you ll do the same thing on exams where those mistakes cost you points You may want to start or join a study group if you re the type of person that benefits from discussing exercises with other students And be prepared to wear down a lot of pencils and use a lot of paper Don t try to conserve paper by writing and drawing so small that you can t see everything Don t Buy a Model Kit Organic chemistry is a very visual subject but most of us have difficulty visualizing a structure in three dimensions Making a 3D model of the structure allows you to detect finer points of conformation steric hindrance and so on Making a model also helps you find those carbons with incorrect numbers of bonds Don t Worry About Falling Behind One surefire way to flunk organic is to get behind Organic is not one of those classes that you can ignore and then cram the night before a test It s best done in small chunks so plan on studying organic six or seven nights a week Don t try to jam it all into two or three study sessions a week How long should you study each time Good question but no pat answer The collegecourse rule of thumb that says you should spend two to three hours studying for every hour you re in class but for organic it s probably more than three for every class hour However you should know from your performance in Organic Chemistry I if this is reasonable for you or if you need even more And remember the quality of your studying is far more important than quantity Chapter 19 Ten Surefire Ways to Fail Organic Chemistry II gU 7 Don t Bother Learnina Reactions Organic chemistry especially Organic 11 is all about the reactions and their mechanisms If you don t know the reactions you can t pass You need to know the reactions by name the reactants the products and the conditions along with the reaction mechanism Practice practice practice Use flash cards Quiz each other in your study groups Write write write Group reactions by product by type and so on Do them forward and backward Know those reactions If your Textbook Confuses 1ou Don t Bother with Additional Resources Sometimes organic textbooks aren t the easiest books to read but no one says you can t use multiple sources You already invested in this For Dummies guide so use it If you re having a difficult time with a particular concept search the Internet and other organic chemistry textbooks until you find an author who explains in a way that makes sense to you Use several sources and compare Yes all of that takes time but it s worth it in the long run Don t Bother Reading the Chapter he fore lttendina Class Going into a lecture cold isn t a good idea Okay so maybe you don t need to read the entire chapter but at least take 30 minutes or an hour to read the part your instructor will be covering that day Don t try to work any exercises just try to familiarize yourself with general concepts Pay particular attention to vocabulary because it will help you get the most from the lecture In other words set the stage before you walk into class Remember the six Ps prior preparation prevents pretty poor performance 1 Part VI The Part of Tens Attend Class Onlg When you Feel Like It Class attendance has a direct correlation to class grade If you don t want to pass don t come to class Most people find that they can t really get organic just by reading the textbook You need to watch your instructor draw structures push electrons around organic teachers are such bullies and so on You ll benefit from the inclass discussions pick up some tricks of the trade and maybe discover the logic of organic chemistry Besides your instructor may give a hint as to what s going to be on the next exam Don t Bother Taking Notes ust Listen When You Aren39t Sleeping or Texting We said you should go to class now we re saying that you should pay attention and take notes A crucial survival skill for Organic Chemistry 11 Learn to listen to your instructor watch the board or screen and take notes all at the same time Have extra pencils handy possibly with different colors If you drop the one you are using just grab another one If you stop to pick up the dropped pencil you may get so far behind that you end up dropping the course especially if you re taking Organic 11 during a summer term Your lecture notes will probably be somewhat messy so recopy them ASAP before you forget what they mean The act of recopying notes within 24 hours of taking them is a great reinforcement of the material and you gotta form those neural pathways And don t even think about texting in organic class unless you plan to repeat the class Don t Bother Asking Questions If you re puzzled about something in class ask about it either during class or during your instructor s office hours Most professors teach because they like it so they re happy to answer your questions But if you go to your instructor for help avoid saying I just don t know anything because what your instructor hears is that you haven t invested the necessary study time to know what you don t know Go in with a list of specific questions and go through them one by one You have to learn the material yourself the instructor is your guide Don t worry about appearing to be stupid The only stupid question is one that you don t ask Better to ask questions than to give proof of your ignorance on the next test Chapter 20 More than Ten Ways to Increase Your Score on an Organic Chemistry Exam In This Chapter Mastering