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Organic SpectroscopyWorkbook

Organic SpectroscopyWorkbookTOM FORREST

JEAN-PIERRE RABINE

MICHEL ROUILLARD

A John Wiley & Sons, Ltd., Publication

This edition first published 2011 2011 John Wiley and Sons Ltd.

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Contents

Preface vii

Acknowledgements ix

How to Use this Workbook 1

Preliminary Observations 2

Centres of Unsaturation 8

Exercises 1–100 10

Infrared Data 210IR Frequencies: Display of Infrared Database by Frequency 210IR Groups: Display of Infrared Database by Functional Group 216

1H NMR Data 221General Tables of Functional Groups 221Comparison of δ of Alkyl Groups Bearing a Substituent in α or β Position 225Alkyl Groups with an α Substituent 226Alkyl Groups with a β Substituent 226

Tables of δ for −CH2− Doubly Substituted in α Position 227Tables of δ for Aromatics 229Coupling Constants 230

13C NMR Data 233General Ranges for CH3−, −CH2−, >CH−. . . 233Chemical Shift Ranges for Functional Groups 234Chemical Shift Estimations 234

Mass Spectrometry Data 240Main Fragments Observed 240Atomic Masses 242Isotopic Abundances in % 243

Internet Resources 254

Glossary of Terms Used in the Exercises 255

Answers 257

Index 261

Preface

This workbook arose out of a long standing project for the development of computer basedprograms to assist students in learning how to deduce the structures of organic compoundsfrom spectroscopic data. The ability to elucidate chemical structures is built largely onimplicit heuristics and informal experience accumulated by solving many examples. Theaccumulation of this experience is greatly facilitated by these computer based programs,http://spectros.unice.fr/. However, the knowledge that is so arduously obtained at the com-puter may evaporate rapidly unless some reinforcement mechanism is in place.

This workbook was originally conceived as a way to provide the computer users with avehicle for making a lasting record of their learning experience. However, we recognizedthat a similar situation occurs for students who prefer to undertake their study without usingcomputer-based exercises. They could also benefit from a workbook where they could getassistance in solving spectral problems and have a record that could be used for review andreinforcement. The resulting workbook was designed so that it can be used with or withoutthe computer based program. The computer exercises, which are freely available online toany user, provide more detailed explanations and supplementary graphical material for thestructural determinations.

For each problem in the workbook we provide four spectra together with a detailedanalysis of the spectra and an explanation of the derivation of the structure of the compound.

The level of detail provided in the solution decreases gradually, until near the end of theset of exercises, very little or no analysis is given. The solutions to all of the exercisesare provided, but no structural formulae are shown on the work area of the exercise, sothat the student may follow along the analysis and draw the structural components as theyare revealed.

There are many advanced techniques available in modern spectrometers that may beemployed to assist in the solution of more difficult structural problems. In these exerciseswe have chosen to give low resolution electron impact mass spectra because these spectra arereadily available, and are rich in structural information. Modern mass spectrometry employsa multitude of other ionization techniques that can be used in particular circumstances toprovide specific information, but these are beyond the scope of an introductory spectralexercise book. In the field of NMR spectroscopy there are numerous sophisticated pulsesequences available that can generate various multidimensional spectra which are used formore difficult structural problems. These techniques are particularly useful for large complexstructures, but they are aids that still require the user to have an understanding of thefundamentals of spectroscopic interpretation, which is the goal of these exercises.

We hope that students will find that using this workbook will prove to be a pleasant andrewarding experience.

Acknowledgements

We have used the resources of our university libraries extensively in preparing this workbook,but have found that we made even more extensive use of the resources available on theinternet. We are indebted to the many people who have taken the time to place their data atthe disposal of the scientific community in this manner. This easy access to spectral data hasbeen of immense assistance to us in the preparation of this workbook. The availability ofsuch a large amount of primary data allowed us to find very good model compounds whichgave more specific support for the assignments than possible from the general correlationtables of secondary publications.

It is impossible to list all of the internet sites that we visited during the preparation of themanuscript; however, we would like to specifically acknowledge the providers of those sitesthat we have found we were consistently using. These are for the most part sites providedby institutions, universities and chemical suppliers:

NIST, National Institute of Standards and Technology, Chemistry Webbook;National Institute of Advanced Industrial Science and Technology (AIST, Japan), Spectral

Database for Organic Compounds;

Sigma-Aldrich Corporation, NMR and IR spectra;Acros Organics, IR spectra;University of Wisconsin, Chemistry Department, H. J. Reich, compilation of NMR chemical

shifts and coupling constants.

