ALL ABOUT SOLVING PROTON-NMR

Whatis1HNMR?1HNMRisNMR(NuclearMagneticResonance)spectroscopyinwhichtheprotonnuclearspinismanipulated.Itrevolvesaroundtheideathattheenergyrequiredtocausenuclearspinflipisafunctionofthemagneticenvironmentofanatomsnucleus.HowcanwefindstructurefromNMR?WefindstructurethroughtheinformationNMRrelaystous.Thesepiecesofinformationinclude:

Numberofsignals[whichisthenumberofnonequivalentprotonsetsinamolecule] Positionofsignals[alsoknownaschemicalshift,whichisthemagneticenvironment

inwhichprotonsarein] Relativeintensityofsignals[alsoknownasintegration,whichistheratioof

equivalentprotontypes] Splittingofsignals[alsoknownasspinspincoupling,whichistheamountofproton

neighborspresent]HowtosolvefornumberofsignalspresentThenumberofsignalsdependsontheequivalencyofprotons.Protonsareequivalentiftheyhavenucleiwiththesamemagneticenvironment.Inotherwords,theyhavetobethesameineverypossibleway.Itiseasiertofindnonequivalentprotons.

http://www.chem.ucla.edu/harding/index.html

Withinthemoleculedepictedabove,therearethreedifferentsignalspresent,inthattherearethreedifferentsetsofprotonswithdifferentequivalencies.Ha,HbandHcareallnonequivalenttooneanother,thusproducingthreeseparatesignalsontheNMRspectrum.Howtosolveforthepositionofsignals/chemicalshiftChemicalshiftisdefinedasthepositionofasignalinanNMRspectrum.Itisdependentondifferentinfluences,suchastheEofthespinstateenergy(whichiscontrolledbythemagneticfieldatthenucleus),andelectronegativity(aschemicalshiftincreases,electronegativityofatomsneartheprotonincreases).ThepositionsofsignalswillbegiventousonatestsoITISNOTMANDATORYTOMEMORIZETHEM!

TheprotonslabeledHaareequivalentbecausetheyallare

connectedtoacarbonthatisconnectedtotwohydrogen

atoms.Theyarethesameineverywaypossible.

Thistablewillbegiventousonanexam.

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ChemicalShiftTrend RCH3

HowtosolveforthesplittingofsignalsSpinspincouplingisdeterminedbythenumberofmagneticfieldsaffectingthenucleusofamolecule.ThiscausessplittingoftheNMRsignals.Thegeneralruleforsplittingisthesignalforaprotonwithnneighborsissplitinton+1lines.Therearesomerulesandrestrictionswemustfollowwhenworkingwithsignalsplitting:

1. Onlynonequivalentprotonscancouple Ifprotonsareequivalent,thentheydonotcouple

2. Protonsthatareseparatedbythreeormoresinglebondsusuallydonotcouple

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Pibondsdonotcounttowardsthebondlimit,butthesplittingconstantJ(whichisthespacingbetweenlinesinasplittingpattern)maybetoosmalltosee,soitmaybethere Forpurposesofcoupling,justpretendthepibondisntthere[ITSAFREE

SPACER]!Tobesafe,adda+1totheamountofbondscountedexcludingthepibond.

Benzeneringscountasonebigfreespacer Allprotonscouplewitheachotherbecauseofresonance Jmaybesmall Thinkofbenzeneasagatedcommunityhydrogenatomsbondedtothe

benzeneringcanonlycouplewithotherhydrogensbondedtothebenzenering.Theycannotbondwithotherhydrogenspresentinthemolecule

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3. SignalsforOHandNHareusuallysinglets

Inthisexample,HaandHddonotcouplebecausetheyarefourbondsapart(seeredbonds).However,HaandHccancouplebecausetheyaretwobondsapart(seegreenbonds).

Thehydrogenatomspresentinthemethylattachedtothebenzeneringcannotcouplewiththehydrogenatomsbondeddirectlytothebenzenering.Thehydrogenatomsbondeddirectlytotheringcanonlybondtootherhydrogensattachedtothering.

