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Structure of Organic Molecules

by: Michael Reilly

Structure of Organic Molecules CHEM 14C

Marketplace > University of California - Los Angeles > Chemistry > CHEM 14C > Structure of Organic Molecules
Michael Reilly
GPA 3.71

S. Hardinger

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S. Hardinger
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This 54 page Class Notes was uploaded by Michael Reilly on Friday September 4, 2015. The Class Notes belongs to CHEM 14C at University of California - Los Angeles taught by S. Hardinger in Fall. Since its upload, it has received 95 views. For similar materials see /class/177971/chem-14c-university-of-california-los-angeles in Chemistry at University of California - Los Angeles.


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Date Created: 09/04/15
Resonance What Is Resonance Example Carbon monoxide CQ versus CEO Which accurately represents the molecular structure Open octet on carbon vs formal charges both decrease stability Double bond longer versus triple bond shorter Chemistry 14C Lecture Supplement Page 16 What Is Resonance The Lewis structure of C0 00 or 00 Laboratory analysis of C039 quot quot H20O OCO carbon monoxide COlength 121ri COlength 12ori COlength 113ri shorter than CO Empirical observations of dipole moment rule out triple bond Conclusion The Lewis structure that accurately represents CO is cb39 CEO Empirical obsenation actual structure is somewhere in between What Is Resonance The Lewis structure of C0 Perhaps an equilibrium between two structures that differin electron distribution c Chemistry 14C Lecture Supplement Page 17 Curved Arrows 19 CQH A curved arrow lndicates electron redistribution Starts at electron source points to electron destination Electron source A Electron destination When arrow starts at bond bond is broken H H H C l ii I 51 H H When arrow points to atom or space between atoms bond is formed H H H Ch il H H What Is Resonance The Lewis structure of C0 Perhaps CO structure is an equilibrium between two structures that differ in electron distribution C2 CQ Equilibrium ruled out by empirical evidence CQ ZC Chemistry 14C Lecture Supplement Page 18 What Is Resonance The Lewis structure of C0 Empin39cal CO structure is intermediate between the Lewis structures c39d CEO quot Adequate structure Inadequate Inadequate What is this between structure Each inadequate structure has some but not all actual features of CO CO cannot be adequately represented by a single Lewis structure Therefore blend or average inadequate structures gt composite structure What Is Resonance A definition Resonance Moecue cannot be adequately represented byjust one Lewis structure Two or more Lewis structures must be used Resonance contributors c39d CEO Chemistry 14C Lecture Supplement Page 19 How Do We Indicate a Molecule Has Resonance co 4 CEO Resonance Accurate depiction of CO c CEO Equilibn39um Not accurate for CO Why Is Resonance Important Common occurrence with Chem 140 atoms CN O F Cl Br I S P Resonance delocalization increases stability electrons charge spread over larger volume T stability Resonance in uences molecular structure causes planarity Examgle An acidbase reaction 0 o H O O H H0 H O g O H20 411 No resonance Has resonance Less stable More stable Keg gt 1 H2804 stronger acid than H2O Chemistry 14C Lecture Supplement Page 20 Drawing Resonance Contributors Key recognize electron pair shift patterns ch c HaC CCH2 3 W H Cit H ccNH H C CNH 3lt We 34W 0 quotz i i Hzgsc iIZ HzgCZQHZ What is electron pair shift pattern Note Not necessary to draW dorboth sides of lt gt Drawing Resonance Contributors i i c cc HZC 4 CH2 H29 v HZ J 9 i i as a HZC EHZ H29 EHZ open octet What is electron pair shift pattern Chemistry 14C Lecture Supplement Page 21 Drawing Resonance Contributors HZE QH H HzcgoiiI walla cng lt gt CH3ZKl CCH3Z P CH3Zcr lt gt Gigi z What is electron pair shift pattern Most common resonance patterns Lone pairpi bond trade places Pi bondopen octet trade places Lone pair open octet switches with pi bond Drawing Resonance Contributors Less common electron pair shift patterns Pi bonds SWitChi 4 Common forbenzene rings Single bond and lone pair switch 7amp2 1B CHg C XCHQZ CHgbg XCHQZ Only rarely important Chemistry 14C Lecture Supplement Page 22 Which Resonance Contributor Represents Reality Once upon a study break Fantasy creatures if Real creature Neither fully represents reality A unicorndragon hybrid The carbon monoxide case c lt gt CEo Neither contributor fully represents CO Resonance hybrid a weighted average or blend of resonance contributors the most accurate representation of the electronic structure of a molecule Drawing the Resonance Hybrid Example Draw the resonance hybrid for acetate ion CH3COZ39 03 6 6 1Draw contributors iv lt x H A CH3 9 CH3 9 CH3 202 2 Draw the features sigma and pi bonds lone pairs formal charges that are the same for all contributors 3 Add features that are not the same for all contributors Resonance hybrid Chemistry 14C Lecture Supplement Page 23 Do All Contributors Have Equal Importance Is a rhinoceros more unicorn or more dragon T contributor stability quot to reality T contribution to hybrid Therefore we need contributor preference stability rules Tnumber andor magnitude of rules violations Jimportance of individual contributor lcontribution to resonance hybrid Resonance Contributor Preference Rules Rule 1 The most important contributor has the maximum number of atoms with full valence shells Example H2COH lt gt H25 H All octets lled Open octet on carbon More important contributor Less important contributor Rule 1 is more in uential than all the other preference rules Rules 26 have no particular order of preference Chemistry 14C Lecture Supplement Page 24 Resonance Contributor Preference Rules Rule 2 The most signi cant contributor has the maximum number of covalent bonds H H Exam Ie Co lt gt c o p H H Four covalent bonds Three covalent bonds More important contn39butor Less important contn39butor Resonance Contributor Preference Rules Rule 3 The most significant contributor has theleast nunber of formal charges H H Exam e c6 lt gt c 6 Pl H H No formal charges Two formal charges More important contributor Less important contributor Chemistry 14C Lecture Supplement Page 25 Resonance Contributor Preference Rules Rule 4 If a contributor must have formal charges the most important contributors has these charges on the atoms than can best accommodate them Negative formal charges best on atoms of high electronegativity Positive formal charges best on atoms of low electronegativity IO 0 Example A H k CH3 CH2 CH3 CH2 Carbon EN 25 Oxygen EN 30 Less important contributor More important contributor Resonance Contributor Preference Rules Rule 5 Resonance interaction ie pi bond is strongest between atoms in the same row of the periodic table Usually CNOF Usually outweighs electronegativity considerations rule 4 F CC391l m H H l F C C 1 Example H More important contributor Less important contributor Even though EN F gt EN Cl Chemistry 14C Lecture Supplement Page 26 Introduction to Aromaticity Historical Background Chemists of the early nineteenth century were puzzled by the structure of benzene an unknown liquid with a pleasant aroma that was isolated from oil gas used to illuminate street lamps Michael Faraday analyzed the substance which was later termed benzene Early chemists knew that benzene had a molecular formula of C6H6 from its molecular mass and CH ratio of 11 which implies high reactivity However benzene was observed to be unusually stable at room temperature and did not undergo the reactions typical of alkenes Based on these properties along with the conclusions that carbon forms four bonds Kekule and that carbon can form multiple bonds Brown some possible structures for Benzene were proposed although these were shown to be incorrect later1 Illa 11 yquot E H3C C E C C 5C CH3 Fulvene Prismane Dewar Benzene 24Hexadiyne In fact under certain conditions both Dewar Benzene and Prismane rearrange to form Benzene below driven by the elimination of ring strain and preference for aromatic stabilization According to Kekule the actual structure of benzene came to him in a dream in which he saw a snake biting its own tail This suggested that benzene is a