We are grateful for the support of:

Universite de Nice Sophia-Antipolis;Universite Numerique Thematique:UNISCIEL (Universite des Sciences en Ligne)Dalhousie University

How to Use this Workbook

The exercises give spectral charts and verbal analyses of the spectra, but the structuraldiagrams that would complete the picture are missing. Although annotated structureswould be very helpful in the spectral assignments, they are deliberately omitted to avoid apremature revelation of the answer. You don’t want to see the solution before you have hadthe opportunity to think about the problem. You may want to use a small sheet of paper tocover the conclusion/summary section as you try to solve each structure. There is adequatewhite space available on the pages to allow you to draw the appropriate structures, and tomake notes on the spectral assignments. Complete structural drawings are available in theAnswers section of the workbook. Each exercise has a second number in brackets whichidentifies the number by which the compound can be located in the Answers section, andin the computer-based exercises. The structures are not presented in the same order asthe exercises, thus lessening the possibility of an unwanted exposure of the answer to thenext exercise.

There is no single method that provides the best way to solve these problems. There isusually an abundance of information that allows many paths to the solution. However, itis unlikely that the complete structure will be revealed intact, so the general approach isto derive the partial structures of the molecule and to connect them to form the completestructure. The order of the spectral analysis provided in each exercise is not presented as thebest way to approach the problem. Many people do a preliminary scan of the spectra to lookfor significant features, identify the responsible structural component, then verify its presenceusing other spectra. For example, a carbonyl group that stands out in the IR spectrum canbe verified by the peak at the appropriate chemical shift in the 13C NMR spectrum. As onemoves from spectrum to spectrum like this, it is imperative that an organized approach bemaintained and that good notes are preserved.

It is recommended that you write down the partial structures as you derive them. Youcan always erase them if you make a mistake or, better still, simply draw a mark through

Organic Spectroscopy Workbook, First Edition. Tom Forrest, Jean-Pierre Rabine, and Michel Rouillard. 2011 John Wiley & Sons, Ltd. Published 2011 by John Wiley & Sons, Ltd.

them, and make notes to ensure that you have them for review and can learn from yourmistakes. Take the opportunity to make detailed notes before you leave each exercise. Asclear as it will be in your mind at that moment, it is amazing how the details evaporatewith time.

All the 13C NMR DEPT spectra are DEPT-135 spectra in which methylene carbons showa negative signal and quaternary carbons show no signal. Methyl and methine carbons showup as normal positive signals.

All the mass spectral charts are low-resolution electron impact mass spectra (EIMS). Thesespectra frequently provide the molecular weight of the compound, and the high degree offragmentation also offers valuable information to those with knowledge of reactivity oforganic compounds. In some examples, the results of high-resolution mass spectra (HRMS)are provided which can be used to obtain the molecular formula directly.

The exercises are organized somewhat in the order of complexity of the molecule andlevel of explanation provided. You will find very simple spectra and detailed explanations atthe beginning, and eventually more complex exercises with less in the way of explanation.For the last 20 exercises, only general hints and suggestions are provided. In these exercisesthe 1H NMR spectral data is provided in standard text format to show the coupling constantswhere appropriate. These data should be compared to the spectral pattern, so that you couldwrite such a verbal description of the spectrum.

The resource material you require to analyse the spectra is provided in tables andalgorithms in separate data sections for each of the spectroscopy types.

The exercises presented in this work book are also available in an on-line program,Multispectroscopy, which you may access at spectros.unice.fr. This program provides moredetailed explanations, as well as elaborate diagrams and graphics, to help to explain thespectral analysis and structure determination. You may wish to take advantage of thisadditional resource, while you are using this workbook.

Preliminary Observations

It is easy to get some information from a quick glance at some spectra. There are sometypical shapes that give a good indication of particular structural features. It is essential thatthese deductions are verified by detailed inspection of the spectrum, and with data fromother spectra.

Infrared

This indicates an alcohol.

This indicates an acid.


This indicates an alkene.

This indicates a terminalalkyne.


This indicates a nitrile.

