Jones M., Fleming S.A. Organic Chemistry

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15.6 NMR Measurements 729

In the general situation, for a given hydrogen, H

, the NMR signal will consist

of n  1 lines,where n is the number of equivalent adjacent (vicinal) hydrogens, H

This observation is called the n ⴙ 1 rule.The lines for the signal are separated from

each other by a frequency of J Hz. The intensity of the lines can be determined by

analyzing the various possible spins of the coupled hydrogens as we have been doing,

or by using a device called “Pascal’s triangle.” In Pascal’s triangle, the value (inten-

sity) of a number is the sum of the two numbers above it. Figure 15.39 explains it

better than these words. Pascal’s triangle allows the rapid determination of the

intensities of lines in NMR signals. Note that we are speaking here of the relative

intensities of the lines within a given signal, not of the intensities of signals for dif-

ferent hydrogens in the molecule (the integral).

1133

11206615 15

Signal for H

(n)

Number of

equivalent

adjacent H

(n + 1)

Number

of lines

in signal

51010 5

15 20 15 6

THE GENERAL CASE

Pascal’s triangle: Each number is the sum of the

numbers above (see examples shown with dashed lines);

the intensities of NMR signals follow this pattern

FIGURE 15.39 In the general case, there will be n  1 lines for a hydrogen

adjacent (vicinal) to n equivalent hydrogens.

730 CHAPTER 15 Analytical Chemistry: Spectroscopy

Multiple Couplings. What happens when a hydrogen is flanked by more than

one kind of hydrogen, a very common situation in complex organic molecules? The

answer is quite simple—just apply the n  1 rule for each set of equivalent adjacent

hydrogens. Consider the situation in Figure 15.40 in which H

is adjacent to two

different hydrogens, H

and H

.The NMR signal for H

will be split into two lines

by H

and each of these lines will be further split into two by coupling to H

.Let’s

say that J

is equivalent to 4 squares and J

is equivalent to 2 squares.The result is

shown in Figure 15.40a: four lines, a doublet of doublets. But does it matter which

way we do it? In Figure 15.40a we started with J

. In Figure 15.40b, we do it the

other way, starting with J

. As you can see, the order makes no difference, we get

the same four lines either way.

= 4

(a)

= 4

(b)

YC C

= 2

FIGURE 15.40 The pattern for a

single hydrogen, H

, is determined by

coupling to two different adjacent

single hydrogens, H

and H

. Note

that the ultimate pattern of lines, a

double doublet, does not depend on

the order in which we write the

coupling constants J

and J

Similarly, Figure 15.41 shows what happens if hydrogen H

has one adjacent

hydrogen H

and two adjacent hydrogens H

,with J

 3 squares and J

 1 square.

A doublet of triplets results.

= 1

= 3

C C

= 1

FIGURE 15.41 The splitting pattern

for a single hydrogen, H

,is

determined by coupling to a single

adjacent hydrogen, H

, and two

equivalent adjacent hydrogens, H

PROBLEM 15.16 What happens in general if hydrogen H

is flanked by one hydrogen

and three hydrogens H

PROBLEM 15.17 What happens to the NMR spectrum of the molecule in Figure

15.41 if J

 2 squares and J

 1 square?

Summary

We have learned about the three important NMR phenomena: the integral, the

chemical shift, and the coupling constant. A signal will appear for each set of

equivalent hydrogens. The integration will tell us the relative number of hydro-

gens in that signal. The chemical shift of the signal is dependent on the chemi-

cal environment in which the hydrogen resides.The number of lines in an NMR

signal results from coupling with vicinal hydrogens. Multiple couplings can be

analyzed by constructing “tree”diagrams such as those in Figures 15.40 and 15.41

in which the different couplings are worked out in sequence.The order in which

one applies the different couplings never matters.

15.6 NMR Measurements 731

The Magnitude of J: The Karplus Curve. There is no observable coupling

between equivalent nuclei. But why is coupling observed only to hydrogens on adja-

cent atoms and not to all other hydrogens in a molecule? Usually, only “two-bond”

or “three-bond coupling” is large enough to be detectable (Fig. 15.42). Two-bond

coupling only occurs when hydrogens on the same carbon are diastereotopic (dif-

ferent).Such coupling can be very small ,and thus hard to detect, or it can

be relatively large (20 Hz). In some cases, long-range coupling can be observed,

especially if the hydrogens are connected in allylic fashion.

