solution manual for understanding nmr spectroscopy

8/14/2019 Solution Manual for Understanding NMR Spectroscopy

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Solutions manual forUnderstanding NMR spectroscopy

second edition

James Keeler and Andrew J. PellUniversity of Cambridge, Department of Chemistry

1

2

1

1

2

2

(a) (b)

Version 2.0 James Keeler and Andrew J. Pell July 2005 and March 2010

This solutions manual may be downloaded and printed for personal use. It may notbe copied or distributed, in part or whole, without the permission of the authors.


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PrefaceWe hope that this solutions manual will be a useful adjunct to Understanding NMR Spectroscopy(2nd edition, Wiley, 2010) and will encourage readers to work through the exercises. The old adagethat practice makes perfect certainly applies when it comes to getting to grips with the theory of NMR.We would be grateful if users of this manual would let us know (by EMAIL to [email protected] )of any errors they come across. A list of corrections will be maintained on the spectroscopyNOW website

http://www.spectroscopynow.com/nmr (follow the link education)

Cambridge , March 2010


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Contents

2 Setting the scene 1

3 Energy levels and NMR spectra 5

4 The vector model 11

5 Fourier transformation and data processing 21

6 The quantum mechanics of one spin 29

7 Product operators 35

8 Two-dimensional NMR 47

9 Relaxation and the NOE 59

10 Advanced topics in two-dimensional NMR 73

11 Coherence selection: phase cycling and eld gradient pulses 89

12 Equivalent spins and spin-system analysis 101

13 How the spectrometer works 123


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iv CONTENTS


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2Setting the scene

2.1

We need Eq. 2.1 on page 6:(ppm) = 106

ref ref

.

For the rst peak(ppm) = 106

500 .135021 500 .134271500 .134271

= 1.50 ppm.

For the second peak the shift is 7.30 ppm.Using Eq. 2.3 on page 8(ppm) = 106

ref rx

,

with rx = 500 .135271 MHz gives the two shifts as 1.50 ppm and 7.30 ppm i.e. identical valuesto three signicant gures. To all intents and purposes it is perfectly acceptable to use Eq. 2.3 onpage 8.

The separation of the two peaks can be converted to Hz using Eq. 2.2 on page 7:

frequency separation in Hz = (1 2) ref (in MHz).So the separation is

(7 .30 1.50) 400 .130000 = 2321 Hz .The conversion to rad s1 is made using Eq. 2.4 on page 17

= 2 = 2 2321 = 14583 rad s1 .

2.2

For J AB = 10 Hz & J AC = 2 Hz, the line positions are 6, 4, + 4, + 6 Hz. For J AB = 10 Hz & J AC = 12 Hz, the line positions are 11 , 1, + 1, + 11 Hz; note that compared to the rst multipletthe two central lines swap positions. For J AB = 10 Hz & J AC = 10 Hz, the line positions are 10, 0,0, + 10 Hz; in this case, the line associated with the spin states of spins B and C being and , andthe line in which the spin states are and , lie of top of one another giving a 1:2:1 triplet.


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Chapter 2: Setting the scene 2

0 10-10 0 10-10 0 10-10

J AB J AB

J AC

J AB

J AC

J AC

J AB = 10 Hz J AC = 2 Hz J AB = 10 Hz J AC = 12 Hz J AB = 10 Hz J AC = 10 Hz

Introducing a thirdcoupling gives a doublet of doublet of doublets. The line positions are1.5, 3.5,6.5, 8.5 Hz. For clarity, only the spin state of the fourth spin, D, are shown by the grey-headedarrows on the last line of the tree.

0 10-10

J AB

J AC

J AD

J AB = 10 Hz J AC = 2 Hz J AD = 5 Hz

2.3

The frequency, in Hz, is 1/period:

= 1

2.5 109 = 4 108

Hz or 400 MHz.

Converting to rad s1gives: = 2 = 2.51 109 rad s1 .

