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Fund of Organic Chemistry I

by: Dominic Kling

Fund of Organic Chemistry I CHEM 3331

Marketplace > University of Houston > Chemistry > CHEM 3331 > Fund of Organic Chemistry I
Dominic Kling
GPA 3.71

Mary Bean

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This 139 page Class Notes was uploaded by Dominic Kling on Saturday September 19, 2015. The Class Notes belongs to CHEM 3331 at University of Houston taught by Mary Bean in Fall. Since its upload, it has received 30 views. For similar materials see /class/208154/chem-3331-university-of-houston in Chemistry at University of Houston.


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Date Created: 09/19/15
VI 1 quotV x LJIl Wm 3 33 Attention Students The pages that follow are copies of the transparencies used in class They are provided to make your note taking easier You are expected to add important information and details during class These pages do not represent a complete set of notes You cannot survive Organic I with these notes alone CHAPTER ONE Introduction and Review Introduction ORGANIC chemistschemistry students nonchemistsphilosophy students TW C Hid Review I Structure of the Atom A Subatomic particles 1 electron 2 proton 3 neutron protons protons neutrons B Models of the Atom where are e P and N 1 Thompson 2 Rutherford 3 Bohr 4 Quantum Mechanical Model Using a model which one shells subshells O gt gtgt orbitals Orbital Shapes C Electron Configuration distribution of electrons among an atom39s orbitals The lowest energy arrangement is called 1 Aufbau principle 2 Pauli exclusion principle 3 Hund39s rule Examples Nitrogen Sodium Why is electron configuration important to an organic chemist octet quotrulequot How can atoms satisfy the octet rule ll Bonding A lonic Bonding Example NaF gtgtgt gtgtcogtgt gtgt B Covalent Bonding The most common type of bonding in organic compounds Simple inorganic example H2 Representing the bonding in organic molecules will be important Mechanisms 1 Lewis symbol F O N C 2 Lewis Structure Electron Dot Structure I2 02 N2 TIPS FOR WRITING LEWIS STRUCTURES 1 Count the total number of valence electrons for each atom in the formula Add or subtract electrons according to charge The completed structure should have the same number of electrons to be correct 2 Determine which atoms are bonded to which Hints a H and F can only form one single bond therefore these atoms always occupy terminal positions always monovalent b Cl Br and I usually form single bonds in organic compounds usually monovalent c S and 0 form single and double bonds usually divalent forms 2 bonds either 2 single or one double d N forms single double and triple bonds usually trivalent e C forms single double and triple bonds and usually does not occupy a terminal position tetravalent with very few exceptions f The central atom is usually the least electronegative 3 Choose the central atom and draw a single bond between it and each of the surrounding atoms 4 Complete the octets of the surrounding atoms by adding lone pairs of electrons except for hydrogen Just do it 5 If the octet rule is not yet satisfied for the central atom draw double or triple bonds between it and surrounding atoms using lone pairs of electrons on appropriate atoms atoms capable of forming double or triple bonds 6 Count the valence electrons in the proposed structure The number of electrons MUST match the number calculated in step one C Electronegativity and Bond Polarity We39ve considered two types of bonds ionic and covalent transfer e39 share complete transfer equal Sharing Consider HF H F electronegativity partial charge negative partial charge assigned to H F H F rules of thumb smaller the difference in EN of the atoms forming a bond greater the difference in EN of the atoms forming a bond dipole moment D Formal Charge The calculation of formal charge allows us to determine which atoms in a molecule bear charge which atoms may be electron rich or electron deficient Importance FC of valence electrons of electrons quotownedquot by the atom quotownershipquot all lone pairs and half of all bonding pairs Example 003239 valence electrons quotownedquot electrons FC E Resonance Theory W 1 Resonance structures exist only on paper Used when a single Lewis structure is inadequate to represent the molecule Only the hybrid exists 2 Resonance structures differ only in the placement of electrons not atoms Move only pi electrons and lone pairs 3 All structures must be proper Lewis structures Atoms of periods 1 and 2 max of 8 electrons 4 All structures must have the same number of unpaired electrons radicals 5 Energy of the actual molecule is lower than any contributing resonance structure Resonance stabilization 6 Equivalent structures make equal contributions to the resonance hybrid and will contribute to resonance stabilization 7 Nonequivalent structures do not make equal contributions to the hybrid The more stable the structure itself the more it contributes to the hybrid 8 To make decisions about the relative importance or stability of resonance contributors follow these guidelines a The more covalent bonds the more stable b Structures in which all atoms have a complete octet of electrons contribute most Io c Charge separation decreases stability Requires energy to separate and charges Uncharged structures contribute more than charged structures d Structures that place negative charge on the more electro negative element and positive charge on the more electro positive element are more stable and contribute more to the hybrid However rule 8b is more important ll Structural Formulas A Lewis Structure shows all atoms bonds and lone pairs EX C3H30 C4H1OO B Condensed Formula shows atoms only not bonds atoms grouped in a manner that indicates bonding in the molecule C LineAngle Formula Skeletal Formula carbon atoms are not drawn and most hydrogen atoms are not drawn 1 Carbon atoms are represented by the ends and intersections of lines 2 Each carbon is understood to have the appropriate number of hydrogens to complete the octet Hydrogen bonded to carbon is not drawn 3 All atoms other than carbon and hydrogen are always drawn 4 If an atom is drawn its hydrogen must be drawn also OH CH32NCH20H3 CH3 g l3 Acids and Bases in Organic Chemistry Many reactons of organic compounds may be classified as acidbase reactions In chapter 6 we will use the more general terms electrophilenucleophile to describe reactions r t acid base Example H Eli39 H Bi gt H Eliw 67 H H conjugate acid conjugate base acidbase pairs AcidBase Strength Strong Acids HI H20 stronger the acid I4 Weak AcidBase 0 ll CHa C O H H20 How StrongWeak Ka 39 pKa 39 smaller the pKa the acid pKE1 of strong acids pKa of organic acids Prediciting Outcomes of AcidBase Reactions AcidBase reactions favor formation of the weaker acid and weaker base Examples use pKa o 0 II II 1 CHa C OoH NaoH quot CHs CO H O H 2 CH33CO H NaoH quot CH33CO H O H What if you don39t have the pKa table Oh no 539 Consider the general reaction HA H Aquot Factors that affect anion stability 1 Electronegativity EN of the atom bearing the charge A more electronegative atom quottoleratesquot negative charge better than a less electronegative one compare acidities of the following H CH3 H NH2 H0H HF 2 Delocalization of charge the more a charge is delocalized spread around the more stable the system a size of the atom the larger the anion the more stable the anion H20 vs H28 b resonance stabilization O I ll CHsCHzo H vs CH3 COH OH O c inductive effects described later G Lewis AcidsBases expands the definition of acids and bases add base includes Bronsted Lowry acidsbases 0 0 9 H H Ci gt H c H Cl I H H F T no F B NH F BNH3 i F F F O F 3 CH30H20 H gt 