16 2 Hz2

2–30

6–8

Identical hydrogens

do not couple

Magnitude of J (Hz)

is two bonds away from H

;

coupling over this distance is

usually observable

is three bonds away from H

;

coupling over this distance is

usually observable

0–1

is four bonds away from H

;

coupling is not usually observable

2–3

H is in the allylic position relative

to H

; coupling is usually observable

FIGURE 15.42 Coupling can be

measured over a three-bond distance,

but is not usually easily detectable

over longer distances.

The coupling constant between hydrogens on adjacent carbons is most sensitive

to the dihedral angle between those hydrogens, and this dependence can often be

used in assigning structure. Figure 15.43a shows the Karplus curve, named for

Eye

Dihedral

angle

J = 12–18 Hz

θ = 180⬚

–1

0⬚ 20⬚ 40⬚ 60⬚

For two adjacent hydrogens on sp

carbons

For two adjacent hydrogens

on sp

carbons

80⬚ 100⬚ 120⬚ 140⬚ 160⬚ 180⬚

J (Hz)

= 6–12 Hz

θ = 0⬚

(a) (b)

FIGURE 15.43 Coupling between

vicinal hydrogens is sensitive to the

dihedral angle between them. (a) The

Karplus curve shows the calculated

relationship between J and the

dihedral angle. (b) Dihedral angles

can be determined by using a

Newman projection and taking the

angle between the bond and

the bond.C

732 CHAPTER 15 Analytical Chemistry: Spectroscopy

PROBLEM 15.18 How many lines do you expect to see in the NMR signal for

the central hydrogen in 2-methoxypropane? How about 1,3-dichloro-2-

methoxypropane? Hint: Be careful! See page 721.

The cis coupling constant, J

cis

,for alkenes and rigid rings is smaller than the trans

coupling constant, J

trans

. In alkenes, J

cis

is in the range 6–12 Hz and J

trans

is in the

range 12–18 Hz. Notice that the ranges slightly overlap. One cannot always assign

stereochemistry from just one stereoisomer. What would you make of a coupling

constant of 12 Hz,for example? But if both isomers are available,the cis stereochem-

istry can be reliably assigned to the one with the smaller J value. Figure 15.44 gives

some typical coupling constants in alkenes and aromatic compounds.

Cl Cl

1,3-Dichloro-2-methoxypropane

OCH

2-Methoxypropane

OCH

CH(CH

)

CH(CH

)

PROBLEM 15.19 The two molecules shown below have coupling constants for the

hydrogens attached to the three-membered ring of 8.4 and 5.3 Hz. Which

compound has which coupling constant?