This indicates acarbonyl.

It may be an aldehyde, ketone, acid, ester, amide, urea...

This indicates analdehyde.

This indicates an amideR–CO–NH2.

Thisindicatesan –NH2.

Thisindicates a >NH.

4 Organic Spectroscopy Workbook

Around 1380 cm −1, this is an isopropyl –CH(CH3)2 or a gem-dimethyl >C(CH3)2.

Around 1380 cm−1,there is a tert-butyl–C(CH3)3.

This indicates the presence of more than4–5 CH2, rocking bandnear 720cm−1.

1H NMR

This indicates an ethylgroup CH3–CH2–X.

This indicates a propyl group CH3–CH2–CH2–X.

This indicates an isopropylgroup (CH3) 2CH–X.

This indicates an allyl group CH2=CH–CH 2–X.


This indicates a benzene ring with one alkylsubstituent, or a benzene symmetrically substituted with alkyl groups.

This indicates a benzene para-disub-stituted by two different groups (not alkyl groups) which gives an AA’XX’ system.

13C NMR

This indicates a quaternary carbon >C<.

This indicates a methine >CH–.

This indicatesa methylene–CH2–.


This indicates a methyl group–CH3.

This indicates an aldehyde.

This indicates a ketone.

Mass SpectraThe molecular ion peak, M, is not generally the highest mass peak observed in the spectrum.Normally one finds higher mass ions due to isotope peaks:

– M+1 is normally due to the presence of the 13C isotope of carbon, 1.1% of M per C, orin the case where silicon is present, it makes a strong contribution to M+1 (5.1% of Mper Si). Sulfur also contributes to M+1 but it is easily detected by M+2.

– M+2 is generally less intense than M+1, but a M+2 that is very intense, perhaps evenmore intense than M, gives a good indication of the presence of chlorine or bromine.

By looking at the isotope patterns in the high end of the spectrum, one can usually detectthe presence of chlorine and bromine, and even the numbers of these atoms in the molecule.

For chlorine, 35Cl (75%) and 37Cl (25%), depending on the number of Cl atomspresent, one has distinctly different profiles of the isotopic peaks (M, M+2, M+4,M+6, etc.).

For bromine, 79Br (50%) and 81Br (50%), the peak pattern (M, M+2, M+4, M+6, etc.) isquite distinctive, depending on the number of bromine atoms in the molecule. Visually, thepattern looks similar to simple NMR splitting patterns.

For chlorine and bromine in the same molecule, the isotope peak profile is morecomplex but one can still see distinctive patterns of isotopic peaks (M, M+2, M+4,M+6, etc.).:


For sulfur, 32S 100% M, 33S 0,8% M+1, 34S 4,4% M+2.

In general, the contribution of sulfur to M+1 is largely masked by the contributionfrom carbon 13C (1.1% per carbon). Only the peak M+2 gives a good indication of thenumber of sulfur atoms, but only in the absence of chlorine or of bromine and only ifthe peak M is sufficiently intense, so that the intensity of peak M+2 can be read withsome precision.

Centres of Unsaturation

If you have a molecular formula for the compound, don’t forget to take the formula intoaccount in all your deductions. The first thing to do is to calculate the number of centres ofunsaturation of the compound.

Number of centres of unsaturation

Knowing the formula CxHyOzNn, it is possible to determine the number of centres ofunsaturation of the compound NCU:

NCU = (2x + 2 + n − y)/2

If the formula contains halogens, count them like hydrogens.If the NCU is less than 4, the compound cannot have a benzene ring.If the NCU is 0, the compound cannot have a double bond, or a carbonyl or a ring.You need an NCU of at least 2 to have a compound containing a triple bond.

CxHy: no oxygen, no nitrogen

The functional group of the compound cannot be alcohol, ketone, aldehyde, phenol, acid,ester, anhydride, amide, amine, nitrile . . .

If the NCU = 0, it is an alkane.If the NCU = 1, this indicates one ring or one double bond (alkene).If the NCU = 2, this indicates two double bond (alkene), or two rings or one double bond

and one ring or one triple bond (alkyne).


If the NCU = 4, this indicates four double bonds, or four rings or one double bondand three rings, or two double bonds and two rings, or three double bonds and one ring(benzene ring).

