Chapter 14 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula

CHAPTER 14

Magnetic resonance
The techniques of ‘magnetic resonance’ probe transitions
between spin states of nuclei and electrons in molecules.
‘Nuclear magnetic resonance’ (NMR) spectroscopy, the focus
of this chapter, is one of the most widely used procedures in
chemistry for the exploration of structural and dynamical properties of molecules of all sizes, up to as large as biopolymers.

14A  General principles
The chapter begins with an account of the principles that govern spectroscopic transitions between spin states of nuclei and
electrons in molecules. It also describes simple experimental
arrangements for the detection of these transitions. The concepts developed in this Topic prepare the ground for a discussion of the chemical applications of NMR and ‘electron
paramagnetic resonance’ (EPR).

14B  Features of NMR spectra
This Topic contains a discussion of conventional NMR, showing how the properties of a magnetic nucleus are affected by
its electronic environment and the presence of magnetic nuclei
in its vicinity. These concepts lead to understanding of how
molecular structure governs the appearance of NMR spectra.

techniques that NMR spectroscopy can probe a vast array of
small and large molecules in a variety of environments.

14D  Electron paramagnetic resonance
The experimental techniques for EPR resemble those used in
the early days of NMR. The information obtained is used to
investigate species with unpaired electrons. This Topic includes
a brief survey of the applications of EPR to the study of organic
radicals and d-metal complexes.

What is the impact of this material?
Magnetic resonance techniques are ubiquitous in chemistry,
as they are an enormously powerful analytical and structural
technique, especially in organic chemistry and biochemistry.
One of the most striking applications of nuclear magnetic
resonance is in medicine. ‘Magnetic resonance imaging’
(MRI) is a portrayal of the concentrations of protons in a
solid object (Impact I14.1). The technique is particularly useful for diagnosing disease. In Impact I14.2 we highlight an
application of electron paramagnetic resonance in materials
science and biochemistry: the use of a ‘spin probe’, a radical
that interacts with biopolymer or a nanostructure and has
an EPR spectrum that reveals its structural and dynamical
properties.

14C  Pulse techniques in NMR
In this Topic we turn to the modern versions of NMR, which
are based on the use of pulses of electromagnetic radiation
and the processing of the resulting signal by ‘Fourier transform’ techniques. It is through the application of these pulse

To read more about the impact of
this material, scan the QR code, or go to
bcs.whfreeman.com/webpub/chemistry/
pchem10e/impact/pchem-14-1.html


14A  General principles
Contents
14A.1 

Nuclear magnetic resonance

The energies of nuclei in magnetic fields
Brief illustration 14A.1: The resonance
condition in NMR
(b) The NMR spectrometer
Brief illustration 14A.2: Nuclear spin populations
(a)

14A.2 

Electron paramagnetic resonance
The energies of electrons in magnetic fields
Brief illustration 14A.3: The resonance condition
in EPR
(b) The EPR spectrometer
Brief illustration 14A.4: Electron spin populations
(a)

Checklist of concepts
Checklist of equations

561
561
563
563
564
564
565
565
566
566

567
567

➤➤ Why do you need to know this material?
Nuclear magnetic resonance spectroscopy is used widely
in chemistry and medicine. To understand the power
of magnetic resonance, you need to understand the
principles that govern spectroscopic transitions between
spin states of electrons and nuclei in molecules.

condition of strong effective coupling when the frequencies of
two oscillators are identical is called resonance. Resonance is
the basis of a number of everyday phenomena, including the
response of radios to the weak oscillations of the electromagnetic field generated by a distant transmitter. Historically, spectroscopic techniques that measure transitions between nuclear
and electron spin states have carried the term ‘resonance’ in
their names because they have depended on matching a set
of energy levels to a source of monochromatic radiation and
observing the strong absorption that occurs at resonance. In
fact, all spectroscopy is a form of resonant coupling between
the electromagnetic field and the molecules; what distinguishes
magnetic resonance is that the energy levels themselves are
modified by the application of a magnetic field.
The Stern–Gerlach experiment (Topic 9B) provided evidence for electron spin. It turns out that many nuclei also
possess spin angular momentum. Orbital and spin angular
momenta give rise to magnetic moments, and to say that electrons and nuclei have magnetic moments means that, to some
extent, they behave like small bar magnets with energies that
depend on their orientation in an applied magnetic field. Here
we establish how the energies of electrons and nuclei depend
on the applied field. This material sets the stage for the exploration of the structure and dynamics of many kinds of molecules
by magnetic resonance spectroscopy (Topics 14B–14D).

➤➤ What is the key idea?
Resonant absorption occurs when the separation of the
energy levels of spins in a magnetic field matches the
energy of incident photons.

➤➤ What do you need to know already?
You need to be familiar with the quantum mechanical
concept of spin (Topic 9B), the Boltzmann distribution
(Foundations B and Topic 15A), and the general features of
spectroscopy (Topic 12A).

14A.1  Nuclear

magnetic resonance

The application of resonance that we describe here depends
on the fact that many nuclei possess spin angular momentum
characterized by a nuclear spin quantum number I (the analogue of s for electrons). To understand the nuclear magnetic
resonance (NMR) experiment we need to describe the behaviour of nuclei in magnetic fields and then the basic techniques
for detecting spectroscopic transitions.

(a)  The energies of nuclei in magnetic fields
When two pendulums share a slightly flexible support and
one is set in motion, the other is forced into oscillation by the
motion of the common axle. As a result, energy flows between
the two pendulums. The energy transfer occurs most efficiently
when the frequencies of the two pendulums are identical. The

The nuclear spin quantum number, I, is a fixed characteristic

property of a nucleus in its ground state (the only state we consider) and, depending on the nuclide, is either an integer or a
half-integer (Table 14A.1). A nucleus with spin quantum number I has the following properties:


562  14  Magnetic resonance
Table 14A.1  Nuclear constitution and the nuclear spin
quantum number*

proportional to its angular momentum. The operators in eqn
14A.2 are then:

Number of protons

Number of neutrons

I

Even

Even

0

Odd
Even

Odd
Odd

Integer (1, 2, 3, …)

Odd

Even

Half-integer ( 1 , 3 , 5 ,…)
2 2 2

µˆ = γ N Iˆ and Hˆ = −γ N B ⋅ Iˆ

Half-integer ( 1 , 3 , 5 ,…)
2 2 2

• An angular momentum of magnitude

{I(I + 1)}1/2.

• A component of angular momentum m I on a
specified axis (‘the z-axis’), where m I = I, I − 1, …, −I.
• If I > 0, a magnetic moment with a constant
magnitude and an orientation that is determined by
the value of m I.

Physical interpretation

* The spin of a nucleus may be different if it is in an excited state; throughout this
chapter we deal only with the ground state of nuclei.

According to the second property, the spin, and hence the magnetic moment, of the nucleus may lie in 2I + 1 different orientations relative to an axis. A proton has I = 12 and its spin may
adopt either of two orientations; a 14N nucleus has I = 1 and its

spin may adopt any of three orientations; both 12C and 16O have
I = 0 and hence zero magnetic moment.
Classically, the energy of a magnetic moment µ in a magnetic field ℬ is equal to the scalar product (Mathematical background 5 following Chapter 9)
(14A.1)



More formally, ℬ is the magnetic induction and is measured
in tesla, T; 1 T = 1 kg s−2 A−1. The (non-SI) unit gauss, G, is also
occasionally used: 1 T = 104 G. Quantum mechanically, we write
the hamiltonian as
Hˆ = − µˆ ⋅B



(14A.2)

To write an expression for μˆ , we use the fact that, just as for
electrons (Topic 9B), the magnetic moment of a nucleus is

(14A.3a)

where γN is the nuclear magnetogyric ratio of the specified nucleus, an empirically determined characteristic arising
from its internal structure (Table 14A.2). For a magnetic field
of magnitude ℬ0 along the z-direction, the hamiltonian in eqn
14A.3a becomes
Hˆ = −γ N B0 Iˆz

(14A.3b)

Because the eigenvalues of the operator Iˆz are mI, the eigenvalues of this hamiltonian are
Em = −γ N B0 mI
I

Energies of a nuclear
spin in a magnetic field (14A.4a)



When written in terms of the nuclear magneton, μN,
μN =

e
= 5.051 ×10−27 J T −1
2mp



Nuclear magneton  (14A.4b)

(where mp is the mass of the proton) and an empirical constant called the nuclear g-factor, gI, the energy in eqn 14A.4a
becomes
Em = − g I µN B0 mI
I

E = − µ ⋅B



gI =

γ N
µN

Energies of a nuclear
spin in a magnetic field (14A.4c)

Nuclear g-factors are experimentally determined dimensionless quantities with values typically between –6 and +6 (Table
14A.2). Positive values of gI and γN denote a magnetic moment
that lies in the same direction as the spin angular momentum
vector; negative values indicate that the magnetic moment and
spin lie in opposite directions. A nuclear magnet is about 2000
times weaker than the magnet associated with electron spin.
For the remainder of our discussion of nuclear magnetic
resonance we assume that γN is positive, as is the case for the
majority of nuclei. In such cases, it follows from eqn 14A.4c

Table 14A.2*  Nuclear spin properties
Nuclide

Natural abundance/%

1n

Spin I

g-factor, gI

Magnetogyric ratio, γN/(107 T−1 s−1)

NMR frequency at 1 T, ν/MHz

1
2

−3.826

−18.32

29.164

1H

99.98

1
2

5.586

26.75

42.576

2H

0.02

1

0.857

4.11

6.536

13C

1.11

1
2

1.405

6.73

10.708

14N

99.64

1

0.404

1.93

3.078

* More values are given in the Resource section.


14A  General principles  
that states with mI > 0 lie below states with mI < 0. It follows that
the energy separation between the lower mI = + 12 (α) and upper
mI = − 12 (β) states of a spin- 12 nucleus, a nucleus with I = 12 , is
∆E = E −1/2 − E +1/2 = 12 γ N B0 − (− 12 γ N B0 ) = γ N B0

γ N B0
2π

Spin- 12 nuclei

Resonance
condition

(14A.6)

is fulfilled (Fig. 14A.1). At resonance there is strong coupling
between the spins and the radiation, and absorption occurs as
the spins flip from the lower energy state to the upper state.
It is sometimes useful to compare the quantum mechanical
and classical pictures of magnetic nuclei pictured as tiny bar
magnets. A bar magnet in an externally applied magnetic field
undergoes the motion called precession as it twists round the
direction of the field (Fig. 14A.2). The rate of precession νL is
called the Larmor precession frequency:
L =

γ N B0
2π

Definition 

Larmor frequency of a nucleus  (14A.7)

β, mI = –½
Magnetic
field off

γNhB0

It follows by comparing this expression with eqn 14A.6 that
resonance absorption by spin- 12 nuclei occurs when the
Larmor precession frequency νL is the same as the frequency of
the applied electromagnetic field, ν.