(a) 90 is one quarter of a rotation so will take 14 2.5 109 = 6.25 1010 s .(b) As 2 radians is a complete rotation, the fraction of a rotation represented by 3/ 2 is

(3/ 2)/ (2) = 3/ 4, so the time is 0.75 2.5 109 = 1.875 109 s .(c) 720 is two complete rotations, so the time is 2 2.5 109 = 5.0 109 s .


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To convert from angular frequency to Hz we need Eq. 2.4 on page 17

= 2

= 7.85 104

2= 12494 Hz.

The period is 1/frequency:T =

1

= 1

12 494 = 8.00 105 s .

2.4

time

x

y

x

y

x

y

= 3 /2 = 135

= 0 or 2

(a) & (c) (b) (d)

y -comp.x -comp.

For = 0 or 2 radians, the x-component is a cosine wave, and the y-component is a sine wave. For = 3/ 2, the y-component is minus a cosine wave, and the x-component is a sine wave.

2.5

We need the identitysin( A + B) sin A cos B + cos A sin B.

Using this we nd:sin( t + ) = sin( t )cos + cos ( t )sin

= sin( t ),where to go to the second line we have used cos = 1 and sin = 0. So the y-component is indeedr sin( t ).


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3Energy levels and NMR spectra

3.1

The expression for H one spin is given by Eq. 3.2 on page 29: H one spin = B0 I z.

We need to work out the effect that H one spin has on 1/2 : H one spin1/2

= B0 I z1/2=

B

0 1

2

1/2

= 12 B01/2 .

To go to the second line we have used Eq. 3.3 on page 30 i.e. that 1/2 is an eigenfunction of I z. The

wavefunction has been regenerated, multiplied by a constant; 1/2 is therefore an eigenfunction of H one spin with eigenvalue 12 B0.

3.2

The Larmor frequency, in Hz, of a nucleus with zero chemical shift is dened by Eq. 3.8 on page 32:

0 = B02

= 6.7283 107 9.42

= 1.01 108 Hz or 101 MHz.To convert to rad s1, we need to multiply the frequency in Hz by 2:

0 = 2 0 = 2 1.01 108 = 6.32 108 rad s1 .In the case of a non-zero chemical shift, the Larmor frequency, in Hz, is:

0 = (1 + 106) B0

2

= 6.7283

107

(1 + 77

10

6)

9.4

2= 1.01 108 Hz .


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Chapter 3: Energy levels and NMR spectra 6

This is an identical value to three signicant gures. We need to go to considerably more guresto see the difference between these two Larmor frequencies. To seven gures the frequencies are1.00659 108 Hz and 1.00667 108 Hz.

3.3

We let H one spin act on the wavefunction + 1/2 :

H one spin+ 1/2 = 0 I z+ 1/2= 1

2 0+ 1/2 ,

where the Hamiltonian has been expressed in angular frequency units. To go to the second line, wehave used the fact that + 1/2 is an eigenfunction of I z with eigenvalue + 12 .In the same way,

H one spin1/2 = 12 01/2 .

Hence, 1/2 are eigenfunctions of H one spin with eigenvalues 12 0.

3.4

Following the approach in section 3.5 on page 35, we let the Hamiltonian act on the productwavefunction:

H two spins, no coupl., 1, 2 = 0,1 I 1 z + 0,2 I 2 z , 1, 2= 0,1 I 1 z, 1, 2 + 0,2 I 2 z, 1, 2= 0,1 I 1 z, 1 , 2 + 0,2, 1 I 2 z, 2 .

To go to the third line, we have used the fact that I 1 z acts only on , 1 and not on , 2. Similarly, I 2 zacts only on , 2.Using Eq. 3.11 on page 35 i.e. that , 1 and , 2 are eigenfunctions of I 1 z and I 2 z, the terms in thesquare brackets can be rewritten:

H two spins, no coupl.

, 1

, 2

= 0,1

I 1 z

, 1

, 2

+ 0,2

, 1

I 2 z

, 2

= 12 0,1, 1, 2 +

12 0,2, 1, 2

= 12 0,1 +

12 0,2 , 1, 2 .

