SolvingSpectroscopyProblems:PuttingitAllTogether
Onceyou’veanalyzedthemassspectrometry,infraredspectrometry,1H‐NMR,and13C‐NMRdata,thereisnoonewaytoputthemtogether.It’sallabouttrialanderror,buthereareafewhelpfultipsandtrickstohelpyougetontherighttrack.First,here’sanoverviewoftheinformationyoushouldhaveobtainedbasedonyouranalysesofthedatafromthe4methodsofspectrometry.MassSpectrometry‐molecularformula,DBEInfraredSpectrometry‐presentfunctionalgroups1H‐NMR‐backbonestructure(C‐H)13C‐NMR‐hydrogensattachedtothecarbons,confirmsfunctionalgroupsfoundinIRandimplicationsdeterminedbyH‐NMRKeythingstorememberasyouproceedwiththepuzzle:
1) Keeptrackofthenumberatomsinyourmolecule.Makesurethesumoftheatomsinyour1H‐NMRimplications+functionalgroupsfromIRspectrummatchesthemolecularformulaobtainedfromthemassspectrum.
2) KeeptrackoftheDBEcount,especiallywhendoingIRanalysis.3) Checkthatyourimplicationsandyourfinalmoleculematch1H‐NMRand13C‐
NMRchemicalshifts.4) Checkthatyourfinalmoleculematchesthenumberofsignals,splittingpattern,
andnumberofhydrogenatomsasthe1H‐NMRand13C‐NMRdataTheinformationfromthemassspectrumisprettystraightforward.UseittokeeptrackofthenumberofeachatomasyouproceedwiththeIRanalysisandthe1H‐NMRspectrum(numberofcarbonsandhydrogens).Here’sahandytablethatwillhelpyououtwithdeterminingthefunctionalgroupassociatedwiththegivencarbonylgroup:
FunctionalGroup IR 1H‐NMR 13C‐NMRAldehyde 2C‐Hpeaks:~2900and
~2700cm‐1C=Ostretch1740‐1720cm‐1
Aldehydeproton9.5‐11ppm
C=Ocarbondoublet180‐220ppm
Ketone C=Ostretch1750‐1705cm‐1
Nocharacteristic1Hchemicalshift
C=Ocarbonsinglet180‐220ppm
Ester C=Ostretch1750‐1735cm‐1
Nocharacteristic1Hchemicalshift
C=Ocarbonsinglet160‐180ppm
Carboxylicacid BroadO‐Hstretch3000‐2500cm‐1
C=Ostretch1725‐1700cm‐1
COOHproton10‐13ppm
C=Ocarbonsinglet175‐185ppm
Amide N‐Hstretch3500‐3300cm‐1(2peaksif1°,1peakif2°,noneif3°)C=Ostretch1690‐1650cm‐1
N‐Hproton5‐9ppm C=Ocarbonsinglet160‐180ppm
Dr.Hardinger’sChemistry14CThinkbook,NinthEditionFirst,listoutallimplications,functionalgroups,andremainingatomsthatyouhavefoundfromtheIR,1H‐NMR,and13C‐NMRdata.Consolidateasmany1H‐NMRimplicationsfromdifferentsignalsaspossible(willbediscussedshortly).Oncethat’ssettled,agoodstartingplaceistodrawoutthefunctionalgroupsfoundintheIRspectrumsuchas: Then,usetrialanderrortoattachtheremainingpieces(1H‐NMRandIRfunctionalgroups).For1H‐NMR,herearesomehintsandpatternsthatoccurfrequently(butnotalways):
Butfirst,here’satableofthemorecommonimplicationsandtheonesthataremostlikelytomakeuppartofthefinalmolecule(i.e.onethatshowupintheThinkbookandpracticeexams).Itdoesnotshowallpossibleimplications.………theseareoftenoverlooked,butkeeptheseimplicationsinmindwhenthereareoxygens(alcoholsmorespecifically)andnitrogensinthemolecularformulaandwhenalcohols,amines,carbonyls,carboxylicacids,andaldehydesarepresentittheIRspectrum……… theseatomsrepresentthecarbonswiththecorrectnumberofhydrogensindicatedbytheintegrationShift Signal Integration Numberof
HydrogensImplications
singlet 123469