CxHyO: one oxygen, no nitrogen

The functional group of the compound cannot be acid, ester, anhydride, amide, amine,nitrile . . .

If the NCU = 0, it is an alcohol or an ether.If the NCU = 1, this indicates one ring or one double bond (>C=C< alkene or >C=O,

ketone, aldehyde).If the NCU = 4, this indicates four double bond, or four rings, or one double bond

(>C=C< or >C=O) and three rings, or two double bonds (>C=C< or >C=O) andtwo rings, or three double bonds (>C=C< or >C=O) and one ring (it may be abenzene ring).

If the NCU ≥ 5, you may have a benzene ring with a double bond (>C=C< or >C=O)or a ring . . .

CxHyO2: two oxygens, no nitrogen

The functional group of the compound cannot be anhydride, amide, amine, nitrile . . .

If the NCU = 0, it is an alcohol or an ether, or both.If the NCU = 1, this indicates one ring or one double bond (>C=C< or >C=O) or an

acid RCOOH or an ester RCOOR’.

Centres of Unsaturation 9

CxHyO3: three oxygens or more, no nitrogen

The functional group of the compound cannot be amide, amine, nitrile . . .

If the NCU = 2, it may be an ahydride RCOOCOR’.

CxHyON: one oxygen, one nitrogen

The functional group of the compound cannot be acid, ester, anhydride . . .

If the NCU = 1, it may be an amide RCON<

CxHyN: no oxygen, one nitrogen

The functional group of the compound cannot be acid, ester, anhydride, amide . . .

If the NCU = 0, it is an amine.If the NCU = 2, it may be a nitrile.

Exercises 1–100Exercise 001 (001)

δ (integration): 1.3 (3), 3.5 (2) δ: 15.19, 65.88


Exercise 001 (001)

Preliminary Observation

The mass spectrum can be used to get the molecular mass of the compound, to detect thepresence of chlorine, bromine or other heteroatoms, and to obtain structural informationfrom the fragmentation pattern. The IR spectrum provides evidence of functional groupsandother structural features. The 13C NMR spectrum gives the number of different carbon in themolecule, as well as their degree of substitution and functionality. The 1H NMR spectrumcan provide a count of the number of protons and information on their environment. The 1Hcoupling patterns give indications of the connectivity of the components of the molecule.

Mass Spectrum

The isotope patterns give no indication of the presence of chlorine or bromine. The molecularion (M+) is usually the highest mass ion (with its associated isotope peaks). An odd-massmolecular ion usually indicates the presence of an odd number of nitrogen atoms. In thisspectrum, the highest mass ion 74 (44%) has a 13C isotope peak at 75 (2%). The naturalabundance of 13C is 1.1%; therefore, each carbon contributes 1.1% to the intensity of theisotope peak. The relative intensity of the M+1 peak in this compound is 4.5% (100 × 2/44),indicating the presence of four carbons. The most intense peak is known as the base peakand is assigned an intensity of 100%. The ion at 31 (100%) indicates the presence of oxygenin the molecule. The highest mass hydrocarbon in that region is 29 (CH3CH2

+), but ifoxygen is present a peak at 31 (CH2OH+) is possible. The peak at 59 corresponds to M−15,suggesting the presence of a CH3 group.

13C NMR

The simplest spectrum, the broad band (BB) decoupled spectrum in which all of the C–Hcoupling is removed, shows signals from only two types of carbon. In the off-resonance(OR) spectrum, the size of the direct C–H coupling is reduced to make thecoupled spectrummore easily readable. In this exercise, thedirect C–H coupling gives a quartet for a methylgroup (CH3), and a triplet for a methylene group (CH2). The DEPT spectrum shows invertedsignals for CH2 carbons, and in this example has one such signal. Carbons bearing an odd

number of protons show upright signals, and here it shows one such signal, that of theCH3 group. Quaternary carbons (bearing no H) give no signal in the DEPT spectrum. Thechemical shift of the lower field carbon, (66 ppm) is in the range of a methylene carbonbonded to oxygen. The molecule must have an element of symmetry, because the massspectrum indicates that it has four carbons.