sixmember ring with alternating sing and double bonds According to the structure proposed by Kekule there should be H a 5333 g H two isomers of l2dichlorobenzene In one structure the chlorine I E atoms are separated by a single bond and in the other they are separated by a double bond However these isomers have never H I H been isolated or detected 2 H Cl Cl atoms separated by Short bond Cl Cl atoms separated by l long bond c1 l Rk jagfquot HH Cl l i I 11 K 12a H 3 Elffa nhh I 1 Fulvene is from htt wwwsteve bcomima esmoleculesarenesfulvene r1 visited 9206 Prismane is from htt wwwan eloedufacult lltboudreamolecule alle 01 alkanes rismane 01 if visited 9206 Dewar Benzene and 24Hexadiyne were constructed on Microsoft Word 2 l2dichlorobenzene structures from httpntp niehsnihgovntphtdocsstructuresZdTRZS5gif site visited on 9306 Kekule suggested that the reason the isomers are not observed is because there may be a very fast equilibrium between the two forms meaning that they interchange so quickly that the individual structures are not even detected But this still does not explain why benzene does not undergo typical alkene reactions Why Benzene is Not a Normal Alkene A normal alkene undergoes an addition reaction when exposed to Brz The three alkene benzene shown above is expected to behave similarly However benzene does NOT react with Brz unless a catalyst FeBr3 is introduced Since it is inert to Brz it cannot be a normal alkene If the true structure of benzene was made up of alternating single and double bonds as Kekule originally suggested then the bonds would also vary in length since double bonds are shorter than single bonds However data obtained through xray crystallography con rms that all CC bonds in benzene are of equal length Electron delocalization through resonance provides a solution to the reactivity inconsistency the bond length observation and why isomers were not observed for 12 dichlorobenzene The resonance contributors of benzene are pure fantasy The resonance hybrid which is a blend of the contributors best represents reality The electrons are spread throughout the ring giving all the bonds partial doublebond character which explains why the bonds are all the same length3 0 Benzene resonance contributors Benzene resonance hybrid A nal piece of information supporting the fact that benzene is not a normal alkene comes from catalytic hydrogenation a method that utilizes a metal catalyst such as Pt to add hydrogen to a double or triple bond This converts the pi bonds to CH sigma bonds so that the energy change can be measured 39 The heat of hydrogenation for cyclohexene is 286 kcal mol391 l 0 0 AH 286 kcal mol391 t I If benzene were simply three alkenes we would expect its heat of hydrogenation to be three times the heat of hydrogenation of cyclohexene since it has three double bonds 3286 kcal mol39l 856 kcal mol39l This value would be a little lower due to conjugation 3 footnote deleted 39 However the experiment shows the heat of hydrogenation of benzene to be AH 498 kcal mol39l This is 36 kcal mol391 less than three times the heat of hydrogenation of cyclohexene 4 This huge difference which corresponds to extra stability is not just explained by conjugation The extra stability is the aromaticity resonance energy Observed AH 498 kcal mol391 Overview Aromaticity also called Resonance Energy is the extra stability observed in molecules such as benzene To be considered aromatic a compound must meet ALL FOUR of the following requirements 1 The molecule must have a continuous ring or loop of 192 orbitals This means that all atoms in the loop must be Sp or sp2 hybridized 2 Atoms forming the loop must be planar so that the 192 orbitals can overlap A massive amount of torsional strain is required to override the amount of stability provided by aromaticity and thereby cause the molecule to lose planarity 3 All of the pi bonds that are counted towards aromaticity must lie within a cyclic structure 4 The molecule must obey Hiickel s Rule the aromatic loop must contain 4n2 pi electrons where n is any integer whole number such as 0 12 In other words the pi cloud must contain an odd number of pairs of pi electrons 5 totaling 2 6 10 14 etc Imgortant Points to Remember All aromatic molecules are conjugated but not all conjugated molecules are aromatic This statement is true because an aromatic molecule must have at least three adjacent parallel p2 orbitals 6 This fits the definition of conjugation However in order for a conjugated molecule to be classi ed as aromatic it must fit a number of other criteria For instance it must also be cyclic planar and contain the correct number of aromatic electrons All aromatic molecules have resonance but not all molecules with resonance are aromatic All aromatic molecules have resonance because they must have at least three overlapping p2 orbitals occupied by a minimum of two pi electrons However not every molecule that has resonance is planar cyclic and obeys Huckel s Rule 4 Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 186 5 Pg 595 of Bruice Organic Chemistry 4th Edition Ch 152 6 Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 191 Aromatic rings must be planar yet they are more stable and less reactive than their nonaromatic counterparts This is true because the stability gained through aromaticity outweighs the strain caused by planarity Classifying Aromatic and NonAromatic Compounds EX Is the molecule shown below aromatic Explain why or why not7 HE gs n DH ANSWER Yes the molecule is aromatic Remember that the entire molecule need not be cyclic for the molecule to be aromatic only the atoms forming the loop of 192 orbitals must be cyclic since these are the ones that will overlap In this case the benzene ring is planar cyclic contains a continuous ring of 192 orbitals and has 4n2 pi electrons 4n2 6 so 11 1 There is not enough torsional strain to override planarity EX Is the molecule furan shown on the right aromatic Why or why is u not8 f ANSWER At rst it may appear that furan has only four sp2 hybridized electrons However the oxygen has two lone pairs so the hybridization can be rearranged to give the correct number of pi electrons If we take the oxygen to be sp2 hybridized instead of 193 then it can contribute a 192 orbital Remember that for sp2 hybridization there will be three lled sp2 orbitals one with the sigma bond one with the lone pair on the outside of the ring and another with the remaining sigma bond and one p orbital that contains the remaining lone pair Thus we will have a continuous loop of 192 orbitals planarity a cyclic ring and 4n2 pi electrons 4n2 6 n1 All of the requirements for aromaticity have been met so furan is aromatic 3 EX Is the molecule shown on the right aromatic Why or why not The negative formal charge is circled9 ANSWER No the molecule is not aromatic The carbon on the bottom right corner has four bonds so it must be 193 hybridized Thus every atom in the ring does not have a p orbital Also it violates Huckel s Rule because it has two pairs of pi electrons if we count the carbon on the top right corner as sp2 hybridized by putting its lone pair electrons into a p orbital 7 Image is a modified form of the one on httpwwwchembaycomupimg200298l4570365 gif site visited 9306 Also the same structure was used in Chem 14C Test 1 Summer 2006 by Professor Steven Hardinger 8 Same question as CFQ 9 on Pg 190 of Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version 9 Structure adopted from Bruice Organic Chemistry 4th Edition Ch 152 Pg 597 Proton Nuclear Magnetic Resonance lHeNMR Spectroscopy Theory behind lVER In the late 1940 s physical chemists originally developed NMR spectroscopy to study different propenies of atomic nuclei but later found it to be useful in determining e molecular structure of organic compounds The theory behind NMR comes from the spin I1 of a nucleus Just as electrons have a l2 l2 spin certain nuclei also experience charged spins that create a magnetic eld called the magnetic moment which allows chemists to study them using NMR Nuclei with even numbers of both neutrons and protons experience NO spin and nuclei with odd numbers of both neutrons and protons have integer spins Nuclei that have the sum of protons and neutrons equal to an odd number like 1H and 13C have halfinteger spins When there is no external