0 0 Na NH2 CH3CH2O H gt O l39l H20 HCI HBI39 HN03 H2804 quot3 OH CH3COH 0 CH30H2OH Li Mgz Br AlCl3 BF3 TICI4 FeCl3 ZnCl2 I B O 9 ll CHscHzOH CHaOCHa CH3CH CHSCCHa 9 9 S C CH3C39CI CH3COH CH3COCH3 C3H3CNH2 CHS39N39CHs CH3SCH3 CH3 Curved Arrow Formalism The Mechanism 1 Curved arrows are used to indicate bonds breaking and forming in a reaction or electron quotflowquot 2 Direction of electron quotflowquot is always from electron rich to electron poor or from nucleophile to electrophile NEVER the reverse 3 Arrow begins with a covalent bond or a lone pair of electrons sometimes with a lone electron REMEMBER Arrows show where electrons flow not where the atoms go See previous acidbase reactions for examples 8 34 C We m 3 53 2 Attention Students The notes that follow are for your use in class They are incomplete You are expected to add important information and details during class You cannot survive Organic 11 with these notes alone CHAPTER 13 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY Structure determination AB gtC IR MS NMR UV Energy required see electromagnetic spectrum p 502 of Wade The Basics of NMR Theory 1 All nuclei are charged 2 Some nuclei have a nuclear spin 3 The spinning charge generates a magnetic field magnetic moment 4 These magnetic fields are randomly oriented and 1 Che m 337 5 If an external magnetic field H0 is applied the orientations of the nuclear magnetic fields are no longer random 9 A EASE 2 C H0 increases F H0 6Energy is required to quotflipquot the nucleus from its lower energy state to its higher energy state When the quotflipquot occurs the nucleus is said to be quotin resonancequot Energy required is proportional to the magnetic field strength NMR continued To obtain an NMR spectrum sample is irradiated with a constant amount of radio frequency rf energy such as 60 MHz by a transmitter the magnetic field strength is varied at the appropriate magnetic field strength for the rt energy the nucleus absorbs energy and quotflipsquot a detector measures the energy absorbed and a peak is recorded at that field strength OJ If examining isolated nuclei protons Nuclei are not isolated circulating electrons generate magnetic fields leffective H0 39 Hi Example naked proton absorbs at 140920 gauss assume shielding of 03 gauss Heff 140920 03 14091 gauss Heff 140923 03 140920 gauss Nuclei in different magnetic environments absorb energy at different magnetic field strengths the more shielded a nucleus the more deshielded a nucleus Example see Fig 139 The Proton NMR Spectrum of Methanol Chemical shift The variations in magnetic field strength necessary to bring nuclei in different magnetic environments in to resonance are actually quite small If field strength in gauss were used as the scale for NMR spectra it would be difficult to distinguish different types of protons Solution 9H3 H30 si CH3 CH3 Structural features that contribute to shieldingdeshielding 1 an electronegative atom H Cl Y examples 0 39 l CH3 C O CH3 CI CHzCli CI Cl 2 1 electrons a aromatic 7t electrons induced field reinforces the applied field b It electrons of double bond induced field reinforces the applied field H C C c R electrons of triple bond induced field opposes the applied field C E C H T74 Fig 1310 Deshielding of Benzene Protons Circulation of electrons induced field reinforces the external field deshielding 39 induced I H0 39 1nduced magnetlc eld Fig 1311 Proton NMR Spectrum of Toluene Hz 600 500 400 300 200 100 O 100 90 80 70 60 50 40 30 20 10 0 5ppm ORGANIC CHEMISTRY 4E 1999 by PrenticeHall Inc by LG Wade Jr Upper Saddle River NJ 07458 T75 Fig 1312 Deshielding of Vinyl Protons 6 induced field reinforces the external field deshielding VHinduceV Fig 1313 Shielding of Acetylenic Protons Hinduced shields the proton H H induced R H induced ORGANIC CHEMISTRY 4E 1999 by PrenticeHall Inc by LG Wade Jr Upper Saddle River NJ 07458 T73 Table 133 Typical Values of Chemical Shifts 7 Type of Proton Approximate 5 Type of Proton Approximate 8 alkane CH3 09 gtCClt 17 CI z kane CHi jL3 3 alkane CH 14 Ph H 7 2 I Ph CH3 23 D R CHO 9 10 CCH3 2391 R COOH 10 12 CECH 2395 R OH variable about 2 5 RCH2X 3 4 Ar OH variable about 4 7 X halogen O R NH2 variable about 15 4 C3 5 6 H Note These values are approximate as all chemical shifts are affected by neighboring substituents The numbers given here assume that alkyl groups are the only other substituents present A more complete table of chemical shifts appear in Appendix 1 Fig 1340 Common Chemical Shifts in the Proton NMR Spectrum Cocl 511512 X X C b O or I i a XOHal I I L I I i I I 8 7 6 5 4 3 2 1 0 5 ppm ORGANIC CHEMISTRY 4E 39 1999 by PrenticeHall inc by LG Wade Jr Upper Saddle River NJ 07458 Equivalent vs Nonequivalent Protons IMPORTANT to recognize If two hydrogens can be replaced in turn one at a time by the same group to obtain a chemically equivalent compound the two hydrogens are chemically equivalent examples C H3 H H H H C H3 H Br H H CH3 0 H C C gt H H Practice 0 H n cmo c om c cm CH3 H H Br H CH3 CH3 H H H Br CH3 H H CH3 H CH3 393 was 7 lama war a mQMdooC q 99 om aunsu 500 400 300 Hz 4H b b H 5b H 1 CH3 CH3 a 200 Hb H b 6H a 100 80 70 60 50 6 Ohm 30 IO Spinspin splitting occurs only between nonequivalent protons does not occur between equivalent protons most splitting is between protons on adjacent carbon atoms also occurs between nonequivalent protons on the same carbon atom cl Hb r Cl C C Cl Ha Hb Hb I Jab 939 Ho Ha Ha l Hal mag field aligned with applied Ha mag field aligned against applied field field effective field or field quotfeltquot by effective field or field quotfeltquot by Hb IS Increased by Ha Hb is decreased by Ha Hb resonates at lower magnetic Hb resonates at higher magnetic field strength downfield field strength upfield observed signal for Hb Jab Jab L i H f f f l l l M Hb2 Hb1 Hb2 Hb1 Hb2 l f Hb1 Hb2 opposing Hb magnetic fields cancel observed signal for Ha The N 1 Rule If a signal is split by N equivalent protons it is split into N 1 peaks Examples Cl T Cl l 1 H Cl c c H CI CI Cl 2 CH3 CH2 Br 0 ll 3 CI CH2 C OCH20H3 5 CH3 CH2 CH2 l I 2 SPLITTING PATTERN Of the 1 PROTONS T60 in nPROPYL IODIDE continued 5 185 chemical shift of Hb JIM Z 7 3 HZ 1 I l Jul 2 68 Hz Y I Y I I 39 l I 1991 by PrenticeHall Inc 0 GAHIC CHEMISTRY A Simon amp Schuster Company R Englewood Cliffs New Jersey 07632 quotdb 73quot quot39 quotsb quot 521 40 quot quot 393F quot3923quot 15quot Now Hz 500 400 300 200 100 o t f 2 H N m K Q 1 g 102mm i i i j i i i I I l 39 i 1 39 I 3 a L I s L I 4 i J 39 r 39 an 10 60 50 40 30 20 10 o Noam Hz 500 am 300 200 1m 0 7 1 003 g 2 z a i 3 a I 1 1 J an 79 so 5 o 40 30 20 10 0 Mom copyrigmom mums Inc FIGURE 1427 94 4 w um 4455 55 6657 9 101112 141 1111 ilzl1llqilllltc1 I 35 r v 1 v 2500 200013901600 140012001000 80 459 30 20 10 50 Wall 80 70 90 100 l5 Carbon13 NMR Theory Important Differences Between 130 and 1H NMR 1 Techniques for obtaining spectra are different Since natural abundance of 13C is low approx 99 of carbon atoms are 120 we would obtain very weak signals using the techniques discussed for 1H NMR spectroscopy Solutions Use computer to average many scans taken in the quotnormalquot continuous wave manner time consuming or use FT NMR spectroscopy 2 Carbon signals occur over much wider range 0 220 ppm 0 10ppm for1H so less signal overlap 3 Peak area is not necessarily proportional to the number of carbons 4 No carbon carbon splitting is observed adjacent 13C atoms unlikely Carbon hydrogen splitting may be observed depending on instrument mode all carbon signals appear as singlets observe a singlet for each type of carbon 0 II CH339C39CHQCH3 Wbsewe splitting of 13C signal by its directly bonded protons only N 1 Rule applies 0 CH3 CH n CH339C39CHQCH3 CH2 0 A C5 HuCl l 5 