Example of how to solve an 1H-NMR Spectroscopy problem

Formula:C6H14OHNMR:3.3ppm(triplet;integral=1) 1.6ppm(sextet;integral=1) 0.90ppm(triplet;integral=1.5)1. ThefirststepistocalculatetheDBE(doublebondequivalency)value,whichwilltellushow

manypibondsarepresentorifthereisapotentialbenzeneringpresentinthestructure.DBE=#ofcarbons(#ofhydrogens2)+(#ofnitrogens2)+1=6(14/2)+(0/2)+1=0(thismeanstherearenopibondswithinthestructure)

2. Knowingthattheintegralisproportionaltotherelativenumberofequivalenthydrogenatomspresentwithinthestructure,wecanusetheintegralinformationgivenintheproblemandthenumberofhydrogens,alsogivenintheproblemthroughthemolecularformula,tofindthefactorinwhichtomultiplytheintegralbysoastodeterminehowmanyhydrogenatomsarepresentwithineachsignal.

Thefirstsignalhadanintegral=1Thesecondsignalhadanintegral=1Thethirdsignalhadanintegral=1.5

Ifyoumultiplyeachofthesenumbersbyafactoroffour,thenyouget:4hydrogenatomsforthefirstsignal4hydrogenatomsforthesecondsignal6hydrogenatomsforthethirdsignal

3. Nowyoucandeterminetheimplicationsforeachintegralandsplittingpattern.

Beforeyoufindeachimplication,youneedtofindthenumberofneighborseachsignalhas.Thisisrevealedthroughthesplittingpatterngivenintheproblem.

Thefirstsignalisatriplet.Fromthegeneralrulestatedabove,thesignalforaprotonwithnneighborsissplitinton+1lines.Becausethereare3splitsinthesignal,thismeansthatthereare31neighbors,whichequalstwoneighbors.Thesecondsignalisasextet.Thismeansthatthereare61neighbors,whichequalsfiveneighbors.

Wecandeducealotfrom1HNMR,especiallythemolecularstructureofamolecule.Hereisanexampleofatypical1HNMRspectroscopyproblemwhenthemolecularformulaisgiven,andthestepsyoutakeinordertosolveit.

Addingalloftheseup,weget14hydrogenatomspresent,whichisequivalenttotheamountofhydrogensgivenintheformula,soweknowwefoundtherightfactortomultiplyby.

Thethirdsignalisatriplet.Thismeansthatthereare31neighbors,whichequalstwoneighbors.

Nowyoucanfindtheimplications,whichdealwiththeCHskeletonofamolecule

(orNH/OHiftheyarepresentintheIRspectrum/molecularformula)FirstSignal:hastwoneighborsmeansthattheunderlinedCHpartoftheskeletonthatwearelookingathastwoprotonsasneighbors 2x:CH2CH2 2x:CHCH2CH

4x:CHCH2 4x:CHCHCHSecondSignal:hasfiveneighbors 2x:CH3CH2CH2 4x:CH3CHCH2 4x:CH3CH(CH)2

ThirdSignal:hastwoneighbors 2x:CH3CH2 3x:CH2CH2 3x:CHCH2CH 6x:CHCH2 6x:CHCHCH

4. Afterfindingtheimplications,youthenlookatallofthemandfindthemostlikelyimplicationthatwouldmakeupamolecularstructure[MOSTLIKELYSTRUCTURE=LEASTNUMBEROFATOMS].Alwayslookforthesimplestimplications!Additionally,lookforimplicationsthatcanbeeliminatedbasedonvalidity(YOUCANNOTHAVEACARBONWITHMORETHANFOURBONDS!).

FirstSignal:2x:CH2CH2 2x:CHCH2CH

SecondSignal2x:CH3CH2CH2 4x:CH3CHCH2

4x:CH3CH(CH)2

ThirdSignal:2x:CH3CH2 3x:CH2CH23x:CHCH2CH 6x:CHCH2 6x:CHCHCH

Twoneighboringprotons

Twototalneighboringprotons

Twoneighboringprotons

Twototalneighboringprotons

**The2xand4xbeforetheCHskeletonrefertothefactorinwhichyoumultiplythenumberofhydrogenatomsunderlinedby.ThiswillgiveyouthetotalnumberofequivalenthydrogensyoucalculatedbeforeinStep2.

Thesecannotbepossibleimplicationsbecausetheyaretheonlyimplicationsthataremultipliedbyafactorof3.Theythereforecanbediscarded.

Thesecannotbepossibleimplicationsbecausetheyformhexavalentcarbons,andcarbonscanonlyformfourbonds.