some study strategies in preparing for an organic chemistry exam Avoiding the mistakes commonly made on exams Remembering that a carbon atom has four bonds n this chapter we focus on those activities and techniques that will help increase your score on an organic exam Many of these can also be used with other courses also Many student freeze on an exam due to lack of selfconfidence but you can build up a confident nature by making sure that you know the material and by having a positive attitude Think of the exam as an opponent you re about to battle and be eager to win that is to show your prof what you know Who knows you may actually grow to like Organic Chemistry Sounds unlikely we know but stranger things have happened Don t Cram the Night he are a Test An organic chemistry test is not a test you can cram for the night before Or even a few days before You probably found that out in Organic Chemistry I You need to study all along a minimum of six days a week Don t try to create review sheets just before an exam either make them all throughout the semester as you study the material Then when exam times rolls around you already have your sheets made and can start reviewing them And don t underestimate the power of positive selftalk Even when you get stuck in your studying never say you can t understand the material or that you re going to fail an exam or the course and so on Keep telling yourself that you can do it I can pass the exam I m going to make an A in the course and pretty soon you ll believe it and with hard work it can come true Part VI The Part of Tens Try Doiny the Problem Sets and Practice Tests Twice When you re doing your problem sets or practice tests work them twice The first time you may have to refer to your notes or book Make sure you understand and master the material that gave you trouble Then work the problems again but the second time try not to use your notes and work them in a random order Sometimes the context of the material gives you clues that may not be present on an exam so you don t want to rely on them Working problems in a random order is an especially powerful study habit when gearing up for the final exam Study the Mistakes you Made on Previous Exams Use your exams to help fix mistakes in your knowledge and reasoning Fully correct your errors on old exams as soon as possible If your teacher thought that material was important enough to put on an exam you will be seeing it again on another exam or the final That corrected exam becomes part of your studyreview material Face it every exam in organic chemistry is cumulative Learn from the mistakes you made and don t make them again Don t worry you ll have the opportunity to make brandnew ones Know Precisely Where Why and How the Electrons Are Motiny When writing and studying reactions pay attention to the mechanism especially where why and how the electrons are moving This is true whether you re doing homework problems or making your practice sheets Be sure to use the right types of arrows double arrows for equilibrium single arrows for reactions curved arrows or curved half arrows for electron movement and so on Also don t try to combine too many mechanistic steps especially on an exam Take it one step at a time and your results will be clearer and easier to grade and this is a very good thing Keep in mind though that you may not be asked for the mechanism on an exam just the reaction In that case only write the reaction You can get yourself into trouble by volunteering extra information Chapter 20 Ten Ways to Increase Your Score on an Organic Chemistry Exam Relax and Get Enough Sleep before the Exam You really have to be able to think when taking an organic chemistry exam If you try pulling an allnighter you won t be able to think So treat it the way athletes do a big match or game Relax and get to bed early Get up in plenty of time to have breakfast before your exam Use positive selftalk You may find that you want to isolate yourself from the other students before an exam so that you don t get into a question and answer dialogue and panic yourself Think Before you Write Before you start to answer a test question stop and think Write down the assumptions that are pertinent to the questionproblem on your scratch paper Make a few notes and maybe even sketch out the reactionanswer before you start answering the question on the exam paper Then work steadily and carefully on the answer Make sure that everything you write down is clear and reasonable Don t write down unnecessary information on the exam It takes time distracts your professor as she grades the answer and gives her more things to count incorrect Pay attention to the question and fully answer it no more no less Include Formal Charges in your Structures When lppropriu te quot Using formal charges sometimes allows you to pick out the most appropriate or likely structure so use them when trying to decide between different Lewis structures Certainly show them on exams when asked If you really feel they re necessary show them even if they weren t requested but be absolutely sure they re correct You don t want to volunteer unnecessary information that could be wrong And be sure to write your structures neatly on the exam because if your instructor can t follow what you have done you re going to lose points If you get into the habit of drawing your structures neatly on your homework you will do the same