i 6 L i i 3 4 i 39 i v 39 190 I80 170 163 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 6npm r if 1 T y BCsHuCl q 39 t39 q 8 CDCI3 L i7 11quotquotii iLJLJJ 190 180 170 160 150 140 130 120 no 100 90 so 7o 60 50 4o 30 20 lo 0 640901 t C 3531101 t q t iiiquot JLJULJU 190 180 170 160150 140 130 120 no 100 90 80 7o 60 so 40 30 2b 10 0 64mm Copyright 0 John Wiley amp Sons Inc 1437 1 g T r r VT73939 quot 7 r a v Q Hi A i quot39 m L L n g l 1 L 1 1 l L 4 1 I 1 l l l J 1 J 1 w o 1 n a a na a gm gm 3 if g 39 1 w 52m 3 331 gift 911 7 llg umz I k is 639 t p L 15quot t 1 L 0 i f 39 I 331quot soNOJ I W 395 b A A 939 n 10 a 0 2 CID leo IR I700 cmquot I Sivan 539 I 1 MR J 7L J 11 2 1 I3 C NMR 01a 3 200 1334 Hm t IBS IZOPM t qo PM 1quot lopw l8 1 O I CQHWO Intensity 39 5H x 10 3 6 4 Chemical shift 5 3g 3350 why4 1 WM 1500 Moo m Z BCNMKS 5130490 34 IS 6201 5904 09pm 0 190 Hz 200 2H b m 20 10 220quot 30 O o m 0 V II 0 I A N 0 I VT UI Tma 5 O 0 Ln Copyright 0 John Wiley amp Sons Inc 6 p39pm J 39 410 50 70 80 FIGURE 1411 87 a L 50M to H Q 4N O m 5 Q0 ltr3 00 L A D V a 5 3 5 H 5 2 0 V I 7 3 6 A Q V O N N I O O m A O V O 239 O 39y C O LO Copyright 6 John Wiley amp Sons Inc J 70 39 80 FIGURE 1424 2d 92 MM Z 100 b 20 quot330 200 a Hz 30 39 50 a C b 80 qx 599 34 Copyright 0 John Wiley 1 Sons Inc 1 500 m 400 3930 suos w MIM uuor w yldoo O H c CH2C1 C OCH2CH3 b a uw 99w 96 6817 L 38091 AIME Hz 200 100 b 39 c a 1L0 5 Wm 3k 20 10 4ch i kamell Chapter 7 Structure and Synthesis of Alkenes omit 711 Important functional groupImportant industrial compounds H20CH2 l Structure compare to alkane H H C Cl In H H H H BDE of CC bond free rotation Physical Properties H H CC H H BDE of CC double bond restricted rotation BDE of a bond Since relatively nonpolar compounds similar to the alkanes of corresponding molecular mass Cis isomers generally more polar than trans isomers ALKENE NOMENCLATURE 1 Parent name is longest chain that contains the double bond Change quotanequot to quotenequot C CC CC CCC CC C l C CC 2 Begin numbering at end nearer the double bond Double bond gets lowest possible number in absence of other functional groups C l C czc c c 3 If double bond has same number from either end number to give substituents lowest possible numbers first point of difference C C CCC 4 Write name according to previous rules numbers commas dashes alphabetical order etc 5 If more than one double bond use prefixes di tri tetra etc diene 2 double bonds triene 3 double bonds etc Choose as parent name the chain that contains the most double bonds possible 393 C C CZC CrC 6 If it is not possible to include all double bonds in parent chain use alkenyl name for side chain containing the double bond Name as complex side chain as for alkanes or may use common alkenyl names on page 306 of Wade cc cC A c cc 7 In cycloalkenes the double bond is always between C1 and C2 Give substituents the lowest numbers possible first point of difference Be sure to include the double bond number in the set If only one double bond in ring not necessary to include the number 1 for the double bond39s location in the name calcug cH3 Geometric lsomerism The at bond restricts free rotation which leads to the possibility of geometric isomers Each isomer must have a unique name H30 CH H CH3 3 CC CC H H H30 H Are these geometric isomers H30 CH3 H CH3 00 00 H CH3 HaC CH3 I I I I I I I II H Br Cl Br 00 00 CI CHa H CH3 1 Assign priority to each group attached to each carbon atom of the double bond according to the CahnlngoldPrelog system learned in chapter 5 Higher atomic gets higher priority lower atomic gets lower priority 2 Look at the positions of the two higher priority groups on each carbon atom of the double bond If both are on the same side of the double bond assign Z stereochemistry If the groups are on opposite sides of the double bond assign E stereochemistry 3 Remember if identical atoms are bonded to the carbon atoms of double bond examine the next sets of atoms Assign priority at the first point of difference 4 Remember groups containing double or triple bonds are assigned priority as if the groups around the bond were doubled or tripled H CH3 HsC H CC cu CH20H3 Br 0 HZCCH H C HC 3 c Alkene Stability Determination of order observed in Ch 6 more substituted more stable Heats of Hydrogenation energy released in the following reaction R R R R H2Pt I 39 CC gt RC C R heat R R 1 39 CH H IPt H lPt 3 HZCCHCHZCH3 Lgt CHSCHZCHZCH3 4 2 CH CCH 3 alkene that evolves the most heat Explanation of order complex 1 hyperconjugation increases the electron density of the pi bond which strengthens the bond more alkyl groups greater the hyperconjugation 2 bond strength greater the 3 character of the hybrid stronger the bonds formed 3 steric factors more stable when alkyl groups are separated as much as possible M vs 5 Cycloalkene stability A QC Q 0 mm a compound that contains two rings each bridge must contain at least one carbon atom CO g BoredjLiBule A bridged bicyclic compound cannot have a double bond at a bridgehead position unless one of the rings contains at least eight carbon atoms co g 61gt Alkene Synthesis A Dehydrohalogenation of AIkyI Halides by E2 Review Ch 6 T f B X Remember strong base required anticoplanar arrangement of X and H trans diaxial if cyclic size of base determines the orientation of the double bond Example CH3 CH33CO 39 Na j CH30 39 Na Br Bulky bases increase E2 C H3CH23N 1 39339 20 CH Nal or KI II acetone 39 Br or Znacetic acid CCH33 Br Nal or KI acetone 391 Br win C Dehydration of Alcohols A 00 H20 OH 3 A want alkene B want alcohol Mechanism 39l T EH H OSO3H T HI step1 c c c c l I l H H oH H I I RDS I Step 2 C C CC I I t 3 I H20 sep 0 0 00 H O I I or HSO439 3 Order of reactivity of alcohols 3 2 1 Examples 3 H OSOaH CHaCHCHQCH3 heat H 0803H CH CH CH OH OH 3 2 2 2 heat Important Tips for Proposing Mechanisms see pp 303306 for others 1 Examine the starting materials and try to indentify potential nucleophiles electron rich and electrophiles electron poor 2 Draw curved arrows to represent electron flow from nucleophile to electrophile 3 Show one step at a time unless certain that steps occur simultaneously such as in 8N2 and E2 4lf reaction conditions involve strong acidsstrong electrophiles avoid forming strong basesstrong nucleophiles NO OK 5 If reaction conditions involve strong basesstrong nucleophiles avoid forming strong acidsstrong electrophiles NO OK 6 If reaction conditions involve initiators such as peroxides or light try a radical process Example 0 H H20 OH Chaim 5 tefaachem siry 359 881 SY M L Simsturai Same ismmia m canneciivity CQN 8 NA Cami bs intercmve ed amass mads are bmken Can be 3333mm E SEREQ SQMERS Same farmuia same cameat v yx diff spatiai awangemam QQNFGRMA Q AL Ragidiy niercanva ed w imm breaking beads Can t be sa ateii usua y 5 i E k g g Mira imagxas hat are newswgerimgmsab e Carmnmama s ngie band miat m swimmers Amine Envafs m Cmmrmers AYEREGMER Ceaiigum mai mamas that are m3 m rm mages inciudas aigiirans somars wentifyi g 23mm cemem gm H i E g waas cg x 3 F F 35 EH3 2 a gc gwgmgmzmEWCHzm sngwmg H HEGN f 3 j H3C MQH3 Hquotquot quot H Assigning 3918 C2132axing0 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wm v iauwfoxmiwwo a 23 665 z a hie 59n gt o iv lt waySaki 53 S V RQQRY 36 I on N 0 AA ller gtsltmlt gtgtltw gt C gt5 x v 5 z 0va mc A273 Y 22 9 k 354 06x Xw 5lt A A Aukvwvwwo w w Q J 43 0 V Amy gt O 29 51 w hymn J 9933 A R 05 g m r 39 A w v 3 24 lt5 99 4 5 gt x I 42 lt 4 5 m m I 1quot 0 ltgt2lt o of 0 lt 39 lt 9 a I lt2 9a ama o v 0 I n 5N0 a 582 mixium 3 wamt amem R am 3 3909 r 393 A 3 45455 a m 53 lt A39r rquotrf AV n l rl 5393 xv w A 44 m x A u lt gt M y 3525 042 x 2ltltgtltgtig 3 1 vlt 7363 Z x98 7 v 553921 3x 3 y 2 51512 39 39 as x quot vs 223 gt W v