(14A.5)

and resonant absorption occurs when the resonance condition
h = γ N B0 or  =

563

Magnetic
field on

α, mI = +½

Figure 14A.1  The nuclear spin energy levels of a spin- 21
nucleus with positive magnetogyric ratio (for example, 1H or
13C) in a magnetic field. Resonance occurs when the energy
separation of the levels matches the energy of the photons in
the electromagnetic field.
z

ms = +½
ms = –½

Figure 14A.2  The interactions between the mI states of a spin1
2 nucleus and an external magnetic field may be visualized
as the precession of the vectors representing the angular
momentum.

Brief illustration 14A.1  The resonance condition in NMR

To calculate the frequency at which radiation comes into resonance with proton (I = 12 ) spins in a 12.0 T magnetic field we
use eqn 14A.6 as follows:
γN

B0

(2.6752 ×10 T s ) × (12.0 T)
= 5.11 × 108 s −1
2π
= 511 MHz

ν=

8

−1 −1

Self-test 14A.1  Determine the resonance frequency for

nuclei, for which γ N
conditions.

 = 1.0841 × 10 8 T−1 s −1,

31P

under the same
Answer: 207 MHz

(b)  The NMR spectrometer
In its simplest form, NMR is the study of the properties of
molecules containing magnetic nuclei by applying a magnetic
field and observing the frequency of the resonant electromagnetic field. Larmor frequencies of nuclei at the fields normally
employed (about 12 T) typically lie in the radiofrequency region
of the electromagnetic spectrum (close to 500 MHz), so NMR
is a radiofrequency technique. For much of our discussion we
consider spin- 12 nuclei, but NMR is applicable to nuclei with any
non-zero spin. As well as protons, which are the most common
nuclei studied by NMR, spin- 12 nuclei include 13C, 19F, and 31P.
An NMR spectrometer consists of the appropriate sources of
radiofrequency radiation and a magnet that can produce a uniform, intense field. Most modern instruments use a superconducting magnet capable of producing fields of the order of 10 T
and more (Fig. 14A.3). The sample is rotated rapidly to average

out magnetic inhomogeneities; however, although sample spinning is essential for the investigation of small molecules, for
large molecules it can lead to irreproducible results and is often
avoided. Although a superconducting magnet (Topic 18C)
operates at the temperature of liquid helium (4 K), the sample
itself is normally at room temperature or held in a variable temperature enclosure between, typically, −150 and +100 °C.
Modern NMR spectroscopy uses pulses of radiofrequency
radiation. These techniques of Fourier-transform (FT) NMR
make possible the determination of structures of very large molecules in solution and in solids. They are discussed in Topic 14C.
The intensity of an NMR transition depends on a number of
factors. We show in the following Justification that
Intensity ∝ (N α − N β )B0

(14A.8a)


564  14  Magnetic resonance
as it is very slow. Therefore, the net rate of absorption is proportional to the difference in populations, and we can write

Superconducting
magnet
Computer

Probe

Preamplifier

Rate of absorption ∝ N α − N β

Receiver Detector

Transmitter

Figure 14A.3  The layout of a typical NMR spectrometer. The
link from the transmitter to the detector indicates that the
high frequency of the transmitter is subtracted from the high
frequency received signal to give a low frequency signal for
processing.

where

The intensity of absorption, the rate at which energy is
absorbed, is proportional to the product of the rate of absorption (the rate at which photons are absorbed) and the energy
of each photon. The latter is proportional to the frequency ν of
the incident radiation (through E = hν). At resonance, this frequency is proportional to the applied magnetic field (through
ν = γ Nℬ0/2π), so we can write
Rate of absorption ∝ (N α − N β )B0
as in eqn 14A.8a. To write an expression for the population
difference, we use the Boltzmann distribution (Foundations B
and Topic 15A) to write the ratio of populations as
∆E

Nα − Nβ ≈

N γ N B0
2kT

Nuclei 

Population difference  (14A.8b)

with N the total number of spins (N = Nα + Nβ). It follows that
decreasing the temperature increases the intensity by increasing the population difference.

For protons γ N = 2.675 × 10 8 T−1 s −1. Therefore, for 1 000 000
protons in a field of 10 T at 20 °C
γN

1000000 × (2.675 ×10 T s ) × (1.055 × 10 Js) × 10 T
Nα − Nβ ≈
2 × (1.381 × 10−23 JK −1 ) × (293 K)
−1 −1

k

≈

1−

γ N B0
kT

The expansion of the exponential term is appropriate for
ΔE = γ Nℬ0 ≪ kT, a condition usually met for nuclear spins. It
follows that
1−γ N B0 /kT

1−γ N B0 /kT

N

1 − (1 − γ N B0 /kT ) γ N B0 /kT
=
≈
2
1 + (1 − γ N B0 /kT )

B0
8

−34

≈1

which is eqn 14A.8b.

T

≈ 35

Even in such a strong field there is only a tiny imbalance of
population of about 35 in a million.
nuclei, γ N  = 6.7283 × 10 7  T−1  s −1 .
Determine the magnetic field that would need to be achieved
in order to induce the same imbalance in the distribution of
13C spins at 20 °C.
Self-test 14A.2  For

e − x =1− x +

B0 /kT

N α − N β N α (1 − N β /N α ) 1 − N β /N α
=
=
N α + N β N α (1 + N β /N α ) 1 + N β /N α

Brief illustration 14A.2  Nuclear spin populations

N

Nβ
= e −γ N
Nα

13 C

Answer: 40 T

Justification 14A.1  Intensities in NMR spectra

From the general considerations of transition intensities in
Topic 12A, we know that the rate of absorption of electromagnetic radiation is proportional to the population of the lower
energy state (Nα in the case of a proton NMR transition) and
the rate of stimulated emission is proportional to the population of the upper state (Nβ). At the low frequencies typical of
magnetic resonance, spontaneous emission can be neglected

By combining eqns 14A.8a and 14A.8b we see that the intensity is proportional to B02, so NMR transitions can be enhanced
significantly by increasing the strength of the applied magnetic
field. The use of high magnetic fields also simplifies the appearance of spectra (a point explained in Topic 14B) and so allows
them to be interpreted more readily. We can also conclude that

absorptions of nuclei with large magnetogyric ratios (1H, for
instance) are more intense than those with small magnetogyric
ratios (13C, for instance).

14A.2  Electron

resonance

paramagnetic

Electron paramagnetic resonance (EPR), or electron spin
resonance (ESR), is the study of molecules and ions containing unpaired electrons by observing the magnetic field at which
they come into resonance with radiation of known frequency.


14A  General principles  
As we have done for NMR, we write expressions for the resonance condition in EPR and then describe the general features
of EPR spectrometers.

(a)  The energies of electrons

in magnetic fields

µˆ = γ e sˆ and Hˆ = −γ eB ⋅ sˆ

(14A.9a)



where sˆ is the spin angular momentum operator, and γe is the

magnetogyric ratio of the electron:
γe =−

g ee
2me

degeneracy is removed: the state with ms = + 12 moves up in energy
by 12 g e μBB0 and the state with ms = − 12 moves down by 12 g e μBB0 .
From eqn 14A.11b, the separation between the (upper)
ms = + 12 (α) and (lower) ms = − 12 (β) levels of an electron spin in
a magnetic field of magnitude B0 in the z-direction is
∆E = E +1/2 − E −1/2 = 12 g e μBB0 − (− 12 g e μBB0 )

The spin magnetic moment of an electron, which has a spin
quantum number s = 12 (Topic 9B), is proportional to its spin
angular momentum. The spin magnetic moment and hamiltonian operators are, respectively,

Electrons 

Magnetogyric ratio  (14A.9b)

with ge = 2.002 319… as the g-value of the electron. (Note that
the current convention is to include the g-value in the definition of the magnetogyric ratio.) Dirac’s relativistic theory, his
modification of the Schrödinger equation to make it consistent with Einstein’s special relativity, gives ge = 2; the additional
0.002 319… arises from interactions of the electron with the
electromagnetic fluctuations of the vacuum that surrounds the
electron. The negative sign of γe (arising from the sign of the
electron’s charge) shows that the magnetic moment is opposite
in direction to the angular momentum vector.
For a magnetic field of magnitude ℬ0 in the z-direction,

565

= g e μBB0

(14A.12a)



The energy separations come into resonance with the electromagnetic radiation of frequency ν when
h = g e μBB0

Electrons 

Resonance condition  (14A.12b)

This is the resonance condition for EPR (Fig. 14A.4). At resonance there is strong coupling between the electron spins and
the radiation, and strong absorption occurs as the spins make
the transition α ← β.
Brief illustration 14A.3  The resonance condition in EPR

Magnetic fields of about 0.30 T (the value used in most commercial EPR spectrometers) correspond to resonance at
ge

ν=

µB

B0

(2.0023) × (9.274 ×10 J T ) × (0.30 T)
6.626 ×10−34 Js
−24

−1

h

= 8.4 ×109 s −1 = 8.4 GHz
which corresponds to a wavelength of 3.6 cm.
Self-test 14A.3  Determine the magnetic field for EPR tran-

(14A.10)

sitions in a spectrometer that uses radiation of wavelength
0.88 cm.

Because the eigenvalues of the operator sˆz are ms with
ms = + 12 (α) and ms = − 12 (β), it follows that the energies of an
electron spin in a magnetic field are

Answer: 1.2 T

Hˆ = −γ eB0 sˆz

Em = −γ e B0 ms
s

Energies of an electron
spin in a magnetic field

(14A.11a)

They can also be expressed in terms of the Bohr magneton, μB, as
Em = g e μBB0 ms
s



Energies of an electron
spin in a magnetic field

(14A.11b)

α, ms = +½
Magnetic
field off

Magnetic
field on
ge µBB0

where
μB =

e
= 9.274 ×10−24 J T −1
2me

Bohr magneton  (14A.11c)

The Bohr magneton, a positive quantity, is often regarded as the
fundamental quantum of magnetic moment.
In the absence of a magnetic field, the states with different values of ms are degenerate. When a field is present, the

β, ms = –½

Figure 14A.4  Electron spin levels in a magnetic field. Note
that the β state is lower in energy than the α state (because the
magnetogyric ratio of an electron is negative). Resonance is
achieved when the frequency of the incident radiation matches
the frequency corresponding to the energy separation.