1H NMR

The 1H NMR spectrum shows only two types of proton, a triplet and a quartet. The smallpeak at 0.0 ppm is that of the tetramethylsilane reference standard. This simple couplingpattern is easily recognized as that of an ethyl group (–CH2CH3), with the methyl protonssplit by the neighbouring CH2 protons into a triplet, and the methylene protons being splitby the neighbouring CH3 protons into a quartet. The integration of the spectrum indicatesthat the two signals have a relative intensity of 2 to 3. A molecule that has no nitrogen orhalogen or other odd-valent atom, must have an even number of protons. Thus, there mustbe an even number of ethyl groups. The low field chemical shift (3.5 ppm) of the CH2 groupindicates that it is attached to an electron-withdrawing group, the oxygen atom in this case.

Infrared

The band at 1140 cm−1 is a clear indication of C–O stretching, most likely that of an etherfunction. Other than the bands due to the alkyl groups, there are no bands that can beassigned to other functional groups. The IR spectrum can be used very effectively to showthat certain functional groups are absent. For instance, in this case one can say with certaintythat there is neither a carbonyl group, nor hydroxyl present in the molecule.

Summary

The 1H and 13C NMR evidence indicates the presence of an ethyl group bonded to oxygen(ethoxy group). The mass spectrum gives a molecular mass of 74, and since an ethoxy grouphas a mass of 45, another ethyl group (mass 29) is required to complete the structure. Thecompound is diethyl ether, CH3CH2–O–CH2CH3.


Exercise 002 (003)

δ (integration): 4.12 (2), 2.04 (3), 1.26 (3) δ: 171.1, 60.4, 21.0, 14.3


Exercise 002 (003)


In a preliminary scan of the spectra there are several points that stand out. In the IR spectrum,there is a very strong peak in the carbonyl region near 1740 cm−1. The mass spectrum appearsto have a peak for the molecular ion, since the highest mass ion has an even mass. The 13CNMR spectrum can be used to determine the number and type of carbons, and the 1HNMR spectrum to get the connections between the carbons. The chemical shifts and theIR spectrum will identify the functional group. The mass spectrum can be used to get themolecular weight of the compound.

13C NMR

The broad band spectrum shows that there are four types of carbon in the molecule. Theoff-resonance spectrum indicates that two of the carbons are methyl groups (quartets, CH3),one is a methylene group (triplet, CH2) and one a quaternary carbon (singlet, no attachedhydrogen). The DEPT spectrum shows one inverted signal corresponding to the CH2 carbon,and the disappearance of a peak at 171 ppm corresponding to the quaternary carbon. Thereare two different CH3 groups, one CH2 group, and a strongly deshielded carbon bearing nohydrogens (quaternary carbon). This spectrum shows that the compound has four carbonsand eight protons or a multiple of these.

1H NMR

The 1H NMR spectrum shows the signals from three different types of proton. The integrationof the spectrum shows a relative intensity of 2:3:3 (4.2, 2.0 and 1.3 ppm), for a total of eightprotons, which agrees with the 13C NMR spectrum that indicated a total of eight protons.One CH3 group is not coupled to any other hydrogen (singlet at 2 ppm), and the other CH3

(triplet at 1.3 ppm) is coupled to the CH2 group. Thus, we have an ethyl group and an isolatedmethyl group. The singlet at 2 ppm is in the chemical shift region of a methyl group attachedto an sp2 carbon, while the CH2 (4.1 ppm) is in the region of a methylene group attached toan oxygen. As the CH2– group is coupled to a methyl group, we have a CH3CH2O– group.

Infrared

This infrared spectrum has two very significant bands of the functional group of the com-pound. The band at 1740 cm−1 clearly shows the presence of a carbonyl group (C=O), andthe band at 1250 cm−1 indicates the presence of a C–O single bond. The frequencies of bothof these bands are compatible with an ester group (C–O–C=O).

Mass Spectrum

The highest even-mass ion in the spectrum is at 88, and is probably that of the molecular ion.The most intense peak (the base peak) is found at 43, which is a fragment most commonlyassociated with an acylium ion (CH3C≡O+). This would arise from the loss of the ethoxyradical by α cleavage at the carbonyl group. The ion at 73 comes from the loss of one ofthe methyl groups. The ion at 70 arises from the loss of a molecule of water, a commonrearrangement fragmentation in esters. The majority of the fragment ions have odd masses,which is typical of compounds without nitrogen. These odd-mass ions have an even numberof electrons, and are generally more stable than the even-mass fragments, as they consistof the less stable odd-electron radical ions. Even-mass ions are generally formed when arearrangement creates a stable small molecule, such as, in this case, the loss of water yieldsthe ion at 70.