Racemic Drama mixtura Ex Hydmgana an a ketone tHnn Et 3 30 W i 3 WW 3 w 1 H33 H H 38mm m e ptica Ey inactive Stamng materia MW 363 E3 nastive pmduc 31hamyim iy37 aaimma ma iyM9 3y5igms 9 a ma ecaie gar Marie 1thth a symme iz icai seniormaiian it is ach s ai was if we a he canfmmations i3 chimi mg m Mama 3 symmetfy Wham determining Chiraiiiy in systems that ex gi as rapidiy iniemmve mg swimmers iaak 05 mamas 3 gymmetyy r ma mast symmetricai egresematian 0 the maieguiev rings CH3 W 2 SH mgt i 3 W 57 acyaiin Br 8 Ma gggiagmmmgmiham mSfiammemej r r CH36HCHCHQCH3 Pass bie a aterea samefs z 2 n m 3 stereacentam 232 Ham 63 MB Ha MB Emmag r 2 CH3 HCZHCH3 43 Sywi m B B CH3 M833 camgaum Naming cemmunda with mere than me stareacamer QHS Brw m 5 i H 3 m8 CH3 ii 5 iSiifEER 39ZJMEQECTEQNS Amaiygimg 313 maiemies m a fiat gufface 1 Main carbmta Chain is veriitai 2 Xe e ica nes repre zimt bon s behiim phi e iik i dasheg 31quot in 33222119 3 Eigrizemai nes represem mags 11 from 0quot pieme 23m wedges 4 intersemam raga356m chiral carbon mama 5 M03 exidized group is uma y places at the 3079 5 Fissrzhezr pragetztizmg must not be removed mm 3216 gimme 0f the paper Ef e 6212131101 be picked up and ipped aver 7 Pre ee ons may Em matted in new ways 2 f The wha ie pregame1 may be mixed it 80 333131 2 One gram aimth m a chiral far39iyon may be new gteaziy arm the Qtth three gmupa may be mated ciackw se or mumerdeckwisa IE gum fomxm Mm 2x x im 0 xi Eim fa magw gem mmam x im 0me 0 Mm ix x im fummm mm 2 Wm mmmm xiTmm mg mmw m0gm xmmm mmm wasng 13 3 mm mmw gma mxw Magnume 3 Emmi ampmms Wimmi Shim Mamas 1 wgiaw es mm are baked 5am asymmeigia mafarma ans 39 if x V 52quot 1 gawk lt 31 I l l X g I s 3 Mm x 39 W x m I gt W I x g 64 K 3 g i 3 wHK u 2 Sumi iumc aiieneg a 3 H364 f3 CmaC ngGwQ CAquot H H33 H Amg uia vg Reiative Qamfigumiim 2H3 2 PB gas ngmmz WC mg CHyQi 3 iv w wCHyB CRECHQ CHZCHS UNSE QHS wwgwcaga imam SQGLKU c f i b Alkynes omit 9 68 39 lo Structure compare to alkene and alkane H H H H Hquot quotIIll H H H H BDE of triple bond BDE of double bond BDE of 00 bond BDE of first 1 bond BDE of second a bond Percent 3 character Importance sp hybrid orbital sp2 hybrid orbital sp3 hybrid orbital Since nonpolar compounds similar to the alkanesalkenes of corresponding molecular mass CommorLNames Acetylenes HCECH terminal alkyne CH3CH2 CEO H internal alkyne CH3 CCCH3 1kChoose longest chain that contains the triple bond Suffix is quotynequot or diynetriyne etc Triple bonds gets priority when it is the only functional group present 9H3 Hac c CH2 CEC CH3 8m 2 Compounds with double and triple bonds in the longest chain are called quotenynesquot Numbering an enyne begins with the end that gives the multiple bonds the lowest possible numbers double or triple may have the lower number 9H3 HC CCHgCHEJHCHQCH CHCH3 CH3 If the numbering is the same from either direction the double bond has priority and receives the lower number HC EC39CHQ39CHZ39CH CH2 3 If the triple bond is not part of the parent name name as an allgmyl side chain j39CHzcEcCH3 4 Cyclic alkynes ring size of at least 8 carbons necessary Triple bond is understood to be between carbons 1 and 2 GHQO Note True bond angle of sp carbon is not reflected in the structure above ALKYNE ACIDITY Compare the pKa values of hydrocarbons H H H CEC H CC HSCH l39 H H H l H l e H CE e H H3oe ZC H H other pKa values H2804 CH3COZH H20 CH3CH20H FORMATION OF ACETYLIDE ANIONS Strong bases such as NaNH2 organolithium reagents and Grignard reagents can deprotonate a terminal alkyne to form an acetylide anion Hydroxide is NOT strong enough Q FlCEC H BEHNHZ e RCEC H QH Reactions of Acetylide Anions A Alkylation of the acetylide anion a substituton reaction 8N2 H R CEC Na R39 CI X gt I H Examples 1 NaNH2 2 CchH2 39 CH2 Br 1NaNH2 CHaCHQ CEC H 2 CH3C392H CH3 Br B Addition to carbonyl groups aldehydes and ketones R R CEC Na CO gt R Exam Ie p 0 H e 9 CH3C CH3 2 H30 Synthesis of Alkynes Dehydrohalogenation of vicinal and geminal dihalides R C C l X X X l I R C C X H HCECCHQCHQCHg CH3C E C39C H20H3 H Base R C C Base heat a x Base X R gt cc ias e r heat R H NaNH2 3 150 C Br Br Cll ig CH CHZCH20H3 Br r Br CHa39 CID CH2 CH20H3 Br Bf39 CH20H2CH3 Br K CH3 CIJH ICH CHZCHg gt Tgt J RCE CFl RC E CR HC E 00 HgCHQC H3 CHgC E CCHQCHa Base Catalyzed Rearrangement of Alkynes H e I e R C CEC R39 B R CID CEC R H H R G H R H G H Rl H 9 2 HB R CEO C lt gt R CCC RI RI REACTIONS OF ALKYNES 1 Addition of H2 A Catalytic Reduction Hydrogenation l Reactive catalysts H2 R CEC R gt Pt Pd or Ni 2 Lindlar39s catalyst H2 PdBaSO4 R C E C R quinoline B Dissolving Metal Reduction LiNH 1 R CEC R 3 2 Mechanism Li o R g H NH2 R a Reactions of Alkynes continued II Addition of H X HCI HBr HI X H l l RCECH AL R C C H l X H l HX THX H X H H R H addition of HBrperoxide l l RCECH HBL gt R c C H peroxude l I Br Ill Addition of X2 Br2 or Clg Z X R X x x x X2 X2 I RCCR gt 00 020 gt R C C R I I R X R R X X Reactivity of Alkyne vs Alkene and Stability of the Vinyl Carbocation CC iL r H H HX 1 sf spa 6 H a 439 z39c CCH3 H 60 7 20 Ca V atquot Muzd m v gt g kwim L l 011 me39fhy 3 F lo IV Addition of HOH Hydration A Mercuric lon catalyzed hydration H H20 H2304 C C H9 4 RCE CH tautomers special constitutional isomers that are interconvertible by a rapid EQ H H HO H I CC C C H l R H R H Mechanism H H 39 H H92 8042 H20 H20 H O RCECH gt Rco gt 00 C C H2804H20 Hg R Hg R Hg HESH2 H H H HO H 0 H20 H O i H O 39 H5H2 czc C C H lt C C H lt gt C CH 4 l gt gt R H R H R H R H Examples CHsCHeCECH Ego H2304 4 H20 st04 CH3CH2CECCH3 Hg 4 l Addition of HOH to alkynes continued B Hydroborationoxidation nonMark first step similar to addition of borane to alkene but must use a hindered borane to prevent double addition R H H BH BH IT HF 2 RCECH 3 gt CC MEI F RCIJ C H i l H BHZ H BH2 Reagent used BH SiazBH disiamylborane R H R H O S39 BH 39 39 ll RCECH 432 CC H202OH oc gtOH RCHzC H H BSiaz H OH 1 Sia BH u EX CH CECH 2 3 2 H202OH39 CH3CH2 C H H2OHZSO4HgSO4 S CH3CCH3 V Oxrdation of aikynes OH OH O O KMnO4HZO 39 2 H2O u n 1de2 RCECR W Fi393 R gt R C CR HO OH KMnO M20 9 S H lt3 9 2Str0ng RCCR OHheat RC C R gt RC CR 1O 0 0 OH HO 3 2 H20 0 O H II R C OH HO O 10 II if terminal RCECH g R C CO2 Similar result With KMnO4 OH Chapter 11 Reactions of Alcohols l Oxidations oxidaLion increases of bonds to O N or halogens or decreases of bonds to H reduction increases of bonds to H or decreases of bonds to O N or halogens neither oxidation or reduction if HX or H20 is gained or lost 0 0 II II RCHa lt R CH20H lt gt R CH R COH A Oxidation of 2 alcohols ox R CHR gt ox 1 chromic acid reagents a NazCl39207 H2304 H20 b 003 HZSO4 H20 lacetone 0 C Jones reagent 2 KMnO4 3 HNO3 4 pyridinium chlorochromate ICHZCIZ written as PCCCHZCIZ or CrO3 pyridine HCI lCH202 B Oxidation of 1 alcohols oxa oxb R CH20H gt gt oxb 13 above oxa PCC ONLY C Oxidation of 3 alcohols ox R IEOH gt R Examples OH CHQOH N a20r207 Na2Cr207 H2 64IH26 W CH3 OH NaZCr207 CH CH CH CH OH 3 2 2 2 W CH H2 4Hz 3 II Formation and uses of the tosylate ester converting the OH group to a better leaving group o lt N H R OH Cl CH3 gt o NaCN acetone li R OH R OTos NaCN pyridine acetone III Conversion of alcohols to alkyl halides The BEST methods A Formation of 3 alkyl halides from 3 alcohols use hydrohalic acids HBr HCl CH3 0H HClether 0 C for 1 and 2 alcohols treatment of alcohol with HX is not the best method of forming alkyl halides addition of catalyst Lucas reagent improves results B Formation of 1 2 alkyl halides from 1 2 alcohols 1 Bromides use PBr3 R OH P Br 2 Chlorides a PC3 Ol39 PC5 b SOCI2 NOTE The mechanism given on page 474 p 449 of 3rd ed occurs only under quotspecial conditionsquot The more typical mechanism is the followmg 3 Iodides use Pl2 P l2 Stereochemistry of Indirect