566  14  Magnetic resonance
(b)  The EPR spectrometer

Microwave
source

Detector

Absorption, A

Signal

It follows from Brief illustration 14A.3 that most commercial EPR spectrometers operate at wavelengths of approximately 3 cm. Because 3 cm radiation falls in the microwave

region of the electromagnetic spectrum, EPR is a microwave
technique.
Both Fourier-transform (FT) and continuous wave (CW)
EPR spectrometers are available. The FT-EPR instrument is
based on the concepts developed in Topic 14C for NMR spectroscopy, except that pulses of microwaves are used to excite
electron spins in the sample. The layout of the more common
CW-EPR spectrometer is shown in Fig. 14A.5. It consists of a
microwave source (a klystron or a Gunn oscillator), a cavity in
which the sample is inserted in a glass or quartz container, a
microwave detector, and an electromagnet with a field that can
be varied in the region of 0.3 T. The EPR spectrum is obtained
by monitoring the microwave absorption as the field is
changed, and a typical spectrum (of the benzene radical anion,
C 6 H6− ) is shown in Fig. 14A.6. The peculiar appearance of the
spectrum, which is in fact displayed as the first-derivative of the

Slope
Slope
Field, B
Derivative
of absorption, dA/dB

Figure 14A.7  When phase-sensitive detection is used, the
signal is the first derivative of the absorption intensity. Note
that the peak of the absorption corresponds to the point where
the derivative passes through zero.

absorption, arises from the detection technique, which is sensitive to the slope of the absorption curve (Fig. 14A.7).
As usual, the intensities of spectral lines in EPR depend on
the difference in populations between the ground and excited

states. For an electron, the β state lies below the a state in
energy and, by a similar argument to that for nuclei,
Nβ − Nα ≈

Sample
cavity
Phase
sensitive
detector

Ng e μBB0
Electrons 
2kT

Population difference  (14A.13)

where N is the total number of spins.

Brief illustration 14A.4  Electron spin populations

Electromagnet

When 1000 electron spins are exposed to a 1.0 T magnetic field
at 20 °C (293 K),

Modulation
unit

N

Figure 14A.5  The layout of a continuous-wave EPR
spectrometer. A typical magnetic field is 0.3 T, which requires
9 GHz (3 cm) microwaves for resonance.

Nβ − Nα ≈

μB

ge

B0

1000 × 2.0023 × (9.274 ×10 J T ) × (1.0 T)
2 × (1.381 × 10−23 JK −1) × (293 K)
−24

k

−1

T

≈ 2.3
There is an imbalance of populations of only about 2 electrons
in a thousand. However, the imbalance is much larger for
electron spins than for nuclear spins (Brief illustration 14A.2)
because the energy separation between the spin states of electrons is larger than that for nuclear spins even at the lower
magnetic field strengths normally employed.

Field strength

Figure 14A.6  The EPR spectrum of the benzene radical anion,
C6H6− , in fluid solution.

Self-test 14A.4  It is common to conduct EPR experiments
at very low temperatures. At what temperature would the
imbalance in spin populations be 5 electrons in 100, with
ℬ0 = 0.30 T?
Answer: 4 K


14A  General principles  

567

Checklist of concepts
☐1.The nuclear spin quantum number, I, of a nucleus is
either a non-negative integer or half-integer.
☐2.Nuclei with different values of m I have different energies in the presence of a magnetic field.
☐3.Nuclear magnetic resonance (NMR) is the observation
of resonant absorption of radiofrequency electromagnetic radiation by nuclei in a magnetic field.
☐4.NMR spectrometers consist of a source of radiofrequency radiation and a magnet that provides a strong,
uniform field.

☐5.The resonance absorption intensity increases with the
strength of the applied magnetic field (as B02 ).
☐6.Electrons with different values of m s have different
energies in the presence of a magnetic field.
☐7.Electron paramagnetic resonance (EPR) is the observation of resonant absorption of microwave electromagnetic radiation by unpaired electrons in a magnetic field.
☐8.EPR spectrometers consist of a microwave source, a

cavity in which the sample is inserted, a microwave
detector, and an electromagnet.

Checklist of equations
Property

Equation

Comment

Equation number

Nuclear magneton

μN = e/2mp

μN = 5.051 × 10−27 J T−1

14A.4b

Energies of a nuclear spin in a magnetic field

EmI = −γ N B0mI

14A.4c

= − g I µNB0mI
Resonance condition (spin- 12 nuclei)

hν = γNℬ0

γN > 0

14A.6

Larmor frequency

νL = γNℬ0/2π

γN > 0

14A.7

Population difference (nuclei)

Nα − Nβ ≈ NγNℬ0/2kT

Magnetogyric ratio (electron)

γe = −gee/2me

Energies of an electron spin in a magnetic field

Ems = −γ e B0ms

14A.8b
ge = 2.002 319

14A.9b
14A.11b

= g e µBB0ms
Bohr magneton

μB = e/2me

Resonance condition (electrons)

hν = geμBℬ0

14A.12b

Population difference (electrons)

Nβ − Nα ≈ NgeμBℬ0/2kT

14A.13

μB = 9.274 × 10−24 J T−1

14A.11c


14B  Features of NMR spectra
Contents
14B.1 

➤➤ What is the key idea?

The chemical shift

Brief illustration 14B.1: The δ scale
Example 14B.1: Interpreting the NMR spectrum
of ethanol

14B.2 

The origin of shielding constants
The local contribution
Example 14B.2: Using the Lamb formula
(b) Neighbouring group contributions
Brief illustration 14B.2: Ring currents
(c) The solvent contribution
Brief illustration 14B.3: The effect of aromatic
solvents
(a)

14B.3 

The fine structure
The appearance of the spectrum
Example 14B.3: Accounting for the fine
structure in a spectrum
(b) The magnitudes of coupling constants
Brief illustration 14B.4: The Karplus equation
(c) The origin of spin–spin coupling
Brief illustration 14B.5: Magnetic fields
from nuclei
(d) Equivalent nuclei
Brief illustration 14B.6: Chemical and magnetic
equivalence

(e) Strongly coupled nuclei
Brief illustration 14B.7: Strongly coupled spectra
(a)

14B.4 Conformational

processes

568
569
569
570
570
570
571
572
573
573
573
573
575
575
576
576
576
577
578
579
579

conversion and exchange

Brief illustration 14B.8: The effect of chemical
exchange on NMR spectra
Checklist of concepts
Checklist of equations

580
580
581
581

The resonance frequency of a magnetic nucleus is
affected by its electronic environment and the presence of
magnetic nuclei in its vicinity.

➤➤ What do you need to know already?
You need to be familiar with the general principles
of magnetic resonance (Topic 14A) and specifically
that resonance occurs when the frequency of the
radiofrequency field matches the Larmor frequency.

Nuclear magnetic moments interact with the local magnetic
field. The local field may differ from the applied field because
the latter induces electronic orbital angular momentum (that
is, the circulation of electronic currents) which gives rise to a
small additional magnetic field δℬ at the nuclei. This additional
field is proportional to the applied field, and it is conventional
to write
δB = −σB0

To make progress with the analysis of NMR spectra and
extract the wealth of information they contain you need to
understand how the appearance of a spectrum correlates
with molecular structure.

Shielding constant  (14B.1)

where the dimensionless quantity σ is called the shielding constant of the nucleus (σ is usually positive but may be negative).
The ability of the applied field to induce an electronic current
in the molecule, and hence affect the strength of the resulting
local magnetic field experienced by the nucleus, depends on
the details of the electronic structure near the magnetic nucleus
of interest, so nuclei in different chemical groups have different shielding constants. The calculation of reliable values of the
shielding constant is very difficult, but trends in it are quite well
understood and we concentrate on them.

14B.1  The
➤➤ Why do you need to know this material?

Definition 

chemical shift

Because the total local field ℬloc is

Bloc = B0 + δB = (1− σ )B0

(14B.2)

the nuclear Larmor frequency (eqn 14A.7 of Topic 14A,
νL = γNℬ/2π) becomes


14B  Features of NMR spectra  

L =

γ N B loc γ N B0
=
(1− σ )
2π
2π


(14B.3)

This frequency is different for nuclei in different environments.
Hence, different nuclei, even of the same element, come into
resonance at different frequencies if they are in different molecular environments.
The chemical shift of a nucleus is the difference between
its resonance frequency and that of a reference standard. The
standard for protons is the proton resonance in tetramethylsilane, Si(CH3)4, commonly referred to as TMS, which bristles
with protons and dissolves without reaction in many solutions.
For 13C, the reference frequency is the 13C resonance in TMS,
and for 31P it is the 31P resonance in 85 per cent H3PO4(aq).
Other references are used for other nuclei. The separation of
the resonance of a particular group of nuclei from the standard increases with the strength of the applied magnetic field
because the induced field is proportional to the applied field;
the stronger the latter, the greater the shift.

Chemical shifts are reported on the δ scale, which is defined as
δ=

 − °
×106
°


The relation between δ and σ is obtained by substituting eqn
14B.3 into eqn 14B.4:
δ=

(1 − σ )B0 − (1 − σ ° )B0
×106
(1 − σ ° )B0

σ ° −σ
=
×106 ≈ (σ ° − σ ) ×106
1− σ °


 =  + ( /106 )δ

RCH3

–COOH
12

A nucleus with δ = 1.00 in a spectrometer where ν° = 500 MHz

(a ‘500 MHz NMR spectrometer’), will have a shift relative to
the reference equal to
 −  = (500 MHz /106 ) ×1.00 = (500 Hz) ×1.00 = 500 Hz
because 1 MHz = 10 6 Hz. In a spectrometer operating at
ν° = 100 MHz, the shift relative to the reference would be only
100 Hz.

RC–CH3
–CO–CH3

–C=CH–

(a)

Ar–H

–CHO
10

8

6

δ

4

–CH –
R–NH 2
–CH– 2

ArC–CH3

ArOH

ROH

2

0

R–C–H
>C=C<
X
–C=C–

(14B.5)

Brief illustration 14B.1 The δ scale

Relation between δ and σ   (14B.6)

The last line follows from σ° ≪ 1. As the shielding constant σ,
gets smaller, δ increases. Therefore, we speak of nuclei with
large chemical shifts as being strongly deshielded. Some typical
chemical shifts are given in Fig. 14B.1. As can be seen from the
illustration, the nuclei of different elements have very different ranges of chemical shifts. The ranges exhibit the variety of
electronic environments of the nuclei in molecules: the higher
the atomic number of the element, the greater the number of
electrons around the nucleus and hence the greater the range of
the extent of shielding. By convention, NMR spectra are plotted

with δ increasing from right to left.

Definition  δ Scale  (14B.4)

where ν° is the resonance frequency of the standard. The advantage of the δ scale is that shifts reported on it are independent of
the applied field (because both numerator and denominator are
proportional to the applied field). The resonance frequencies
themselves, however, do depend on the applied field through

569

(b)
300

–C=C<
C–X in ArX
–
R–C=N
R–CHO
R–COOH
R2C=O
R=C=R
R3C+
200
100
δ

R3C–

0

Figure 14B.1  The range of typical chemical shifts for (a) 1H
resonances and (b) 13C resonances.

