-- nmr interpretation guide

8/13/2019 -- NMR Interpretation Guide

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NMRInterpretation Guide

Therearethreethingstolookforwhenyouseeaspectrum.First,thesplittingpatternwhichtellsyou

whatisnexttotheprotongivingthesignal.Second,theintegrationwhichtellsyouhowmany

equivalentprotonsaregivingthesignal.Finally,theshiftwhichtellsyouthetypeofprotonthatisgiving

thesignal(wecallthisthechemicalenvironment).Whenyoulookataspectrum,itcanbevery

overwhelmingsoweuseatabletoorganizeallofthedata(belowisanexampleofatableandthe

analysisofacompound).

TheIntegration

Theintegrationoftheareaunderthepeakisdirectlyproportionaltothenumberofequivalentprotons

givingthatsignal.Therefore,ifyoucalculatethearea,you*know*howmanyprotonsthereare.Usually,

weneedtoknowsomethingaboutthecompoundtobeabletousethisknowledgebecausethefirst

integrationis

set

to

1proton

and

everything

else

is

relative

to

that

area

under

the

curve.

If

you

know

thatacertaintripletis3protons,youcansetthatareatobeequalto3protonsandeverythingelsewill

berelativetothatknown.

Importanttothisdiscussionisthetermequivalentprotons.Therearetwoclassificationsofprotons.

Theyareeitherhomotopicorheterotopic.Homotopicmeanstheprotonsareexactlythesameand

indistinguishablefromoneanother(wesubstitutedeuteriumforaprotonandseeifthereisachangein

thestereochemistry).

Heterotopicmeanstheprotonsarenotquitethesamebuttherearedifferentdegreesofdifference.

First,theycanbecompletelydifferent.Thesearetheeasiesttospotandwillgivecompletelydifferent

signalsandsplittingpatterns(whatwecallspinsystems).Anexampleisshownforethanolwherethe


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CH2protonsarecompletelydifferentfromtheCH3protons(differentconnectivity).Iftheprotonsare

notcompletelythesame(homotopic)orcompletelydifferent,theyareeitherenantiotopicor

diastereotopic.Thisismoredifficulttoputintowordssoanexampleisshownbelow.

Intheaboveexamplewithethanol,themethylprotonsarecompletelydifferentfromtheCH2protons

and

therefore,

heterotopic

because

of

connectivity

(they

are

not

connected

to

the

same

type

of

carbon).

TheCH2protonsareheterotopicbecauseoftheirspatialarrangement(theyseethingsalittle

differently).

Fordiastereotopicprotons,therehastobeastereocenteralreadyinthemolecule(otherwise

substitutingadeuteriumforaprotonwillonlyresultintheenantiomer).Belowisanexampleof

homotopicanddiastereotopicprotons.

ProvetoyourselfthattheotherCH2protonsarediastereotopicandbothmethylgroupprotonsare

homotopic.

Sohomotopicandenantiotopicprotonsgivethesamesignalandtheintegrationwillreflectthenumber

ofprotons.

SplittingPattern

Thesplittingpatternisdependentonthenumberofequivalentprotonsonthecarbonnextdoor

(enantiotopicorhomotopic).Iftheprotonsnextdoorarenonequivalent,thenmorecomplexsplitting


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patternsoccur(forexample,doubletoftriplets).Thistakespracticetovisualizebutcanbeprettyeasy.

Belowisanexampleofanaldehydewithadoubletoftriplets.

Thesetwospinsystemsarerelatedandtheyaresaidtobecoupled(Jcoupled).Ifwereallywantedto

makesuretheywerecoupledwewoulddeterminethedistancebetweenthepeaksandseethatthey

werethesame(thesmallsplitting).

TheShift

Theshifttellsyouwhatkindofprotonitis.Intheaboveexamplethetripletat9.77ppmisaprotonon

thecarbonyl(aldehydicproton).Thedoubletoftripletsareprotonsonacarbonnexttoacarbonyl

(alphatothecarbonyl).Theshiftisprettyconsistentthroughoutorganiccompounds.Therearetablesin

books

for

all

of

the

different

shifts

that

you

can

use

to

help

identify

which

carbons

are

giving

specific

signals.

SpinWorks 2.5: 1D Proton

PPM 9.86 9.84 9.82 9.80 9.78 9.76 9.74 9.72 9.70

file: C:\Documents and Settings\Wei tu\Desktop\Alkemyst NMR\cs- nonanalsm\1\fi d expt:

transmitter freq.: 300.131853 MHz

time domain size: 32768 points

width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt

number of sc ans: 16

freq. of 0 ppm: 300.130000 MHz

processed size: 32768 complex points

LB: 0.000 GB: 0.0000

x 1.924

2.48 2.46 2.44 2.42 2.40

R

H2C

H H

O

H

Sees two neighbors on this sideso splits into a triplet.

Sees one neighbor on this sideso splits into a doublet.