or applied magnetie eld By the nuclear spins orient mndomly however when there is an applied magnetic eld the nuclei orient themselves with or against the larger applied eld The 1 spin state is parallel to the applied force and has lower energy than the greater the strength of the applied magnetic eld the greater is the 7 E between the a and 39n states2 The 7 E between the a and i spin state is N002 cal mol39l which lies in tl e radio frequency region The emitted energy in this region produces an NMR signal lt5 P k J 7 Absorbewvgy 7 75 lt5 Reluxnnon I release energy alpha rspm state betarsmn state lqwel eneruv mum 2quotng Mr Diagram 13 Diagram 2 Addition of energy results in a spin ip nuclei shed excess energy in a process called relaxation Brspm state A Re 31va Energy 3 E espinstate Apphad Magnetlc Fweld Strength gt increases 1 2p 527 PY Bruice Organic Chemistry 4 3 th facul kutztownedubettshtmlNMR htm 1 1705 HTM 1 1 nm Dmgum 3 mun pptnunrpmn nuumnms so damn energy ntwnn ts mugspn one ntp two cnmmymgmpn cnmmmuunaucuotcs um NMR 82 ectnlm The NMR spcctrum 15 u ptot ottm mtcnsrty af MR sxgqals versm tnc mnpncucnctuttrcnucncy t t 5mm mrctcrcncctoms The nensw nuctcr tcnsrty 15 mcusurcd by mm m Lhunteg cruncncn unucr cs downnuu upnu Frequency 5 Interpreting H NMR s2 cctra Four types aflnfmmananfmm NMR 1 Num 2r n signas Protons wrtnrn a compound cnpcncncc drtrcrcnt magncuc cnvrronmcnts wnrcn grvc a scparatc srgna1 m tnc N39MR spcctrum Fnui alum 391 I A t V cqurvalcnt Symmctnca1 compounds such as bcnzcnc arc also cqurva1cnt nowcvcr smce many compounds arc not symmctnca1rtrs rmportant to know how to rdcntrfy noncqurvalcnt protons Protons that arc dffcrcnt m any way even m tncrr stcrcocncmrstry arc not cqurva1cnt and Wm absorb at ddfferent frcqucncrcs grvc a scparatc srgna1 on tnc N39MR spcctra Enamplcs5 All of tncsc compounds have 2 srgna1srntncrr NMR spcctrum htg www on nsu cunarcuscmrnuumtppccup tnnrnnn nm 1 mumo 521w urucc Wm mm a wwwcemmsuen1xemcthuaszx395 ecu tnnrnnn hunOllEID j it pp Lg mfg p0 a f 3 T ch cquot EGG f 3 3 3 cl CHIPS Cl 3 4quot b z i at 2 U 57 TMS L TMS TMS I I I I I I I I I I I I I I I l I I I I I I I I 7654321061098 qazlo loe ire432166 A B C 4 H s are on a plane of symmetry The CH2 H s are attached to a C One CH3 is attached to an O with each other CH3 are also on that s attached to a C1 The CH3 while the other CH3 is attached a plane of symmetry so 2 signlas H s are attached to the same C to a C and have the same neighbors 10 393 S 2 Position of signals chemical shift The position on the horizontal frequency scale at which the equivalent proton signals occur E is called a chemical shift7 measured in ppm Protons generally sense 3 different magnetic elds magnetic eld of the Earth the NMR spectra and different protons in the molecule Since the magnetic elds of the Earth and NMR spectra are felt similarly by all the protons in the molecule the chemical shift depends only on the varying local magnetic elds from the neighboring protons Reference compound Tetramethylsilane TMS CH1 In order to standardize the NMR spectra the chemical shifts are I positioned in relation to a reference proton set at 000 ppm H39IC Si CHT Tetramethylsilane CH34Si is the standard for H1 NMR TMS is A practical as a reference compound because of its inert quality that prevents it from reacting with the sample and its highly volatile nature that makes it easy to evaporate out of samples Few compounds have a lower frequency reading than TMS and it has 12 equivalent protons that read strongly on the NMR spectra What in uences a chemical shift Shielding effects Under an applied magnetic eld circulating electrons in the electron cloud produce a small opposing magnetic eld ultimately decreasing the effective magnetic eld felt by the proton shifting the signal to the right or up eld This effect in which the electron cloud shields the proton from the applied magnetic eld is called local diamagnetic shieldingg Electronegativity and deshielding H s that are attached to more electronegative atoms experience higher chemical shifts Electronegative atoms also remove electrons from the 6 p 779 l p lIrl1rin Tonhninnoc in nrnnnin Fhomicfmz 7ndl electron cloud which decreases their density and results in less shielding hence electronegative atoms are said to derhr39eld the proton and che e ect h atom increases refer to NMR Spectrum diagram above cause it to have a higher al shi moving it to the lett or doomem The magnitude ofttie dnhizlding a i i Examlr i cu ere CHZBr CHZCHzBr CHZCHZCHzBr CHZCHZCHZCHzBr 2 69 ppm 166 ppm l ppm n 93ppm Magnetic Anisotropy Lesystem ring currentsg In molecules containing 7 orhitals anisotrth having a different effect along symmetrical Characteristic Chemical Shirts tables oftypieal chemical shitts can be made since similar functional groups experience similar chemical shitts however due to the variability between molecules at s mew mu ushering r in time one cannot assume a certain frequency corresponds to one type ofproton An exception I n rr 3 Relative Intensity of Signals Integration inlzgmn39onn corresponds to signal Therefore the relative intensities ofthe sign relative number of proton equivalents It is imponan integration only provides ratios of protons not the absolute number convenience in calculating the relative 39 39 39 1 dtl th 39 The ar ea under the sign e number of protons responsible for that al are rop 39 For signal strengths the smallest 7 SEE F Y Bruiee Otgams alumina xhttp huuman chem geurgetuwn edunmrinteraetiunehemshfhtm 1 llKU y pszn F Y Bruiee Otgams alumina Ell 31n5 31 l 20mm NMR l Signal 3 I39ll 7quot 39 trength mm 1 39C39ll L 1 4 ppm Diagram 412 gtl lntegrals appear as lines on the spectra above the signals in which their heights correspond to the integration ratios In this spectra the OH H is correctly determined to be in a 31 ratio with the 3 CH3 H s 4 Splitting of signals spinspin coupling NMR signals are not usually single triangles but a complex pattern of split triangles labeled as doublets 2 peaks triplets 3 peaks quartets 4 peaks etc The distance between the split peaks are called coupling constants denoted by J13 The interaction between nearby protons produce different spin ip energies E as they can orient themselves in a pattern of parallel or antiparallel to the applied magnetic force This phenomenon where the spin of the nucleus of one proton is close enough to affect the spin of another is called spinspin coupling Splitting is always reciprocated between the protons if Ha splits Hb then Hb must split Ha and provides information on the neighbors of a proton within the molecule N1 Rule For a proton with n neighbors its signal will be split into n1 lines Intensity ratios Spin do ublet F attain f Np Cpu pl ed CC CH triplet 1 Hydrageng f r C A z Singlet quartetl1 J V 1 H ll Cine Coupled m i pH Hydrogen 39 r a Hquot C I NV 31 uu net H H l 1 ll J Two coupled In J Hydrogens HECH msgl lw lamp Pascal39s Triangle Diagram 513 Diagram 614 gtl Pascal s triangle gives the intensity ratios 1 neighbor 9 doublet 2 neighbors 9 triplet 3 neighbors 9 quartet between the split signals Splitting patterns that are too difficult to analyze are called multiplet 10p58 Prof Steven Hardinger Chem 14C Thinkbook version 4 ll p538 PY Bru1ce Organic Chemistry 4th 12 l1tth39nnxmv r l IPmiqtrV lln llPl l l Parlrimmnakllvlp lllTpinl l lTlVl l l 106 sputung results tron the ureterent energes nan sprn patterns 3 Rulesst 39e 39 nsfnr Prntnn chug 39 g 1 Nucler wth the sarne chemlcal shrlt ixachmnaux do not couple wth each er the protons rnustbe nonequryalent n oroler to couple 2 Vicirml protons protons separated by 3 bonds can H H couple wrth each other Protons that are rnore than 3 l bonols away cannot because the slgnal they feel from cigic ic therr nelghborls too srnall to affect therr 5pm There ls c also gemhml couplingicouplmg through 2 bonds and WW Hmmggng allyb39r caupung Leouphng through 4 bonols lf one ls p bo otf l th 3 bonol rule because the electron denslty d ls hlgher then m a 5m le bonol henee lbonds e E39 8 benzene nng only couple wrth each other and not wrth H atorns eyen lfthey are wrthrn the 3 bonol max lrrnrt coupllng club Hyolrogens bonoleolto aNltxogen or Oxygen usually do not couple wth other protons and appear as srnglets on the NMR spectra quotgt17 5quot V P of none rst oroler splrttrng