Hence, , 1, 2 is an eigenfunction of H two spins, no coupl. with eigenvalue 12 0,1 + 12 0,2.

Letting the coupling term act on the product wavefunction:

J 12 I 1 z I 2 z, 1, 2 = J 12 I 1 z, 1 I 2 z, 2= J 12 12 , 1 12 , 2= 1

4 J 12 , 1, 2 .

, 1, 2 is indeed an eigenfunction of the coupling term, with eigenvalue 14 J 12 : this corresponds tothe energy.


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H two spins, no coupl. and the coupling term share the same eigenfunctions (a result of the fact that thetwo terms commute). Since the Hamiltonian for two coupled spins can be represented as the sumof these two terms,

H two spins = H two spins, no coupl. + 2 J 12 I 1 z I 2 z,

it follows that it must also have the same eigenfunctions. Hence, , 1, 2 is an eigenfunction of H two spins with energy eigenvalue 12 0,1 + 12 0,2 + 14 J 12 , i.e. the sum of the individual eigenvalues of H two spins, no coupl. and J 12 I 1 z I 2 z.

3.5

Reproducing Table 3.2 on page 38 for 0,1 = 100 Hz, 0,2 = 200 Hz and J 12 = 5 Hz:number m1 m2 spin states eigenfunction eigenvalue/Hz

1 + 12 + 12 , 1, 2 + 12 0,1 + 12 0,2 + 14 J 12 = 148 .752 + 12 12 , 1 ,2 + 12 0,1 12 0,2 14 J 12 = 48 .753 12 + 12 ,1, 2 12 0,1 + 12 0,2 14 J 12 = 51 .254 12 12 ,1 ,2 12 0,1 12 0,2 + 14 J 12 = 151 .25

The set of allowed transitions is:

transition spin states frequency/Hz1 2 E 2 E 1 = 197 .503 4 E 4 E 3 = 202 .501 3 E 3 E 1 = 97 .502 4 E 4 E 2 = 102 .50

80 100 120 140 160 180 200 220

frequency / Hz

24 3413 12

spin 1 flipsflipsspin 2

spin 1spin 2


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If J 12 = 5 Hz, the table of energies becomes:number m1 m2 spin states eigenfunction eigenvalue/Hz

1 + 12 + 12 , 1, 2 + 12 0,1 + 12 0,2 + 14 J 12 = 151 .252 + 12 12 , 1 ,2 + 12 0,1 12 0,2 14 J 12 = 51 .253 12 + 12 ,1, 2 12 0,1 + 12 0,2 14 J 12 = 48 .754 12 12 ,1 ,2 12 0,1 12 0,2 + 14 J 12 = 148 .75

80 100 120 140 160 180 200 220

frequency / Hz

13 1224 34

spin 1 flipsflipsspin 2

spin 1spin 2

The spectrum in unchanged in appearance. However, the labels of the lines have changed; the spinstate of the passive spin for each line of the doublet has swapped over.

3.6

The allowed transitions in which spin two ips are 12, 34, 56 and 78. Their frequencies are:

transition state of spin one state of spin three frequency/Hz12 0,2 12 J 12 12 J 23 = 19334 0,2 + 12 J 12 12 J 23 = 20356 0,2 12 J 12 + 12 J 23 = 19778

0,2 + 12 J 12 +

12 J 23 = 207

The multiplet is a doublet of doublets centred on minus the Larmor frequency of spin two.

There are two lines associated with spin three being in the state, and two with this spin being inthe state. Changing the sign of J 23 swaps the labels associated with spin three, but leaves thoseassociated with spin one unaffected.