123469
CH,OH,NH,C=O,CHO,COOHCH2,CHx2CH3CHx3CH2x2,CHx4CH3x2,CH2x3,CHx6CH3x3,CHx9
Doublet 123469
123469
CHCHCHCH2CHCH3
CHCH2,CHCH3x2CHCH3x3
Triplet 123469
123469
CH2CH,CHCHCHCH2CH2,,CHCH2CHCH2CH3,CHCH3CHCH2CH2x2,CHCH2CHx2CH2CH3x2,CHCH3CHx2CH2CH3x3,CHCH3CHx3
Quartet 123469
123469
CH3CH,CH2CHCHCH3CH2,,CH2CH2CHCH3CH3,CH2CH3CHCH3CH2x2,CH2CH2CHx2CH3CH3x2,CH2CH3CHx2CH3CH3x3,CH2CH3CHx3
Pentet 123469
123469
CH3CHCH,CH2CHCH2CH3CH2CH,,CH2CH2CH2
CH3CH3CH,CH2CH3CH2
CH3CH2CHx2,CH2CH2CH2x2CH3CH3CHx2,CH2CH3CH2x2CH3CH3CHx3,CH2CH3CH2x3
Sextet 12
12
CH3CHCH2,CH2CHCH2
CH3CH2CH2,CH2CH2CH3
3469
3469
CH3CH3CH2,CH2CH3CH3
CH3CH2CH2x2CH3CH3CH2x2CH3CH3CH2x3
Septet 123469
123469
CH3CHCH3
CH3CH2CH3
CH3CH3CH3
CH3CH2CH3x2CH3CH3CH3x2CH3CH3CH3x3
Multiplet 5 5 C6H5monosubstitutedbenzenering 2
Doublets4 4 C6H4disubstitutedbenzenering
Multipletsand“doubletsofdoublets”areusuallybenzeneringsforourpurposes.ImplicationsoftheformCH3x2or3areoftenpresentinmoleculeswith2or3CH3groupsattachedtothesameatomorshowsymmetryinotherwaysacrossthemolecule. i.e.CH3x2: CH3x3:Oftentimesyoucancoupletogetherimplicationsforseveralsignalsiftheycoincidewitheachother. i.e.atripletCH2CH3andaquartetCH3CH2canbeputtogetherintoonesingle piece:CH2CH3Whenyouneedtoseparateimplicationsfordifferentsignalssothattheydon’tcouple,tryinsertingtheseatoms/moleculesinbetweenthem: O C=O N i.e.CH3CH2CH2CH2CH2 Gives:2triplets,2pentets,1sextet whereas CH3CH2OCH2CH2CH2 Gives:3triplets,1quartet,and1pentetNowlet’sputitalltogetherwithsomepracticeproblems!13C‐NMR,2D‐NMR,andMRIOWL#4.SuggestthestructureforamoleculeofformulaC10H14OwiththefollowingNMRdata.Formula:C10H14O DBE=10–(14/2)+(1/0)+1=4(probablybenzenering)
1H‐NMR:7.12ppm(doubletofdoublets;integral=2),6.82ppm(doubletofdoublets;integral=2),3.74ppm(singlet;integral=3),2.84ppm(septet;integral=1),and1.21ppm(doublet;integral=6).13C‐NMR:157.9ppm(singlet),141.1ppm(singlet),127.3ppm(doublet),113.9ppm(doublet),55.2ppm(quartet),33.4ppm(doublet),and24.2ppm(quartet)AftercompletionoftheIRandNMRanalyses,youshouldhavethefollowingpieces: 1H‐NMR: C6H4disubstitutedbenzenering CH3singlet CH(CH3)2 CH3CHx2 13C‐NMR: 157.9ppmsinglet–benzeneringC 141.1ppmsinglet–benzeneringC 127.3ppm,113.9ppm–2CHonbenzenering 55.2ppmquartet–CH3(probablydirectlyattached tobenzenering) 33.4ppmdoublet–CH(probablydirectlyattachedto benzenering) 24.2ppmquartet–CH3KEEPTRACKofthenumberofatomsinyourmolecule! C6H4+CH(CH3)2+CH3=C10H14 What’sleft:O
1) startbydrawingthebenzenering:
2) Sincethebenzeneringisisubstiutedthereare2carbonsonitthatcanconnecttothe2remainingpiecesofthepuzzle.These2piecescanbearrangedin3possibleways:
Sincethereare413C‐NMRsignalsattachedtothebenzenering.Onlythepara isomergivesthisarrangementfitsourstructure.
3) Sincewestillhaveanoxygenleft,itcanattachtoeithertheCHortheCH3singlet.Lookatthe1H‐NMRchemicalshiftandyouwillseethattheCH3singlethasthehigherchemicalshift,soitisprobablyattachedtotheoxygen.So,attachtheCH(CH3)2ononepositiononthebenzeneringandtheCH3Oyoujustmadeontheotherposition.
