Physical Chemistry Third Edition

1022 24 Magnetic Resonance Spectroscopy

EXAMPLE24.15

Find the ratio of the populations of the two spin states of protons in a magnetic field of

4.6973 T at 298 K.

Solution

(Population ratio)e−∆E/kBTe−gpβNB/kBT

exp

(

−(5.5857)(5. 05079 × 10 −^27 JT−^1 )(4.6973 T)

(1. 3807 × 10 −^23 JK−^1 )(298 K)

)

e−^3.^2205 ×^10

− 5

 0. 9999678

NMR Spectroscopy with Other Nuclei

Although proton NMR is the most common type of NMR spectroscopy, other nuclei

are also used. It turns out that nuclei withIgreater than 1/2 give broad lines that are not

likely to be resolved well enough to give much useful information, so these nuclei are

not used in NMR. The most prominent NMR nucleus after the proton is^13 C, although

NMR of^19 F and^31 P is fairly common. The natural abundance of the^13 C isotope is

approximately 1%, so that a^13 C nucleus is unlikely to have another^13 C nucleus in

close proximity in a substance with the natural isotopic composition. In this case the

(^13) C– (^13) C spin–spin coupling can be ignored.
Almost every^13 C nucleus will be close to protons, so the spin–spin splitting between
protons and^13 C nuclei is important in^13 C NMR. The technique ofspin decoupling
is used to simplify the spectra. The sample is strongly irradiated with radiation at the
resonant frequency of some or all of the protons. This causes the^13 C multiplets to
collapse to singlets. The simple explanation is that the irradiation causes the protons
to make rapid transitions between the spin-up to the spin-down states so that they do
not remain in a single spin state for long enough to provide the splitting. This expla-
nation is similar to the explanation of the lack of spin–spin splitting in an alcohol with
water present. There is a more complicated explanation in terms of precession of the
spins.^4

EXAMPLE24.16

Calculate the resonance frequency of^13 C nuclei at the magnetic field that produces resonance

with protons at 200.00 MHz.

Solution

From Example 24.11,Bz 4 .6973 T. From Table A.24 of Appendix A,gN 1 .4048.

ν

gNβNBz

h



(1.4048)(5. 050824 × 10 −^27 JT−^1 )(4.6973 T)

6. 6261 × 10 −^34 Js

 5. 0300 × 107 s−^1  50 .300 MHz

(^4) J. B. Lambert and E. P. Mazzola,Nuclear Magnetic Resonance Spectroscopy, Pearson, Upper Saddle
River, NJ, 2004, p. 144ff.