on an exam Part VI The Part of Tens Check That You Haven39t Lost lnu Carhon Atoms Losing carbon atoms not accounting for all your carbon atoms is easy to do when writing a multistep mechanism Being sure to account for all your carbons is especially important in those reactions where you might be losing carbon dioxide or another small carboncontaining molecule Even though many times your instructor won t require a balanced chemical equation he will be upset if you lose carbons Include EZ RS cistrans Prefixes in Naming Organic Structures Forgetting to use prefixes is a common mistake that students make in the midst of an exam Make sure you indicate the appropriate stereochemistry in your nomenclature especially if your instructor takes time to indicate the specific stereochemistry of a compound Again get into the habit of doing this when you re working on homework exercises Think of Spectroscopy Especially NMR As a Puzzle Spectroscopic data can be very useful on an exam but think of it as individual pieces of a puzzle Write down each absorption and assign a structural value to it and then step back and look at the overall picture and try to see how all those individual pieces fit together This is an especially valuable tip when faced with NMR and IR data since many times this data is far more detailed than UVvis spectra or mass spectra Make Sure That Each Carhon Atom Has Four Bonols We believe that more points have been lost on organic exams due to this one mistake more any other Be sure that every carbon atom has four bonds T his might not be true for ions Leaving off a bond commonly to a hydrogen atom is like waving a red flag in front of the grader She gets that small smile shakes her head and down comes the red pen Remember Carbon makes four bonds Appendix Named Reactions Lots of chemical reactions take place in the diverse world of organic chemistry Some are named after people some are named after reactants or products If you want to pass your Organic Chemistry 11 class you need to know the following named reactions V Acetoacetic Ester Synthesis The formation of a substituted acetone through the basecatalyzed alkylation or arylation of a Bketo ester V Aldol Cyclization An internal aldol condensation V Aldol Reaction The formation of an aldol Bhydroxy carbonyl com pound through the catalyzed condensation of an enolenolate with a carbonyl compound V Cannizzaro Reaction The formation of an acid and an alcohol through the basecatalyzed disproportionation of an aliphatic or aromatic aldehyde with no ochydrogen atoms V Claisen Condensation The formation of a Bketo ester through the basecatalyzed condensation of an ester containing an ochydrogen V ClaisenSchmidt Reaction The production of an ocBunsaturated aldehyde or ketone from an aldehyde or ketone in the presence of strong base V Cope Elimination The pyrolysis of an amine oxide to produce a hydroxylamine and an alkene V Crossed Aldol Condensation An aldol condensation involving different carbonyl compounds V Crossed Cannizzaro Reaction A Cannizzaro reaction involving two different aldehydes V Crossed Claisen Condensation A Claisen condensation utilizing a mixture of two different esters V Curtius Rearrangement Similar to a Hofmann degradation with an azide replacing the amide V Dieckmann Condensation The intramolecular equivalent of a Claisen condensation where dicarboxylic acid ester undergoes basecatalyzed cyclization to form a Bketo ester Organic Chemistry II For Dummies V DielsAlder Reaction The reaction of an alkene dienophile with a conjugated diene to generate a sixmembered ring V FriedelCrafts Reaction The Lewis acid catalyzed usually AlCl3 alkylation or acylation of an aromatic compound V Gabriel Synthesis The reaction of an alkyl halide with potassium phthalimide to form after hydrolysis a primary amine V Grignard Reaction The reaction of an organomagnesium compound typically with a carbonyl compound to produce an alcohol although it may be used in other situations V HellVolhard Zelinsky Reaction A method for forming ochalo acid V Hofmann Elimination Converts an amine into an alkene V Hofmann Rearrangement A useful means of converting an amide to an amine V Hunsdiecker Reaction A freeradical reaction for the synthesis of an alkyl halide V Knoevenagel Condensation A condensation of an aldehyde or ketone with a molecule containing an active methylene in the presence of an amine or ammonia V Malonic Ester Synthesis Synthesis involving a malonic ester or a related compound with a strong base such as sodium ethoxide The ultimate product is a substituted carboxylic acid V Mannich Reaction The reaction of a compound with a reactive hydrogen with aldehydes non enol forming and ammonia or a primary or secondary amine to form a Mannich base aminomethylated compound V Michael Addition Condensation Reaction The addition of a carbon nucleophile to an activated unsaturated system V Reformatsky Reaction A reaction leading to formation of Bhydoxy esters using an organozinc intermediate V Robinson Annulation The addition of a methyl vinyl ketone or deriva tive to a cyclohexanone to form an 0cBunsaturated ketone containing a sixmembered ring V Sandmeyer Reaction A reaction utilizing a diazonium salt to produce an aryl halide The process begins by converting an amine to a diazonium salt V Schiemann Reaction A means of preparing aryl fluorides V Stork Enamine Synthesis A reaction leading to the formation