Physical Chemistry Third Edition

24.4 Nuclear Magnetic Resonance Spectroscopy 1015

The selection rule for magnetic dipole transitions of nuclear spins is

∆MI± 1 (24.4-1)

so that the energy of the photon absorbed or emitted is

Ephotonhν|gN|βNBzγNhB ̄ z

γNhBz

2 π

(24.4-2)

For the frequencyνthe magnetic field must have the value

BzBres

hν

gNβN



2 πν

γN

(24.4-3)

wheregNis the nucleargfactor for the nucleus andγNis its magnetogyric ratio.

EXAMPLE24.11

Find the value of the magnetic field necessary for protons to absorb at a frequency of

200.00 MHz.

Solution

Bz

hν

gNβN



(6. 6261 × 10 −^34 J s)(200. 00 × 106 s−^1 )

(5.5857)(5. 050787 × 10 −^27 JT−^1 )

 4 .6973 T46973 gauss

Diamagnetic Shielding

If the frequency of the radiation is fixed and if the externally applied field were the

only contribution to the magnetic field at the nucleus, every proton would absorb

at the same value of the magnetic field, every^13 C nucleus would absorb at another

value of the magnetic field, and so on. However, there are two contributions to the

magnetic field at a given nucleus in addition to the externally applied magnetic field.

The first is due to the fact that the externally applied magnetic field induces a net

current in the electrons of the molecule. This produces a magnetic field in the opposite

direction to the externally applied field. This phenomenon is calleddiamagnetismand

is said to provideshielding. The diamagnetic contribution to the magnetic field at a

given nucleus is proportional to the applied field and depends on the electron density

around the nucleus. For nucleus numberjin a given molecule, this diamagnetic

contribution is

Bjdiamagnetic−σjB 0 (24.4-4)

whereσjis called theshielding constantfor nucleus numberjand whereB 0 is the

externally applied magnetic field. The shielding constant has a larger value when the

probability of finding electrons around the nucleus is larger. Typical values ofσrange

from 15× 10 −^6 to 35× 10 −^6 (15 to 35 parts per million).

If we ignore other factors, the magnetic field at thejth nucleus is

Bj(1−σj)B 0 (24.4-5)