Frrst oroler splrttrng prooluees norrnal splrttrng Tvalues w A l sprneeouples wth 2 or rnore sets ofnearby nuclel thathaye dlfferent Jyalues Exam le ofFlrst Orders llttln and Noanlrst Order quartet uoublet af daublets relateye rntensrtees l 3 3 l relateye rntensrtees l l l l equal yalues merentl yalues 239quotI Order Cnllpling Effects Frrst oroler theory ls sumerent when descnblng N39MR slgnals ofcoupled nuclel that are far enough apart from each other however when the 3an www cam rnsu eWNXeLwnhVmualTex peckpymrmxl htrn quotm2 www cam rnsu eWNXemuthnualTex gecgg imxnmxl htrn pol me Stevenllarunger chem 14c man Vernon 4 5 P m2 JR Mohng Techquotqu m Organic chemm 2 quotp52 me Stevenllarunger chem 14c man Vernon 4 chemical shifts come closer other effects second order coupling effects begin to take over Two things occur as a result of these weaker interactions spin state energies shift and the intensities of the peak change and no longer adhere to the pascal triangle pattern Hydrogen a and b are both doublet of doublets due to 2 d order coupling effects QUICK SUMMARY Major Pieces of Info from NMR19 39 Number of signals number of sets of nonequivalent protons 39 Integration relative number of nonequivalent protons 39 Chemical shift info about proton environment and presence of benzene rings 39 Splitting info about neighboring protons Technigue for solving NMR Problems Calculate DBE first NMR data is interpreted after the molecular formula and functional groups are found from mass and IR spectrometry so this information is already known When the molecular formula is provided it can be used to determine the DBE or double bond equivalents DBE provides the degree of unsaturation double bonds and reveals the presence of rings in the molecule 20C number of carbons DBE C 7 M M 1 H number onydrogens 2 2 h number of halogens N number ofNitrogens Organize Sgectral Information Looking at all of the NMR Spec data at once can get too overwhelming so it is a good idea to organize all of the information into a table and work with the pieces individually A table with the following headings is recommended 39 Chemical Shift ppm 39 Splitting 39 Integration 39 Hydrogens 39 Possible Fragment Structure Once the information is organized add up the number of hydrogens carbons and DBE s to make sure that everything has been used up Perform an atom check and a DBE check Checking the individual atoms before forming them into pieces will prevent mistakes in later stages that may become frustrating After the pieces are gathered all you have to do is put them together like pieces of a puzzle IghttpwwwchemistryuoguelphcadriguanaMIR2NDORDERHTM 112006 19 P276 IR Mohrig Techniques in Organic Chemistry 2 CASE STUDY cuntzins atoms mher than just hydmgen and carbon Interpret the NMR spectra and flnd the flnal structure Molecular formula CAHQBrO Hr 49 ppm sactet rntegrall 3 73 ppm tnplet rntegral z 3 20 ppm srnglet lntegal 1 l 95 ppm quartet lntegal2 and l 70 ppm doublet lntegal 3 Step 1 The molecular formula ls grven to be ctHaBro so calculate the DBE DBE4cr9H 1h01 ODBE Z alkanes Step 2 Flll ln all othe glvm lnformauon rnto the table 51m Splitting Integratinn x H Pnsslble Fragment sunmne 4 49 sextet l l 3 7 3 trrplet 2 2 3 20 srnglet l l l 95 quartet 2 2 l 7 o doublet 3 3 All ofthe lnformatlon above was glvm ln the problem Before golng on you should see Analyzln WW k k through the problem rather than Lrylng to flll ln the gaps at the very end Andy As seen above LheNMR spectrum rndrcatesthat there are 5 srgaals whlch means 39 Also 39 A AAMAM lulat ulla all mum StevEnHardlnger Chem 140 nmzdmok Wm0Y1 4 2 p zluR Muhng Tachmqulz m organc Chamer 2 21powmr StevEnHardlnger Chem 140 nmzdmok Wm0Y1 4 m whic an only be oxygen because oxygen ean have 2 attaelunents Whereas lunent ato h e a halogen like bromine only has 1 center ofattac t u u Sh sullmug lumgmion 11 rumble magnum Smlrnno 449 sextet 1 1 CH in CH CH 373 iplet 2 2 CH2 in CH CHQQCH 320 singlet 1 1 0H 195 quartet 2 2 CH2 in CH CHCH2CH2 171 doublet 3 3 CH3 in CHE Analysxs e canwn e1 all analysis Hydrogenbonded molecules like NH and 0H are usually singlets so w 39t 39t down as a fragment with con dence 0 be careful 39 quot 39 39 39 39 you have to remember to leave enough openings to attach them Also you know equally considered Choose your pieces and che k 31 SIIIming Lunganon 11 Posuble 17 nglnenr Slnlrnue 4 49 sailet l 1 CH In CHz cHg 3 73 mp let 2 2 CH2 1n gm CHcHaCH 3 20 slnglet l l l 95 quartet 2 2 1 7o doublet 3 3 Totnls 9 11 CHCH2OHCH2CH CngO Atom check CAHgBrO CAH0 Br DBE check There were no DBE s to account for so no need to check Analysxs ofa carbon later on Now all you have to do is add Br to the total number ofmolecular pieces you have and attaen them togetlnen 1339C NMR The 13C NMR is generated in the same fundamental was as proton NMR spectrum Only 11 of naturally occurring carbon is 13C and actually an advantage because of less coupling Requirement for NMR Spin quantum I i 0 Meaning 9 must be an odd number andor neutrons EX 1H 2H 13C 19F etc Thz39nkbook How do we nd this Look at the atomic number The 13C NMR is directly about the carbon skeleton not just the proton attached to it a The number of signals tell us how many different carbons or set of equivalent carbons b The splitting of a signal tells us how many hydrogens are attached to each carbon N1 rule c The chemical shift tells us the hybridization sp3 spz sp of each carbon d Integration Not useful for 13C NMR Proton coupled spectrum shows splitting of the carbon signal only by protons attached to that carbon itself 13CH coupling not 13C 13C H or not 13C713C7 13CiH or not 12C713C couplingoccurs but very low No coupling due to low abundance 11 X111 12 I Thinkbook Thinkbook Thus for each carbon the multiplicity of the signal depends upon how many protons are attached to it Note Due to low natural abundance13C NMR spectra do not ordinarily show carbon carbon splitting two 13C being next to other is 11 x 110012 because 12C does not have a magnetic moment it cannot split the signal of an adjacent 13C and are thus enormously simpli ed Thinkbook Proton Decoupled Spectrum shows no splitting at all it consists of a set of single peaks one for each carbon or each set of equivalent carbons in a molecule Even for very complicated molecules such a spectrum is amazingly simple because overlapping multiplets very difficult to interpretmost commonly run spectrum for structural analysis and will list the multiplicity of the peaks in the upper lefthand corner Bruice Chemical Shift in 13C NMR spectrum arises in the same way as in the proton NMR spectrum Each carbon nucleus has its own electronic environment different from the environment of other nonequivalent nuclei it feels a different magnetic field and absorbs at different applied fields strength 39239 Electronegative atoms and pi bonds cause downfield shifts Thinkbook 39239 13C chemical shift range 0250 ppm Thinkbook In 13C NMR spectrum the more electronegative group bonded to carbon atom 9 deshielding increases This table demonstrates this effect I Br 25 28 Sp carbon 96 256 How many signals are in the 13C NMR spectrum H3JHZJHZJHZJHZJHZJH H2 141 229 321 293 291 341 139 114 sp2 9 larger chemical shift Eight signals no equivalent carbons 291 156 0 f CH2 CH3 6 7 7 CH27C707CH27CH3 1442 1279 b b g 142 ppm b129 ppm c c e 414 ppm c 127 ppm 1257 1284 f 606 ppm d 126 pm Equivalent carbons 8quot 13539ppm On benzene ring 6 signals Note there are two methyl groups and one corresponding to 7CH2 downfield 606 ppm is attached to 0 cause deshielded Benzyl CH2 411 ppm Aromatic ring carbons have resonance over range from 126 ppm to 135 ppm Determine the structure from this formula C4H802 in 13 spectrum 1799 ppm triplet 515 ppm quartet 275 ppm triplet and 92 ppm quartet Thinkbook PP12 1799 ppm corresponding to ester or ketone carbonyl group 515 ppm is down eld must be close to carbonyl or oxygen OCH3 275 ppm CH2 and 92 ppm quartet H3 group further away from carbonyl group in up eld region These are three structures possibilities O CH3iCiOCHziCH3 CH3 CH2C O CH3 CH3iCiCHziOiCH3 Methyl propionate a b c d Note b this C must downfield and probably over 200 ppm o o o o o o o o o o 2 o 180 160 140 120 100 80 60 40 20 o usrnnrn4 ppm Actual strnomre MethylPropionate quotC spectrum above 2HzJHz H 137 221 m 183 845 6 