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190 195 200 205 210 190 195 200 205 210

frequency / Hz

12 56 34 78

spin 1spin 3

56 12 78 34

spin 1spin 3

0,2 0,2

J 12

J 12 = 10 Hz J 23 = 4 Hz J 12 = 10 Hz J 23 = -4 Hz

J 23

3.7

The six zero-quantum transitions have the following frequencies:

transition initial state nal state frequency23 0,1 + 0,2 12 J 13 + 12 J 2367 0,1 + 0,2 + 12 J 13 12 J 2335 0,1 0,3 +

12 J 12

12 J 23

46 0,1 0,3 12 J 12 + 12 J 2325 0,2 0,3 + 12 J 12 12 J 1347 0,2 0,3 12 J 12 + 12 J 13

1

2 3

4

5

6 7

8


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The six transitions can be divided up into three pairs: 23 and 67 in which spins one and two ip, and spin three is passive,

35 and 46 in which spins one and three ip, and spin two is passive,

25 and 47 in which spins two and three ip, and spin one is passive.

Each pair of transitions is centred at the difference in the Larmor frequencies of the two spins whichip, and is split by the difference in the couplings between the two active spins and the passive spin.


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4The vector model

4.1

z

x

eff

1

The offset of the peak is 5 ppm. This can be converted to Hz using Eq. 2.2 on page 7:

2= 106 ref = 106 5 600 106 = 5 600 = 3000 Hz or 3 kHz.

From the diagram,tan =

1

= 25 103 2

3 103 2=

253

= 8.33 ,

so = 83 .For a peak at the edge of the spectrum, the tilt angle is within 7 of that for an on-resonance pulse;the B1 eld is therefore strong enough to give a reasonable approximation to a hard pulse over thefull shift range.

For a Larmor frequency of 900 MHz, the peak at the edge of the spectrum has an offset of 4.5 kHz,so the tilt angle is 80 . The larger offset results in the same B1 eld giving a poorer approximationto a hard pulse.


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Chapter 4: The vector model 12

4.2

From Fig. 4.16 on page 58, the y-component of the magnetization after a pulse of ip angle is M 0 sin . The intensity of the signal will, therefore, vary as sin , which is a maximum for = 90.

(a) If = 180 , the magnetization is rotated onto the z-axis. As sin 180 = 0, the signal intensityis zero.(b) If = 270 , the magnetization is rotated onto the y-axis. As sin 270 = 1, the signal will havenegative intensity of the same magnitude as for = 90.

4.3

From Fig. 4.16 on page 58, the intensity of the signal is proportional to sin , where the value of theip angle is given by Eq. 4.5 on page 58:

= 1 t p.

The pulse lengths of 5 and 10 s correspond to ip angles below 90. Increasing t p further causes to increase past 90, and so the value of sin (and hence the signal intensity) decreases. The null at20 .5 s corresponds to = 180 .

From the expression for the ip angle, it follows that = 1 t 180 . Therefore,

1 = t 180

=

20 .5 106 = 1.5 10

5 rad s1 or 2.4 104 Hz .

Another way to answer this question is to see that since a 180 pulse has a length of 20.5 s, acomplete rotation of 360 takes 41 .0 s. The period of this rotation is thus 41 .0 s, so the frequencyis

141 .0 106

= 2.4 104 Hz .This frequency is 1 / 2, the RF eld strength in Hz.

The length of the 90 pulse is simply half that of the 180 pulse:

t 90 = 12 20 .5 = 10 .25 s.The further null occurs at a pulse length that is twice the value of t 180 . This corresponds to a ipangle of 360 , for which the magnetization is rotated back onto the z-axis.


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4.4

x

-y

x

-y x

-y

x

-y

x

-y

x

-y

2 180 pulse

about y

startingposition

finalposition

resolved into x - andy - components

components after180 pulse

The vector has been reected in the yz-plane, and has a nal phase of 2, measured anti-clockwisefrom the y-axis.

4.5

x

-y

time

p h

a s e

,

0

0

/2

3 /2 2

2

1 8 0 o ( y

) p u

l s e

2

/2

3 /2

=

= 2

2

The spin echo sequence 90( x) 180 ( x) results in the magnetization appearing along they-axis. In contrast, the 90( x) 180 ( y)sequence results in the magnetization appearing alongthe y-axis. Shifting the phase of the 180 pulse by 90 thus causes the phase of the magnetizationto shift by 180 .A 180( x) pulse rotates the magnetization in the opposite sense to a 180 ( x) pulse, but the net effectis still to reect the magnetization vectors in the xz-plane. The sequence 90( x) 180 ( x) will, therefore, have the same effect as the 90( x)

180 ( x)

sequence i.e. the vector appears

on the y-axis at the end of the sequence.


