BIOINORGANIC CHEMISTRY A Short Course Second Edition

NUCLEAR MAGNETIC RESONANCE 99

nucleus fi nds itself. This variation results in the “ chemical shift effect ” (to be
explained further below), one of the most useful results of nuclear magnetic
resonance spectroscopy.
Nuclei possessing even numbers of both protons and neutrons have angular
momentums equal to 0 ( I = 0) and are magnetically inactive. Two major
isotopes in organic and biological species,^12 C and^16 O, belong to this group,
a fact that simplifi es the NMR spectra of molecules containing them. Odd
numbers of either (or both) protons or neutrons lead to nonzero spins,
with the actual number of spin states being dependent on the number of
possible arrangements. Because s bonding electrons have charge density
within the nucleus and will perturb nuclear spin states, ultimately the NMR
technique leads to information about molecular arrangements surrounding
the nucleus under study. This is the information being sought in the NMR
experiment.
Placing the nucleus with I = 1 and a positive magnetogyric ratio in a particu-
lar magnetic fi eld B 0 (aligned along a z axis) leads to 2 I + 1 arrangements at
differing potential energies split by ( m μ /I) B 0. as shown in Figure 3.9A.
Alignment of vectors relative to B 0 for each value of m results in vectors
of length [ I ( I + 1)] 1/2 and a z component [ I ( I + 1)] 1/2 , where cos θ = m [ I ( I + 1)] 1/2
as shown Figure 3.9B. Differing energy states lead to interaction with electro-
magnetic radiation of the correct frequency according to ΔE = hν. For nuclear
magnetic resonance observations, equation 3.14 must hold because the selec-
tion rule for energy transitions between adjacent states ( Δm = ± 1) operates and
in terms of the magnetogyric ratio, one fi nds equation 3.15.

h

I

νμ=

B 0

(3.14)

γ

πμ

=

2

hI

(3.15)

Figure 3.9 (A) Nuclear spin energy for a nucleus having I = 1 (e.g.,^14 N) plotted as a
function ofB 0. (B) Nuclear vectors relative to B 0 with vector length [ I ( I + 1)] 1/2 and z
component m so that cos θ = m [ I ( I + 1)] 1/2. (Adapted with permission of Nelson Thornes
Ltd. from Figures 1a and 1b of reference 21 .)

Energy

μBo

μBo

m = -1

m = 0

m = +1

Field Strength Bo Bo

m = -1

m = 0

m = +1

θ

hm

h[I(I+1)]1/2

I

I

A

B

ћ ћ

ћ

ћ