of an ocalkyl or occarbonyl compound from an alkyl or aryl halide reacting with an enamine Index 00 absorption 71 acetal 148150 acetaldehyde 139 acetic acid 189 194 acetic anhydride 201 acetoacetic ester synthesis 267268 347 acetone 139 acid anhydride 207 acid catalysis 182 acid chloride 206 acid dissociation constant 194 acid hydrolysis 210 212 acidcatalyzed dehydration 41 acidic carbonyl 127 acidic cleavage of epoxides 50 of ethers 49 acidity acetic acid 194 carboxylic acids 194195 acids BronstedLowry 125 combining with alcohols 203205 combining with bases 206208 protonation 23 reaction of alcohols as 44 activation 107108 195 acyl chloride 126 191192 acyl group carbonyl 125 acyl halide 199200 acylation FriedelCrafts 100 110 acylium ion 100 adipic acid 190 alcohols acids 44 boiling point 3334 classifying 32 45 combining with acids 203205 conversion to alkyl halides 44 conversion to esters 43 dehydration 41 density 34 Grignard reagents and production of 3839 hydration of alkenes 35 hydrogen bonding 29 infrared and proton NMR data 50 Lucas test 45 melting point 3334 nomenclature 3233 oxidation 4243 oxymercurationdemercuration reactions with alkenes 3537 physical properties of 3334 preparation by reduction of carbonyls 38 primary 32 reactions 4045 secondary 32 solubility 34 structure and nomenclature of 3233 synthesis of 3439 tertiary 32 aldehydes addition reactions 173177 BaeyerVilliger reaction 158159 boiling point 139 carbonyl 124125 CH stretch 131 ClaisenSchmidt reaction 173175 Grignard reagents 144145 IR spectroscopy 160 mass spectroscopy 160 melting point 139 NMR spectroscopy 160 nomenclature 137138 nucleophilic attack 147 oxidation 140141 156158 196 oxygencontaining nucleophile 148 physical properties 139 reactions 143146 reduction reactions 142 solubility 139 UVvis spectroscopy 160 alditol 290 aldol addition 168 aldol condensation 168169 350 Organic Chemistry II For Dummies aldol cyclization reaction 171 347 aldol reactions de ned347 enolates 169 enols 169 aldose sugars 293 alkanes 29 alkenes catalytic hydration 35 diols from reactions with 38 hydration 35 hydroborationoxidation reactions with 3637 oxidation 196 oxymercurationdemercuration reactions with 3537 peroxidation 48 alkoxides 33 166 alkyl benzenes 197 alkyl chloroformate 214 alkyl halides 44 alkylated aromatics 101 alkylated benzene 101 alkylation FriedelCrafts 99 110 allylic radical 54 allyllic bromination 59 amides acids and bases to form 206208 boiling point 194 carbonyl 126127 CH stretch 131 dehydration 212 melting point 194 nomenclature 192 reactions 211212 amines primary 151 222 quaternary 223224 secondary 151 223 tertiary 223 amino acids as building block of protein 301 glycine 303 physical properties 302304 resolution 308 structure of 302 synthesis 304308 zwitterion 302 304 amino sugars 298 ammonia 206 amphoteric 302 amylose 297 anhydride acetic 201 acid 207 asymmetric 200 cyclic 202 nomenclature 191 removing water to form 200202 symmetric 200 anion 11 anomer 283 anti addition 37 anylopectin 297 aromatic compound alkylated 101 benzene 8186 branches of 89 de nition 15 group in uencing reactivity of 103 heteroatoms 8990 Hiickel rule 8586 IR spectroscopy 91 Kekul ring structure 8283 mass spectroscopy 92 nitrogen 90 NMR spectroscopy 91 nomenclature 87 oxygen 90 sulfur 90 synthetic strategies for making 115116 systems mistaken for 86 UVVis spectroscopy 91 aromatic substituent 108109 aromatic systems 114115 arrow curved 1920 doubleheaded curved 18 equilibrium 18 head point 18 reaction 18 resonance 17 83 singleheaded curved 18 my branch 89 asymmetric anhydride 200 Index 35 I ate acid 188 atom bonding 10 030 BaeyerVilliger reaction 158159 barbital 279 barbiturate 279280 base combined with acids 206208 base hydrolysis 210 212 basicity de nition 25 Grignard reagents 250251 nitrogen compounds 226 nucleophilicity difference 25 benzaldehyde 139 benzene alkylated 101 aromatic compound 8186 boiling point 85 chlorination of 95 derivatives of 88 halogenation of 9596 Kekul ring structure 8283 melting point 85 metadistributed 91 nitration 9697 orthodistributed 91 paradistributed 91 physical properties 85 resonance 8384 stability 8485 structure of 81 sulfonation of 9798 benzophenone 139 Biochemistry For Dummies Moore and Langley 9 224 281 biomolecule 281 blood sugar 282 boiling point alcohols 3334 aldehydes 139 amides 194 benzene 85 carboxylic acids 193 cyclohexane 85 esters 193 ethers 45 ketones 139 bonding atom 10 breaking and forming of bond 19 delocalized 83 double bond 6869 Fischer project formula 14 hydrogen 10 29 single bond 89 triple bond 69 bondtobond movement 21 bond tolone pair movement 21 23 bromine free radical 29 BronstedLowry acids 125 butadiene 55 butanoic acid 189 butyric acid 189 0C0 CA conjugate acid 251 Cannizzaro reaction 181182 347 carbocation 23 carbohydrate glycoside 285286 monosaccharide 282 mutarotation 282285 oxidative cleavage 289 photosynthesis 281 carbon atom 12 carbon tetrachloride 215 carbonation 198 carboncontaining nucleophile 154156 carbon13 spectra 77 carbonyl compound 261 carbonyls acid anhydride 126 acidic 127 acyl chlorides 126 acyl group 125 aldehyde 124125 amides 126127 carboxylic acids 125 esters 126 functional groups 124127 group 124 IR spectroscopy 130131 ketones 124125 mass spectroscopy 134135 Organic Chemistry II For Dummies carbonyls continued NMR spectroscopy 132133 polarity 128 preparation of alcohols by the reduction of 38 reactivity 130 