= 6–12 Hz

= 6–8 Hz J

= 1–3 Hz J

= 0–1 Hz

= 12–18 Hz

= 0–3 Hz

= 4–10 Hz

= 2–3 Hz

~~~

FIGURE 15.44 Some coupling

constants for hydrogens attached to

double bonds.

Martin Karplus (b. 1930) of Harvard University, which shows the dependence of J

on the dihedral angle for two hydrogens on adjacent carbons. Figure 15.43b shows

the dependence of the coupling constant on the angle between hydrogens attached

to sp

hydridized carbons.

15.6 NMR Measurements 733

Chemical Exchange. The NMR splitting patterns of alcohols and amines are

often not what we would expect. Consider the common alcohol,ethyl alcohol, whose

H NMR spectrum is shown twice in Figure 15.45. Notice that these two spectra,

taken from different catalogs in the chemical literature, show a different chemical

shift for the OH signal. Remember there are several factors that can influence the

hydrogen bonding, and thus the chemical shift of a hydrogen on oxygen. These

include concentration, pH, solvent, and temperature.The methyl and methylene sig-

nals are in the same place in the two spectra.The spectra show the expected triplet

for the methyl group, but the methylene and hydroxyl hydrogens do not appear as

our analysis would predict.The most accurate description of them might be “blobs.”

What is going on?

543210

(ppm)

543210

(ppm)Chemical shift

(δ)

Chemical shift

(δ)

abc

FIGURE 15.45 Two

H NMR spectra of ethyl alcohol show different shifts for the OH signal.

Here the issue is one of chemical exchange. In the presence of either an

acidic or a basic catalyst, or even under neutral conditions, alcohols and amines

exchange the OH or NH hydrogens. Figure 15.46 gives the mechanism for base-

catalyzed exchange of an alcohol and acid-catalyzed exchange of an amine, using

deuterated materials in order to detect the exchange reaction. This exchange

reaction can be used to locate the signals for and groups.

The NMR spectrum is taken, then a drop of D

O is added to the sample tube.

–

—

HOD

FIGURE 15.46 The base-catalyzed

exchange reaction of alcohols, and the

acid-catalyzed exchange reaction of

amines.

734 CHAPTER 15 Analytical Chemistry: Spectroscopy

The tube is shaken and the spectrum taken again. Hydrogens on carbon will not

exchange in this experiment, so their signals will not shift significantly. However,

the hydrogens exchange and the molecule becomes .The signals

for deuterium, which has a different magnetogyric ratio γ from hydrogen, will not

be observed because they appear in a different NMR range, so the signals for the

exchangeable hydrogens effectively vanish from the spectrum and are replaced by

an HOD signal.

DRO

In IR spectroscopy, the absorption of energy is essentially instantaneous.The IR

spectrometer takes a stop-action picture of the molecule. Nuclear magnetic resonance

spectroscopy is different. Acquisition of the signal is much slower, about 10

3

s, and

the exchange reaction is fast compared to the time it takes to make the NMR meas-

urement. During this time the alcohol oxygen of a given molecule will be bonded

to many different hydrogens.This is why OH and the NH signals are often broad.

In addition, the two methylene hydrogens of ethyl alcohol “see” only an average of

all spins of the hydroxyl hydrogen, not the two different 

and 

states.

Exchange can be so rapid that there is no coupling with the OH hydrogen.The same

phenomenon serves to remove coupling between alkyl hydrogens that are vicinal to

the NH hydrogens of primary and secondary amines.

If this explanation is correct, the coupling should be detectable in scrupulously

purified ethyl alcohol because the exchange reaction will be stopped or, at least,

greatly slowed. It is very hard to remove the last vestiges of acidic or basic catalysts

from these polar molecules, but if the effort is made, the coupling returns, as it

should. Absolutely pure ethyl alcohol does give the anticipated triplet for the OH

hydrogen and a well-defined doublet of quartets for the methylene signal (Fig.15.47).

10 9 8 7 6 5 4

3210

(ppm)Chemical shift (δ)

C OH

FIGURE 15.47 The

H NMR

spectrum of absolutely pure ethyl

alcohol. See Problem 15.60.

This section gives the first hint that NMR spectra are dependent on the rates

of reactions. If the rate of the exchange reaction were slow on the NMR time scale,

we would not see the averaged spectra that we do for alcohols and amines. We will

return to this notion in Section 15.9.

PROBLEM 15.20 Write a mechanism for acid-catalyzed exchange of ethyl alcohol

and the base-catalyzed exchange of diethylamine.

15.6 NMR Measurements 735

Summary

In the best of all worlds (which means in this case that your budget includes

enough money to buy the highest magnetic field spectrometer available), we will