Example 14B.1  Interpreting the NMR spectrum of

ethanol

A note on good practice  In much of the literature, chemical shifts are reported in parts per million, ppm, in
recognition of the factor of 106 in the definition; this is
unnecessary. If you see ‘δ = 10 ppm’, interpret it, and use it
in eqn 14B.5, as δ = 10.

Figure 14B.2 shows the NMR spectrum of ethanol. Account
for the observed chemical shifts.

Self-test 14B.1  What is the shift of the resonance from TMS of
a group of nuclei with δ = 3.50 and an operating frequency of
350 MHz?

Answer   The spectrum is consistent with the following

Answer: 1.23 kHz

Method  Consider the effect of an electron-withdrawing atom:

it deshields strongly those protons to which it is bound, and
has a smaller effect on distant protons.

assignments:

• The CH3 protons form one group of nuclei with δ = 1.


570  14  Magnetic resonance
The empirical approach supposes that the observed shielding constant is the sum of three contributions:
CH3CH2OH

4.0

CH3CH2OH

3.6

δ

CH3CH2OH

σ = σ (local) + σ (neighbour) + σ (solvent)

(14B.7)

The local contribution, σ (local), is essentially the contribution
of the electrons of the atom that contains the nucleus in question. The neighbouring group contribution, σ (neighbour), is
the contribution from the groups of atoms that form the rest
of the molecule. The solvent contribution, σ (solvent), is the
contribution from the solvent molecules.
1.2

Figure 14B.2  The 1H-NMR spectrum of ethanol. The bold

letters denote the protons giving rise to the resonance peak,
and the step-like curve is the integrated signal.
• The two CH2 protons are in a different part of the
molecule, experience a different local magnetic field,
and resonate at δ = 3.
• The OH proton is in another environment, and has a
chemical shift of δ = 4.
The increasing value of δ (that is, the decrease in shielding)
is consistent with the electron-withdrawing power of the O
atom: it reduces the electron density of the OH proton most,
and that proton is strongly deshielded. It reduces the electron
density of the distant methyl protons least, and those nuclei
are least deshielded.
The relative intensities of the signals are commonly represented as the height of step-like curves superimposed on the
spectrum, as in Fig. 14B.2. In ethanol the group intensities are
in the ratio 3:2:1 because there are three CH3 protons, two CH2
protons, and one OH proton in each molecule.
Self-test 14B.2  The NMR spectrum of acetaldehyde (ethanal) has lines at δ = 2.20 and δ = 9.80. Which feature can be
assigned to the CHO proton?
Answer: δ = 9.80

14B.2  The origin of shielding constants
The calculation of shielding constants is difficult, even for small
molecules, for it requires detailed information (using the techniques outlined in Topic 10E) about the distribution of electron density in the ground and excited states and the excitation
energies of the molecule. Nevertheless, considerable success
has been achieved with small molecules such as H2O and CH4
and even large molecules, such as proteins, are within the scope
of some types of calculation. However, it is easier to understand
the different contributions to chemical shifts by studying the
large body of empirical information now available.

(a)  The local contribution
It is convenient to regard the local contribution to the shielding
constant as the sum of a diamagnetic contribution, σd, and a
paramagnetic contribution, σp:
σ (local) = σ d + σ p


Local contribution to the shielding constant  (14B.8)

A diamagnetic contribution to σ (local) opposes the applied
magnetic field and shields the nucleus in question. A paramagnetic contribution to σ (local) reinforces the applied magnetic
field and deshields the nucleus in question. Therefore, σd > 0
and σp < 0. The total local contribution is positive if the diamagnetic contribution dominates, and is negative if the paramagnetic contribution dominates.
The diamagnetic contribution arises from the ability of the
applied field to generate a circulation of charge in the groundstate electron distribution of the atom. The circulation generates a magnetic field that opposes the applied field and hence
shields the nucleus. The magnitude of σd depends on the electron density close to the nucleus and can be calculated from the
Lamb formula:1
σd =

e 2 µ0 1
12πme r

Lamb formula  (14B.9)

where μ0 is the vacuum permeability (a fundamental constant, see inside the front cover) and r is the electron–nucleus
distance.
Example 14B.2  Using the Lamb formula

Calculate the shielding constant for the proton in a free H

atom.
Method   To calculate σ d from the Lamb formula, calculate the expectation value of 1/r for a hydrogen 1s orbital.
Wavefunctions are given in Table 9A.1.
Answer  The wavefunction for a hydrogen 1s orbital is
1 

For a derivation, see our Molecular quantum mechanics (2011).


14B  Features of NMR spectra  

571

1/2

e − r /a0

so, because dτ = r2dr sin θ dθdφ, the expectation value of 1/r is
2π

1
ψ *ψ
1
=
dτ = 3
r
r
πa0

∫

=

∫

2π

0

dφ

IntegralE.1

2

∫

π

0

sin θ dθ

∫

∞

re −2r /a0 dr

0

1
a02 1
3 × 4π × 4 = a
πa0
0

Chemical shift relative to CH4, δ

1

 1 
ψ = 3 
 πa0 

CH3CH2X
2

CH3CH2X
3

4

Therefore,


kgm2 s−2
(1.602 ×10−19 C)2 ×  4 π ×10−7 J s2 C −2 m −1 



e 2 µ0
σd =
=
−
31
−
11
12πme a0
12π × (9.109 ×10 kg) × (5.292 ×10 m)
= 1.775 ×10−5
Self-test 14B.3  Derive a general expression for σd that applies
to all hydrogenic atoms.
Answer: Ze2μ0/12πmea0

The diamagnetic contribution is the only contribution in
closed-shell free atoms. It is also the only contribution to the
local shielding for electron distributions that have spherical or
cylindrical symmetry. Thus, it is the only contribution to the
local shielding from inner cores of atoms, for cores remain
nearly spherical even though the atom may be a component of
a molecule and its valence electron distribution is highly distorted. The diamagnetic contribution is broadly proportional
to the electron density of the atom containing the nucleus of
interest. It follows that the shielding is decreased if the electron
density on the atom is reduced by the influence of an electro­
negative atom nearby. That reduction in shielding as the electronegativity of a neighbouring atom increases translates into
an increase in the chemical shift δ (Fig. 14B.3).
The local paramagnetic contribution, σp, arises from the ability of the applied field to force electrons to circulate through the
molecule by making use of orbitals that are unoccupied in the
ground state. It is zero in free atoms and around the axes of linear molecules (such as ethyne, HC ≡ CH) where the electrons
can circulate freely and a field applied along the internuclear

axis is unable to force them into other orbitals. We can expect
large paramagnetic contributions from small atoms (because the
induced currents are then close to the nucleus) in molecules with
low lying excited states (because an applied field can then induce
significant currents). In fact, the paramagnetic contribution is
the dominant local contribution for atoms other than hydrogen.

(b)  Neighbouring group contributions
The neighbouring group contribution arises from the currents
induced in nearby groups of atoms. Consider the influence

I
2

Br

Cl

2.5
3
3.5
Electronegativity of halogen, χ

F
4

Figure 14B.3  The variation of chemical shielding with
electronegativity. The shifts for the methyl protons agree with
the trend expected with increasing electronegativity. However,
to emphasize that chemical shifts are subtle phenomena,

notice that the trend for the methylene protons is opposite
to that expected. For these protons another contribution (the
magnetic anisotropy of C–H and C–X bonds) is dominant.

of the neighbouring group X on the proton H in a molecule
such as H–X. The applied field generates currents in the electron distribution of X and gives rise to an induced magnetic
moment proportional to the applied field; the constant of proportionality is the magnetic susceptibility, χ (chi), of the group
X: μinduced = χℬ0. The susceptibility is negative for a diamagnetic
group because the induced moment is opposite to the direction
of the applied field. The induced moment gives rise to a magnetic field with a component parallel to the applied field and
at a distance r and angle θ (1) that has the form (The chemist’s
toolkit 14B.1):
θ

r

µinduced

1

Blocal ∝

µinduced
(1− 3 cos2θ )
r3


Local dipolar field  (14B.10a)

The chemist’s toolkit 14B.1  Dipolar fields

Standard electromagnetic theory gives the magnetic field at a
point r from a point magnetic dipole μ as

B=

µ0 
3(µ⋅r)r 
µ−
4 πr 3 
r 2 

where μ0 is the vacuum permeability (a fundamental constant
with the defined value 4π × 10−7 T2 J−1 m3). The electric field due
to a point electric dipole is given by a similar expression:


572  14  Magnetic resonance
1

where ε0 is the vacuum permittivity, which is related to μ0 by
ε0 = 1/μ0c2. The component of magnetic field in the z-direction is

Bz =

µ0 
3(µ⋅r )z 
µ −
4 πr 3  z
r 2 

with z = r cos θ, the z-component of the distance vector r. If
the magnetic dipole is also parallel to the z-direction, it follows that
µ⋅r

µz

Bz =

z

µ0
µµ0
3(µr cosθ )(r cos θ)
µ−
=
(1 − 3 cos 2 θ)
4 πr 3
4 πr 3
r2

We see that the strength of the additional magnetic field experienced by the proton is inversely proportional to the cube of the
distance r between H and X. If the magnetic susceptibility is independent of the orientation of the molecule (is ‘isotropic’), the local
field averages to zero because 1 − 3 cos2 θ is zero when averaged
over a sphere (see Problem 14B.7). To a good approximation, the
shielding constant σ(neighbour) depends on the distance r
 1− 3 cos2 θ 
σ (neighbour) ∝ ( χ  − χ ⊥ )

r3


θ
H



Neighbouring
group
contribution

(14B.10b)

r
χ||
χ⊥

X
2

–

+

1 – 3 cos2 θ

1 
3(µ⋅r )r 
µ−
E=
4 πε 0r 3 

r 2 

0

–2

Figure 14B.5  The variation of the function 1 − 3 cos2 θ with the
angle θ.

where χ∥ and χ⊥ are, respectively, the parallel and perpendicular components of the magnetic susceptibility, and θ is
the angle between the XeH axis and the symmetry axis of
the neighbouring group (2). Equation 14B.10b shows that the
neighbouring group contribution may be positive or negative according to the relative magnitudes of the two magnetic
susceptibilities and the relative orientation of the nucleus with
respect to X. If 54.7° < θ < 125.3°, then 1 − 3 cos2 θ is positive,
but it is negative otherwise (Figs. 14B.4 and 14B.5).
Brief illustration 14B.2  Ring currents

A special case of a neighbouring group effect is found in aromatic compounds. The strong anisotropy of the magnetic susceptibility of the benzene ring is ascribed to the ability of the
field to induce a ring current, a circulation of electrons around
the ring, when it is applied perpendicular to the molecular
plane. Protons in the plane are deshielded (Fig. 14B.6), but
any that happen to lie above or below the plane (as members of
substituents of the ring) are shielded.