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4.6

From section 4.11 on page 67, the criterion for the excitation of a peak to at least 90% of itstheoretical maximum is for the offset to be less than 1.6 times the RF eld strength. The Larmorfrequency of 31P at B0 = 9.4 T is:

0 = B02

= 1.08 108 9.4

2= 1.62 108 Hz or 162 MHz.

If the transmitter frequency is placed at the centre of the spectrum, the maximum offset isapproximately 350 ppm. In Hz, this is an offset of

2= 350 162 = 5.66 10

4 Hz or 56.6 kHz.

According to our criterion, the RF eld strength must be at least 56 .6/ 1.6 = 35 .3 kHz, from whichthe time for a 360 pulse is simply 1/ (35 .3 103) = 28 .28 s. Thus, the 90 pulse length is 14 28 .28 =7.07 s .

4.7

The ip angle of a pulse is given by Eq. 4.5 on page 58:

= 1 t p

So, 1 =

t p

.

For a 90 pulse, = / 2, so the B1 eld strength in Hz is:

12

= (/ 2)2 t p

= 1

4 10 106 = 2.5 104 Hz or 25 kHz.

The offset of 13C from 1H is 300 MHz, which is very much greater than the B1 eld strength. The13C nuclei are therefore unaffected by the 1H pulses.


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4.8

From Eq. 4.4 on page 57,eff = 21 + 2 .

If we let = 1, eff can be written

eff = 21 + 2 21 = 1 1 + 2 . (4.1)If t p is the length of a 90 pulse, we have 1 t p = / 2 and so

1 = 2t p

,

and hence substituting this into Eq. 4.1 on page 15 gives

eff = 2t p

1 + 2 .

Therefore the angle of rotation about the effective eld, eff t p, is given by

eff t p =

2t p 1 + 2 t p

=

2

1+

2

.

The null condition is when there is a complete rotation about the effective eld i.e. eff t p = 2:

2 = 2 1 + 2 .

Rearranging this gives

4 = 1 + 2 i.e. = 15 or = 15 1 ,which is in agreement with Fig. 4.28 on page 68.

The next null appears at eff t p =

4 i.e. two complete rotations; this corresponds to =

63 .For large offsets, 1, so 1 + 2 . The general null condition is eff t p = 2n, where n =1, 2, 3, . . . Combining these two conditions gives

2n = 2 1 + 2

2

,

for which we nd = 4n.


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4.9

In section 4.11.3 on page 70, it was demonstrated that, on applying a hard 180 pulse, the range of offsets over which complete inversion is achieved is much more limited than the range over whicha 90 pulse gives signicant excitation. Therefore, only peaks with small offsets will be invertedcompletely. Peaks with large offsets will not exhibit a null on the application of the 180 pulse.

4.10

The initial 90( x) pulse rotates the magnetization from the z-axis to the y-axis; after this theevolution in the transverse plane is as follows:

x

-y

x

-y 90(+ x )delay

x

-y

The x-, y- and z-components after each element of the pulse sequence are:

component after rst 90( x) after after second 90( x) x 0 M 0 sin M 0 sin

y M 0 M 0 cos 0 z 0 0 M 0 cos

The nal pulse is along the x-axis, and so leaves the x-component of the magnetization unchanged,but rotates the y-component onto the z-axis. The overall result of the sequence is M y = 0 and M x = M 0 sin .

0 /2 3 /2 2

-M 0

M 0

M x

A null occurs when M x = 0, i.e. when = n, where n = 0, 1, 2, . . .


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4.11

The initial spin echo sequence refocuses the offset, and aligns the magnetization along the y-axis.