Substitution of ROH TsCl 4 IV Acidcatalyzed dehydration to alkene first discussed in Ch 7 Mechanism E1 Intermediate C Major product formed according to Saytzeff39s rule more stable alkene is major product Example H HOSO H CHgCHgCHgCHZOH lt 3 gt CH30H20HCH2IDH gt I Acidcatalyzed dehydratons of 1 alcohols do NOT give good yields of terminal alkenes Phosphorous oxychloride POCI3 a good dehydrating agent for 1 2 and 3 alcohols 9 N CH CH CH CH OH P 39 3 2 2 2 ClC39Cl 0 C V Formation of Ethers A Symmetrical ethers through intermolecular dehydration R OH HO R H ROR HoH H20 H so EX 20H3CH2OH W CHgCHg OCH20H3 HOH Mechanism H I ll FtOH gt ROH R OR gt RO R R OH NOT for unsymmetrical ethersl CH30H2 0H CHaO39H 44108 B Williamson Ether Synthesis symm or unsymmetrical ethers alkoxide alkyl halide gt ether NaorK ROH order of reactivity of ROH in alkoxide formation Me gt 1 gt 2 gt 3 Williamson Ether Synthesis continued R O Na R39 CH2X gt ROCHZR39 9H3 CH3 O CHCH3 a VI Ester Formation 4n A Fischer Esterification intermolecular dehydraton mechanism next semester 0 II II R OH HOCR39 lt gt R OCR HOH 9 H EX CH3CH2 OH HO 39CCH3 B Nucleophilic acyl substitution of acid chloride 0 0 II ROH Cl C R39 gt R OCquotR39 HCI l r9 r0 l ROCR39 gt ROCR I Cl 0 EX CH3CH2 0H IsCH3 Vll Unique reactions of 12 diols APeriodic acid cleavage OH OH I HO4 gt c c gt C0 Chapter 8 Reactions of Alkenes omit 85a and 816 Electrophiiic Additions addition of E writsr THIS MATERME MAY BE Pam rmrn m BGPYRIGHT LAW quotTITLE 37 US CUHFquot General iquot CC E Nu gt C C I l Addition of HX HCI HBr HI H H 39f39 39l339 CC H Br C C H H I Mechanism H H 00 H Br H H Addition to unsymmetrical alkenes a CH2CHCH3 HBr H30 H Regiospecific product is formed from only one of two possible orientations of addition Markovnikov39s Rule observation The addition of HX to the double bond of an alkene results in a product with the acidic proton bonded to the carbon that already has the greater number of hydrogens WHY What is RDS CH3CHCH2 reaction coordinate 39 Mark39s Rule extended In an electrophilic addition E adds in a way that generates the most stable intermediate Example CH3IHCHCH2 ii CH3 Free Radical Addition of HBr A Puzzle HBr Peroxide Effect Only with HBr Step 1 ROOR 2 BO 63 Step2 R O H Br ROH Br Step 3CH3CH quot CH2 Br 39 CHacHCHZBr Step 4 H3 Br CH3CHCHzBr CH3CHZCHZBr Br without peroxide with peroxide ll Addition of HOH A Hydraton acid catalyzed addition of H20 across double bond reverse of dehydration of alcohols H OH W I I CC H20 c c H30 l l Mechanism if H u H 0H n H OH CC HBH H I 6 H20 H20 l l 2 u n 39 I I I I I I Example CH3 CH 3 H30 B Indirect hydration product is alcohol but not from direct addition of H20 1 Oxymercurationdemercuration H CC g39 2NaBH4 I I OH 0 ll ll II lI CHa C39O39HQ39O39C39CHs CH3COHg ocCH3 Mechanism CH30HCH2 My CH30H70H2 Hg II l CH3COHg 0Ac IDH CH30HH2 H Example D 1 2 NaBH4 CH3 39 H bquot HIl 3 39gt CH30H2H2 HgOAc lHab39 oH Ja BHJ OHSCHcH2 HgOAc useful variation alkoxymercurationdemercuration use an alcohol instead of H20 to form ethers mechanism stereochemistry and regiochemistry same as above Example D 1 HgOAc2CH30H 2 NaBH4 CH3 fmmz N 03220va F IoNoNI m IIIIIIIII IIHamIm F 50 meamem mIomIome mIoononl mINIoNIomIo 31 IomIomIomIo Io quotIomIo I mIomIomIo I m INIomIomIo I NIo quotIomIo Q I I I j I Tm NIolIIo Tu NIohomIo Allll NIoqumIo HEwEmcomZ IoNoNIm III Almlql H o o EEIm o 0 cozmhonegl m IV Addition of XX halogens Clg Br2 sometimes l2 x 06 X2CCI4 I3 39 l I 5 C x Example Br2 CC4 5 C Mechanism X2H20 l CC C 3 I OH Example CH3 Cl2H20 Mechanism CH3 Cl Cl gt H20 VI Addition of Carbenes and Carbenoid Species G r l ene a C 0 Example CH2N2 gt heat B SimmonsSmith Reagent produces carbenoid species CH212 ZnCu gt 1 CH2 ZnI Example CH3 gt ZnCu NOTE The stereochemistry of the alkene is always maintained in carbenecarbenoid additions H30 CH3 CH2N2 C C gt H quotH heat CH3 CH 212 C C gt HaC s 9 ZnCU C Dibromocarbene and Dichlorocarbene Formation of carbene C39 039 Cl 39 KOH I Cl C H gt Cl C e gt I 4 lt Cl Cl Cl Br Br I KOH I 339 Br C H Br C 9 gt lt lt Br Br Br Example HSC CH3 C 0 CHCI3 gt CH3CH2 H KOH Cle BIG Oxidation of Alkenes A Epoxidation with peroxyacids ll CC R C O OH gt R C OH Mech a concerted process H H i 3 X 0 c O H g Stereochemistry of the alkene is maintained H CH3 MCPBA H O CC g S i i CHs H30 H quotI Acid catalyzed ring opening of epoxides as H 392 OH 1 OH 39 o 0111 OH Hzo CCH Halt2 I ll 9 quotI 39 I H H lgH OH H Isolation of epoxide vs diol Strongly acidic peroxyacids in aqueous solution give diol epoxide not isolated Weakly acidic peroxyacids in nonaqueous solution give epoxide B Hydroxylation quot00 0quot ox 1 COLD basic KMnO4 2 OsO4IH202 I quotcc SYN Add39t non o O O Example CH3 Q 0504m202 39 gt CH3 MCPBACCN C Oxidative Cleavage ll C C 0X ox 1 WARM concentrated KMnO4 acidic or neutral conditions oH IDH C CIIIIIII 4 ts lCC V H20OH gt O O gt Mn 0 0 H30 fragments that contain oxygen 2 Ozonolysis 1O3 2 CH3ZS or Znacetic acid IO quot390 c f HO OH Ozonolysis cc C O General examples R R 10 CH Cl CC 2 CH328 R R R R 1 0 CH Cl CC 2 CH32S R H R R 1 0 CH CI CC 2 CH328 H H R H 10 ICH Cl CC 2 CH32S H H C O O carbonyl compounds CH32S m if carbon has H giggle if carbon has no H Warm KMnOiI cleavage gives ketone if carbon has no H gives carboxylic acid if carbon has H KMnO4 oxidizes aldehydes to carboxylic acids and formaldehyde to CO2 H20 R R KMnO C C warm conc R R R R KMnO C C warm conc R H R R KMnO C C warm conc H H R H KMnO C C warm conc H H R co Fl R CO R R c HO O R co Ho R OC R R OC OH R OC OH co2 H20 Alkene Reaction Summary 1BH3THF H202fOH 1J xOA 2 bO 2NaBH4 H30 CCI4 HX X2 H20 4 1 g HiPt Pd or Ni 2 CH3ZS CHgCHCH2 r KMnO4warm CH2N2 heat or light cone KMnO4 cold b 03quot CHZCIZ oso4m202 or other RCO3H sz3 KOH CHAPTER 6 Alky Halides Nucleophilic Substitution and Elimination No sections omitted 39 39 I r rvrr squot39 r31 r i Tli E39 71158 IAs h EE39ih fquot F3171 u an 1m 6 I Types of Organohalogen Compounds WYWH my Tm E W M Common Uses section 3 F F l Cl C F F c H DDT F Cl Nomenclature No new rules halogen has no priority For name as substituent drop quotidequot add quot0quot Example chloride becomes chloro new terms vicinal dihalide two halogens on adjacent carbons geminal dihalide two halogens on the same carbon Structure 1 2 3 determined by the type of carbon the halogen is bonded to Example In a 1 halide the halide is bonded to a 1 carbon atom H quot Physical Properties types of intermolecular attractions 1 2 M F Cl Br M I Preparation 1 Free radical halogenatlon Caution mixtures of products Best results when a single major product is produced Rememberzbromine is more selective than chlorine O Br2 ight l Br2 light W Br2 light 2 Allylic brominaton O Br2 Ight Reason for selectivity Consider first step of propagation H H gt Better reagent Nucleophilic Substitution Reactions H N X H H General Reaction 9 Nu R X gt R Nu X Two specific reactons G A CN CH30I gt CH3CN are e e B CN CH330CI gt CH33CCN Cl Both reactions are nucleOphiIic substitution reactions but A and B proceed through differentmechanisms Before proposing a mechanism for each we consider two kinds of experimental evidence 1 The kinetics of the reaction Examine rate changes in relation to changes in the concentrations of the reactants 2 The stereochemistry of the products Reaction A 1 Kinetics Observations if cone of Nu is doubled if cone of substrate is doubled rate The Mechanism 2 Stereochemistry Nucleophile Nu begins to form bond to electrophilic carbon while leaving roup LG is still bonded Nu donates electron pair to the small back lobe of the sp hybrid that forms bond to LG This is called quotbackside attackquot and results in inversion of carbon39s configuration Walden Inversion Carbon configuration is turned quotinside