Six signals no equivalent carbons o o o o o o o o o 20 180 160 140 120 100 80 6O 40 20 O rrrrrrrr 9 ppm EHgl leszlezj g note there are sixearbons but 13C a a b b e e NMRshowedoI y3signals a1416pp1n b3187pp1n 02289pp1n duetomoleeulansymnnetry r r r r 20 180 160 140 120 100 80 so 40 20 o r rlli p i H a i i i rpm a r Note i l l l i i H diastereomers spectra are NOT identical Typical 13C illJR Chemical Shift ranges 070 100160 160210 SplitingPattern NH rule for each carbon the multiplicity of the signal depends upon how many protons are attached to it o no nion Hon 7 no proton oneproton two protons three protons singlet doublet triplet quartet 2D llldll lnteraction of nuclear spins plotted in two dimensions ConelationSpectroscopy COST 39139 Two axes conespond to the single isotope Thahkhook l 39139 The interaction indicates with H s are coupled gives better understanding of structure Thialbook Carbohydrates Notes All earbohydrates have the general formula Cannon A m belreved p Ho F F p l h repeaung unrt groups aldoses andketoses al dosesr earbohydrates that eontarn an aldehyde 0 project baekbone of aldoses rn Frseher all aldo b tt cllu lf the OH group attaehedto the see the saeeharrde ls denoted wrth an L ln front of the m lf the OH group a a d the saeeh the number of earbons m th rfthere ar th e ear th the 0 see Carbohydrate you don arrde ls denoted wrth a D m fr on ses have the CHO at the top and the CHon at the o om e van ble gene pornts squotmustbean Lhe number ofhonzontal lrnes ln between ean b a between 14 andrndreate earbons at the eonver e Lhe attaehments ofeach of these mlddle earbon OH on one srde and an H on the opposrte dlfferent monomers have dlfferent onentauons and orders of eaeh of these attaehments OHVH vs Heo Lhe seeondto last c ls the mostrmportant ln terms othe attaehment orderbeeause rt dlfferentlates between aD aldose andL aldose ondto last earbon atom ls on the left name to the seeondto last earbon atom ls on the nght ont ofthe monomername e atoms depends on the monomer bonsrtnose tt ehe ehyde sur earbons and a OH on the nght of us amonome h an ald seeond to last earbon atom would be named Draldohexose Le ture Supplement to see all the varlous 39t eedto memorrze eaeh ofthemijustbe famrlrarwrth the DrAldoses n baekbone stxucture three aldoses thatmay be bene clal to memorlze are glucose galactose and nbose CHO CHD 4 all I o CHO Ho H Ho H H OH H OH Ho H quot OH H OH H OH H OH HIOH HsOH HlOII Drrglucose Drrga1actose Drnbose ketosesr carbohydrates that contaln aketose o backbone ofketoses ln Flscher prolecuon CEO N all ketoses have the cHaOH at the top and a carbonyl at the 39 second carbon thls ls actually lust aketone but ln Flsher ns they are shown ln thls way they all have the cHaOH at the bottom Lhe same rules apply to ketoses as those of aldoses as far as the number of mlddle carbons onentatlon of attachment groups andD vs Lmthe only real dlfference between aldoses and ketoses ls there functlonal group and aldehyde vs aketone thus a merWlth aketose flve carbons and a OH on the le of the second to last carbon atom would be named eretopenthexose For the rest of our dscusslons of carbohydrates we wlll contlnually use glucose as our example bec se thls ls the mostlmportantmonomer and most abundant monomer We wlll see the reason forthls ln allttle blt The nul lhrlum on39t really 39 ln thls form the aldehyde doesn39t emst but ls rather turnedlnto an ether When the rlng formed by a monomerhas lye members lt ls called afuranose When the rlng formed by a monomerhas sllr members ltls called apyranose Uslhg glucose as an ocample ofllnearvs mg equlllbhum CHUH o 3920 OH Hgtc 0 CHIOH oll lt 011 v caaoll cu u cltou t V o thalamus mloll mam Mammoth OH HO Ho ande xhlcuse 0H laaomtpmtem r t a t to form 15 called an ahomenc carbon o Deglucose acycllc then forms a new bond Wthln ltself and becomes Dr glucopyranose the mechahlsm 15 hot lmp oner Just memonze the glucose structure and glucopyranose structure H mall l CH20H group l pll uual are onmted below the rug equatonal posltlohs Thls means that Lhae IS a mmlmum amount of stress k a to alpla Conformatlon and beta Conformation o TF Vh CH20H attachment and anomalc hydroxyl the OH attached the ahomehc carb on are 115 to each oLha then the mg 15 alpha Dlsacchandes e t u a The lmpo znt ohsacchaholes to remember are o maltose l 4 alpha glycoslde llhks between glucose o lactose 14 beta gl coslde llnksbetwem glucose and galactose o sucrose 111 alpha glycoslde llnksbetween glucose and 39uctose Look h thlnkb ook lecmre supplements to see how the bonds look Molecular Structure Introduction and Review Organic chemistm What is it 39 The study of molecules containing carbon Why all this fuss about carbon 39 Carbon is unique in its ability to form stable chains and rings gt millions of molecules known from small set of elements CHON most important 39 Carbon compounds basis for life as we know it Why should I study organic chemistry 39 Broadly applicable to other fields biochemistry pharmaceuticals biology etc 39 Skills learned useful elsewhere information organization critical analytical thinking etc How often should I study organic chemistrV 39 At least one hour on each day of the week that ends in day In Chemistry 14C we expand our knowledge of organic molecular structure by exploring 39 Selected topics in structural theory resonance conjugation aromaticity stereochemistry 39 Laboratory determination of structure spectroscopy 39 Structure controls properties physical chemical biological 39 Reaction chemistry transforming one substance into another covered in Chem 14D Molecular structure electron distribution in bonds in molecule and positions of atoms in space 4 Lecture Supplement Molecular Structure Introduction and Review Molecular Representations How do we represent molecular structure is a bond electron pair shared by two atoms is a lone nonbonded electron pair 39 Carbons do not always have to be shown 39 Hydrogens can be omitted only if carbon not written as C 39 All other atoms must always be shown 39 Lone pairs do not always have to be shown 39 Formal charges must always been shown 39 Three dimensional geometry does not always have to be shown indicates the bond is coming out of the plane of the paper towards the Viewer mIIII indicates the bond is retreating from the plane of the paper away from the Viewer Molecular structure samples H Methane H IZ H or HIIII Major component of natural gas J1 H H Paclitaxel Taxol Anticancer drug omquot E Your Molecular Model Kit A Fun To and a Ve Good Habit 39 Models useful to Visualize and manipulate structure in three dimensions 39 Compare molecular model of molecules in this review with their paper structures 39 Models can be used on exams Do I have to memorize these or any structures 39 The more often you see a molecule the more important it probably is and therefore there is a greater chance you might need to know its structure 39 Examples Methane and glucose are common paclitaxel is not Lecture Supplement Molecular Structure Introduction and Review 5 What do I have to know about svstem for naming 39 You should be able to name or draw from the name simple molecules such as 2 chlorobutane or 3 methylcyclohexanol Review Bruice sections 21 27 The Electron Count Counts Review Lewis structures tutorial at course web site Valence shell electron count 39 H full shell 2 2 same as He 39 CNOF 2 d row elements full shell 2 8 same as Ne T Octet rule Eight electrons and four bonds max C exceptions very rare none in Chem 14C H H Pentavalent carbon very bad 39 3rd row elements easily violate octet rule P often has 10 S often has 12 Formal charge 39 Definition the charge on an atom in a Lewis structure if the bonding was perfectly covalent and the atom has exactly a half share of the bonding electrons The difference between the number of electrons owned by a covalently bonded atom versus the same atoms without any bonds ie a free atom of the same element 39 Significance indicates electron excessdeficiency desire to gainlose electrons electrostatic interaction between regions of charge 39 How to assign formal charges Review formal charge tutorial at course web site Self test Use the following molecules to test your ability to count electrons Verify the formal charges assume full or expanded octets NH2 quotN H2N lt I N O g u u if N H9 9H