BothoftheseimplicationshaveCH3CH2inthem,sowecanassumetheyarepartofthemolecularstructure.

BecausetheimplicationscircledgreenbothhaveCH3CH2inthemandcanthereforebeconfirmedaspartofthestructure,wenowneedtofindafinalimplicationthatwillcoincidewiththem.Becausethisimplicationisverysimilartotheoneswehavealreadypickedout,andbecauseitcontainsaCH2CH2,justliketheimplicationfromthesecondsignal,wecanthereforeconcludethatthisisthefinalimplicationthatwillmakeupourmolecularstructure.

5. Nowweknowthatourmolecularstructureiscomposedofthreeimplicationsthatmakeup2CH3CH2CH2structures.ButifyoulookatthecircledimplicationfromtheFirstSignal,thereisanattachmentonthefirstCH2(CH2CH2).Thismeansthatsomethingattachestothiscarbon.

Ifwelookattheoriginalformulagiventous,weseethatwehaveaccountedforeverythingexcepttheoxygen[thereare7hydrogenatomsx2=14hydrogenatomstotal;thereare3carbonatomsx2=6carbonstotalthisishowmanycarbonsandhydrogenswereinthegivenmolecularformula].Nowwehavetoaccountfortheoxygenatom.BecausethereisanattachmenttotheCH2,wecanattachtheoxygentothiscarbon.Ourstructurewillthereforelooklikethis:

CH3CH2CH2OCH2CH2CH3Andtheresyourmolecularstructureforthemolecule!

Note:Makesuretocheckyourwork!Checktheformula,integrals,numberofsignals,splitting,etc.andmakesuretheyareallcorrect.Alwaysperformanatomcheckandcompareittotheformulagiven.Additionally,checktoseeifyourstructureagreeswiththeDBEforthemolecule.

Massspectrum:m/z=154(M;100%),m/z=155(11.23%),andm/z=156(0.26%)IRspectrum:

1HNMR:2.1ppm(singlet;integral=2),1.9ppm(singlet;integral=1),and1.1ppm(singlet,integral=6)

Nowletsdoanotherexampleinvolvingmassspectroscopyandinfraredspectroscopy.

1. Wefirstneedanalyzewhattheinformationthemassspectroscopyistellingus.Thefirstm/zvalueistheMpeak,whichtellsushowmanynitrogenatomsarepresentinthemolecularformula.Ifthem/zvalueiseven,thentherearezerooranevenamountofnitrogenatoms.Ifthem/zvalueisodd,thenthereareanoddamountofnitrogenatoms.

Inthisexample,theMpeakhasanintensityof154.Thisisanevennumber,sothereare0oranevenamountofnitrogensinthestructure.

ThesecondvalueistheM+1peak,whichtellsushowmanycarbonsarepresentinthestructure.HowyoufindthisisbydividingtherelativeabundanceoftheM+1peakby1.1%.MAKESUREYOUDONOTDIVIDEBY1%;THISWILLNOTGIVEYOUTHERIGHTANSWER.

Inthisexample,theM+1peakrelativeabundanceis11.23%.Youdividethisby1.1%andyouget10.209%.Thismeansthatthestructurehaseither10or11carbons.

ThethirdvalueistheM+2peak,whichtellsushowmany(ifany)bromine,chlorine,orsulfuratomsarepresentwithinthestructure.Iftherelativeabundanceisapproximately4%,thensulfurispresent.Iftherelativeabundanceisapproximately33%,thenchlorineispresent.Iftherelativeabundanceisapproximately100%,thenbromineispresent.

Inthisexample,theM+2peakrelativeabundanceis0.26%.Thisdoesnotcorrespondtoanyoftheabovestatedelements,andistoosmalltobesulfur,sothereisnochlorine,bromineorsulfuratomspresentinthismolecule.

Wecannowusethisinformationtofindamolecularformulaforthisstructure.C10Possibility154(12amuperCx10C)=34amuleftforhydrogen,nitrogenandoxygen

Inthiscase,wecanruleoutC10H6N2becausetheIRspectrumshowsthatthereisacarbonylinZone4,thusrevealinganoxygenintheformula.C10H6N2lacksanoxygen.YoucanalsoruleoutC10H202becauseitdoesnotcorrespondwiththeintegralsgiveninthe

1HNMR.Theintegralequals9(2+1+6),andH2isnotaproportionateratio.ThatleavesuswithC10H180asourformula.C11Possibility154(12amuperCx11C)=22amuleftforhydrogen,nitrogenandoxygen

O N H Formula Doesitwork?0 0 34 C10H34 No,violatesHRule1 0 18 C10H180 Acceptable2 0 2 C10H202 Acceptable0 2 6 C10H6N2 Acceptable

O N H Formula Doesitwork?