Together it is a doublet of triplets


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TheTable

Shift SplittingPattern Numberof

Neighbors

Numberof

Protons

(Integration)

ChemicalEnvironment

Where

isthe

setof

peaks

Whattypeof

peaks(s,d,t,

quart,quint,)

n1,where

nisthe

numberof

peaks

Justthe

numberof

protonsgiving

thepeaks

Fromthebooks

Hereisasimpleexample.

Thefullspectrumisaboveandtheexpandedspectraarebelow.


PPM 7.2 6.8 6.4 6.0 5.6 5.2 4.8 4.4 4.0 3.6 3.2 2.8 2.4 2.0 1.6 1.2 0.8 0.4

2.

999

0.

393

0.

392

0.

891

3.

025

6.

006

1.

005

0.

246

1.

004

1.

010

1.

004

1.

984

1.

992

8.

137

file: C:\Documents and Settings\Wei tu\Desktop\AlkemystNMR\cs- 3-18-pnmr\ 1\fid expt:



width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt



processed size: 32768 complexpoints

LB: 0.000 GB: 0.0000


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PPM 7.36 7.32 7.28 7.24 7.20 7.16 7.12 7.08 7.04 7.00 6.96 6.92

1.

984

1.

992

8.

137

7.3

874

7.3

817

7.3

589

7.3

564

7.3

400

7.3

345

7.3

274

7.3

124

7.2

861

7.2

803

7.2

771

7.2

713

7.2

621

7.2

558

7.2

364

7.2

288

7.2

142

7.1

149

7.1

091

7.0

896

6.9

571




width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt




LB: 0.000 GB: 0.0000


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PPM 5.04 5.00 4.96 4.92 4.88 4.84 4.80 4.76 4.72 4.68 4.64 4.60 4.56 4.52 4.48

1.

004

1.

010

1.

004

5.0

178

4.8

307

4.7

771

4.6

113

4.5

577




width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt




LB: 0.000 GB: 0.0000


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PPM 4.24 4.20 4.16 4.12 4.08 4.04 4.00 3.96 3.92 3.88 3.84 3.80 3.76 3.72 3.68

1.

005

0.

246

4.1

567

4.1

329

3.8

781

3.8

717

3.8

545

3.8

481




width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt




LB: 0.000 GB: 0.0000


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PPM 2.68 2.64 2.60 2.56 2.52 2.48 2.44 2.40 2.36 2.32 2.28 2.24 2.20 2.16 2.12 2.08 2.04 2.00 1.96

0.

392

0.

891

3.

025

6.

006

2.6

718

2.3

153

2.2

407

2.0

655




width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt




LB: 0.000 GB: 0.0000


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Shift

Splitting

Pattern

Numberof

Neighbors

Numberof

Protons

(Integration)

ChemicalEnvironment

1.061 d 1 3 CH3CH

1.29 t 2 0.39 Unknown(maybeCH3CH2)

2.07 s 0 0.39 Unknown(maybeCH3C=O)

2.2 broads 0 1 ProbablyHOC(broadsingletshereare

usuallyOHprotons)

2.3 s 0 3 Oddbutprobablyamethyl

2.7 s 0 6 Oddbutprobably2equivalentmethyls

3.86 dquart 1and3 1 CH3CHCH

4.14

quart

3

0.24

Unknown(probably

CH3

CH2

O

4.58 d 1 1 OddbutlookslikeaABpattern

4.80 d 1 1 OddbutlookslikeaABpattern

5.02 Broads 0(?) 1 NotreallyanOHsounknown

6.96 s 0 2 Aromatic

7.067.13 mult ? 2 Aromatic

7.187.42 mult ? 8 Aromatic


PPM 1.52 1.48 1.44 1.40 1.36 1.32 1.28 1.24 1.20 1.16 1.12 1.08 1.04 1.00 0.96 0.92 0.88 0.84 0.80

2.

999

0.

393

1.3

090

1.2

852

1.2

614

1.0

731

1.0

494




width: 6172.84 Hz = 20.567092 ppm = 0.188380 Hz/pt




LB: 0.000 GB: 0.0000


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First,wecanthrowoutthesignalsthatdonothaveintegrationsthatlookgood(youshouldseethe

ethylacetatepeaksasbeingthe0.39,0.39and0.24integrations).Nowthetableiscompleteandshould

helpyourealizethereareatleasttwoaromaticrings(aromaticprotonsequal12),atwoCHgroups,an

OHand 4methylgroups(threenonequivalent).Thiswouldbeagoodstarttoidentifyingthe

compoundandtherearepeoplewhocanlookatthisdataandcomeupwiththecorrectcompoundbut

wereally

need

more

information.

This

is

the

product

from

the

following

reaction:

HN(R) (S)

OH

PhSO2

Ph Br

N(R) (S)

OH

PhSO2

Ph

Youshouldbeabletoidentifythesignalsfromthetableandthereactionwiththeexceptionofthe

broadsingletat5.02ppmbutyoucanusetheprocessofeliminationtoidentifyeventhat.