Ring
current
+

µ

π

½π
θ

0

B
Magnetic
field

–

Figure 14B.4  A depiction of the field arising from a point
magnetic dipole. The three shades of colour represent the
strength of field declining with distance (as 1/r3), and each
surface shows the angle dependence of the z-component of
the field for each distance.

Figure 14B.6  The shielding and deshielding effects of the
ring current induced in the benzene ring by the applied
field. Protons attached to the ring are deshielded but a
proton attached to a substituent that projects above the
ring is shielded.


14B  Features of NMR spectra  

Self-test 14B.4  Consider ethyne, HC ≡ CH. Are its protons

shielded or deshielded by currents induced by the triple bond?

Answer: Shielded

(c)  The solvent contribution
A solvent can influence the local magnetic field experienced by
a nucleus in a variety of ways. Some of these effects arise from
specific interactions between the solute and the solvent (such as
hydrogen-bond formation and other forms of Lewis acid–base
complex formation). The anisotropy of the magnetic susceptibility of the solvent molecules, especially if they are aromatic,
can also be the source of a local magnetic field. Moreover, if
there are steric interactions that result in a loose but specific
interaction between a solute molecule and a solvent molecule,
then protons in the solute molecule may experience shielding
or deshielding effects according to their location relative to the
solvent molecule.

spectrum. It arises because each magnetic nucleus may contribute to the local field experienced by the other nuclei and
so modify their resonance frequencies. The strength of the
interaction is expressed in terms of the scalar coupling constant, J. The scalar coupling constant is so called because the
energy of interaction it describes is proportional to the scalar
product of the two interacting spins: E ∝ I1⋅I2. As explained in
Mathematical background 5, a scalar product depends on the
angle between the two vectors, so writing the energy in this way
is simply a way of saying that the energy of interaction between
two spins depends on their relative orientation. The constant
of proportionality in this expression is written hJ/2 (so E = 
(hJ/2)I1⋅I2): because each spin angular momentum is proportional to , E is then proportional to hJ and J is a frequency
(with units hertz, Hz). For nuclei that are constrained to align

with the applied field in the z-direction, the only contribution
to I1⋅I2 is I1zI2z, with eigenvalues m1m22, so in that case the
energy due to spin–spin coupling is
Em m = hJm1m2
1

2

Brief illustration 14B.3  The effect of aromatic solvents

An aromatic solvent like benzene can give rise to local currents that shield or deshield a proton in a solute molecule. The
arrangement shown in Fig. 14B.7 leads to shielding of a proton
on the solute molecule.

B

Self-test 14B.5  Refer to Fig. 14B.7 and suggest an arrangement

that leads to deshielding of a proton on the solute molecule.

Answer: Proton on the solute molecule coplanar
with the benzene ring

14B.3  The

fine structure

The splitting of resonances into individual lines by spin–spin
coupling shown in Fig. 14B.2 is called the fine structure of the

Spin–spin coupling energy  (14B.11)

(a)  The appearance of the spectrum
In NMR, letters far apart in the alphabet (typically A and X)
are used to indicate nuclei with very different chemical shifts;
letters close together (such as A and B) are used for nuclei with
similar chemical shifts. We shall consider first an AX system, a
molecule that contains two spin- 12 nuclei A and X with very different chemical shifts in the sense that the difference in chemical shift corresponds to a frequency that is large compared to J.
For a spin- 12 AX system there are four spin states: αAαX,
αAβX, βAαX, βAβX. The energy depends on the orientation of
the spins in the external magnetic field, and if spin–spin coupling is neglected
Em

Figure 14B.7  An aromatic solvent (benzene here) can give
rise to local currents that shield or deshield a proton in a
solute molecule. In this relative orientation of the solvent
and solute, the proton on the solute molecule is shielded.

573

A mX

= −γ N (1− σ A )B0 mA − γ N (1− σ X )B0 mX
= −h A mA − h X mX



(14B.12a)

where νA and νX are the Larmor frequencies of A and X and
mA and mX are their quantum numbers (mA = ± 12 , mX = ± 12 ).
This expression gives the four lines on the left of Fig. 14B.8.
When spin–spin coupling is included (by using eqn 14B.11),
the energy levels are
Em

A mX

= −h A mA − h X mX + hJmA mX



(14B.12b)

If J > 0, a lower energy is obtained when mAmX < 0, which is
the case if one spin is α and the other is b. A higher energy is
obtained if both spins are α or both spins are b. The opposite
is true if J < 0. The resulting energy level diagram (for J > 0) is
shown on the right of Fig. 14B.8. We see that the αα and bb


574  14  Magnetic resonance

No spin–spin βΑαX ¼hJ
coupling
–½hνL(A) + ½hνL(X)
αΑβX
¼hJ

βΑαX
αΑβX

With
spin–spin
coupling

αΑαX
–½hνL(A) – ½hνL(X)

αΑαX ¼hJ

Figure 14B.8  The energy levels of an AX system. The four levels
on the left are those of the two spins in the absence of spin–spin
coupling. The four levels on the right show how a positive spin–
spin coupling constant affects the energies. The transitions
shown are for β ← a of A or X, the other nucleus (X or A,
respectively) remaining unchanged. We have exaggerated the
effect for clarity. In practice, the splitting caused by spin–spin
coupling is much smaller than that caused by the applied field.

states are both raised by 14 hJ and that the aβ and βa states are
both lowered by 14 hJ .
When a transition of nucleus A occurs, nucleus X remains
unchanged. Therefore, the A resonance is a transition for which
ΔmA = +1 and ΔmX = 0. There are two such transitions, one in
which βA ← aA occurs when the X nucleus is a , and the other in
which βA ← aA occurs when the X nucleus is β. They are shown
in Fig. 14B.8 and in a slightly different form in Fig. 14B.9. The

energies of the transitions are
∆E = h A ± 12 hJ

(14B.13a)

Therefore, the A resonance consists of a doublet of separation J centred on the chemical shift of A (Fig. 14B.10). Similar
remarks apply to the X resonance, which consists of two
βΑβX

βΑαX
αΑβX
αΑαX

Figure 14B.9  An alternative depiction of the energy levels
and transitions shown in Fig. 14B.8. Once again, we have
exaggerated the effect of spin–spin coupling.

X resonance

½hνL(A) – ½hνL(X)

A resonance

βΑβX

βΑβX ¼hJ
Energy

½hνL(A) + ½hνL(X)

J

J

δA

δX

Figure 14B.10  The effect of spin–spin coupling on an AX
spectrum. Each resonance is split into two lines separated by
J. The pairs of resonances are centred on the chemical shifts of
the protons in the absence of spin–spin coupling.

transitions according to whether the A nucleus is a or β (as
shown in Fig. 14B.9). The transition energies are
∆E = h X ± 12 hJ

(14B.13b)

It follows that the X resonance also consists of two lines of the
same separation J, but they are centred on the chemical shift of
X (as shown in Fig. 14B.10).
If there is another X nucleus in the molecule with the same
chemical shift as the first X (giving an AX2 species), the X
resonance of the AX2 species is split into a doublet by A, as
in the AX case discussed above (Fig. 14B.11). The resonance
of A is split into a doublet by one X, and each line of the doublet is split again by the same amount by the second X (Fig.
14B.12). This splitting results in three lines in the intensity
ratio 1:2:1 (because the central frequency can be obtained in
two ways).

X resonance
in AX2

X resonance
in AX

J

δX

Figure 14B.11  The X resonance of an AX2 species is also a
doublet, because the two equivalent X nuclei behave like a
single nucleus; however, the overall absorption is twice as
intense as that of an AX species.


14B  Features of NMR spectra  

575

Example 14B.3  Accounting for the fine structure in a

spectrum
Account for the fine structure in the NMR spectrum of the
C–H protons of ethanol.
Method  Consider how each group of equivalent protons (for
instance, three methyl protons) split the resonances of the
other groups of protons. There is no splitting within groups of
equivalent protons. Each splitting pattern can be decided by
referring to Pascal’s triangle.

δA

Figure 14B.12  The origin of the 1:2:1 triplet in the A resonance
of an AX2 species. The resonance of A is split into two by
coupling with one X nucleus (as shown in the inset), and then
each of those two lines is split into two by coupling to the
second X nucleus. Because each X nucleus causes the same
splitting, the two central transitions are coincident and give rise
to an absorption line of double the intensity of the outer lines.

Three equivalent X nuclei (an AX3 species) split the resonance of A into four lines of intensity ratio 1:3:3:1 (Fig. 14B.13).
The X resonance remains a doublet as a result of the splitting
caused by A. In general, N equivalent spin- 12 nuclei split the
resonance of a nearby spin or group of equivalent spins into
N + 1 lines with an intensity distribution given by Pascal’s triangle (3). Successive rows of this triangle are formed by adding
together the two adjacent numbers in the line above.

1

1

1
5

1
4

1
3
10

1
2
6

1
3
10

1
4

1
5

1

1

3

δA

Figure 14B.13  The origin of the 1:3:3:1 quartet in the A
resonance of an AX3 species. The third X nucleus splits each of
the lines shown in Fig. 14B.11 for an AX2 species into a doublet,
and the intensity distribution reflects the number of transitions
that have the same energy.

Answer  The three protons of the CH3 group split the resonance of the CH 2 protons into a 1:3:3:1 quartet with a splitting J. Likewise, the two protons of the CH 2 group split the

resonance of the CH3 protons into a 1:2:1 triplet with the same
splitting J. The OH resonance is not split because the OH protons migrate rapidly from molecule to molecule (including
molecules of impurities in the sample) and their effect averages to zero. In gaseous ethanol, where this migration does not
occur, the OH resonance appears as a triplet, showing that the
CH2 protons interact with the OH proton.
Self-test 14B.6  What fine structure can be expected for the protons in 14 NH4+ ? The spin quantum number of nitrogen-14 is 1.
Answer: 1:1:1 triplet from N

(b)  The magnitudes of coupling constants
The scalar coupling constant of two nuclei separated by N bonds
is denoted NJ, with subscripts for the types of nuclei involved.
Thus, 1JCH is the coupling constant for a proton joined directly
to a 13C atom, and 2JCH is the coupling constant when the same
two nuclei are separated by two bonds (as in 13CeCeH). A typical value of 1JCH is in the range 120 to 250 Hz; 2JCH is between
10 and 20 Hz. Both 3J and 4J can give detectable effects in a
spectrum, but couplings over larger numbers of bonds can generally be ignored. One of the longest range couplings that has
been detected is 9JHH = 0.4 Hz between the CH3 and CH2 protons in CH3C b CeC b CeC b CeCH2OH.
As remarked (in the discussion following eqn 14B.12b), the
sign of JXY indicates whether the energy of two spins is lower
when they are parallel (J < 0) or when they are antiparallel
(J > 0). It is found that 1JCH is often positive, 2JHH is often negative, 3JHH is often positive, and so on. An additional point is that
J varies with the angle between the bonds (Fig. 14B.14). Thus, a
3J
HH coupling constant is often found to depend on the dihedral
angle φ (4) according to the Karplus equation:
H φ
H

4

Spin–spin coupling constant, J/Hz

576  14  Magnetic resonance
A nucleus with spin projection mI gives rise to a magnetic
field with z-component Bnuc at a distance R, where, to a good
approximation,

14
12 HCCH
10
8

Bnuc = −

6

HNCH

2
π/2
Angle, φ

0

π

Figure 14B.14  The variation of the spin–spin coupling constant
with angle predicted by the Karplus equation for an HCCH
group and an HNCH group.