C11H22andC11H8Ncanberuledoutbecausetheylackanoxygen.C11H60canalsoberuledoutbecauseitdoesnotcorrespondtotheintegralgiventousonthe1HNMR.Therefore,ourfinalformulaisC10H18O.ThelaststepistocalculatetheDBE:10(18/2)+(0/2)+1=2(possiblytworingsor2pibondsoraringandapibond)

2. NowwehavetoanalyzetheIRspectrum.Wedothisbylookingateachzoneandseeingiftherearepeaksthatcorrespondtodifferentfunctionalgroups.

Zone1: AlcoholOH:absent;nopeakpresentAmine/AmideNH:absent;nopeak,nonitrogeninformulaTerminalAlkyneCH:absent;nopeakinzone3

Zone2: Aryl/Vinylsp2CH:absent;nopeak>3000cm1Alkylsp3CH:present;peaks

AlkeneC=C:absent;nopeakFromtheIRspectrum,weseethatthereisaketoneandalkylsp3CHbondswithinthestructure.

3. Nowwehavetoanalyzethe1HNMRgiventous.First,putallgiveninformationintoatable:

ppm Neighbors Integration #H Implications2.1ppm Singlet Integration=2 1.9ppm Singlet Integration=1 1.1ppm Singlet Integration=6

Nowwehavetocalculatehowmanyhydrogenatomsarepresentforeachsignal.Whenyouaddupeachintegralvalue,yougetatotalof9.Thereare18hydrogenatoms(asseeninthemolecularformula),soitisa1:2ratio.Thismeansthatyoumultiplyeachintegralby2togetthenumberofhydrogens:

ppm Neighbors Integration #H Implications2.1ppm Singlet Integration=2 2x2=4 1.9ppm Singlet Integration=1 1x2=2 1.1ppm Singlet Integration=6 6x2=12

Nowwehavetofillintheimplications:

ppm Neighbors Integration #H Implications2.1ppm Singlet Integration=2 2x2=4 2x:CH24x:CH1.9ppm Singlet Integration=1 1x2=2 CH22x:CH1.1ppm Singlet Integration=6 6x2=12 12x:CH4x:CH3

6x:CH2

Thecircledpiecesaretheskeletonsthatwecanseearepartofthestructure.Theywereeachchosenbecausetheywerethemorelikelyoftheimplicationsassignedfortheirspecificsignal.

4. Nowwehavetoputthepiecesalltogether.

WeseefromtheDBEthatwecouldhavetwopibonds,tworings,orapibondandaring.TheketoneusesoneDBE,sowehaveoneleft(eitheraringorapibond).

WeseefromtheIRspectrumthatwehaveaketoneandalkylsp3CHbonds. Weseefromthe1HNMRthatwehave4x:CH3,2x:CH2andaCH2.The2xCH2will

bemoredeshielded,whichisevidentbytheirhigherppmvalue.Thismeansthattheywillbeclosertoanelectronegativeatom,suchtheoxygenintheketone.

Andheresourmolecularstructure!

Note:Makesuretocheckyourwork!Checktheformula,integrals,numberofsignals,splitting,etc.andmakesuretheyareallcorrect.Alwaysperformanatomcheckandcompareittotheformulagiven.Additionally,checktoseeifyourstructureagreeswiththeDBEforthemolecule.

TrialanderrorishugeinsolvingprotonNMRproblems.Themorepracticeyoudo,thebetteryouwillbe.PRACTICE,PRACTICE,PRACTICE!Goodluck,andIhopethishelped!WorksCitedDr.HardingersChemistry14CLectureSupplementDr.HardingersChemistry14CLectureSupplementPowerPointCDDr.HardingersChemistry14CThinkbookhttp://www.chem.ucla.edu/harding/index.html[IllustratedGlossary]

TheblueCH2groupsareclosesttotheketonebecausetheyaredeshielded,asseenintheirppm,thussuggestingthattheyareclosetoanelectronegativeatom,likeoxygen.