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

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

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

prop-erties of molecules of all sizes, up to as large as biopolymers

14A general principles

The chapter begins with an account of the principles that

gov-ern spectroscopic transitions between spin states of nuclei and

electrons in molecules It also describes simple experimental

arrangements for the detection of these transitions The

con-cepts developed in this Topic prepare the ground for a

dis-cussion of the chemical applications of NMR and ‘electron

paramagnetic resonance’ (EPR)

This Topic contains a discussion of conventional NMR,

show-ing 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

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

trans-form’ techniques It is through the application of these pulse

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

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 ful for diagnosing disease In Impact I14.2 we highlight an

use-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

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

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

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 electromag-netic field generated by a distant transmitter Historically, spec-troscopic 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 dence 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 elec-trons 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

evi-we establish how the energies of electrons and nuclei depend

on the applied field This material sets the stage for the tion of the structure and dynamics of many kinds of molecules

explora-by magnetic resonance spectroscopy (Topics 14B–14D)

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 logue of s for electrons) To understand the nuclear magnetic

ana-resonance (NMR) experiment we need to describe the

behav-iour of nuclei in magnetic fields and then the basic techniques for detecting spectroscopic transitions

(a) The energies of nuclei in magnetic fields

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

property of a nucleus in its ground state (the only state we sider) and, depending on the nuclide, is either an integer or a half-integer (Table 14A.1) A nucleus with spin quantum num-

con-ber I has the following properties:

Contents

14a.1 Nuclear magnetic resonance 561

(a) The energies of nuclei in magnetic fields 561

brief illustration 14a.1: the resonance

brief illustration 14a.2: nuclear spin populations 564

14a.2 Electron paramagnetic resonance 564

(a) The energies of electrons in magnetic fields 565

brief illustration 14a.3: the resonance condition

brief illustration 14a.4: electron spin populations 566

Checklist of concepts 567

Checklist of equations 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.

➤

➤ 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).

• 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

According to the second property, the spin, and hence the

mag-netic moment, of the nucleus may lie in 2I + 1 different

orien-tations relative to an axis A proton has I =1

2 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

mag-netic field B is equal to the scalar product (Mathematical

back-ground 5 following Chapter 9)

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

To write an expression for ˆμ , we use the fact that, just as for

electrons (Topic 9B), the magnetic moment of a nucleus is

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

where γN is the nuclear magnetogyric ratio of the

speci-fied 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

JT

m



2 5 051 10. 27 1 nuclear magneton (14A.4b)

(where mp is the mass of the proton) and an empirical

con-stant called the nuclear g-factor, g I, the energy in eqn 14A.4a becomes

E m I = −g I µN m I g I=γ µN

N

Nuclear g-factors are experimentally determined

dimension-less quantities with values typically between –6 and +6 (Table

14A.2) Positive values of g I 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

Odd Odd Integer (1, 2, 3, …)

Even Odd Half-integer ( , , , ) 1

2 32 52 … Odd Even Half-integer ( , , , ) 1

2 32 52 …

* 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.

Nuclide Natural abundance/% Spin I g-factor, g I Magnetogyric ratio, γN /(10 7 T−1 s−1 ) NMR frequency at 1 T, ν/MHz

(14A.4c)

energies of a nuclear spin in a magnetic field

that states with m I > 0 lie below states with m I < 0 It follows that

the energy separation between the lower m I= +1

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= γ2NB0

π Definition larmor frequency of a nucleus (14A.7)

It follows by comparing this expression with eqn 14A.6 that resonance absorption by spin-1

2 nuclei occurs when the Larmor precession frequency νL is the same as the frequency of the applied electromagnetic field, ν

(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 electromag-netic 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-1 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-1

2 nuclei include 13C, 19F, and 31P

An NMR spectrometer consists of the appropriate sources of radiofrequency radiation and a magnet that can produce a uni-form, intense field Most modern instruments use a supercon-ducting 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 spin-ning 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 tem-perature 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 mol-ecules 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

Brief illustration 14A.1 The resonance condition in NMR

To calculate the frequency at which radiation comes into nance with proton (I=1)

reso-2 spins in a 12.0 T magnetic field we use eqn 14A.6 as follows:

Self-test 14A.1 Determine the resonance frequency for 31P

nuclei, for which γN = 1.0841 × 108 T−1 s−1, under the same conditions

Answer: 207 MHz

Magnetic

field off

Magnetic field on

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

m s = +½

m s = –½

spin-1 nucleus and an external magnetic field may be visualized

as the precession of the vectors representing the angular

momentum

(14A.6)