Brief illustration 14B.5  Magnetic fields from nuclei

The z-component of the magnetic field arising from a proton
(mI = 12 ) at R = 0.30 nm, with its magnetic moment parallel to
the z-axis (θ = 0) is
γN

3

(14B.15)

The angle θ is defined in (1); we saw a version of this expression
in eqn 14B.10a.

4

0

γ N µ0
(1− 3 cos2 θ )mI
4 πR 3


J HH = A + B cos φ + C cos2φ

Karplus equation  (14B.14)

with A, B, and C empirical constants with values close to +7 Hz,
−1 Hz, and +5 Hz, respectively, for an HCCH fragment. It follows that the measurement of 3JHH in a series of related compounds can be used to determine their conformations. The

coupling constant 1JCH also depends on the hybridization of the
C atom, as the following values indicate:
1J

CH/Hz

sp

sp2

sp3

250

160

125

Brief illustration 14B.4  The Karplus equation

The investigation of HeNe CeH couplings in polypeptides
can help reveal their conformation. For 3JHH coupling in such
a group, A = +5.1 Hz, B = −1.4 Hz, and C = +3.2 Hz. For a helical
polymer, φ is close to 120°, which would give 3J HH≈ 4 Hz. For
the sheet-like conformation, φ is close to 180°, which would
give 3JHH ≈ 10 Hz.
Self-test 14B.7  NMR experiments reveal that for HeCeCeH
coupling in polypeptides, A = +3.5 Hz, B = −1.6 Hz, and
C = +4.3 Hz. In an investigation of the polypeptide flavodoxin,
the 3J HH coupling constant for such a grouping was determined to be 2.1 Hz. Is this value consistent with a helical or

sheet conformation?
Answer: Helical conformation

(c)  The origin of spin–spin coupling
Spin–spin coupling is a very subtle phenomenon and it is better to treat J as an empirical parameter than to use calculated
­values. However, we can get some insight into its origins, if not
its precise magnitude—or always reliably its sign—by considering the magnetic interactions within molecules.

µ0

(2.821 ×10 J T ) × 4 π ×10 T 2 J −1 m 3
Bnuc = −
×
3
4 π × (3.0 × 10−10 m)
−26

−1

−7

(1−3cos2 θ ) mI

(−1)

R

= 1.0 × 10

−4

T = 0.10 mT

A field of this magnitude can give rise to the splitting of
resonance signals in solid samples. In a liquid, the angle θ
sweeps over all values as the molecule tumbles, and the factor
1 − 3 cos2 θ averages to zero. Hence the direct dipolar interaction between spins cannot account for the fine structure of the
spectra of rapidly tumbling molecules.
Self-test 14B.8  In gypsum, CaSO 4⋅2H 2O, the splitting in the
H2O resonance can be interpreted in terms of a magnetic field
of 0.715 mT generated by one proton and experienced by the
other. With θ = 0, what is the separation of the protons in the
H2O molecule?
Answer: 158 pm

Spin–spin coupling in molecules in solution can be explained
in terms of the polarization mechanism, in which the interaction is transmitted through the bonds. The simplest case to
consider is that of 1JXY, where X and Y are spin- 12 nuclei joined
by an electron-pair bond. The coupling mechanism depends
on the fact that the energy depends on the relative orientation
of the bonding electrons and the nuclear spins. This electron–
nucleus coupling is magnetic in origin, and may be either a
dipolar interaction or a Fermi contact interaction. A pictorial
description of the latter is as follows. First, we regard the magnetic moment of the nucleus as arising from the circulation of a
current in a tiny loop with a radius similar to that of the nucleus
(Fig. 14B.15). Far from the nucleus the field generated by this
loop is indistinguishable from the field generated by a point
magnetic dipole. Close to the loop, however, the field differs
from that of a point dipole. The magnetic interaction between
this non-dipolar field and the electron’s magnetic moment is

14B  Features of NMR spectra  

577

Hund

Pauli

C

Pauli

Fermi

Fermi
X

Figure 14B.15  The origin of the Fermi contact interaction.
From far away, the magnetic field pattern arising from a ring
of current (representing the rotating charge of the nucleus,
the pale grey sphere) is that of a point dipole. However, if an
electron can sample the field close to the region indicated by
the sphere, the field distribution differs significantly from that
of a point dipole. For example, if the electron can penetrate the
sphere, then the spherical average of the field it experiences is
not zero.

the contact interaction. The contact interaction—essentially

the failure of the point-dipole approximation—depends on the
very close approach of an electron to the nucleus and hence can
occur only if the electron occupies an s orbital (which is the
reason why 1JCH depends on the hybridization ratio). We shall
suppose that it is energetically favourable for an electron spin
and a nuclear spin to be antiparallel (as is the case for a proton
and an electron in a hydrogen atom).
If the X nucleus is α, a b electron of the bonding pair will
tend to be found nearby, because that is an energetically favourable arrangement (Fig. 14B.16). The second electron in the
bond, which must have α spin if the other is b (by the Pauli
principle; Topic 9B), will be found mainly at the far end of
the bond because electrons tend to stay apart to reduce their
mutual repulsion. Because it is energetically favourable for the
spin of Y to be antiparallel to an electron spin, a Y nucleus with
b spin has a lower energy than when it has α spin. The opposite is true when X is b, for now the α spin of Y has the lower
energy. In other words, the antiparallel arrangement of nuclear
spins lies lower in energy than the parallel arrangement as a
Fermi
X

Pauli

Fermi
Y

Y

Figure 14B.17  The polarization mechanism for 2 JHH spin–spin
coupling. The spin information is transmitted from one bond
to the next by a version of the mechanism that accounts for the

lower energy of electrons with parallel spins in different atomic
orbitals (Hund’s rule of maximum multiplicity). In this case, J < 0,
corresponding to a lower energy when the nuclear spins are
parallel.

result of their magnetic coupling with the bond electrons. That
is, 1JCH is positive.
To account for the value of 2JXY, as for 2JHH in HeCeH,
we need a mechanism that can transmit the spin alignments
through the central C atom (which may be 12C, with no nuclear
spin of its own). In this case (Fig. 14B.17), an X nucleus with
α spin polarizes the electrons in its bond, and the α electron is
likely to be found closer to the C nucleus. The more favourable
arrangement of two electrons on the same atom is with their
spins parallel (Hund’s rule, Topic 9B), so the more favourable
arrangement is for the α electron of the neighbouring bond
to be close to the C nucleus. Consequently, the b electron of
that bond is more likely to be found close to the Y nucleus,
and therefore that nucleus will have a lower energy if it is α.
Hence, according to this mechanism, the lower energy will be
obtained if the Y spin is parallel to that of X. That is, 2JHH is
negative.
The coupling of nuclear spin to electron spin by the Fermi
contact interaction is most important for proton spins, but it is
not necessarily the most important mechanism for other nuclei.
These nuclei may also interact by a dipolar mechanism with the
electron magnetic moments and with their orbital motion, and
there is no simple way of specifying whether J will be positive
or negative.

(d)  Equivalent nuclei
X

Y

Figure 14B.16  The polarization mechanism for spin–spin
coupling (1JHH). The two arrangements have slightly different
energies. In this case, J is positive, corresponding to a lower
energy when the nuclear spins are antiparallel.

A group of nuclei are chemically equivalent if they are related
by a symmetry operation of the molecule and have the same
chemical shifts. Chemically equivalent nuclei are nuclei that
would be regarded as ‘equivalent’ according to ordinary chemical criteria. Nuclei are magnetically equivalent if, as well as
being chemically equivalent, they also have identical spin–spin
interactions with any other magnetic nuclei in the molecule.


578  14  Magnetic resonance
Brief illustration 14B.6  Chemical and magnetic

equivalence
The difference between chemical and magnetic equivalence is
illustrated by CH2F2 and H2C = CF2. In each of these molecules
the protons are chemically equivalent: they are related by symmetry and undergo the same chemical reactions. However,
although the protons in CH 2F2 are magnetically equivalent,
those in CH 2 = CF2 are not. One proton in the latter has a cis
spin-coupling interaction with a given F nucleus whereas
the other proton has a trans interaction with it. In contrast,
in CH2F2 both protons are connected to a given F nucleus by

identical bonds, so there is no distinction between them.
Self-test 14B.9  Are the CH3 protons in ethanol magnetically

inequivalent?

Answer: Yes, on account of their different interactions
with the CH2 protons in the next group

Strictly speaking, CH3 protons are magnetically inequivalent. However, they are in practice made magnetically equivalent by the rapid rotation of the CH3 group, which averages out
any differences. Magnetically inequivalent species can give very
complicated spectra (for instance, the proton and 19F spectra of
H2C = CF2 each consist of 12 lines), and we shall not consider
them further.
An important feature of chemically equivalent magnetic
nuclei is that, although they do couple together, the coupling has no effect on the appearance of the spectrum. The
qualitative reason for the invisibility of the coupling is that all
allowed nuclear spin transitions are collective reorientations of
groups of equivalent nuclear spins that do not change the relative orientations of the spins within the group (Fig. 14B.18).
Then, because the relative orientations of nuclear spins are
α

φ

not changed in any transition, the magnitude of the coupling
between them is undetectable. Hence, an isolated CH3 group
gives a single, unsplit line because all the allowed transitions of
the group of three protons occur without change of their relative orientations.
To express these conclusions more quantitatively, we first
need to establish the energy levels of a collection of equivalent
nuclei. As shown in the following Justification for an A2 system,

they have the values depicted on the right of Fig. 14B.19.
Justification 14B.1  The energy levels of an A system
2

Consider an A 2 system of two spin- 12 nuclei. First, consider the
energy levels in the absence of spin–spin coupling. There are
four spin states that (just as for two electrons) can be classified
according to their total spin I (the analogue of S for two electrons) and their total projection M I on the z-axis. The states
are analogous to those for two electrons in singlet and triplet
states (Topic 9C):
Spins parallel, I = 1:

Spins paired, I = 0:

αα

MI = 0

(1/21/2){αβ + βα}

MI = −1

ββ

MI =0

(1/21/2){αβ − βα}

The sign in αβ + βα signifies an in-phase alignment of spins
and I = 1; the − sign in αβ − βα signifies an alignment out of

phase by π, and hence I = 0. The effect of a magnetic field on
these four states is shown in Fig. 14B.19: the energies of the

ββ
No
spin–spin
coupling
αβ + βα
αβ – βα

α

MI = +1

α

+ 14 J

+ 14 J
– 34 J

φ

φ

(a)
α

β

β

αα

β

α

β
α

(b)

β
Two intermediate
orientations

β

Figure 14B.18  (a) A group of two equivalent nuclei realigns
as a group, without change of angle between the spins, when
a resonant absorption occurs. Hence it behaves like a single
nucleus and the spin–spin coupling between the individual
spins of the group is undetectable. (b) Three equivalent
nuclei also realign as a group without change of their relative
orientations.