outquot H H A 0C6N Nu OCltI C Cl gt NC C H quotW H H H e H C3H C Br L gt HO CH3 MUCH CH30H2 CH20H3 8N2 is a stereospecific reaction a particular stereoisomer reacts to give one specific stereoisomer of the product even though another is possible 8N2 summary 1 rate kNu substrate 2 bimolecular TS 3 proceeds with 100 inversion Reaction B 9 CN CH33CCI gt CH33CCN Cl 1 Kinetics Observations if conc of substrate is doubled if conc of Nu is doubled rate The Mechanism 2 Stereochemistry Nu attacks after LG has left Nu attacks a planar carbocation and can attack from either side Attack from the opposite side of the LG39s original position results in inversion of the carbon39s configuration Attack from the same side as the LG39s original position results in retention of the carbon39s configuration This result is known as racemization of stereochemistry The 8N1 process is not stereospecific 55 I 55 39 co 39 H30 CHa Racemization Example CHSCHQCHQ CH W C Br HZOacetone 3 heat CH30H2 SN1 summary 1 rate ksubstrate 2 unimolecular TS for RDS 3 proceeds with racemization may not get a 5050 mixture due to ion pair formation We39ve looked at two reactions two mechanisms for nucleophilic substitution of alkyl halides 8N1 and 8N2 We are left with a BIG QUESTION What determines which mechanism is taking place in the reaction of a given alkyl halide Simple Answer Rate faster pathway wins Factors that Affect Rate 1 Substrate Structure 2 Nature of the Leaving Group 3 Concentration and Reactivity Strength of the Nucleophile 3N2 only 4 Solvent Substrate Structure vs Reaction Rate 3N2 Process Relative Reaction Rates for Alkyl Halides H H H H H 39C I C X H C H I H C X le H H I c H H H H u X IlIIII C C H H H l H Hnlt C HH H C 039 I l I x H Substrate Structure vs Reaction Rate 3N1 Process Relative Reaction Rates for Alkyl Halides Allyl C HQCCH CH2 lt gt HZC CHCH2 Benzyl C OCHQ gt QCHZ lt gt GCH2 lt gt GCHQ Order of 0 stability 3 allylicbenzylic gt 3 2 allylicbenzylic gt 2 1 allylicbenzylic gt 1 gt methyl Illlquot X C C X 8N1 ISN2 Competition CH3X 1 2 3 Nature of the Leaving Group Same influence on rate in 8N1 and 8N2 only the timing of the departure is different LG must be electron withdrawing must polarize the C X bond must put 6 charge on C LG must form a stable species a stable anion a neutral molecule a weak base The weaker the base formed Order of Reactivity of Alky Halides by 8N1 or 8N2 RI RBr RCI RF How is reaction rate affected by the LG I I I ll Table 6 4 gives other good LG Example Strong bases make bad leaving groups in 8N1 and 8N2 processes Example Nu R OH gt But under acidic conditions R OH gt Concentration and Strength of Nucleophile a concentration 8N1 8N2 b strength nucleophilicity affinity for the 6 carbon 8N1 8N2 strong Nu weak Nu Io H Trends in Nucleophilicig Af nity for a positive or 6 carbon atom 1 A negatively charged nucleophile is stronger than its conjugate acid 2 Nucleophilicity decreases from left to right in the periodic table Electronegative elements quothold onquot to lone pairs more tightly therefore lone pairs are less available to form new bonds 3 N ucleophilicity increases down a group in the periodic table as size and polarizability increase Larger atoms hold outer electrons more 1005er so electrons move more freely toward a positive charge This contributes to earlier and stronger bonding in the transition state 4 Bulky groups on the nucleophile decrease nucleophilicity I2 Strength of the Nucleophile and the SN1SN2 Choice for 2 Substrate m KN 3sz Ni 05 3N1 N39 0 Strong Nu necessary Weak Nu OK Strong Nu favors 8N2 Weak Nu favors 8N1 CAUTION Does not mean a 1 substrate with a weak Nu will proceed by SN1 or a 3 substrate with a strong Nu will proceed by 8N2 Example CH3 CH3 039 NaI 039 AL gt heat IH acetone I Solvent influence on rate 8N2 1 Table 63 lists nucleophiles in order of strength in a particular type of solvent The trends in nucleophilicity were also determined in this type of solvent Example I39 gt Br gt Cl39 gt F39 Consider how the protic hydroxylic solvent interacts with Nu G 006 I3 Now consider nucleophilicity in a polar aprotic nonhydroxylic solvent Examples 0 0 II g CH3 3 H N C C so HQC CH3 H H3C CH3 CH3 polar aprotic Using a polar aprotic solvent greatly increases the rate of the 8N2 process Switching from a hydroxylic solvent to an aprotic solvent can increase the rate as much as one million times protic solvent aprotic solvent Example Br 0 KFHgO gt Br KF CH CN W Solvent effect on SW gt R X gt R X gt R x9 m 3H3 CHg ID Cl CH3 The more polar the solvent gt Higher the dielectric constant the more polar the solvent dielectric const relative rate of tBuCI ionization H20 MeOH EtOH Acetone Ether Hexane Protic solvents Igt Solvent Summary 8N2 Solvent affects strength of Nu Aprotic solvents destabilize Nu Nu is not solvated well so increase the strength of Nu Aprotic solvent is not required 8N1 Solvent facilitates ionization by stabilizing the TS the carbocation and the LG Protic solvent is required for significant ionizaton Summary of SN1SN2 choices CH3X 1 30 2 allyl benzyl m0 mo of MIA u foiwl wnfo 0 mo mT w fo hxo mIo MO 3 4 Iomromzo 0in mo mIO foiol ohfo mo 00 Illllll sxowzomzo Io m Iofo Alql n 0 o 10 0 0 mo foiw m mIO foiol oufo 5 0 0 foozomo m 355 69me 8 ES gtclt H EmEmmcmtmwm 2250850 lb Elimination Reactions H H I I gt f Nu X Two pathways timing of bond breaking and forming different as in SN1SN2 E1 elimination unimolecuiar T8 of RDS involves a single molecule the substrate rate kRX Mechanism H I l stept c c c c 39 l l 39i step2 C C V CC Consider the reaction and mechanism 3 CH3 CC 95EtOHH20 l 65 C CH3 lacs Order of reactivity by E1 3 2 1 E2 elimination bimolecular TS of RDS involves two molecules the substrate and Nu rate kRX Nu Mechanism H H B c Ros I O O a quot39 I B X Consider the reaction and mechanism H H3044 39 CH CH 039 Na H3CKC C39WH 3 2 l gt CH3CH20H Br H Order of reactivity by E2 3 2 1 Orientation of Elimination and Alkene Stability H CH3 H H CH33CO Na CH3393H5 39H2 939Na l Br Saytzeff39s Rule in an elimination the more highly substituted alkene more stable alkene predominates order of P R R H R H R R R H alkene 00 00 00 C C CC stability R R R R H R H H H H Stereochemistry of E2 A stereospecific reaction 1 All four substrate atoms involved in the reaction must be coplanar Two conformations meet this requirement H x R C39 C R D R x R R k B o 4 A R 39 c c R I I I 9 I4 x Rquot R R R 2 Preferred conformation Exception to preferred conformation base 8 I Experimental Evidence for Antiperiplanar Coplanar Geometry in E2 Process front carbon H H o B H C C Ph NaNH2 r Fgth H Br kr et anol CC CC Ph Ph Ph back carbon Ph A Br 3 Antiperiplanar result H H H Ph H Ph OR Br Ph Br Ph Synperiplanar result OR gt Br Ph Ph H 20 Experimental Evidence for 12 Diaxial Requirement in E2 Process on Cyclohexyl Substrates H30gtltgtHIICHCH32 H3C uICHCH32 0 C CI A neomenthyl chloride 3 menthyl Chloride H Hsc QCHCH3gt2 H CHCH32 H EtO Na H30 EtOH Cl H H3C ltgtHIICHCH32 A neomenthyl chloride H CH3 H CHCH32 CI Cl H H30 H H H H EtO Na H B menthyl chloride EtOH CHCH32 WCQHHHCWCHQQ 21 Substitution vs Elimination Wade summaries sections 616 and 622 Nu strong base 39OH HO 39NH2 nose bulky bases particularly favor E2 tBuO 39 Nu weak base A fsN2 ksN1 E1 Ezt CH3 1 2 allylicbenzylic 3 2 allylicbenzylic Strong Nu but weak quot ggirnNusolvem base such as iodide g is necessary HS 39 RS or 39 ON 8N1 usually majorprocess faster rate in aprotic high T increases E1 solvents Exception 1 halide with a strong unhindered base 8N2 gives major product Examples Emil 1 CH3CH28r CH30H2039Na 5T 9 CH30HZOCHZCH3 CH2CH2 0070 512 10 6E2 2 CH3CH2150H2CHQBf CH33COK CH3CH215CHQCHQOCCH33 CH3CH215CHCH2 15 612 9670 62 6 quot E 0H 1 3 CH3CH20Na 3939 39 CH2 I v d CACE l39o 4 CH30HCH3 