Summary

The structural components that have been identified are: an ethyl group, a methyl group anda quaternary carbon, the mass of these adding to 56, which is 32 less than the molar mass of88 that was found from the mass spectrum. The missing mass is equivalent to two oxygenatoms (2 × 16), exactly the number required for the ester group that was identified in the IRspectrum. The NMR chemical shifts of the CH2 group indicate that it is attached directly tothe oxygen of the ester. The chemical shifts of the uncoupled methyl group indicate that itis attached to a polarized sp2 carbon, which would place it on the carbonyl (C=O) carbon.This gives us the complete structure CH3(C=O)OCH2CH3, ethyl acetate.


Exercise 003 (010)

δ (integration): 3.37 (2), 1.59 (2), 0.93 (3) δ: 72.84, 23.34, 10.60


Exercise 003 (010)


The infrared spectrum frequently gives the first clue as to the functional group in themolecule. When it is not obvious from the IR, then the chemical shifts in the NMR spectramay give some clues to the functional groups. The NMR spectra give the best informationon the substructures within the molecule. The mass spectra may have a lot of information inthe fragmentation patterns, but these must be used with caution as they arise from reactionsof ions in very high energy states.

Mass Spectrum

The peak at 102 is a good candidate for the molecular ion. It has an even mass and the spec-trum contains many odd-mass fragment ions: at 87 (M−15), at 73 (M−29) and at 59 (M−43).Note the fragment at 31, which usually indicates the presence of oxygen (CH2OH+). Themolecule is possibly a mono-oxygenated saturated compound (C6H14O) or a dioxygenatedunsaturated compound (C5H12O2). Note that there is a peak for the loss of a methyl group(M−15), and for the loss of an ethyl group (M−29), so one might be tempted to concludethat the molecule contains a methyl group and an ethyl group, but this should always beverified in the 1H and 13C NMR spectra.

1H NMR

The spectrum has three signals with an integration ratio of 2:2:3. The coupling pattern ofa triplet, sextet and triplet suggests the presence of –CH2CH2CH3, theterminal CH2 splitby two neighbours, the middle CH2 split by five neighbours, and the terminal CH3 split bytwo neighbours. In view of the chemical shift of the terminal CH2 group at 3.4 ppm, onewould expect this carbon to be attached to an oxygen. The mass spectrum of the compoundrequires a higher mass than an oxygen and one propyl group. As there are no peaks for otherprotons, there are probably two equivalent propyl groups in the molecule.

Infrared

The strong band at 1120 cm−1 is in the region of C–O stretching. As there is no carbonylband or OH band, this band belongs to an ether. There is little else that one can see in the

spectrum other than the bands from the alkyl chains. There is a doublet of equal intensity near1380 cm−1 which normally indicates the presence of a gem-dimethyl group. However thispair of peaks is deceptive, as the NMR spectra shows clearly that there is no gem-dimethylgroup in the compound.

13C NMR

There are three types of carbon in the compound, two CH2 (triplets in off-resonance andinverted in DEPT), and one CH3 (quartet in off-resonance). The chemical shift of the carbonat 73 ppm is in the correct region for a carbon attached to the oxygen of an ether.

Summary

The molecule has a propyl group attached to oxygen and a molecular mass of 102. The massof one propyl group (CH3CH2CH2) is 43, and one oxygen is 16, for a total of 59. This is43 less than the molecular mass of 102. One more propyl group is needed to complete thestructure. The compound is dipropyl ether, CH3CH2CH2–O–CH2CH2CH3. Notice that thepeaks of the two triplets in the 1H NMR spectrum are not symmetrical, particularly that ofthe methyl group. Each one is distorted, having a larger peak on the side towards the groupto which it is coupled. This is typical of systems in which the chemical shift difference isnot large compared to the size of the coupling constant. If the spectrum were run in a higherfield spectrometer (say, 400 MHz, instead of this one, 60 MHz), one would not expect to seesuch distortion, as the chemical shift difference in Hertz would be much larger, while thecoupling constant would not change. The ratio of the chemical shift to coupling constantbecomes much larger, consequently the distortion decreases. Compare the coupling in thiscompound with that of the compound in Exercise 004.

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