Tutorial 49
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ALL ABOUT SOLVING PROTON-NMR What is 1 H‐NMR? 1 H‐NMR is NMR (Nuclear Magnetic Resonance) spectroscopy in which the proton nuclear spin is manipulated. It revolves around the idea that the energy required to cause nuclear spin flip is a function of the magnetic environment of an atom’s nucleus. How can we find structure from NMR? We find structure through the information NMR relays to us. These pieces of information include: ‐ Number of signals [which is the number of nonequivalent proton sets in a molecule] ‐ Position of signals [also known as chemical shift, which is the magnetic environment in which protons are in] ‐ Relative intensity of signals [also known as integration, which is the ratio of equivalent proton types] ‐ Splitting of signals [also known as spin‐spin coupling, which is the amount of proton neighbors present] How to solve for number of signals present The number of signals depends on the equivalency of protons. Protons are equivalent if they have nuclei with the same magnetic environment. In other words, they have to be the same in every possible way. It is easier to find nonequivalent protons. http://www.chem.ucla.edu/harding/index.html Within the molecule depicted above, there are three different signals present, in that there are three different sets of protons with different equivalencies. H a , H b and H c are all nonequivalent to one another, thus producing three separate signals on the NMR spectrum. How to solve for the position of signals/chemical shift Chemical shift is defined as the position of a signal in an NMR spectrum. It is dependent on different influences, such as the ΔE of the spin state energy (which is controlled by the magnetic field at the nucleus), and electronegativity (as chemical shift increases, electronegativity of atoms near the proton increases). The positions of signals will be given to us on a test so IT IS NOT MANDATORY TO MEMORIZE THEM! The protons labeled H a are equivalent because they all are connected to a carbon that is connected to two hydrogen atoms. They are the same in every way possible.
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Transcript of Tutorial 49
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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.
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Thistablewillbegiventousonanexam.
Chemistry14CLectureSupplementPowerPointCD
ChemicalShiftTrend RCH3
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HowtosolveforthesplittingofsignalsSpinspincouplingisdeterminedbythenumberofmagneticfieldsaffectingthenucleusofamolecule.ThiscausessplittingoftheNMRsignals.Thegeneralruleforsplittingisthesignalforaprotonwithnneighborsissplitinton+1lines.Therearesomerulesandrestrictionswemustfollowwhenworkingwithsignalsplitting:
1. Onlynonequivalentprotonscancouple Ifprotonsareequivalent,thentheydonotcouple
2. Protonsthatareseparatedbythreeormoresinglebondsusuallydonotcouple
Chemistry14CLectureSupplementPowerPointCD
Pibondsdonotcounttowardsthebondlimit,butthesplittingconstantJ(whichisthespacingbetweenlinesinasplittingpattern)maybetoosmalltosee,soitmaybethere Forpurposesofcoupling,justpretendthepibondisntthere[ITSAFREE
SPACER]!Tobesafe,adda+1totheamountofbondscountedexcludingthepibond.
Benzeneringscountasonebigfreespacer Allprotonscouplewitheachotherbecauseofresonance Jmaybesmall Thinkofbenzeneasagatedcommunityhydrogenatomsbondedtothe
benzeneringcanonlycouplewithotherhydrogensbondedtothebenzenering.Theycannotbondwithotherhydrogenspresentinthemolecule
Chemistry14CLectureSupplementPowerPointCD
3. SignalsforOHandNHareusuallysinglets
Inthisexample,HaandHddonotcouplebecausetheyarefourbondsapart(seeredbonds).However,HaandHccancouplebecausetheyaretwobondsapart(seegreenbonds).
Thehydrogenatomspresentinthemethylattachedtothebenzeneringcannotcouplewiththehydrogenatomsbondeddirectlytothebenzenering.Thehydrogenatomsbondeddirectlytotheringcanonlybondtootherhydrogensattachedtothering.
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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.
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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.
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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.
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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?
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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
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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.
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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.
