Physical Chemistry Third Edition

24.4 Nuclear Magnetic Resonance Spectroscopy 1021

Impurity and Solvent Effects

If a substance contains impurities or if it is dissolved in a solvent, its proton NMR

spectrum can be different from that of the pure substance. Figure 24.6a shows the

proton NMR spectrum of ethanol containing a trace of water. Figure 24.6b shows the

spectrum of carefully purified ethanol, and Figure 24.6c shows the spectrum of a dilute

solution of ethanol in deuterated chloroform. We first consider the chemical shifts. In

all three spectra the –CH 3 protons have aδvalue near 1 ppm, as expected, since they

are most distant from the electronegative O atom. The –CH 2 – protons have aδvalue

between 3 and 4, because they are closer to the O atom than the –CH 3 protons. In the

first spectrum, the spectral line of the –OH proton is atδ 4 .8 ppm and in the second

spectrum this peak is atδ 5 .3 ppm. This is reasonable, because this proton is bonded

to an electronegative O atom. However, in the solution with deuterated chloroform

this spectral line is atδ 2 .3 ppm. Ethanol molecules form hydrogen bonds with other

ethanol molecules or with water molecules. When a hydrogen bond forms, the proton

is attracted to the other molecule and is at a greater distance from the oxygen nucleus.

It lies in a region of lower electron density and gives a spectral line at a greater value

ofδ. In the solution in deuterated chloroform, the ethanol molecules are distant from

each other and cannot form hydrogen bonds. In the absence of hydrogen bonding the

O–H bond distance, approximately 96 pm, is so small that the proton is imbedded

in the electrons close to the oxygen and is more highly shielded, giving a smaller

value ofδ.

3 2 1 0

(a)

6 5 4

6 5 4 3 2 1 0

6 5 4 3 2 1 0

/ ppm

(b)

(c)

OH CH 2

CH 3

Figure 24.6 The Proton NMR Spec-

trum of Ethanol.(a) With a trace of

water present. Because of exchange of

the hydroxyl proton with water protons,

the hydroxyl proton shows no spin–

spin splitting. From G. W. Castellan,

Physical Chemistry, 3rd ed., Addison-

Wesley, Reading, MA, 1983, p. 606.

(b) Highly purified ethanol. In the

absence of water, the spin–spin split-

ting of the line from the hydroxyl proton

is split into three lines by the methylene

protons. From I. N. Levine,Molecular

Spectroscopy, Wiley, New York, 1975,

p. 353. (c) In deuterated chloroform. In

this spectrum, the solvent interaction

with chloroform moves the hydroxyl

line to near 2.6 ppm. The splitting into

three lines is not quite resolved. From

C. J. Pouchert and J. R. Campbell,

The Aldrich Library of NMR Spectra,

The Aldrich Chemical Co., 1974, Vol. I,

p. 79.

We now consider the spin–spin splitting. As expected the –CH 3 protons produce

a triplet in all three spectra because of the splitting due to the –CH 2 – protons. In the

spectrum of the carefully purified ethanol the line of the –OH proton exhibits spin–

spin splitting, being split into three lines by the two protons of the –CH 2 – group. The

–CH 2 – protons produce a set of eight lines (not all completely resolved) from the

four lines because of the –CH 3 protons, each of which is split into two lines by the

–OH proton. In the solution spectrum this splitting is also present although poorly

resolved. In the presence of a trace of water there is no spin–spin splitting in the

–OH proton line. This is attributed to exchange of the proton with protons on water

molecules. If the average time for this exchange process is shorter than the period of

oscillation of the NMR radiation, the proton NMR spectrum will be an average of

the spectrum of the proton that is leaving and the proton that is arriving, and these

protons have equal probabilities of being spin up and spin down. An average line

with no spin–spin splitting is observed for the –OH proton and the –CH 2 – protons

produce only the quartet expected from the splitting by the three protons on the –CH 3

group.

Saturation of a Signal

The difference in energy between the two states of protons in a typical magnetic field is

much smaller thankBT, so that the populations of the two states are nearly equal. If the

populations are exactly equal, there would be no signal, because the stimulated emission

and the absorption would cancel each other. It is possible with a strong signal to produce

enough transitions to equalize the populations and cause the signal to disappear (this is

called “saturating” the signal). In certain kinds of experiments, the effect of a spin–spin

coupling can be removed by equalizing the populations of some nuclei (this is called

spin decoupling).