resonance 129 stability 123 UVvis spectroscopy 131132 carboxylic acids acetic acid 189 acidity 194195 acyl chlorides 191192 acyl halide reactions 210 acyl halides 199200 amide nomenclature 192 anhydride 191 200202 anhydride reactions 210 boiling point 193 butanoic acid 189 butyric acid 189 carbamic acid 214 carbonic acid derivatives 213 carbonyl 125 CH stretch 131 chemical test 218 decarboxylation 214 dicarboxylic acids 189190 ethanoic acid 189 formic acid 189 HellVolhard Zelinsky reaction 209 Hunsdiecker reaction 215216 IR spectroscopy 218 melting point 193 methanoic acid 189 molecular weight 218 naming 188189 natural and synthetic compound 187 nomenclature 188192 pentanoic acid 189 physical properties 193194 propanoic acid 189 propionic acid 189 reactions 208209 Reformatsky reaction 216 spectroscopy 217218 structure of 188 synthesis 196198 valeric acid 189 catalytic hydration 35 catenation 10 cation 11 CB conjugate base 251 cellobiose 296 cellulose 298 CH stretch aldehyde 131 amide 131 carboxylic acid 131 ester 131 guidelines 69 IR spectroscopy 6970 ketone 131 chemical shift 7475 chemical test carboxylic acids 218 Chemistry Essentials For Dummies Moore 15 Chemistry For Dummies Moore 4 15 chiral compound 1213 chlorination 95 chlorine free radical 2729 cis isomer 12 202 cisglycol 38 cistrans isomer 12 Claisen condensation 262264 266 347 ClaisenSchmidt reaction 173175 347 class attendance 342 cleavage 288 combustion 42 condensation aldol 168169 Claisen 262264 266 347 crossed aldol 171 347 crossed Claisen 265 347 Dieckman 264 347 Knoevenagel 273274 348 conjugate acid CA 251 conjugate base CB 251 conjugated unsaturated system allylic radical 54 butadiene 55 de ned15 DielsAlder reaction 6264 electrophilic addition 5961 Index 353 examples of 5354 reactions 5764 resonance rules 5657 stability of 57 substitution reaction 257259 conservation of charge 23 conversion alcohols to alkyl halides 44 alcohols to esters 43 Cope elimination 243 347 coupling 7576 coupling 11 1 rule 75 cramming before exam 343 crossed aldol condensation reaction 171 347 crossed Claisen condensation reaction 265 347 Curtius rearrangement reaction 347 curved arrow 1920 cyanide ion 180181 cyanohydrin 198 cyclic anhydride 202 cyclohexane 85 oo Daldose family 293 data See spectroscopy deactivation 107108 195 deamination reaction 238 decarboxylation 214 dehydration acidcatalyzed 41 alcohols 41 amides 212 delocalized bonding 83 density alcohols 34 dextrorotatory 12 dextrose 282 dialkyl carbonate 214 diastereomer 13 dicarboxylic acids 189190 Dieckman condensation reaction 264 347 DielsAlder reaction conjugated unsaturated system 6264 de ned348 exam questions 6566 stereochemistry 6364 dipoledipole force 10 disaccharides cellobiose 296 maltose 296 sucrose 295 double bond 6869 doubleheaded curved arrow 18 doublet 76 0E0 E isomer 152 E1 mechanism 27 E2 mechanism 27 EDG electrondonating groups 63 electromagnetic spectrum 68 electron movement 1819 electrondonating groups EDG 63 electronwithdrawing groups EWG 63 electrophilic substitution reaction activation and deactivation 107108 FriedelCrafts reaction 99101 limitations of 109110 mechanism 94 monosubstituted benzene 101 elimination Cope 243 347 Hofmann 241242 348 rate of reaction 25 elimination reaction 11 eliminationaddition reaction 114115 enantiomer 1213 endoproduct 64 energy barrier 61 enolates acid catalysis 182 addition reactions 180 aldol cyclization 171 aldol reactions 169 Cannizzaro reaction 181182 carbonyl groups 161 crossed aldol condensation 171 haloform reactions 166168 Michael addition reaction 176177 nitriles 179 nitroalkanes 178 reactions 166 Organic Chemistry II For Dummies enolates continued Robinson annulation 184 structure of 162 synthesis 164166 enols acid catalysis 182 addition reactions 180 aldol cyclization 171 aldol reactions 169 Cannizzaro reaction 181182 carbonyl groups 161 crossed aldol condensations 171 haloform reactions 166168 Michael addition reaction 176177 nitriles 179 nitroalkanes 178 reactions 166 Robinson annulation 184 structure of 162 synthesis 164166 epimer 290 epoxides 50 equilibrium arrow 18 esters acetoacetic ester synthesis 267268 347 boiling point 193 carbonyl 126 CH stretch 131 conversion of alcohols to 44 Fischer esteri cation 203 hydrolysis 210 malonic ester synthesis 269272 348 methyl 205 nomenclature 190 physical properties 193 transesteri cation 205 uniting acids and alcohols to make 202205 ethanoic acid 189 ethanol 74 76 ethers acidic cleavage 49 boiling point 45 infrared and proton NMR data 50 melting point 45 nomenclature 46 physical properties of 4647 reactions 4951 solubility 4647 structure and nomenclature of 46 sulfuric acid 50 synthesis 4748 William ether synthesis 47 EWG electronwithdrawing groups 63 exam tips additional resource importance 341 class attendance 342 cramming the night before 343 falling behind 340 familiarization of general concepts 341 formal charges 345 homework 340 knowledge of reactions 341 losing carbon atoms 346 memorization 339 model kit 340 note taking 342 old exam corrections 344 positive selftalk 345 pre x mistakes 346 question and answer session 342 repeating practice sets 344 resting before exam 345 study group 340 study time 340 thinking before answering 345 exoproduct 64 CFO fats lipid 281 melting point 299 physical properties of 299 polyunsaturated 299 saponi cation 300 structure of 299 unsaturated 299 Fischer esteri cation 203 Fischer projection 14 282 