CH(CH3)2 HereI’vematchedtheprotonandcarbonNMRimplicationsbyhighlightingthemwiththesamecolor.Youcanseethat13C‐NMRconfirmsthe1H‐NMRimplications
Makesuretocheckyouhavealltherequiredatomsandcheckforthecorrect numberofNMRsignalsandthecorrectchemicalshifts. Andthereyouhaveit!Let’stryaharderone:13C‐NMR,2D‐NMR,andMRIPracticeProblem#16:Deducethestructurethatcorrespondstothespectraldata.MassSpectrum:m/z242(M:100%),m/z243(17.87%),andm/z132(0.50%) Solveforformula:C16H18O2
DBE=8IR:alcoholO‐H,aryl/vinylsp2C‐H,alkylsp
3C‐H,benzenering1H‐NMR:7.24‐7.14ppm(multiplet;integral=5),3.62ppm(triplet;integral=1),2.50ppm(singlet;integral=1)1.68ppm(pentet;integral=1),and1.51ppm(triplet;integral=1)13C‐NMR:142.3ppm(singlet),128.4ppm(doublet),128.3ppm(doublet),125.7ppm(doublet),78.5ppm(singlet),62.0ppm(triplet),39.5ppm(triplet),and27.2ppm(triplet)Pieces: 2xC6H5monosubstitutedbenzenering CH3CH2CHorCH2CH2CH2 CH2CH2
2xOH CH2CH2leftovers:O,C
1) Sincethe2benzeneringsareequivalent(samesignal),drawoutthebenzeneringsnexttoeachothersotheycansomehowbeequivalent.
2) Itseemsyoucan’tattachthemwiththeremainingpiecesinbetweenthem,
linkingthemtogetherastheCH2CH2CH2,C,andOcannotformasymmetricchainonitsown.TheleftoverCmustattachthetworings.
3) TheleftoverChas2benzeneringsattached,soithas2morebondstoform.OnecanbetheCH2CH2CH2chain.
(thisonefitsbetterwiththeotherimplications‐CH2CH2)
Putthese3togetherCH2CH2CH2
4) TheothermustbeanOHasthatisallwehaveleft.
5) TheotherOHmustcapofftheendoftheCH2CH2CH2chaintocompletethemolecule.
Andsince“ThetwoendsofthisCH2CH2CH2piecearenotequivalent”the moleculemustbeconstructedthisway.(Dr.Hardinger’sChemistry14C Thinkbook,NinthEdition) Rememberagaintocheckyourwork‐molecularformula,IRfunctionalgroups, NMRsignals,integrals,andchemicalshifts.References:Dr.Hardinger’sChemistry14CThinkbook,NinthEdition(pgs.217,221‐222,229,233‐235)http://www.google.com/imgres?q=functional+groups+with+carbonyl+group&um=1&hl=en&client=safari&rls=en&biw=1139&bih=680&tbm=isch&tbnid=aOKDVE2NLZKKqM:&imgrefurl=http://chemcases.com/nutra/nutra1b.htm&docid=1MbiE9tMRQOpmM&imgurl=http://chemcases.com/nutra/images/fig1‐06.jpg&w=602&h=721&ei=VgbAT8GgLcWTiQKJuPWZCA&zoom=1http://www.google.com/imgres?q=para+meta+ortho+benzene&um=1&hl=en&client=safari&sa=N&rls=en&biw=1139&bih=680&tbm=isch&tbnid=KArztwsBDmPf3M:&imgrefurl=http://www.chemistry.ccsu.edu/glagovich/teaching/311/content/nomenclature/aromatic/aromatic.html&docid=gnnv_DZHQ3IZsM&imgurl=http://www.chemistry.ccsu.edu/glagovich/teaching/311/content/nomenclature/aromatic/images/sysnameb1.gif&w=575&h=181&ei=JHLAT6GjJuXdiAKXjfHyBw&zoom=1&iact=hc&vpx=294&vpy=193&dur=15&hovh=126&hovw=401&tx=136&ty=95&sig=114357793350899697903&page=1&tbnh=75&tbnw=238&start=0&ndsp=15&ved=1t:429,r:1,s:0,i:75http://www.google.com/imgres?q=benzene&um=1&hl=en&client=safari&rls=en&biw=1139&bih=680&tbm=isch&tbnid=ptp3zroofLsWhM:&imgrefurl=http://bond‐ashleyandlauren.blogspot.com/2011/05/alicylics‐aromatics.html&docid=_seNoTRwc7M1pM&imgurl=http://4.bp.blogspot.com/‐Y6KOQloTUtw/TdyYzb5AC‐I/AAAAAAAAAEw/oHGYdE‐xszc/s1600/71432.gif&w=444&h=500&ei=6XTAT4GVNabhiALm‐Mj8Bw&zoom=1&iact=hc&vpx=127&vpy=324&dur=402&hovh=233&hovw=208&tx=114&ty=166&sig=114357793350899697903&page=1&tbnh=141&tbnw=130&start=0&ndsp=18&ved=1t:429,r:6,s:0,i:137