Spin­ 1 nuclei resonance condition

kT

α− β≈ γNB0

2 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

increas-ing the population difference

By combining eqns 14A.8a and 14A.8b we see that the sity is proportional to B0, 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 appear-ance 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)

resonanceElectron paramagnetic resonance (EPR), or electron spin resonance (ESR), is the study of molecules and ions contain-

ing unpaired electrons by observing the magnetic field at which they come into resonance with radiation of known frequency

Brief illustration 14A.2 Nuclear spin populations

For protons γN = 2.675 × 108 T−1 s−1 Therefore, for 1 000 000

K)

T



≈

Even in such a strong field there is only a tiny imbalance of

population of about 35 in a million

Self-test 14A.2 For 13C nuclei, γN = 6.7283 × 107 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

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

electromag-netic 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

popula-tion of the upper state (Nβ) At the low frequencies typical of

magnetic resonance, spontaneous emission can be neglected

as it is very slow Therefore, the net rate of absorption is portional to the difference in populations, and we can writeRate of absorption∝Nα−Nβ

pro-The intensity of absorption, the rate at which energy is absorbed, is proportional to the product of the rate of absorp-tion (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

fre-quency is proportional to the applied magnetic field (through

ν = γNB0/2π), so we can writeRate of absorption∝(Nα−Nβ)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

The expansion of the exponential term is appropriate for

ΔE = γNB0 ≪ kT, a condition usually met for nuclear spins It follows that

−

N

kT N

1//

/

1 γ B0

 1

β/ α

//

N N

Probe Computer

Preamplifier Receiver Detector Transmitter

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

As we have done for NMR, we write expressions for the

reso-nance condition in EPR and then describe the general features

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

magnetogyric ratio of the electron:

e

= −2g e m 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

defini-tion of the magnetogyric ratio.) Dirac’s relativistic theory, his

modification of the Schrödinger equation to make it

consist-ent 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,

Because the eigenvalues of the operator ˆs z are m s with

m s= +1( )α and m s= −1( )β , it follows that the energies of an

electron spin in a magnetic field are

2 9 274 10. 24 1 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

differ-ent values of m s are degenerate When a field is present, the

degeneracy is removed: the state with m s= +1

m s= +1( )α and (lower) m s= −1( )β levels of an electron spin in

a magnetic field of magnitude B0 in the z-direction is

electro-h=geμBB0 Electrons resonance condition (14A.12b)

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

reso-Brief illustration 14A.3 The resonance condition in EPR

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

which corresponds to a wavelength of 3.6 cm

Self-test 14A.3 Determine the magnetic field for EPR sitions in a spectrometer that uses radiation of wavelength 0.88 cm

tran-Answer: 1.2 T

Magnetic field off

Magnetic field on α,

β,

m s = +½

m s = –½

geµB B 0

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

(14A.11a)

energies of an electron spin in a magnetic field

(14A.11b)

energies of an electron spin in a magnetic field

(b) The EPR spectrometer

It follows from Brief illustration 14A.3 that most

commer-cial EPR spectrometers operate at wavelengths of

approxi-mately 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

spec-troscopy, 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 H6 6−) is shown in Fig 14A.6 The peculiar appearance of the

spectrum, which is in fact displayed as the first-derivative of the

absorption, arises from the detection technique, which is tive to the slope of the absorption curve (Fig 14A.7)

sensi-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 α state in energy and, by a similar argument to that for nuclei,

kT

β− α≈ eμBB0

2 Electrons Population difference (14A.13)

where N is the total number of spins.

Brief illustration 14A.4 Electron spin populations

When 1000 electron spins are exposed to a 1.0 T magnetic field

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 trons is larger than that for nuclear spins even at the lower magnetic field strengths normally employed

elec-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

Phase sensitive detector

spectrometer A typical magnetic field is 0.3 T, which requires

9 GHz (3 cm) microwaves for resonance

Field, B Derivative

of absorption, dA/dB

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

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

ener-gies in the presence of a magnetic field

☐ 3 Nuclear magnetic resonance (NMR) is the observation

of resonant absorption of radiofrequency

electromag-netic radiation by nuclei in a magelectromag-netic field

☐ 4 NMR spectrometers consist of a source of

radiofre-quency 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 B0)

☐ 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

observa-tion of resonant absorpobserva-tion of microwave netic radiation by unpaired electrons in a magnetic field

electromag-☐ 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 E m

N N

B B

0 0

14A.4c

Resonance condition (spin- 1 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 14A.8b