force 10 formaldehyde 139 formic acid 189 fragmentation 73 288 free radical bromine 29 chlorine 2729 Index X halogenation of alkenes 29 Hunsdiecker reaction 215 free radical mechanism 2728 free radical substitution 118 FriedelCrafts reactions acylation 100 110 alkylation 99 110 de ned348 ketones 144 fuming sulfuric acid 97 functional group carbonyls 124127 de nition 11 furanose 284 0amp0 Gabriel synthesis reaction 229 348 glucose acetal 285 common names for 282 different representations to 284 mutarotation in 282283 structure of 283 glutaric acid 190 glycine 303 glycogen 297 glycoside 285286 grape sugar 282 Grignard reagents about this book 4 aldehypes and ketones 144145 basicity 250251 carbonation 198 de ned348 limitations 259 nucleophilicity 251254 preparation of 250 productions of alcohols 3839 reactions of 250255 group G 102 group study 340 0H0 haloform reaction 166168 halogen 199200 halogenation of benzene 9596 Haworth projection 282 HellVolhard Zelinsky reaction 209 348 hemiacetal 148150 heteroatoms aromatic compound 8990 doublebonded 89 singlebonded 89 heterocyclics 224 highest occupied molecular orbital HOMO 71 Hinsberg test 247 Hofmann elimination reaction 241242 348 Hofmann rearrangement reaction 232 348 homework 340 HOMO highest occupied molecular orbital 71 Hiickel rule 8586 Hunsdiecker reaction 215216 348 hydration alkenes 35 hydrazone 152153 hydroborationoxidation reaction 3637 hydrogen atom 273 hydrogen bonding 10 29 hydrogen ion 162 165 hydrogenation reaction 117 hydrolysis acid 210 212 base 210 212 of cyanohydrin 198 hydroxylamine 152 0 I 0 ic acid 188 inductive effect 194195 infrared spectroscopy See IR spectroscopy intermediate 2425 intermolecular force 10 International Union of Pure and Applied Chemistry IUPAC 88 iodide ion 236 iodoform test 157158 ion acylium 100 cyanide 180181 hydrogen 162 165 iodide 236 metal 286 molecular 7273 phenolate 106 356 Organic Chemistry II For Dummies ionic bonding 128 ionic interaction 11 IR infrared spectroscopy alcohols and ethers 50 aldehydes 160 aromatic compound 91 carbonyls 130131 carboxylic acids 218 CH stretch 6970 double bonds 6869 ngerprint region 68 functional group identi cation 11 ketones 160 NH stretch 69 nitrogen compounds 246 OH stretch 69 triple bonds 69 isomer chiral compound 1213 cis 12 202 E 152 tautomers 163 trans 12 Z 152 IUPAC International Union of Pure and Applied Chemistry 88 ojo jasmone 171172 Jones reagent 43 0K0 Kekul ring structure 8283 ketones BaeyerVilliger reaction 158159 boiling point 139 carbonyls 124125 CH stretch 131 ClaisenSchmidt reaction 173175 FriedelCrafts reactions 144 Grignard reagents 144145 IR spectroscopy 160 mass spectroscopy 160 melting point 139 NMR spectroscopy 160 nomenclature 137138 nucleophilic attack 147 oxidation 140141 156158 physical properties 139 reactions 143146 reduction reactions 142 solubility 139 UVVis spectroscopy 160 KilianiFischer synthesis 291 kinetic energy 61 kJmol kilojoules per mole 8485 Knoevenagel condensation reaction 273274 348 oo Langley Richard H Biochemistry For Dummies 9 224 281 levorotatory 12 Lewis structures 56 lipid biomolecule 281 fats and fatty acids 299 fats and oils 281 lithium aluminum hydride 38 lodoform 158 London dispersion force 10 lone pairto bond movement 2223 Lucas test alcohols 45 LUMO lowest unoccupied molecular orbital 71 0M0 magnetic eld NMR spectroscopy 73 malonic acid 190 malonic ester synthesis 269272 348 maltose 296 Mannich reaction 275276 348 Markovnikov addition mechanism 35 118 mass spectroscopy aldehydes 160 aromatic compound 92 carbonyls 134135 fragmentation 73 ketones 160 molar mass 11 molecular ion 7273 MCAT 2 me masstocharge ratio 72 Index p6 mechanism conservation of charge 23 de nition 14 electrophilic substitution reactions 94 elimination 25 27 freeradical 2728 general knowledge of 29 Grignard reaction 3940 importance of understanding the 20 intermediates 2425 Markovnikov addition 35 118 multistep synthesis problem 29 nucleophilic substitution process 112113 protonation 23 substitution 2526 tips for working with 29 Meisenheimer complex 112 melting point alcohols 3334 aldehydes 139 amides 194 benzene 85 carboxylic acids 193 cyclohexane 85 ethers 45 fats and oils 299 ketones 139 memorization 339 meta position 102103 108 metadistributed benzenes 91 metal ion 286 methanoic acid 189 methyl esters 205 methyl ketones 197 micelle 300 Michael addition reaction 176177 348 model kit 340 molecular ion 7273 molecular weight 218 monosaccharide carbohydrate 282 furanose 284 KilianiFischer synthesis 291 oxidation 286289 reactions of 286290 reduction of 290 Ruff degradation 292 Moore John T Biochemistry For Dummies 9 224 281 Chemistry Essentials For Dummies 15 Chemistry For Dummies 15 multistep synthesis description of 312 mistaking with mechanism 29 nitrogen compounds 244246 regiochemistry 313 retrosynthetic analysis 313321 323324 stereochemistry 313 mutarotation anomer 283284 Fischer and Haworth projection 292 in glucose 283 straight chain 285 ONO nanometer nm 131132 neutralization reaction 218 NH stretch 69 nitration of benzene 9697 nitric acid 288 nitriles 179 nitroalkanes 178 nitrogen aromatic compound 90 nitrogen compounds basicity 226 Cope elimination 243 Curtius rearrangement 232 deamination reaction 238 description of 221 elimination reactions 241243 formation of ethers and phenols 237 Gabriel synthesis 229 heterocyclics 224 Hofmann elimination 241242 Hofmann rearrangement 232 IR