o Q T Q T P 9 quot Ho OH H9 29H quot quot Sulfuric acid NAD Very strong acid Coenzyme in biological oxidation reactions 6 Lecture Supplement Molecular Structure Introduction and Review Electrons in Bonds 39 Electron distributionsharing can be even covalent bond or uneven polar covalent bond 39 Electronegativity the power of an atom in a molecule to attract electrons to itself High electronegativity 2 strong attraction low electronegativity 2 weak attraction Electronegativity values for elements important to Chem 14C Pauling scale H21 C25 N30 035 F40 P21 225 Cl30 Br28 125 Electronegativity j away from fluorine right gt left top gt bottom diagonally Must I memori e quot itv values Electronegativity values are useful for many things so they must be learned You can avoid their memorization by doing lots of problems and absorb them by exposure Polar Covalent Bonds 39 Uneven electron distribution leads to partial charges If EN X lt EN Y 6 X Y 6 39 Magnitude of bond dipole influenced by Electronegativity difference T EN difference T 6 and 6 ie more polar bond Length T Increasing bond length T 6 and 6 ie more polar bond Example C H bonds almost nonpolar even though AEN 04 due to short length 39 Consequences of bond polarity Electrostatic interaction with other ions or molecules influences physicalchemicalbiological properties Lecture Supplement Molecular Structure Introduction and Review 7 Bond polarity example Polar X H bond gt hydrogen bonding gt DNA base pairs 9quot Hm gt N N 55 N l 67 a NmmmH 539 5 N N Hmmmo CH3 H Adenine Thymine Functional Groups 39 Functional group A characteristically bonded group of atoms that determines molecular properties regardless of what molecule contains it 39 Similar functional groups 2 similar properties I CH and CC bonds and the carbonyl group CO are not considered to be functional groups by themselves 39 Table on page 12 You are responsible for all the functional groups on this table Atomic Positions and Molecular Geometry 39 Atoms 2 balls of electrons gt repel each other gt get as far away as possible without rupturing bond Magnitude of repulsion in uences bond angle 39 When comparing atoms or groups attached to the same atom ie X and Z in X Y Z relative repulsion size H F lt lone pair Cl Br I lt groups of atoms May not apply to other cases ie X Y Y Z Methane H El Four things around central atom tetrahedral 10950 El Equal repulsion between all hydrogens causes all H C H bond Humm H angles to be equal 1095 H Water H El Four things around central atom tetrahedral quot m104 El Lone pairlone pair repulsion gt hydrogenhydrogen repulsion 39 H so H O H bond angle lt 1095 8 Lecture Supplement Molecular Structure Introduction and Review Bonds Molecular Geometry and Orbitals 39 Covalent bond formed by orbital overlap 39 Orbital Mathematical equation that describes a volume of space in which there is a certain probability of finding an electron of a certain energy We can draw ie graph an orbital but it has no physical reality 39 Carbon atomic orbitals s pl py pl The p orbitals are mutually orthogonal do not allow for 1095 CH4 bond angles Building methane using carbon s atomic orbitals gives the wrong bond angles pxIZgt T I71 HVCVH bond angle 90 39 Solution Pauling 1931 Carbon uses combinations of atomic orbitals called hybrid orbitals to give observed bond angles Hybridization schemes Atom with four electron groups lone pairs count Tetrahedral bond angles 1095 need four hybrid orbitals spxpypZ gtsp3sp3sp3sp3 3 H SP 1095O H H39C H Atom with three electron groups H 5 72 H Trigonal planar bond angles 120 need three hybrid orbitals N 1200 sppypZ gtsp2sp2sp2 pZ H H Atom with two electron groups Sp Linear bond angle 180 need two hybrid orbitals HCECH V SPxPyPzquotgt 517 SP PxPz 180 Summary To achieve lowest energy the molecule seeks to minimize destabilizing effects electron repulsion strain etc and maximize stabilizing effects resonance conjugation aromaticity etc The molecule adopts whatever geometry is necessary to achieve these goals Hybridization is used to achieve this geometry Stability causes geometry geometry causes hybridization Lecture Supplement Molecular Structure Introduction and Review Bond Rotation and Molecular Conformations Alkanes 39 Sigma bond overlap of two orbitals along bond axis orbital overlap does not change with rotation so rotation allowed Single bonds can allow significant exibility in molecular structure Example butane conformations CH3 H H H H H H H H CH3 H H H H H H I H CH I ch M H M ch CH3 Anti staggered Eclipsed Staggered Eclipsed Lowest e39 repulsion Most e39 repulsion Lowest torsional strain Most torsional strain Most stable Least stable Strain decrease in molecular stability due to electron repulsion or deviation from ideal geometry Torsional strain strain caused by electron repulsion as atoms move past each other Steric strain strain caused by electron repulsion from atoms that are fixed in space Bond Rotation and Molecular Conformations Cycloalkanes 39 Cyclic molecules can have significant torsional steric and angle strain when flat Nonplanar conformations reduce this strain Verify with models Angle strain strain caused by deviation from ideal bond angles mull C clo ro ane cannot escape planarity CCC bond angle H00 inescapgbk steric and angle strain unavoidable 50 mus log5 H HVH eclipsmg H H Cyclobutane some strain relieved in puckered conformation A Cyclopentane very little strain in envelope conformation Q Cyclohexane very little strain in chair conformation m 10 Lecture Supplement Molecular Structure Introduction and Review Substituted Cyclohexanes 39 Ring ip interchanges axial and equatorial positions 39 Equilibrium favors conformation with equatorial substituent less torsional strain w X is equatorial X is axial Less torsional strain More torsional strain More stable Less stable Bond Rotation and Molecular Conformations Pi Bonds 39 Pi bond results from overlap of parallel adjacent p orbitals 39 Pi bond has lobe on each side of molecular plane 39 Rotation removes p orbital overlap therefore double triple bonds fairly rigid H109 OH H1110 d 1 UH 1 H Ethylene p orbital After 90 rotation Benzene planar due overlap forms pi bond pi bond lost to p orbital overlap Lecture Supplement Molecular Structure Introduction and Review Mass Spectrometry Overview Mass Spectrometry is an analytic technique that utilizes the degree of de ection of charged particles by a magnetic eld to find the relative masses of molecular ions and fragments2 It is a powerful method because it provides a great deal of information and can be conducted on tiny samples Mass spectrometry has a number of applications in organic chemistry including Determining molecular mass Finding out the structure of an unknown substance Verifying the identity and purity of a known substance 1 Providing data on isotopic abundance How a Mass Spectrometer Works 2 heater to vapajurige sample J i Emmi beam imises Ehmgg pml k hm f sample inject sample h quot i V V s z r V r Wiggles r field Figure A3 Step 1 The sample is vaporized and then ionized by being bombarded by a beam of high energy electrons usually at 70 eV The electron beam knocks out an electron from the molecule of the injected sample creating a molecular ion which is also a radical cation because it has an unpaired electron and a positive charge Losing an electron weakens the bond while the collision gives it extra kinetic energy These factors make it more likely for the molecular ion to break into fragments as it travels through the mass spectrometer Step 2 There is a pair of oppositely charged plates in the ionization chamber The positively charged one causes the positively charged radical cation to accelerate into an analyzer tube 1 Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 41 2 Bibliographical source Pg 702 of Atkins and Jones Chemical Principles 3rd Edition and Pg 484 of Bruice Organic Chemistry 4Lh Edition 3 Figure A from httpwwwchemucalgarycacourses351CareyChl 31 334 gif Site visited on 82306 Step 3 The analyzer tube is surrounded by a curved magnetic eld which causes the path of the radical cation to be de ected in proportion to its mass to charge ratio mz The ight path of the ion depends on its molecular mass its charge and the strength of the magnetic eld Thus at a given magnetic eld strength ions of only one speci c mass collide