Magnetogyric ratio (electron) γe = −gee/2me ge = 2.002 319 14A.9b

Energies of an electron spin in a magnetic field E m

e

e B

B B

0 0

14A.11b

Bohr magneton μB = e/2me μB = 9.274 × 10 −24 J T −1 14A.11c

Resonance condition (electrons) hν = geμBB0 14A.12b

Population difference (electrons) Nβ − Nα ≈ NgeμBB0/2kT 14A.13

14B Features of nmr spectra

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 = −σ B0 Definition shielding constant (14B.1)

where the dimensionless quantity σ is called the shielding stant of the nucleus (σ is usually positive but may be negative)

con-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 ent 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

Because the total local field Bloc is

the nuclear Larmor frequency (eqn 14A.7 of Topic 14A,

νL = γNB/2π) becomes

Contents

example 14b.1: Interpreting the nmr spectrum

14b.2 The origin of shielding constants 570

brief illustration 14b.3: the effect of aromatic

example 14b.3: accounting for the fine

brief illustration 14b.4: the karplus equation 576

brief illustration 14b.5: magnetic fields

brief illustration 14b.6: chemical and magnetic

brief illustration 14b.7: strongly coupled spectra 579

14b.4 Conformational conversion and exchange

➤ Why do you need to know this material?

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.

➤

➤ What is the key idea?

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.

L=γN2Bloc=γ2NB0 1−σ

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

molec-ular 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

tetramethylsi-lane, 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

stand-ard 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 Definition δ scale (14B.4)

where ν° is the resonance frequency of the standard The

advan-tage 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

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 differ-ent 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.

Brief illustration 14B.1 The δ scale

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

 − = (5 MHz 100 / 06) ×100=(5 Hz 100 ) × 00=5 Hz00

because 1 MHz = 106 Hz In a spectrometer operating at

ν° = 100 MHz, the shift relative to the reference would be only

100 Hz

A note on good practice In much of the literature,

chemi-cal 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.

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

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

Answer The spectrum is consistent with the following assignments:

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

RCH3

–CH2– R–NH2–CH–

RC–CH3

ArC–CH3–CO–CH3

ROH –C=CH–

ArOH Ar–H –CHO –COOH

0 2 4 6 8 10 12

R3C –

R3C +

R–C–H

>C=C<X –C=C– –C=C<

C–X in ArX R–C=N–R–COOH R–CHO

resonances and (b) 13C resonances

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

tech-niques outlined in Topic 10E) about the distribution of

elec-tron 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

The empirical approach supposes that the observed ing constant is the sum of three contributions:

shield-σ shield-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

ques-tion 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

(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 σd+ 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

paramag-netic contribution to σ (local) reinforces the applied magparamag-netic field and deshields the nucleus in question Therefore, σd > 0

and σp < 0 The total local contribution is positive if the netic contribution dominates, and is negative if the paramag-netic contribution dominates

diamag-The diamagnetic contribution arises from the ability of the applied field to generate a circulation of charge in the ground-state electron distribution of the atom The circulation gener-ates a magnetic field that opposes the applied field and hence

shields the nucleus The magnitude of σd depends on the tron density close to the nucleus and can be calculated from the

where μ0 is the vacuum permeability (a fundamental

con-stant, 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 σd from the Lamb formula,

calcu-late 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

• 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

repre-sented 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

(etha-nal) has lines at δ = 2.20 and δ = 9.80 Which feature can be

assigned to the CHO proton?

Answer: δ = 9.80

1.2 3.6

C H3CH2OH

CH3C H2OH

CH3CH2O H

letters denote the protons giving rise to the resonance peak,

and the step-like curve is the integrated signal

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

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

dis-torted 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

elec-tronegativity 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

abil-ity 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

lin-ear 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

of the neighbouring group X on the proton H in a molecule

such as H–X The applied field generates currents in the

elec-tron distribution of X and gives rise to an induced magnetic moment proportional to the applied field; the constant of pro-

portionality is the magnetic susceptibility, χ (chi), of the group

X: μinduced = χ B0 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 netic field with a component parallel to the applied field and

mag-at a distance r and angle θ (1) thmag-at has the form (The chemist’s

toolkit 14B.1):