Na ie CH3C 2HCH3A3910070 3amp2 I EH3 lCHa W CHsICOCHQCHg CH2C 3 2 H 1 CH3 9H3 C 3 3706511 777492 5 CH3IC Br 4 CH3 9H3 CH3 lCtiaQ 29 c3H3cOCH20H3 CH2C 50 C CH3 CH3 802 3076 El CHAPTER 10 Structure and Synthesis of Alcohols omit 1012 Nomenclature 1 Longest chain containg the OH is the parent chain Chain is numbered to give OH the lowest possible Suffix is quotolquot CH3 C3H20H I CH30HCHQCHCHQCH3 2 Alcohols have priority over alkenes and alkynes OH CH3 CH2 CHCHzCHOH I CE 3 In cyclic alcohols OH is always on carbon 1 CH3 OH CH3 4 When the OH group cannot be incorporated into the parent chain it is named as a quothydroxyquot substituent OH IDH CH2CHCH3 5 Alcohols with two OH groups are named as diols Add quotdiolquot to the alkanealkenealkyne name OH oH C3H3 CHZCHaCHCHQCHOH OH 6 In phenols the OH group is always on carbon 1 OOH QOH OOH Br CH3 PHYSICAL PROPERTIES OF ROH Boiling Point CH3CHQCH3 CH3OCH3 CH3CH20H HOCH2CH20H Solubility H OH VS 0 CHEMICAL PROPERTIES ACIDITYBASICITY can serve as weak acid or weak base As a Base 0 R po H H X gt As an Acid e 9 Na H Structure vs Acidity F CHgCHgO H VS F JCHgo H F EHB VS CH3 CH3 Acidity of Phenols OH O OH OH 02 Nio W G OH NaOH gt SYNTHESIS OF ALCOHOLS I From Alkyl Halides Ch 6 heat R X OH gt R o H OR R X HOH gt R O H II From Alkenes Ch 8 A Hydration acid catalyzed B OxymercurationDemercuration C Hydroboration D Epoxidation followed by acidic hydrolysis E Hydroxylation III From Acetylide Anion and Carbonyl Compound Ch 9 R R CEC Na CO th R H3O IV Organometallic Reagents Carbonyl Compounds gt ROH A Preparation of Reagents 1 Grignard Reagent ether R X Mg gt R MQX 2 Organolithium Reagent pentaneor R X 2L1 gt R LI LIX hexane Examples CHSCHQ Br t39eman H M CHSCH C Br j er P B Mechanism of Grignard Reaction mechanism of organolithium reaction is similar at o r ether l HOH2 C R x gt R gt C R C Grignard Reactions RLi reactions are similar H 1 with formaldehyde R ng CO 1 r 2 H30quot H RI 2 with aldehydes R Mg x 30 1 t r 2 H3O H Rquot 3 with ketones R Mg x 00 1 t r 2 H3O RI Examples H30 CHscHgMgBr co 1emergt 2 H3O H H H30 CHSCH C CO MgBr 2 H3O H30 4 with esters or acid halides CI R MgX CO tether R 2 H30 Example of RMgX acid chloride or ester 0 MgBr C O O OCH3 ether gt o 5 with ethylene oxide R MgX i 1 t r 2 H30 Example 0 CHs CH MgBr 4mg 3 D Limitations on Grignard Organolithium reactions Use caution when selecting the halide compound that forms the Grignard or organolithium reagent and the carbonyl compound Observe the following rules 1 NO ACIDIC HYDROGENSI NO OH NH SH C02H or CECH Examples OH H2 CHTBr Mgether ll R MgX R C O H gt 2 NO functional groups that react with Grignard organolithium reagents besides the intended carbonyl component 0 0 NO I I II II N C C S C N Notice the general structural feature a multiple bond in which one atom is strongly partial Examples 0 II CH3 C CHZCHQCHQCHzBr m 0 O H II 1 ether CH3 Mg Br CHa C CHQCHz CH 2 H30 3 Dubious and Worthwhile Uses of a Side Reaction RX Mgether Rng H20 Mth R X ii R MgX D20 SYNTHESIS INTERLUDE CH3 l CH30H2 C OH CHQCHQCHa A Slightly Related Reaction of an Organolithium Reagent The CoreyHouse Reaction forms a new carboncarbon bond a multiple step synthesis 1 RX 2 Li 39 gt RLi LiX 2 2 RLi Cul gt RgCuLi Lil 3 RZCuLi R39X gt RR39 RCU LiX Mechanism Not well understood but NOT a typical substitution reaction I R Example Br CHaBr L39 CHaLi Qi 39 O gt V Synthesis of Alcohols Reduction of a Carbonyl The quotorganicquot definitions of oxidation and reduction oxidation reduction A Hydride Reductions 0 OH 0 OH ll H I H H 1 RC H gt RQ H 2 R C R gt Rg R H H OH 0 OH ll H I H H l 3 R COH gt R c H H20 4 R COR39 gt R c H R39OH H H OH H H I 5 R C Cl ii Rc H HCI H Reducing agents NaBH4 the milder choice usually chosen for aldehydes and ketones may use H20 or alcohol as solvents very slow reduction of esters will not reduce acid chlorides or carboxylic acids LiAlH4 more reactive than NaBH4 reduces aldehydes ketones esters acid halides and carboxylic acids CAUTION reacts violently with H20 and alcohols Must use ethers as solvents Mechanism 0 0 O u H AlH Li HESH I R C H gt3 Fl CliH 3 9 I H H II H AlH Li I n 393 HoH R C OH i p n c H gt RCH W R C H 39 2 I OH H l l R39OH RCIJH Examples NaBH4 61 CH30H2 C CH3 EtOH 1 LiAH4 II b CchHgCOCHCH32 2H30 B Catalytic Hydrogenation of Aldehydes and Ketones CC CO catalytic reduction of carbonyl is slower than reduction of double bond Raney Ni is best catalyst Examples gt aCHaCH2 c H RaCNI 0 Hz b CchHQ C CHQCH CH2 NaBH4EtOH CHAPTER 12 Infrared Spectroscopy and Mass Spectrometry quotReal Worldquot Example Biologically Active Compounds heroin morphine codiene demerol methadone These compouds have useful analgesic properties and harmful side affects addiction HO C C 0 N CH3 0 C C C C I ll C CGHS I N C H3 Q CH3 Framework necessary HO for biological activity Morphine Rule Methadone Morphine Drug Design Maintain positive properties eliminate negative properties A B gt C A Puzzle The puzzle pieces composition gives of each element leads to empirical formula MS gives molecular mass leads to molecular formulaunsaturation also gives other structure clues IR gives functional groups NMR gives carbonhydrogen framework Infrared Spectroscopy Sample is bombarded with radiation in the IR region of the electromagnetic spectrum Fig 12 1 p 502 This is enough energy to affect molecular vibrations Important Terms and Relationships frequency complete wave cycles that pass a fixed point in 1 second directly proportional to energy wavelength distance between peaks or troughs of the wave inversely proportional to frequency and energy wavenumber of cycles of the wave in 1 cm it is the reciprocal of the wavelength it is directly proportional to frequency and energy measured in reciprocal centimeters cm39 Example Longer the wavelength Lower the frequency Lower the energy Shorter the wavenumber Lower the frequency Lower the energy Bombard sample with IR radiation 100 T A A v N AAAAAAAAAAAAAA A B WWW Igt C VVVVVVVVVVVVVV v A vAV v AAAAAAAAAAAAA v I D vvvvvvvw 0 T samp39e 4000 cm391 600 cm391 IMPORTANT QUESTION What determines which frequencies are absorbed and which are transmitted Answer IR uses only enough energy to affect molecular vibrations When vibrations caused by IR radiation match the natural vibrations of the molecule energy is absorbed Natural Vibrations of Molecules I Bond Stretching stretching vibration Frequency of the natural stretching vibration depends on two factors 1 size of atom heavy atoms vibrate more slowly than light atoms 2 strength stiffness of the bond stronger bonds require more force to stretch and compress Bond Stretching Frequencies Bond Bond Energy kcal Strectching Freq cm 391 C H 100 3000 C D 100 2100 C C 83 1200 C C 83 1200 C 0 146 1660 0 EC 200 2200 C N 73 1200 C N 147 1650 C EN 213 2200 C O 86 1 100 C 0 178 1700 Examples Consider the stretching frequencies of the following bonds 393 I H c C 0 0 H a I H b H30 F H3O H3O Br Conjugation lowers frequency 0 0 ll Molecular StretchingBending Vibrations 151 lll IR Active vs IR Inactive Vibrations Bonds with a dipole moment polar bonds are IR active An electromagnetic wave has a component that consists of a rapidly reversing electric field This electric field can cause bond and molecular vibrations by alternately stretching and compressing polar bonds If the vibration caused by the IR radiation matches the frequency of a natural vibration of the compound the radiation is absorbed Peak observed Bonds with no dipole moment nonpolar bonds are IR inactive Electric field component cannot stretch and compress a nonpolar bondNo peak observed INFRARED SPECTRAL REGIONS T54 wavelength um 25 4 45 5 55 6 65 7 8 9 101112 1416 100 80 percent transmlttance 1 60 40 20 0 1m 4000 3 500 2 5 00 2000 l 800 l 600 1 400 l 