see a different chemical shift for every different hydrogen in our molecule, and

we will be able to determine the number of equivalent adjacent hydrogens from

the n  1 rule. In practice, things may not be quite so simple. We can see some

possible complications right away. Perhaps the signals will not be sufficiently

separated, and the spectrum may consist of complicated overlapping multiplets.

Here is an example that shows why chemists are so anxious to spend taxpayers’

money on ever higher field spectrometers.The higher the field, the better the dis-

persion of the signals (recall Problem 15.9, p. 718).Remember that chemical shifts

(in hertz) are dependent on the strength of the applied field and the higher the

field, the more separated the signals.

109876543210

(ppm)Chemical shift (δ)

COOH

FIGURE 15.48 Two different

hydrogens on adjacent carbons will

give rise to a pair of doublets as long

as the hydrogens are very different.

The signal at δ 12 ppm for the

COOH hydrogen is not shown.

Remember that a doublet should appear as two lines in a 1:1 ratio. The sig-

nals in Figure 15.48 are indeed doublets. But the doublets are not simple 1:1 pairs

of lines; the doublet at δ 7.4 ppm is leaning to the right and the doublet at

δ 6.2 ppm is leaning to the left. In fact, we see the simple spectrum of 1:1 dou-

blets only when H

and H

are very different from each other: when they have

very different chemical shifts. The more alike the two hydrogens, the more the

spectrum deviates from first order, the spectrum predicted by the n  1 rule. As

the chemical environments of the two hydrogens become more similar, the chem-

ical shifts for the corresponding NMR signals of course also become more alike.

15.6d More Complicated NMR Spectra Spectra with coupling that

clearly fits the n  1 rule are called first-order spectra. Even at high field, NMR

spectra are often not the simple first-order collections of lines we might expect.

A classic example of this phenomenon occurs in the two-spin system shown in

Figure 15.48. In the NMR spectrum for this molecule, we expect to see a pair of

doublets because each hydrogen is different, and coupled to only one hydrogen

on the adjacent carbon. The hydrogen is not shown in this spectrum

(it is at δ 12 ppm).

COOH

Predicted spectrum for H

, four lines

2-Furoic acid

COOH

WEB 3D

Here is the real spectrum of 2-furoic acid; the COOH hydrogen is off-scale

736 CHAPTER 15 Analytical Chemistry: Spectroscopy

The more alike the hydrogens are,the closer the signals are and the more pronounced

the leaning of the peaks toward the center (Fig. 15.49).

In the limit, when H

and H

become the same, they have the same chemical

shift. Because there is no coupling between equivalent hydrogens, the signal for them

collapses to a single sharp line. Figure 15.49 shows the gradual transition between

the case in which the two hydrogens are very different and that in which they are

identical. In the language of NMR, the very different system is called AX, the iden-

tical system A

, and an intermediate case either AB or AM.

As we saw earlier (p. 730) it is often simple to derive the expected multiplicity of

an NMR signal from the n  1 rule. For example, in a molecule such as 2-furoic acid

(Fig. 15.50), the two different hydrogens on C(3) and C(5) (H

and H

) are each

coupled to the single hydrogen on C(4) (H

). This system of three very different

hydrogens is called an AMX system.We would predict the signals for H

and H

be a pair of doublets. The actual signals are more complicated because H

and H

are coupled to each other through long-range coupling (Fig. 15.44). Hydrogen H

in furoic acid is coupled differently to H

and H

by the coupling constants J

and J

. How can we figure out the pattern for H

that should be observed in the

Two very

different

hydrogens

Intermediate

cases

Two similar

hydrogens

Two identical

hydrogens

–30 –20 –10 0

2010 30

(Hz)

FIGURE 15.49 As two hydrogens in a

molecule become more similar, the

inner lines grow at the expense of the

outer lines.The signals increasingly

lean toward each other. In the limit,

the two hydrogens become identical,

and there is only a single line.

FIGURE 15.50 In 2-furoic acid, we expect H

and H

to appear as doublets. Each

is strongly coupled to a single adjacent hydrogen, H

. However, both hydrogens

are further split by long-range coupling (J

) into doublets of doublets.

15.6 NMR Measurements 737

spectrum? The first-order pattern can be determined by working out the number of

lines from each coupling one at a time (p. 730). We first determine the number of

lines produced by coupling H

to H

, and then apply the coupling of H

to H