θ r

1

µinduced

Blocal∝µinduced − cosθ

r3 (1 3 2 ) 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

2 0

Self-test 14B.3 Derive a general expression for σd that applies

to all hydrogenic atoms

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

We see that the strength of the additional magnetic field

expe-rienced by the proton is inversely proportional to the cube of the

distance r between H and X If the magnetic susceptibility is

inde-pendent 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

σ(neighbour) (∝ χ χ− ⊥)1 3− cos2θ

3

r

θ r

χ||

χ⊥

2

X H

where χ∥ and χ⊥ are, respectively, the parallel and

perpen-dicular 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 tive according to the relative magnitudes of the two magnetic susceptibilities and the relative orientation of the nucleus with

nega-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)

where ε0 is the vacuum permittivity, which is related to μ0 by

ε0 = 1/μ0c2 The component of magnetic field in the

z-direc-tion is

z r

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

–

–

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

Brief illustration 14B.2 Ring currents

A special case of a neighbouring group effect is found in matic compounds The strong anisotropy of the magnetic sus-ceptibility of the benzene ring is ascribed to the ability of the

aro-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

B

Ring current

Magnetic field

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

(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

suscepti-bility 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

The splitting of resonances into individual lines by spin–spin

coupling shown in Fig 14B.2 is called the fine structure of the

spectrum It arises because each magnetic nucleus may tribute to the local field experienced by the other nuclei and

con-so modify their recon-sonance frequencies The strength of the

interaction is expressed in terms of the scalar coupling

con-stant, 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 propor-

tional 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 I 1z I 2z , with eigenvalues m1m22, so in that case the energy due to spin–spin coupling is

E m m1 2=hJm m1 2 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-1

2 nuclei A and X with very ferent chemical shifts in the sense that the difference in chemi-

dif-cal shift corresponds to a frequency that is large compared to J.

For a spin-1 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 cou-pling is neglected

where νA and νX are the Larmor frequencies of A and X and

mA and mX are their quantum numbers (mA= ±1,mX= ± )

2 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

E m mA X = −h mA A−h mX X+hJm mA X (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 β A higher energy is obtained if both spins are α or both spins are β 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 ββ

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

shielded or deshielded by currents induced by the triple bond?

Answer: Shielded

Brief illustration 14B.3 The effect of aromatic solvents

An aromatic solvent like benzene can give rise to local

cur-rents 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

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

B

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

states are both raised by 1

4hJ and that the αβ and βα states are both lowered by 1

4hJ.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 ← αA occurs when the X nucleus is α, and the other in

which βA ← αA 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

Therefore, the A resonance consists of a doublet of

separa-tion J centred on the chemical shift of A (Fig 14B.10) Similar

remarks apply to the X resonance, which consists of two

transitions according to whether the A nucleus is α or β (as shown in Fig 14B.9) The transition energies are

∆E h= X±1hJ

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 blet 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)

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 ← α 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

βΑαX

αΑαX

αΑβX

βΑβ X

and transitions shown in Fig 14B.8 Once again, we have

exaggerated the effect of spin–spin coupling

A resonance X resonance

J J

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

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

Three equivalent X nuclei (an AX3 species) split the

reso-nance 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-1

2 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

tri-angle (3) Successive rows of this tritri-angle are formed by adding

together the two adjacent numbers in the line above

(b) The magnitudes of coupling constants

The scalar coupling constant of two nuclei separated by N bonds

is denoted N J, 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 typ-ical 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 erally be ignored One of the longest range couplings that has been detected is 9JHH = 0.4 Hz between the CH3 and CH2 pro-tons in CH3C b CeC b CeC b CeCH2OH

gen-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 nega-tive, 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

3JHH coupling constant is often found to depend on the dihedral

angle φ (4) according to the Karplus equation:

φ

H H

Answer The three protons of the CH3 group split the nance of the CH2 protons into a 1:3:3:1 quartet with a split-

reso-ting J Likewise, the two protons of the CH2 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

pro-tons migrate rapidly from molecule to molecule (including molecules of impurities in the sample) and their effect aver-ages 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 tons in 14

pro-4

NH+? The spin quantum number of nitrogen-14 is 1

Answer: 1:1:1 triplet from N

δA

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

δA

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

J = +A B φ+C φ 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

fol-lows that the measurement of 3JHH in a series of related

com-pounds 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:

(c) The origin of spin–spin coupling

Spin–spin coupling is a very subtle phenomenon and it is

bet-ter to treat J as an empirical paramebet-ter 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

consider-ing the magnetic interactions within molecules

A nucleus with spin projection m I gives rise to a magnetic

field with z-component Bnuc at a distance R, where, to a good

Spin–spin coupling in molecules in solution can be explained

in terms of the polarization mechanism, in which the

inter-action is transmitted through the bonds The simplest case to consider is that of 1JXY, where X and Y are spin-1 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 netic 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

mag-Brief illustration 14B.4 The Karplus equation

The investigation of HeNeCeH 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 3JHH≈ 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 3JHH coupling constant for such a grouping was

deter-mined to be 2.1 Hz Is this value consistent with a helical or

sheet conformation?