200 l 000 800 600 x fr wavenumber cm l w H15 equot y 1991 by Premi gl A Simon 8 Schuster Compan En Iewood Cl39ff Ne Jerse 07632 ORGANIC CHEMISTRY g 39 S w y IR SPECTRUM of nHEXANE T29 wavelength Hm 25 3 35 4 45 5 55 6 65 7 8 9 101112 1416 f I I I T T T I I I 1 I In 80 J I 53 E 60 E G i C H bendmg 3 40 E B a a P I 20 it C I I stretch CH3CH24CH3 0 0 AL I I I L I 1 I I I 4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 wavenumber cm 1 IR SPECTRUM of lI39IEXEI IE wavelength Hm 25 3 35 4 45 5 55 6 65 7 8 9 101112 1416 100 f I I r r f I I I I I T I I Q 80 Q I g E 60 g CH3CH23 CHCH2 g 3080 E 40 e CH 1645 Ink C H bendlng 1 E Stretch CZC Stretch z 20 alkene a F C H stretch alkane 0 I I I I g L 1 I I I 4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 wavenumber cm I 1 IR SPECTRUM of CiS4OCTE1 1E wavelength urn 25 3 35 4 45 5 55 6 65 7 8 9 101112 1416 100 I f I I r j F 7 I I I I I r I I Q 80 E S E 60 E E E 40 alkene 3 C II stretch b H H a 20 3020 z d H C CH CH CH alkane C H stretch 3 2h 2 3 0 I I 4 I I I 1 I L 4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 wavenumber cm quot1 1991 by PrenticeHall Inc A Simon amp Schuster Compan ORGANIC CHEMISTRY Englewood Cliffs New Jersey 07632 IR SPECTRUM of 1OCTY1 1E T50 wavelength pm 25 3 35 4 45 5 55 6 65 7 8 9 10 1112 1416 17 1 1 I I I I I I 5 o C 00 O 0 C 2120 4 O H CECCH35CH3 percent transmittanc E C H stretch N O 3 320 V O I 1 I 39 1 I 1 l I 4000 3 S00 3000 2500 2000 1 800 l 600 1 400 1200 1000 800 600 wavenumber cm 1 IR SPECTRUM of 2OCTY1 1E I wavelength hm 25 3 35 4 45 5 55 6 65 7 8 9 101112 1416 100 l I 139 7 17 I r I I I I I 1 I l 80 J E g no no CEC stretch E 60 stretch ViSlble E 4o 93 g 20 alkane CH3 C2C CH24CH3 C H stretching I A I l 4 1 I 4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 wavenumber em 1 IR SPECTRUM 0f 1OCTEl lE wavelength um 25 3 35 4 45 5 55 6 6 7 8 9 101112 1416 100 I I 139 f r I 7 T I I I r T r I q 80 if g g 60 E A g 401 5 U y 20 V I I 1 1 J l I I l I g l 4000 3500 3000 2500 2000 1800 1600 1400 1200 1000 800 600 wavenumber cm quot l 1991 by PrenticeHaH Inc A Simon amp Schuster Compan ORGHm C EMISTRY Englewood Cliffs New Jersey 07632 Wavelength u 3 l 5 0 7 8 9 l0 2 I5 U i V I I I 1 I r I r I 1 I I T I 1 IT 00 IO H 1 gt rlrelch g E OIHI C O 140 a fire 391 a TINllquot OH 80 1 I 1 1 L 1 J 1 1 1 39 1 1 1 1 1 l 1 I 1 I 1 1 1 l 1 l 1 351 10m 2500 2000 III woo 1400 1200 I 300 an I09 K Frequency cmquot Wivelength mm 25 v 4 s o 7 s 9 1o 12 1 5 l Wavenumber cm 39 Wavelength mm 25 3 4 5 6 7 v8 19 IVZH VlS O u 2 CHJCCHZCH3 00 quot 350039 39 3000 2500 2000 1300 1600 1400 1200 1000 80o WavenumbeHcm 39 Wavelength pm lt 3 4 5 o 7 gs 7 9 10 12 15 0 0 39 W3300 I800 1600 140039 1200 m8 00 Wavenumbcr 1cmquot Wavelength turn 5 3 4 5 6 7 IO 1239 IS u J C33CH2COH CO 9000 3500 3000 2500 2000 1800 1600 1400 1200 800 Wavenumber cmquot 1 Wavelength um 3395 3 4 5 6 7 3 9 10 12 1s O n CH3CH23COCH2CH3 00 393500 3000 2500 2000 l800 1600 1400 1200 100039 800 Wavenumber cm l Vluelcn muun 25 J 6 S b 7 a 9 IO I I5 01011012an m 3500 m 2500 2a 1800 lm I m 1200 I m Vannumbet cm quot 739 lemmm l 2 s 3 4 5 0 7a 9 I0 I21 I5 01301201212101 M 3500 3000 2500 2000 1300 1600 1400 39 1200 1000 300 Vamumbet cm 39 Human 0239s 3 5 6 Z 2 10 120 15 c113c112c1121314 39 39 39 quot 500 3500 3000 2500 2000 1000 1600 1200 391000 quot000 Vammnbc cmquot Ovenones of monosubstituted benzene I Aromatic stretch c H bending 000 3600 3200 2800 2400 2000 1amp0 1600 1400 1200 1000 800 650 Wavenumber cm 1 Wavelength u J l s s 7 I 9 10 I I5 r r r If Y T r I i I I I T7 T m 4 no CL L r 10 gt SIVPILII 8 2 239 0 H 0 a my 01on 8 1 CO mnno 9 subml llllllllll39l LLLIIJII 1 1 1 I L 14 L1 1 1 Muf e 39 mo soon 2500 zoao woo woo mo mo loan sun can r 985 K Frequency cmquot Wavelength um 3 4 5 6 7 8 9 IO 127IS Y I Y I 1 w v 0 3500 3000 2500 2000 39 1300 1600 A 1400 I200 IOOO 800 Wavenumber cmquot Wavelength um 035 3 i A 7 X 77 3 10 13 395 0 m 7 vd 0 V 4mm l 392506 quot1530quot quot1660 7 300 13de 866 Wavenumbcr cm 39 ll 3 Mass Spectrometry Another piece of the puzzle mass of the compoun mass of fragments formula structural features especially from high resoultion MS Destructive technique or an39c highE 9 39 Li LQClLOrl phadical cation electrons molecule high energy causes molecule to break apart fragments neutral fragments radicals or small neutral molecules Summary 1 In the mass spectrometer 6 fragments are deflected by a magnetic field to a detector 2 Neutral fragments are not deflected and crash into walls so not detected 3 Magnetic field strength is varied so only fragments of a particular mass to charge ratio mz enter the dector Charge 1 so mz mass Example H CH3 H CH4 gt 1e H51H mZ15 H CH2T 2H gtCH H mz mZ14 mz13 molecular ion parent ion base peak the strongest peak in the mass spec represents the most abundant ion assigned an intensity of 100 other peak intensities are proportional to the base peak not necessarily the molecular ion peak often represents the most stable fragment Transparency 43 Figure 127 Page 418 Mass spectra of a methane and b propane 2 Copyright 1992 by Wadsworth Inc All rights reserved A 100 r r b3 5 W2 11 53 80r 5 8 60 vquot E quotS 40 39 Q N O E 20 2 g 0 III AL I l I I I l I I I I I I 20 39 4O 60 80 100 120 140 mz gt 39 a A 100 5o OS 80 P 6 o 8 6O 5 quotU 5 40 393 cu mz44 E 20 5 Q m 0 I I IIIII quot 39 I l l I I I I I I I 20 40 60 80 100 120 140 mz gt 39 b gt CH3CH2 39CH3 CH3CH2CH3 I 539 High Resolution Mass Spec HRMS While mass spec rounds masses to the nearest whole number HRMS determines an exact mass or a mass to several Significant figures The exact mass allows fairly acurate determination of the molecular formula Example data for an unknown compound Molecular ion peak from mass spec 44 possible formulas C3H8 02H4O C02 CN2H4 calculated exact mass 4406260 4402620 4398983 4403740 Molecular ion peak from HRMS 44063 The unknown compound is Tables of calculated exact masses and formulas are available Isotope Peaks observed for any peak large enough ion is abundant enough small peaks at M mass 1 and sometimes M 2 Example see methane mass spec Because a small 111 of methane molecules will have a13C atom and a very small will have a H atom very few would have both we observe a small peak at mz 17 This peak is called the M 1 isotope peak Some elements are readily identified by their isotope peaks Br gt 2 isotopes 7QBT505 o and 81Br495 rgt M and M 2 almost equal size Ci gt 2 isotopes 35CI755 and 37C245 rgt M 2 is 13 the size of M SIgt3 isotopes3289503380834S42igt M 2 is small but larger than usual I gt 127I100 o observe 1 at mz127 and a 127 unit gap IQ T98 Table 124 Element M M1 M2 hydrogen 1H 1000 carbon 12C 989 13c 11 nitrogen 14N 996 15N 04 oxygen 160 998 180 02 sulfur 32 950 33s 08 34s 42 chlorine 35a 755 37c1 245 bromine 79Br 505 81Br 495 iodine 1271 1000 Page 524 Mass Spectrum of 1bromopropane abundance 10 20 3O 40 50 60 70 80 90 100 110 120 130 140 150 160 quotdz Page 524 Mass Spectrum of 2chloropropane 100 80 C1 60 H3C CH CH3 40 abundance 20 C 37C1 0 39 10 20 30 40 50 6O 7O 80 90 100 110 120 130 140 150 160 n z Organic Chemistry 5e 2003 by PrenticeHall Inc by LG Wade Jr 39 A Pearson Education Company Upper Saddle River New Jersey 07458 1739 T99 Page 524 Mass Spectrum of ethyl methyl sulfide 100 60 S 1 B 40 V M 2 larger than usual 20 0 10 20 30 40 50 6O 70 80 90 100 110120130 140150 160 mz Prob 127b Mass Spectrum of iodoethane 100 80 d 8 60 S a 3 40 Cd 20 0 10 20 30 40 50 6O 70 80 90 100110120130140150160 mZ Prob 1220 Mass Spectrum of 1bromobutane 100 80 g 60 E g 40 20 0 10 20 30 4o 50 60 70 80 9o 100110120 130140150160 mz Organic Chemistry 5e 2003 by PrenticeHall Inc by LG Wade Jr 39 A Pearson Education Company Upper Saddle River New Jersey 07458


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