with the detector and are recorded Step 4 The strength of the magnetic eld is varied in increments to produce a mass spectrum which is a plot of mz on the X axis against relative abundance on the y axis If we assume that all ions have a charge of 1 then the peaks give the mass ratios and their heights give the proportions of ions of different masses Aston and Thomson used a process similar to the one outlined above to acquire the mass spectrum of neon in 1919 They observed that neon s mass spectrum had two peaks one at mz 20 90 abundance and one at mz 22 10 but no peak at the atomic weight ofNeon 2020 which is the weighted average Since Neon is inert and lacks chemical bonds the peaks did not correspond to masses of fragments of the molecular ion but to the masses of different isotopes of Neon The eXperiment s conclusion was that neon has two natural isotopes 20Ne 90 abundance and 22Ne 10 abundance 4 This was the rst time that an isotope had been detected for a nonradioactive substance Note that modern spectrometers actually indicate that neon has a third isotope Intergreting a Mass SQectmm 5 J Flgure B 339 39 4 Base CH3CH2CHzBr Peak w Relative ion abundance J ngl ng ngl M2 M mm 3939 39II N g I Masstocharge ratio mz 4 Quoted from Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 215 5 Mass spectrum from httpwww chem ncaloarv vi arev hl 3 ch 3ms html labels have been modified in Adobe Photoshop Site visited on 82406 The mass spectrum of 1bromopropane shown on the previous page illustrates some characteristic features The Base Peak is the peak with the greatest intensity usually set to 100 relative abundance in the mass spectrum corresponding to the most abundant ion M and the base peak are only the same if many of the molecular ions make it to the detector without breaking into fragments Since M1 and M2 are always less intense than M they can never be the base peak M is the molecular ion unfragmented molecule minus one electron made up of isotopes with the lowest mass numbers H1 C etc 6 This peak will most likely be indicated on an exam but is usually the highest peak with the largest mz When we assume that 21 charge is usually 1 M provides the mass in amu of the compound with the lowest mass isotopes To nd the mz for the M peak of a compound refer to the table of natural isotopes Chem 14C Thinkbook Summer 2006 Pg 42 and multiply the isotopic mass not the atomic weight of each atom by the number of atoms of each in the formula EX We expect the M peak of CH3CH2CHzBr should be at 31271179122 This is verified by the mass spectrum above Note that we use the isotopic mass not the atomic weight of each atom M1 is nextdoor to M on the mass spectrum and corresponds to the molecular ion with mass one higher than M due to presence of one atom that is a heavier isotope EX For 1bromopropane the M peak is at 122 so the M1 peak should be at 1221123 Only 13C15N and 33S contribute significantly to the M1 peak 13C is most important If M is scaled to 100 then the M1 intensity1 1 gives an estimate for the number of carbons in the compound BE CAREFUL TO AVOID MATH ERRORS HERE Note about rounding If remainder is 04 or less then round down If remainder is 07 or more then round up It is important to be conservative in mass spectrometry so if remainder is between 04 07 then consider two different carbon counts for formula candidates Some of these may be rejected later based on other data EX If the relative abundance of the M1 peak relative to M is 495 how many carbons are there in the formula ANSWER 49511 45 There are 4 or 5 carbons M2 is nextdoor to M1 on the mass spectrum and has a mass two higher than M EX For 1bromopropane the M peak is at 122 so the M2 peak should be at 1222124 For our purposes the elements that make significant contributions to this peak are 34S 37Cl and 81Br The MM2 ratio indicates the presence of S10044 Cl 100319 or Br100972 in the compound 6 Quoted from Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 214 EX Notice that the M and M2 peaks for lbromopropane are almost the same height revealing that Br is a component of the molecule Getting the Formula from the Mass Sgectrum7 The Nitrogen Rule states that if mz for M is odd then the molecular formula must have an odd number of nitrogens If mz for M is even then the molecular formula must have an even number of nitrogens this includes 0 EX For lbromopropane mz for M122 The even number is in accordance with the even number of nitrogens in the formula zero The Hydrogen Rule states that the maximum number of hydrogens in the molecular formula is 2CN2 In the formula C of carbons N of nitrogens EX For CH3CH2CHzBr there are three carbons so the maX of hydrogens is 2328 Getting the 39 39 Structure from the Formula One Double Bond Equivalent DBE also known as a degree of unsaturation is one pi bond or one ring A triple bond counts as 2 DBE Having 4 DBE indicates the possibility of a benzene ring since benzene has three pi bonds plus one ring The formula for DBE is the following C of carbons DBE C HZ NZ 1 H of hydrogens or halogens N of nitrogens Formula8 Important Note DBE can never be negative and can never be fractional Beware of math errors EX For CH3CH2CHzBr the DBE equals 38202l0 No pi bonds no rings Putting it All Together EX Use the given mass spectrometry data to determine the molecular formula M 59 100 Ml 60 385 ANSWER 7Nitrogen and Hydrogen Rules are outlined in the Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 4445 8 Formula from Chem 14C Course Thinkbook by Professor Steven Hardinger Summer 2006 Version Pg 4445 Stereochemist The Basics of Solvin Problems Important Terms 1 Optically Active optically active compounds rotate plane polarized light Figure 11 Figure 11 Nonpolarized light hitting a cwstal j 39 L which is optically active and becoming polarized 1 light lying in the same plane 2 Dextrorotatory compound that rotates plane polarized light clockwise Figure 12 A Figure 12 a dextrorotatow compound being rotated clockwise 3 Levorotatory compound that rotates plane polarized light counterclockwise Figure 13 A Figure 13 A levorotatory compound being rotated counterclockwise 4 Stereocenter generally a Carbon with four different attachments Figure 14 F Cl Figure 14 The molecule on the left is a F C stereocenter since it has FOUR different C attachments while the one on the right has C only THREE different attachments H Br H H 5 Stereoisomer isomers that differ only in the position of atoms in space and cannot be interconverted by rotation around a sigma bond 6 Enantiomer a pair of molecules that are non super imposable mirror images 7 Diastereomer any two molecules that are not enantiomers 8 Chiral an optically active molecule that is not super imposable with its mirror image chiral chiral enantiomer 9 Racemic Mixture an equimolar mixture of enantiomers Note Do not use the two terms configuration and confirmation interchangeably there is a difference Configurations take into consideration of whether the attachments in a molecule are going clockwise or counterclockwise R or S while a conformation quotare the different positions that a molecule can twist into 2 How to find a stereocenter It s pretty simple to locate a stereocenter Only one rule should be kept in mind and that is that the Carbon atom has to have FOUR DISTINCT ATTACHMENTS As long as you keep that in mind you should be fine now here are a few examples to go through Cl This Carbon is not a stereocenter because it has two propyl attachments which violates the rule of having four different attachments Diagram taken from Chemistry 14C Thinkbook Steven Hardinger page 32 2 Organic Chemistry as a Second Language David R Klein page 134 Cl c This Carb 3 stereocenter since It has four dlstlnct attachments b CH3 How to determine configuration of a stereocenter O I CahnIngoldPrelog Priority Rules Rules that assign priority to atoms or groups for various purposes such as labe ing the absolute configuration of a stereocenter Lowest priority is assigned to the atom or group wit the lowest atomic number3 The three basic rules of finding a configuration 39 stereocen er 2 Assign priorities through highest atomic numbers CahnIngoldPrelog Priority Rule 3 Put number four to the back Note it is especially helpful to use a model kit when determining stereocenters and finding configurations 0H 4 