Answer: Helical conformation

Brief illustration 14B.5 Magnetic fields from nuclei

The z-component of the magnetic field arising from a proton

(m I=1) at R = 0.30 nm, with its magnetic moment parallel to the z-axis (θ = 0) is

R

mI

(1 3cos ) 2θ

00 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

interac-tion between spins cannot account for the fine structure of the spectra of rapidly tumbling molecules

Self-test 14B.8 In gypsum, CaSO4⋅2H2O, 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

with angle predicted by the Karplus equation for an HCCH

group and an HNCH group

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 β electron of the bonding pair will

tend to be found nearby, because that is an energetically

favour-able arrangement (Fig 14B.16) The second electron in the

bond, which must have α spin if the other is β (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

β spin has a lower energy than when it has α spin The

oppo-site is true when X is β, 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

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 β 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

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 chemi-

cal 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

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

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

Fermi Fermi

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

Strictly speaking, CH3 protons are magnetically

inlent However, they are in practice made magnetically

equiva-lent 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

cou-pling 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

rela-tive orientations of the spins within the group (Fig 14B.18)

Then, because the relative orientations of nuclear spins are

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 rela-tive 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

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

sym-metry and undergo the same chemical reactions However,

although the protons in CH2F2 are magnetically equivalent,

those in CH2 = 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

β

β β (a)

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

With spin–spin coupling

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

Justification 14B.1 The energy levels of an A2 system

Consider an A2 system of two spin-1

2 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 trons) and their total projection M I on the z-axis The states

elec-are analogous to those for two electrons in singlet and triplet states (Topic 9C):

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

Spins parallel, I = 1: M I = +1 αα

M I = 0 (1/21/2){αβ + βα}

M I = −1 ββ

Spins paired, I = 0: M I =0 (1/21/2){αβ − βα}

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 ΔM I = ±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

equiva-lent nuclei and to think of the groups of nuclear spins as

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

‘sec-ond-order spectra’) and are much more difficult to analyse

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 chemi-cal shift difference) has a first-order spectrum An AB sys-tem, 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 cies because A and X are nuclei of different elements (such

frequen-as 13C and 1H), in which case they form a heteronuclear spin system AX may also denote a homonuclear spin system in

Brief illustration 14B.7 Strongly coupled spectra

Figure 14B.20 shows NMR spectra of an A2 system (top) and

an AX system (bottom) Both are simple ‘first-order’ spectra

At intermediate relative values of the chemical shift ence and the spin–spin coupling, complex ‘strongly coupled’ spectra are obtained Note how the inner two lines of the bot-tom 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

differ-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

ν°Δδ << J

ν°Δδ >> J

ν°Δδ ≈ J

ν°Δδ ≈ J

AX system (bottom) are simple ‘first-order’ spectra

two states with M I = 0 are unchanged by the field because they

are composed of equal proportions of α and β 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

For parallel spins, I = 1 and E= +1hJ

4 ; for antiparallel spins

I = 0 and E= −3hJ

4 , 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

which the nuclei are of the same element but in markedly

dif-ferent environments

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

spec-trum 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

con-formation gives rise to a linewidth that is comparable to the

dif-ference 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)

Brief illustration 14B.8 The effect of chemical exchange

on NMR spectra

The NO group in N,N-dimethylnitrosamine, (CH3)2NeNO

(5), rotates about the NeN bond and, as a result, the magnetic

A similar explanation accounts for the loss of fine ture in solvents able to exchange protons with the sample For example, hydroxyl protons are able to exchange with water

struc-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 cal 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

chemi-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

environments of the two CH3 groups are interchanged The two CH3 resonances are separated by 390 Hz in a 600 MHz spectrometer According to eqn 14B.16,

τ = ×23901 2/ −1 =1 2( ) .

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 tion of a single line from the same molecule in a 300 MHz spectrometer?

observa-Answer: Conformation lifetime less than 2.3 ms

H C

N

O

conformation to another, the positions of its protons are

interchanged and the protons jump between magnetically

distinct environments

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

contribu-tion, a neighbouring group contribucontribu-tion, and a solvent

contribution

☐ 3 The local contribution is the sum of a diamagnetic

con-tribution and a paramagnetic concon-tribution

☐ 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

indi-vidual 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

allo-cated to definite groups

☐ 12 Coalescence of two NMR lines occurs when a tional interchange or chemical exchange of nuclei is fast

conforma-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) × 10

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 − 3cos

2θ)/r3 } The angle θ is defined in (1) 14B.10b

Karplus equation 3JHH = A + B cos φ + C cos 2φ A, B, and C are empirical constants 14B.14

Condition for coalescence of two NMR lines τ = 21/2 /πδν Conformational conversions and

exchange processes 14B.16