each of the lines in a “tree diagram.” For example, coupling of H

to H

produces a

doublet, separated by J

. Each of these two lines is further split into two by cou-

pling to H

. The result is four lines, a doublet of doublets. Figure 15.50 shows the

theoretical prediction for H

and the real spectrum of 2-furoic acid.

First-order spectra are quite easy to predict. For example, according to the n  1

rule, the two different methylene groups of a generic molecule

should appear as a pair of 1:2:1 triplets. 2-Chloropropanoic acid fits the pattern and, as

Figure 15.51 shows,satisfactorily produces the expected pair of methylene triplets in its

H NMR spectrum at δ 2.89 and 3.78 ppm.However,when the molecules become more

complex we can anticipate complicated NMR spectra, even for first-order systems.

109876543210

(ppm)Chemical shift (δ)

COOH

(not shown)

δ~12.0

FIGURE 15.51 In 2-chloropropanoic

acid, each methylene group appears

as a simple triplet. Notice that the

signals for these coupled hydrogens

lean toward each other.

These hydrogens are split into a triplet by the two

adjacent methylene hydrogens, and into a

quartet

by the three adjacent methyl hydrogens;

the result

is a sextet at δ 1.66, an overlapping set of 12 lines

Methyl butanoate

Triplet at

δ 0.94

Triplet at

δ 2.30

Singlet at

δ 3.68

OCH

43210

(ppm)Chemical shift (δ)

FIGURE 15.52 The

multiplet at δ 1.66 ppm

results from the splitting of

one methylene group by the

adjacent methyl and

methylene groups.The

result is 12 lines that overlap

in a complex pattern to give

a sextet.

Methyl butanoate is a simple molecule, and we can predict its first-order spectrum

quite easily (Fig. 15.52). The methyl group attached to oxygen has no neighboring

hydrogens and appears as a singlet at δ 3.68 ppm. The other methyl group is adja-

cent to a pair of methylene hydrogens and appears as a triplet at δ 0.94 ppm. The

methylene group adjacent to the carbon–oxygen double bond is flanked by the same

methylene group. It, too, is a triplet, this time at δ 2.30 ppm. But what about the

other methylene group? It is split by three methyl hydrogens, and then split again

738 CHAPTER 15 Analytical Chemistry: Spectroscopy

by two methylene hydrogens. The result should be a quartet of triplets, or, equiva-

lently, a triplet of quartets. Either way you work it out, the result is a 12-line pat-

tern, a dodectuplet! Fortunately, the coupling constants are nearly identical for a

freely rotating alkane and the signal collapses to a sextet (1:5:10:10:5:1). We can

analyze the coupling by noting that this methylene sees five very nearly equivalent

hydrogens and therefore the n  1 rule correctly predicts the sextet.

First-order spectra are only observed when all the hydrogens in a molecule are very

different from each other. As they become less different, more complicated spectra

appear. If we imagine carrying the process of making the hydrogens more and more

alike to its logical conclusion, we get a single signal for equivalent hydrogens.But when

are hydrogens “very different” and when are they not? In a practical sense, for two

hydrogens H

and H

to give rise to first-order spectra,the difference in hertz in their

chemical shifts (¢δ) must be at least 10 times greater than their coupling constant, J

(15.5)

This requirement is another reason why high-field NMR spectrometers are so much

more useful than lower field instruments.Although the coupling constant J does not

depend on the field strength, the value (in hertz) of the chemical shifts does (p. 717).

The greater the field strength, the greater the chemical shift difference (¢δ) in hertz,

and therefore the greater the likelihood that the condition of Eq. (15.5) will be sat-

isfied. Figure 15.53 shows the

H NMR spectrum of 1-bromobutane at 60, 90, and

300 MHz. Even the modest change from 60 to 90 MHz results in substantial

improvement in the spectrum. You get what you pay for.

¢δ 7 10 J

43210

(ppm)Chemical shift (δ)

43210

(ppm)Chemical shift (δ)

60 MHz

43210

(ppm)Chemical shift (δ)

90 MHz

300 MHz

FIGURE 15.53 The

H NMR spectra of 1-bromobutane taken at 60, 90, and 300 MHz.

PROBLEM 15.21 In Figure 15.52, the expected dodectuplet has simplified to six

lines. The methylene group giving rise to that signal is coupled to both the

adjacent methyl and methylene groups. Use a “tree” diagram such as the one in

Figure 15.50 to show that, if the coupling constants J to the methyl group and the

methylene are fortuitously the same, the observed six lines should appear. Graph

paper is very useful!

PROBLEM 15.22 Explain why the methylene group adjacent to the carbonyl group in

methyl butanoate is downfield of the methylene group adjacent to the methyl group.