First Locate the stereocenter Cl 2 Now assign priorities to the different attachm ts b 39 it an Chlorine would e num er one sI ce has the atomic 39 h 2 n on number then Oxygen wit 4 OH Once the priorities are assigned make sure the Lowest one 4 is sent c to the back The hydrogen which has the number 4 priority is already Cl in the back with the dashed wedge 1 3 Z 4 H OH Once priorities have been determined follow these rules to determine if it is in the s or R configuration Since this molecule is going clockwise in a 123 Cl AA pattern it is in the R configuration 3 R Configuration lJnvigulauon 59 E a Clockwi l uuhiun m kmse 3 Chemistry 14c Thinkbook Steven Hardinger page 194 Drawing Enantiomers and Diastereomers Enantiomers these are basically two compounds that are nonsuper imposable mirror images It means no matter how you place them they will never be imposed on top of each other Just think of it as your hands they are mirror images of each other But when you try to stack one hand on top of another it doesn t fit That s the gist of it Note When drawing an enantiomer of a molecule with wedges the solid wedges become dashed wedges and vice versa quotI ll0H a Draw the entantiomer of Wm Remember that when drawing an enantiomer the wedges turn into dashes and dashes turn into wedges So the H which was originally wedges turns into a dash and the OH which was originally dashed turns into a wedge Think of the blue box as the mirror itself Diastereomers Diastereomers are different form enantiomers because enantiomers only come in twos while a whole group ca be considered diastereomers A molecule has to have more than one stereocenter to have diastereomers Only one stereocenter in a molecule is limited to only an enantiomer a Draw a diastereomer of OH H Remember the quotone stereocenter rule This molecule has only one stereocenter and therefore can only have an 3 enantiomer which means no diastereomers o v pKa Values of Selected Compounds 39 In structures with more than one proton type proton corresponding to given pKE is in bold 39 Values in parentheses are alternate pKE values given by some sources Inor anic ineral Acids HSbF6 lt 12 H00 10 H1 10 H2804 pKul 399 395 PKuz 20 HBr 9 58 HCl 7 39 HNO3 13 H3PO4 pKM 21 pKHZ 72 pKu3 123 HF 32 H20 157 Oxonium Ions O H 39 H H H O Co 430 o H OCH3 OH 8 73 65 61 9 43 9 9 CH9on GH30H22OH CH3OH2 CH3CH20H2 38 36 25 24 430 H3O NH2 17 00 Carboxylic Acids BCOQE CF3COZH c13cc02H HOZGCOZH CIZCHCOZH 02 064 pKnl 12 pan 42 13 COZH A H3N COZH FCHZCOZH ClCHZCOZH COZH pKal 19 pKaz 61 22 27 28 Lecture Supplement pKn Values of Selected Compounds 15 c1 ACOZH 283 HOZCHZCOZH pKn1 29 pKE 57 O CH3 O COZH 35 CICHZCHZCOZH 398 HOZCCHZ3COZH pKn1 43 pan 54 HOZC cozH pKn1 35 pan 46 BrCHZCOZH 29 ICHZCOZH 32 HOZCOCOZH pKn1 35 pan 48 42 H3COQCOZH 45 OCOZH c02H pKn1 29 pKE 54 HOZCCHZZCOZH pKn1 42 pan 56 CH3COZH 478 CH3CHZ8COZH CH3CHZ5COZH CH3CHZ14COZH HZC03 484 489 65 pK1 65 pK2 102 PhenolsgArOHL No2 No2 N02 OZN OH 02N070H OH OZNGOH No2 078 40 72 72 c1 1 11 a OH OH OZN OH c1 OH 81 83 88 93 16 Lecture Supplement pKn Values of Selected Compounds 03011 96 CH3 Gm H2 102 9 100 N 3 103 Aminium 1 B3NH and Iminium 1 B2CNH Ions Cam 9 H3C NH3 HZN NH3 81 e HOCHZCHZNH3 95 OZN NH3 10 Pr H 47 e Mq w 639 Z 3 H 52 HgCNH 6 00 N z C gt9 w e G OZCCHZNH3 97 Br NH3 lt2 39 4 3 e Z H3CO NH3 6 H3 NACOZCH3 775 e CHZNH3 93 6 CH33NH 98 e c1 lt gt NH3 e NH2 80 NH4 94 Xe 1045 Lecture Supplement pKn Values of Selected Compounds 3 NH3 NHs 1049 1056 CHH3 vNH3 1065 1067 a 6 CH3CH2NH3 CH30H2gt2NH2 1076 1098 w 1 12 Alcohols QBOHL O L OOH CF3CHZOH 82 124 CH3OH H20 155 157 OOH X OH 18 18 19 H3 1061 09 NH3 1067 9 CH3CH2CH22NH2 1098 HCECCHZOH 135 CH3CHZOH 16 17 Carbon 1 Com ounds xDe rotonation to Form Enolate O O O G CH3CH2CH23NH 1064 gt NH3 1073 lt NH2 1115 CH33NCH2CHZOH 139 ya 17 O O MOCHZCH3 CH3CHZOMOCH2CH3 110 H H 133 Lecture Supplement pKn Values of Selected Compounds o O o H PhiCH3 HsciH H 155 159 17 o o o ykocm H3cJOCHZCH3 H3CJkNCH32 H H 22 23 245 30 Amides 10CNH o o o 9 15 16 Amines QB2NH ij E5 QH E 11 89 144 17 H N Y Y NH3 CH3NH2 36 36 38 40 Sulfur and Selenium Com ounds o O F3C S OH Gig OH CH3SH2 o 391 16 65 to 06 68 HZSe H28 08H 389 70 78 s o H II CH3SH gtlt lt S H H3c SCH3 107 3 1 3 13 19 20 CNH 36 if H3C OH 0 CH3CHZSH 105 Lecture Supplement pKn Values of Selected Compounds Introduction and Review 1 A Brief Introduction source Lecture Supplement Introduction and Review pg 1 a Organic Chemistryfocuses on carbon containing molecules b Special Properties of Carbon 39 Able to form stable chains and rings 39 Basis of life important to biology 2 Molecular Structure source Lecture Supplement Introduction and Review pg 2 Molecular structure has two components 39 Electron distribution bonds 39 Positions of atoms in space Representing Molecular Structure Drawing Diagrams 39 Carbons are not drawnin 39 Hydrogens are omitted if carbons are not shown 39 Any other atom is shown by its symbol 39 Lone pairs are not always shown 39 Formal Charges are always writtenin 39 Solid wedges indicate that the bond is pointing out of the page 39 Hatched wedges indicated that the bond is pointing into the page Example 1 a Label all of the missing hydrogens in Vitamin E b Count the lone pairs c Give the formula CHsH H H30 H Vitamin E without hydrogens a Vitamin E with hydrogens b 4 lone pairs two pairs on each oxygen c Formula C29H5002 3 Counting Valence Shell Electrons 39 H full shell has 2 electrons 39 2quotd row elements CNOF full shell has 8 electrons o Octet rule atoms can have 8 valence shell electrons resulting in a maximum of four bonds 0 Watch out for PENTAVALENT carbons 39 3rd row elements easily violate the octet rule ie P10 812 4 Assigning Formal Charge source Bruice Section 14 pg 13 Formal Charge the dz rence between the number of valence electrons of a free atom and the number of electrons belonging to that atom when it is bonded EC VL12 B Vvalence electrons Llone electrons BBonding electrons Example 2 Calculate the Formal Charge on oxygen that is part of the ether group in Vitamin E see Example 1 FC VL 7 B v 6 L4 134 FC 64 12 4 FC 0 5 Electron Distribution sources Bruice Section 13 pg 10 and Lecture Supplement Introduction and Review Hardinger pg 4 39 Even sharing of electronscovalent bond 39 Uneven sharingpolar covalent bond Electronegativity an atom s ability to attract electrons from other atoms in the same molecule Using the Pauling Scale we have FOClNBrISCPH F40 035 Cl30 Br28 125 P2l N30 825 H2l C25 39 Fluorine is the most electronegative atom on the periodic table 39 Electronegativity decreases from right to left and from top to bottom on the periodic table 6 Bond Polarity Lecture Supplement Introduction and Review Hardinger pg 45 39 Bond Polarity An uneven sharing of electrons yielding partial charges 7 or 7 39 More polar bonds larger electronegativity difference between atoms results in a dipole of higher magnitude The bond length between the atoms ampli es the affect of electronegativity thus increasing the magnitude of the dipole Bond polarity in uences the physicalchemicalbiological properties of a molecule 7 Molecular Geometry Orbitals Hybridization Bond Rotation Newman Projections Lecture Supplement Introduction and Review Hardinger pg 57 a Hybridization by Pauling involves COMBINATIONS of atomic orbitals 39 Stability causes geometry geometry causes hybridization Lecture Supplement pg 6 39 Hybridization is a RESULT of geometry NOT a CAUSE of geometry 39 The repulsion of electrons affects bond angle 39 Lone pairs count in hybridization schemes Shape Hybridization Number Bond Angle of electron groups Linear sp sp px pZ 2 1807 Trigonal Planear sp2 sp2 sp2 pZ 3 120 Tetrahedral sp3 sp3 sp3 sp3 4 1095 b Strajn a decrease in the stability of a molecule caused by 39 Angle Strain a deviation from ideal bond angles which results in strain on the molecule 39 Torsional Strain strain resulting from electron repulsion when groups of atoms get too close to one another 0 Cyclic molecules reduce torsional stain by adopting nonplanar conformations For example the chair conformation of cyclohexane c Chair Conformation Cyclohexane 39 Ring ip axial 7 equatorial 39 Equatorial groups are preferred over aXial groups because they have less torsional strain and therefore provide an increase in stability


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