spectroscopy 246 multistep synthesis 244246 nitro reductions 229230 NMR spectroscopy 246 nucleophilic substitution reactions 227229 physical properties 225226 primary amines 222 reactions with nitrous acid 233234 reactions with sulfonyl chlorides 239240 358 Organic Chemistry II For Dummies nitrogen compounds continued reductive amination 230231 replacement by iodide ion 236 replacement reactions 235238 Sandmeyer reaction 235 secondary amines 223 synthesis 227232 tertiary amines 223 nitrogen rule 72 nitrogencontaining nucleophile 151 nitrous acid reactions with 233234 nm nanometer 131132 NMR nuclear magnetic resonance spectroscopy alcohols and ethers 50 aldehydes 160 aromatic compound 91 carbon13 spectra 77 carbonyls 132133 characteristic ratio of intensity 76 chemical shift 11 7475 coupling 7576 ethanol 76 integration 75 ketones 160 magnetic eld 73 nitrogen compounds 246 proton spectra 7476 spin ipping 74 splitting pattern 11 nomenclature alcohols 3233 aldehydes 137138 amide 192 aromatic compound 87 esters 190 ethers 46 ketones 137138 note taking 342 nuclear magnetic resonance See NMR spectroscopy nucleophile carboncontaining 154156 intermediate 2425 nitrogencontaining 151 oxygencontaining 148 nucleophilic substitution process description of 111112 eliminationaddition reactions 114115 features 26 hydrogenation reaction 117 Markovnikov addition mechanism 118 mechanisms 112113 Meisenheimer complex 112 nitrogen compounds 227229 rate of reaction 25 synthetic strategies 116 nucleophilicity basicity difference 25 Grignard reagents 251254 000 observed rotation 13 OH stretch 69 oic acid 188 onestep synthesis 312 orbital 20 5354 86 89 Organic Chemistry I For Dummies Winter 19 67 331 organization about this book 34 organolithium compounds 256 organometallic compounds 257258 ortho position 103 108 orthodistributed benzenes 91 osazone 290 oxalic acid 190 oxidation alcohols 4243 aldehydes 140141 156158 196 alkenes 196 alkyl benzenes 197 combustion reaction form 42 hydroborationoxidation reaction 3637 ketones 140141 156158 methyl ketones 197 monosaccharide 286289 primary alcohol 196 oxygen 90 oxymercurationdemercuration reaction 3537 ozone 141 ozonolysis 141142 0P0 para position 103 paradistributed benzenes 91 Index P6 pentanoic acid 189 periodic acid 288 peroxidation 48 phenlate ion 106 phosgene 213214 phosphorus ylide 155 photosynthesis 281 physical properties alcohols 3334 aldehydes 139 amino acids 302304 benzene 85 carboxylic acids 85 esters 193 ethers 4647 fats 299 ketones 139 nitrogen compounds 225226 planar system 85 polarity carbonyls 128 polysaccharides cellulose 298 description 296 glycogen 297 starch 297 porbital 5354 86 89 practice problems 328334 prediction 335 pre x mistake 346 primary alcohols 32 196 primary amines 151 222 product identi cation 6566 product prediction 335 propanoic acid 189 propanone 139 propionic acid 189 protein 281 301 proton NMR 7476 protonation 23 pyridine 201 ago quaternary amines 223224 question and answer session 342 0R0 racemix mixture 13 reaction alcohols 4045 aldehydes 143146 aldol 347 aldol cyclization 171 347 amides 211212 BaeyerVilliger 158159 Cannizzaro 181182 347 carboxylic acids 208209 ClaisenSchmidt 173175 347 conjugated unsaturated system 5764 crossed aldol condensation 347 crossed Claisen condensation 265 347 Curtius rearrangement 347 deamination 238 Dieckman condensation 264 347 DielsAlder 6265 348 electrophilic substitution 94 99101 107110 elimination 11 eliminationaddition 114115 enolates 166 enols 166 epoxides 50 ethers 4951 FriedelCrafts 348 Gabriel synthesis 229 348 Grignard reagents 250255 348 haloform 166168 HellVohhard Zelinsky 209 348 Hofmann elimination 241242 348 Hofmann rearrangement 348 Hunsdiecker 215216 348 hydroborationoxidation 3637 hydrogenation 117 identi cation 66 ketones 143146 Knoevenagel condensation 273274 348 malonic ester 348 Mannich 275276 348 Michael addition 176177 348 monosaccharides 286290 neutralization 218 x Organic Chemistry II For Dummies reaction continued oxymercurationdemercuration 3537 product identi cation 6566 redox 181 reduction 142 Reformatsky 216 348 Robinson annulation 348 Sandmeyer 235 348 Sarett s 42 Schiemann 236 348 stork enamine synthesis 277278 348 substitution 11 Wittig 155 reaction arrow 18 redox reaction 181 reducing sugars 286 reduction reactions 142 reductive amination 230231 308 Reformatsky reaction 216 348 regiochemistry 313 resonance allylic radical 54 benzene 8384 carbonyls 129 stabilization 194 resonance arrow 17 83 resonance structure curved arrows 19 Lewis structures 56 rules 5657 retrosynthetic analysis description of 313 problem examples and solution to 314324 ring system vemembered 171 Hiickel rule 85 sixmembered 62 171172 threering system 5051 roadmap problem how to approach 328 practice problems 328334 Robinson annulation reaction 184 348 rotation 1213 RS notation 13 Ruff degradation 292 050 SA strong acids 250 salt 200 salting out 47 Sandmeyer reaction 235 348 saponi cation 210 300301 Sarett s reagent 42 SB strong conjugate bases 250 Schiemann reaction 236 348 scienti c notation 1213 secondary alcohols 32 secondary amines 151 223 sia group 143144 side chain 117118 sigma complex 94 100 103 single bond 89 singleheaded curved arrow 18 SN1 reaction 26 SN2 reaction 27 soap 300 sodium salt 200 sodium stearate 300 solubility alcohols 34 aldehydes 139 ethers 4647 ketones 139 spectroscopy See also IR spectroscopy mass spectroscopy NMR spectroscopy UVVis