+ 14 J

I = 1, MI = +1

With
spin–spin
coupling
I = 1, MI = 0

I = 0, MI = 0
I = 1, MI = –1

Figure 14B.19  The energy levels of an A2 system in the
absence of spin–spin coupling are shown on the left. When
spin–spin coupling is taken into account, the energy levels
on the right are obtained. Note that the three states with
total nuclear spin I = 1 correspond to parallel spins and give
rise to the same increase in energy (J is positive); the one
state with I = 0 (antiparallel nuclear spins) has a lower energy
in the presence of spin–spin coupling. The only allowed
transitions are those that preserve the angle between the
spins, and so take place between the three states with I = 1.
They occur at the same resonance frequency as they would
have in the absence of spin–spin coupling.


14B  Features of NMR spectra  
two states with M I = 0 are unchanged by the field because they
are composed of equal proportions of α and b spins.
The spin–spin coupling energy is proportional to the scalar
product of the vectors representing the spins, E = (hJ/ 2)I1⋅I2 .
The scalar product can be expressed in terms of the total
nuclear spin I = I1 + I2 by noting that
I 2 = (I1 + I 2 )⋅(I1 + I 2 ) = I12 + I 22 + 2I1 ⋅ I 2

579

reorienting relative to each other. The spectra that result are
called first-order spectra.
Transitions cannot be allocated to definite groups when the
differences in their chemical shifts are comparable to their
spin–spin coupling interactions. The complicated spectra that
are then obtained are called strongly coupled spectra (or ‘second-order spectra’) and are much more difficult to analyse.

rearranging this expression to
I1 ⋅ I 2 = 12 {I 2 − I12 − I 22 }
and replacing the magnitudes by their quantum mechanical
values:
I1 ⋅ I 2 = 12 {I (I + 1) − I1(I1 + 1) − I 2 (I 2 + 1)}2
Then, because I1 = I 2 = 12 , it follows that
E = 12 hJ {I (I + 1) − 23 }
For parallel spins, I = 1 and E = + 14 hJ ; for antiparallel spins
I = 0 and E = − 43 hJ , as in Fig. 14B.19. We see that three of the
states move in energy in one direction and the fourth (the
one with antiparallel spins) moves three times as much in the
opposite direction.

Brief illustration 14B.7  Strongly coupled spectra

Figure 14B.20 shows NMR spectra of an A 2 system (top) and
an AX system (bottom). Both are simple ‘first-order’ spectra.
At intermediate relative values of the chemical shift difference and the spin–spin coupling, complex ‘strongly coupled’
spectra are obtained. Note how the inner two lines of the bottom spectrum move together, grow in intensity, and form the
single central line of the top spectrum. The two outer lines

diminish in intensity and are absent in the top spectrum.

ν°Δδ << J
ν°Δδ ≈ J

We now consider the allowed transitions between the states
of an A2 system shown in Fig. 14B.18. The radiofrequency field
affects the two equivalent protons equally, so it cannot change
the orientation of one proton relative to the other; therefore, the
transitions take place within the set of states that correspond
to parallel spin (those labelled I = 1), and no spin-parallel state
can change to a spin-antiparallel state (the state with I = 0). Put
another way, the allowed transitions are subject to the selection
rule ΔI = 0.This selection rule is in addition to the rule ΔMI = ±1
that arises from the conservation of angular momentum and
the unit spin of the photon. The allowed transitions are shown
in Fig. 14B.19: we see that there are only two transitions, and
that they occur at the same resonance frequency that the nuclei
would have in the absence of spin–spin coupling. Hence, the
spin–spin coupling interaction does not affect the appearance
of the spectrum.

(e)  Strongly coupled nuclei
NMR spectra are usually much more complex than the
foregoing simple analysis suggests. We have described the
extreme case in which the differences in chemical shifts are
much greater than the spin–spin coupling constants. In such
cases it is simple to identify groups of magnetically equivalent nuclei and to think of the groups of nuclear spins as

ν°Δδ ≈ J

ν°Δδ >> J

Figure 14B.20  The NMR spectra of an A2 system (top) and an
AX system (bottom) are simple ‘first-order’ spectra.
Self-test 14B.10  Explain why, in some cases, a second-order
spectrum may become simpler (and first-order) at high fields.
Answer: The difference in resonance frequencies increases with
field, but spin–spin coupling constants are independent of it

A clue to the type of analysis that is appropriate is given
by the notation for the types of spins involved. Thus, an AX
spin system (which consists of two nuclei with a large chemical shift difference) has a first-order spectrum. An AB system, on the other hand (with two nuclei of similar chemical
shifts), gives a spectrum typical of a strongly coupled system.
An AX system may have widely different Larmor frequencies because A and X are nuclei of different elements (such
as 13C and 1H), in which case they form a heteronuclear spin
system. AX may also denote a homonuclear spin system in


580  14  Magnetic resonance
which the nuclei are of the same element but in markedly different environments.

14B.4  Conformational

conversion
and exchange processes
The appearance of an NMR spectrum is changed if magnetic
nuclei can jump rapidly between different environments.
Consider a molecule, such as N,N-dimethylformamide, that
can jump between conformations; in its case, the methyl
shifts depend on whether they are cis or trans to the carbonyl

group (Fig. 14B.21). When the jumping rate is low, the spectrum shows two sets of lines, one each from molecules in each
conformation. When the interconversion is fast, the spectrum
shows a single line at the mean of the two chemical shifts.
At intermediate inversion rates, the line is very broad. This
maximum broadening occurs when the lifetime, τ, of a conformation gives rise to a linewidth that is comparable to the difference of resonance frequencies, δν and both broadened lines
blend together into a very broad line. Coalescence of the two
lines occurs when
τ=

21/2
πδ

Condition for coalescence of two NMR lines  (14B.16)

O
C

H

N

Figure 14B.21  When a molecule changes from one
conformation to another, the positions of its protons are
interchanged and the protons jump between magnetically
distinct environments.
Brief illustration 14B.8  The effect of chemical exchange

on NMR spectra
The NO group in N,N-dimethylnitrosamine, (CH3)2N eNO
(5), rotates about the N eN bond and, as a result, the magnetic

environments of the two CH 3 groups are interchanged. The
two CH 3 resonances are separated by 390 Hz in a 600 MHz
spectrometer. According to eqn 14B.16,
τ=

21/2
= 1.2 ms
π × (390 s −1)

It follows that the signal will collapse to a single line when the
interconversion rate exceeds about 1/τ = 830s−1.

5 N,N-Dimethylnitrosamine

Self-test 14B.11  What would you deduce from the observation of a single line from the same molecule in a 300 MHz
spectrometer?
Answer: Conformation lifetime less than 2.3 ms

A similar explanation accounts for the loss of fine structure in solvents able to exchange protons with the sample. For
example, hydroxyl protons are able to exchange with water
protons. When this chemical exchange occurs, a molecule
ROH with an α-spin proton (we write this ROHα) rapidly
converts to ROHβ and then perhaps to ROHα again because
the protons provided by the solvent molecules in successive
exchanges have random spin orientations. Therefore, instead
of seeing a spectrum composed of contributions from both
ROHα and ROHβ molecules (that is, a spectrum showing a
doublet structure due to the OH proton) we see a spectrum
that shows no splitting caused by coupling of the OH proton

(as in Fig. 14B.2 and as discussed in Example 14B.3). The effect
is observed when the lifetime of a molecule due to this chemical exchange is so short that the lifetime broadening is greater
than the doublet splitting. Because this splitting is often very
small (a few hertz), a proton must remain attached to the
same molecule for longer than about 0.1s for the splitting to
be observable. In water, the exchange rate is much faster than
that, so alcohols show no splitting from the OH protons. In
dry dimethylsulfoxide (DMSO), the exchange rate may be
slow enough for the splitting to be detected.


14B  Features of NMR spectra  

581

Checklist of concepts
☐1.The chemical shift of a nucleus is the difference
between its resonance frequency and that of a reference
standard.
☐2.The shielding constant is the sum of a local contribution, a neighbouring group contribution, and a solvent
contribution.
☐3.The local contribution is the sum of a diamagnetic contribution and a paramagnetic contribution.
☐4.The neighbouring group contribution arises from the
currents induced in nearby groups of atoms.
☐5.The solvent contribution can arise from specific molecu­
lar interactions between the solute and the solvent.
☐6.Fine structure is the splitting of resonances into individual lines by spin–spin coupling.

☐7.Spin–spin coupling is expressed in terms of the spin–
spin coupling constant J and depends on the relative

orientation of two nuclear spins.
☐8.The coupling constant decreases as the number of
bonds separating two nuclei increases.
☐9.Spin–spin coupling can be explained in terms of the
polarization mechanism and the Fermi contact
interaction.
☐10. Chemically and magnetically equivalent nuclei have
the same chemical shifts.
☐11. In strongly coupled spectra, transitions cannot be allocated to definite groups.
☐12. Coalescence of two NMR lines occurs when a conformational interchange or chemical exchange of nuclei is fast.