spectroscopy aromatic compound 9092 carboxylic acids 217218 description of 67 spin ipping 74 splitting pattern 11 stability benzenes 8485 carbonyls 123 conjugated unsaturated system 57 starch 281 297 stereochemistry DielsAlder reaction 6364 multistep synthesis 313 steric hindrance 25 109 steroid 281 Index 5 I stork enamine synthesis reaction 277278 348 Strecker synthesis 306307 strong acids SA 250 strong conjugate bases SB 250 structure acyl chloride 126 amino acids 302 benzene 81 carboxylic acids 188 enolates 162 enols 162 ethers 46 fats 299 glucose 283 study group 340 study habit See exam tips substitution electrophilic 94 99101 107110 free radical 118 nucleophilic 2526 111118 227229 substitution reaction 11 257259 succinic acid 190 sucrose 295 sugars amino 298 nitrogencontaining 298 sulfonation of benzene 9798 sulfur aromatic compound 90 sulfuric acid ethers 50 fuming 97 symmetric anhydride 200 syn addition 37 synthesis acetoacetic ester 267268 347 of alcohols 3439 amino acid 304308 carboxylic acids 196198 description of 311 enolates 164166 enols 164166 ethers 4748 Gabriel 229 348 KilianiFischer 291 malonic ester 269272 348 multistep synthesis 29 312 nitrogen compounds 227232 onestep 312 stork enamine 277278 348 Strecker 305 307 of urea 10 0T0 tautomers 163 tertiary alcohols 32 tertiary amines 223 test Hinsberg 247 iodoform 157158 Lucas 45 theory Vital force 10 thionyl chloride 199200 Tollen s reagent 157 trans isomer 12 transesteri cation 205 transglycol 38 triple bonds 69 triplet 76 Qua unsaturated system allylic radical 54 butadiene 55 de ned 15 DielsAlder reaction 6264 electrophilic addition 5961 examples of 5354 reactions 5764 resonance rules 5657 stability of 57 substitution reaction 257259 urea 10 214 UVvis ultraviolet and Visible spectroscopy absorbance and wavelength 71 aldehydes 160 aromatic compound 91 carbonyls 131132 ketones 160 spectral range 7071 362 Organic Chemistry II For Dummies o0 valeric acid 189 Vibration 68 Visible spectroscopy See UVVis spectroscopy Vital force theory 10 own WA weak acids 250 water dehydration 41 212 hydrolysis 198 210 212 wavelength 71 WE weak conjugate bases 250 Williamson ether synthesis 47 Winter Arthur Organic Chemistry For Dummies 67 331 Wittig reaction 155 WolffKishner reduction 153 155 aye ylide carboncontaining nucleophile 154 phosphorus 155 sulfur 156 0Z0 Z isomer 152 ZaitseV s rule 242 zwitterion 302 304 BusinessAccounting amp Bookkeeping Bookkeeping For Dummies 9780764598487 eBay Business AllinOne For Dummies 2nd Edition 9780470385364 Job Interviews For Dummies 3rd Edition 9780470177488 Resumes For Dummies 5th Edition 9780470080375 Stock Investing For Dummies 3rd Edition 9780470401149 Successful Time Management For Dummies 9780470290347 Computer Hardware BlackBerry For Dummies 3rd Edition 9780470457627 Computers For Seniors For Dummies 9780470240557 iPhone For Dummies 2nd Edition 9780470423424 Laptops For Dummies 3rd Edition 9780470277591 Macs For Dummies 10th Edition 9780470278178 Cooking amp Entertaining Cooking Basics For Dummies 3rd Edition 9780764572067 Wine For Dummies 4th Edition 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Dummiescom Dummies products I make life easier DVDs 0 Music 0 Games 0 DIY 0 Consumer Electronics 0 Software 0 Crafts 0 Hobbies 0 Cookware 0 and more For more information go to Dummiescom and search the store by category iDLtt iWE 3 r i Etewtt new 0 D Starter Ptqck Earn ism 3 Tire to A ve ni5hi39ng it 39 Dur itttsrj 0 EnnuibJnIir I1I13939ruJaJ i E9IIuIjI39 l39IBuai139 With more than 200 million books in print and over 1600 unique titles Dummies is a global leader in how to information Now you can get the same great Dummies information in an AppWith topics such as Wine Spanish Digital Photography Certification and more you39ll have instant access to the topics you need to know in a format you can trust To get information on all our Dummies appsvisit the following wwwDummiescomgomobile from your computer wwwDummiescomgoiphoneapps from your phone ScienceChemistryOrganic Chemistry Your plainEnglish guide to understanding chemical reactions and tackling tests with ease Open the book and find Whether you39re confused by carbocations or baffled by biomolecules this straightforward easy to read guide demystifies Organic Chemistry II From appreciating aromatic compounds to comprehending carbonyls you39ll discover what you need to know about organic reactions in order to master the course and score high on your exams Get comfortable with the course follow your coursework alongside the book as it corresponds to a typical second semester collegelevel organic chemistry class Refresh your knowledge of key topics from mechanisms to alcohol and ethers from conjugated unsaturated systems to electrophiles and nucleophiles get prepped for Organic II Grasp carbonyl group basics get an overview of structure reactivity and spectroscopy before delving deeper into aldehydes and ketones enols and enolates and carboxylic acids and their derivatives Ace advanced topics take a closer look at nitrogen compounds organometallic compounds the Claisen Condensation and its variations and biomolecules such as carbohydrates lipids andproteins Go to Dummiescom for videos stepbystep examples howto articles or to shop For Dummies A Branded Imprint of WILEY 1999 US 2399 CN 1499 UK ISBN I76El39l7EILBl5 John T Moore EdD Richard H Langley PhD
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