Checklist of equations
Property

Equation

Comment

Equation number

δ Scale of chemical shifts

δ = {(ν − ν°)/ν°} × 106

Definition

14B.4

Relation between chemical shift and shielding
constant

δ ≈ (σ ° − σ) × 106

14B.6

Local contribution to the shielding constant

σ(local) = σd + σp

14B.8

Lamb formula

σd = (e2μ0/12πme)〈1/r〉

14B.9

Neighbouring group contribution to the
shielding constant

σ(neighbour)∝(χ∥ − χ⊥){(1 − 3cos2θ)/r3}

Karplus equation

3J
HH = A + B

Condition for coalescence of two NMR lines

τ = 21/2/πδν

cos φ + C cos 2φ

The angle θ is defined in (1)

14B.10b

A, B, and C are empirical constants

14B.14

Conformational conversions and
exchange processes

14B.16


14C  Pulse techniques in NMR
Contents
14C.1 

The magnetization vector
The effect of the radiofrequency field
Brief illustration 14C.1: Radiofrequency pulses
(b) Time- and frequency-domain signals
Brief illustration 14C.2: Fourier analysis
(a)

14C.2 

Spin relaxation
Longitudinal and transverse relaxation
Brief illustration 14C.3: Inhomogeneous
broadening
(b) The measurement of T1 and T2
(a)

14C.3 
14C.4 

Spin decoupling
The nuclear Overhauser effect

14C.5 

Two-dimensional NMR

Brief illustration 14C.4: NOE enhancement
Example 14C.1: Interpreting a two-dimensional
NMR spectrum
14C.6 

Solid-state NMR
Brief illustration 14C.5: Dipolar fields in solids

Checklist of concepts
Checklist of equations

582
583

583
584
585
585
585
587
587
588
589
590
590
591
592
592
593
593

➤➤ Why do you need to know this material?
To understand how nuclear magnetic resonance
spectroscopy is used to study large molecules and even
diagnose disease, you need to understand how spectral
information is obtained by analysing the response of
nuclei to the application of strong pulses of radiofrequency
radiation.

➤➤ What is the key idea?
Fourier-transform NMR spectroscopy is the analysis of
the radiation emitted by nuclear spins as they return to
equilibrium after stimulation by one or more pulses of
radiofrequency radiation.

➤➤ What do you need to know already?
You need to be familiar with the general principles of
magnetic resonance (Topic 14A), the features of NMR
spectra (Topics 14B), the vector model of angular
momentum (Topic 8B), the magnetic properties of

molecules (Topic 18C), and Fourier transforms (Topic 12A
and Mathematical background 7). The development makes
use of the concept of precession at the Larmor frequency
(Topic 14A).

The common method of detecting the energy separation
between nuclear spin states in NMR spectroscopy is more
sophisticated than simply looking for the frequency at which
resonance occurs. One of the best analogies that have been
suggested to illustrate the preferred way of observing an NMR
spectrum is that of detecting the spectrum of vibrations of a
bell. We could stimulate the bell with a gentle vibration at a
gradually increasing frequency, and note the frequencies at
which it resonated with the stimulation. A lot of time would
be spent getting zero response when the stimulating frequency
was between the bell’s vibrational modes. However, if we were
simply to hit the bell with a hammer, we would immediately
obtain a clang composed of all the frequencies that the bell
can produce. The equivalent in NMR is to monitor the radiation nuclear spins emit as they return to equilibrium after the
appropriate stimulation. The resulting Fourier-transform
NMR (FT-NMR) spectroscopy gives greatly increased sensitivity, so opening up much of the periodic table to the technique.
Moreover, multiple-pulse FT-NMR gives chemists unparalleled
control over the information content and display of spectra.

14C.1  The

magnetization vector

Consider a sample composed of many identical spin- 12 nuclei.
By analogy with the discussion of angular momenta in Topic
8C, a nuclear spin can be represented by a vector of length
{I(I + 1)}1/2 units with a component of length m1 units along
the z-axis. As the uncertainty principle does not allow us to
specify the x- and y-components of the angular momentum, all
we know is that the vector lies somewhere on a cone around
the z-axis. For I = 12 , the length of the vector is 12 31/2 and when
mI = + 12 it makes an angle of arccos( 12 /( 12 31/2 )) = 55° to the z-axis
(Fig. 14C.1).
In the absence of a magnetic field, the sample consists of
equal numbers of α and β nuclear spins with their vectors lying
at random angles on the cones. These angles are unpredictable, and at this stage we picture the spin vectors as stationary.


14C  Pulse techniques in NMR  

the z-direction and with a length proportional to the population difference (Fig. 14C.2b).

z

(a)  The effect of the radiofrequency field

½

½√3

Figure 14C.1  The vector model of angular momentum
for a single spin- 21 nucleus. The angle around the z-axis is
indeterminate.
M

α
β

(a)

583

(b)

Figure 14C.2  The magnetization of a sample of spin- 21
nuclei is the resultant of all their magnetic moments. (a) In the
absence of an externally applied field, there are equal numbers
of α and β spins at random angles around the z-axis (the field
direction) and the magnetization is zero. (b) In the presence
of a field, the spins precess around their cones (that is, there
is an energy difference between the a and β states) and there
are slightly more a spins than β spins. As a result, there is a net
magnetization along the z-axis.

The magnetization, M, of the sample, its net nuclear magnetic
moment, is zero (Fig. 14C.2a).
Two changes occur in the magnetization when a magnetic
field of magnitude ℬ0 is present and aligned in the z-direction:

• The energies of the two orientations change, the α spins
moving to low energy and the b spins to high energy
(provided γ N > 0).
At 10 T, the Larmor frequency for protons is 427 MHz, and in
the vector model the individual vectors are pictured as precessing at this rate (Topic 14A). This motion is a pictorial representation of the difference in energy of the spin states (it is not
an actual representation of reality but is inspired by the actual
motion of a classical bar magnet in a magnetic field). As the
field is increased, the Larmor frequency increases and the precession becomes faster.
• The populations of the two spin states (the numbers of a
and β spins) at thermal equilibrium change, with slightly
more a spins than β spins (see Topic 14A).
Despite its smallness, this imbalance means that there is a net
magnetization that we can represent by a vector M pointing in

Now consider the effect of a radiofrequency field circularly
polarized in the xy-plane, so that the magnetic component of
the electromagnetic field (the only component we need to consider) is rotating around the z-direction in the same sense as
the Larmor precession of the nuclei. The strength of the rotating magnetic field is ℬ1.
To interpret the effects of radiofrequency pulses on the magnetization, it is useful to imagine stepping on to a platform, a
so-called rotating frame, that rotates around the direction
of the applied field. Suppose we choose the frequency of the
radio­frequency field to be equal to the Larmor frequency of the
spins, νL = γNℬ0/2π; this choice is equivalent to selecting the resonance condition in the conventional experiment. The rotating
magnetic field is in step with the precessing spins, the nuclei
experience a steady ℬ1 field, and precess about it at a frequency
γNℬ1/2π (Fig. 14C.3). Now suppose that the ℬ1 field is applied
in a pulse of duration ∆τ = 14 × 2π /γ N B1, the magnetization tips
through an angle of 14 × 2π = π /2 (90°) away from the vertical
z-direction and we say that we have applied a 90° pulse, or a
‘π/2 pulse’ (Fig. 14C.4a).

Brief illustration 14C.1  Radiofrequency pulses

The duration of a radiofrequency pulse depends on the
strength of the ℬ1 field. If a 90° pulse requires 10 µs, then for
protons

B1 =

π
= 5.9 ×10−4 T
2 × (2.675 ×10−8 T −1 s −1) × (1.0 ×10−5 s)
γN

∆τ

or 0.59 mT.
Self-test 14C.1  How long would a 180° pulse require for

protons?

Answer: 20 µs

Now imagine stepping out of the rotating frame. To a fixed
external observer (the role played by a radiofrequency coil), the
magnetization vector is now rotating at the Larmor frequency in
the xy-plane (Fig. 14C.4b). The rotating magnetization induces
in the coil a signal that oscillates at the Larmor frequency and
which can be amplified and processed. In practice, the processing takes place after subtraction of a constant high frequency
component (the radiofrequency used for ℬ1), so that all the signal manipulation takes place at frequencies of a few kilohertz.
As time passes, the individual spins move out of step (partly

because they are precessing at slightly different rates, as we


584  14  Magnetic resonance
M

M

Signal

νL

ν = νL

Figure 14C.3  (a) In a resonance experiment, a circularly
polarized radiofrequency magnetic field ℬ1 is applied in the
xy-plane (the magnetization vector lies along the z-axis). (b) If
we step into a frame rotating at the radiofrequency, ℬ1 appears
to be stationary, as does the magnetization M if the Larmor
frequency is equal to the radiofrequency. When the two
frequencies coincide, the magnetization vector of the sample
rotates around the direction of the ℬ1 field.
M
B0

90° pulse

Detecting
coil

M
(a)

B1

νL

(b)

Figure 14C.4  (a) If the radiofrequency field is applied for a
certain time, the magnetization vector is rotated into the
xy-plane. (b) To an external stationary observer (the coil), the
magnetization vector is rotating at the Larmor frequency, and
can induce a signal in the coil.

explain later), so the magnetization vector shrinks exponentially with a time constant T2 and induces an ever weaker signal
in the detector coil. The form of the signal that we can expect is
therefore the oscillating-decaying free-induction decay (FID)
shown in Fig. 14C.5. The y-component of the magnetization
varies as
M y (t ) = M 0 cos(2πL t )e − t /T

2



Time, t

B1

(b)

Figure 14C.5  A simple free-induction decay of a sample of
spins with a single resonance frequency.

(b)  Time- and frequency-domain signals
We can think of the magnetization vector of a homonuclear
AX spin system with spin–spin coupling constant J = 0 as
consisting of two parts, one formed by the A spins and the
other by the X spins. When the 90° pulse is applied, both magnetization vectors are rotated into the xy-plane. However,
because the A and X nuclei precess at different frequencies,
they induce two signals in the detector coils, and the overall
FID curve may resemble that in Fig. 14C.6a. The composite
FID curve is the analogue of the struck bell emitting a rich
tone composed of all the frequencies (in this case, just the two
resonance frequencies of the uncoupled A and X nuclei) at
which it can vibrate.
The problem we must address is how to recover the resonance frequencies present in a free-induction decay. We know
that the FID curve is a sum of decaying oscillating functions,
so the problem is to analyse it into its components by carrying out a Fourier transformation. The analysis of the FID curve
is achieved by the standard mathematical technique of Fourier
transformation, which is explained more fully in Mathematical
background 7 following chapter 18.

Free-induction decay  (14C.1)

We have considered the effect of a ℬ1 pulse applied at exactly
the Larmor frequency. However, virtually the same effect is
obtained off resonance, provided that the pulse is applied close
to νL. If the difference in frequency is small compared to the

inverse of the duration of the 90° pulse, the magnetization will
end up in the xy-plane. Note that we do not need to know the
Larmor frequency beforehand: the short pulse is the analogue
of the hammer blow on the bell, exciting a range of frequencies.
The detected signal shows that a particular resonant frequency
is present.

(a)
Signal

B1

(a)

Time, t

(b)

Frequency, ν

Figure 14C.6  (a) A free-induction decay signal of a sample of
AX species and (b) its analysis into its frequency components.