If the nal pulse is about the y- or y-axis, then it has no effect on the magnetization as the vectoris aligned along the same axis as the B1 eld. The magnetization remains along y.If the nal pulse is about the x-axis, then it rotates the magnetization from the y-axis to the z-axis.Overall, the sequence returns the magnetization to its starting position.

If the nal pulse is about the x-axis, then the magnetization is rotated from the y-axis to thez-axis. Overall, the magnetization has been inverted.

4.12

The initial 90( x) pulse rotates the magnetization from the z-axis to the y-axis. For on-resonancepeaks, = 0, so the magnetization does not precess during the delay . The nal 90( x) thensimply undoes the rotation caused by the rst pulse. Overall, the magnetization is returned to itsstarting position.

= / 2. During the delay, the magnetization rotates to the x-axis and is therefore not affected bythe nal 90(

x) pulse. The net result is that the magnetization appears along the x-axis.

= . During the delay, the magnetization rotates onto the y-axis. The nal pulse rotates themagnetization onto thez-axis. Theequilibrium magnetization is inverted: no observable transversemagnetization is produced.

x

-y

x

-y 90( -x )

90( -x )

90( -x )

delay

x

-y

x

-y

x

-y delay

x

-y

x

-y

x

-y delay

x

-y

= 0

= /2

=

/2

The overall effect of the sequence is to produce x-magnetization which varies as M 0 sin( ).


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0 /2 3 /2 2

-M 0

M 0

M x

To suppress a strong solvent peak, it is placed on-resonance. The delay is then chosen so that av = / 2, where av is the average value of the offset of the peaks we wish to excite.

4.13

The initial 90 pulse rotates the equilibrium magnetization to the y-axis; from there the magneti-zation precesses about the z-axis through an angle of . The 90( y) pulse rotates the x-componentof the magnetization onto the z-axis.

x

-y

x

-y 90( y )delay

x

-y

The y-component of the magnetization varies as M 0 cos :

0 /2 3 /2 2

-M 0

M 0

M y

The nulls are located at = (2n + 1)/ 2, where n = 0, 1, 2, . . .

To suppress the solvent peak, the transmitter frequency is placed in the middle of the peaks of interest, and then is chosen so that = / 2, where is the offset of the solvent. With such achoice, the solvent will not be excited.


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4.14

Line A is on-resonance, so its magnetization does not precess during the delay . The pulse sequenceis, effectively, a 180 ( x) pulse, and so the magnetization is inverted.For line B, the x-, y- and z-components of the magnetization after each element of the sequence are:

component after rst 90( x) after after second 90( x) x 0 M 0 sin M 0 sin

y M 0 M 0 cos 0 z 0 0

M 0 cos

The nal pulse is along the x-axis, so leaves the x-component of the magnetization unchanged.Substituting in the values of and we nd (note that the offset of 100 Hz has to be converted torad s1):

M x = M 0 sin(2 100 5 103) = M 0 sin = 0 M z = M 0 cos(2 100 5 103 ) = M 0 cos = M 0 .

The magnetization is therefore returned to the z-axis.

The 90 pulse rotates the equilibrium magnetization onto they-axis. During the delay , the vectorprecesses about z to give the following x- and y-components: M x = M 0 sin M y = M 0 cos .

For line A, offset 50 Hz:

M x = M 0 sin(2 50 5 103) = M 0 sin(/ 2) = M 0 M y = M 0 cos(2 50 5 103) = M 0 cos( / 2) = 0.

For line B, offset 50 Hz: M x = M 0 sin(2 50 5 103) = M 0 sin(/ 2) = M 0 M y = M 0 cos(2 50 5 10

3) = M 0 cos(/ 2) = 0.The two magnetization vectors rotate at the same rate in the opposite sense. After a delay of = 5 ms, they are both aligned along the x-axis, but pointing in opposite directions.


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5Fourier transformation and data processing

5.1

One desirable feature of the dispersion lineshape is that it crosses the frequency axis at the frequencyof the transition. This allows for a more accurate measurement of the chemical shift than might bepossible for the absorption lineshape, especially in the case of broad lines.

In a spectrum containing many peaks, the following features of the dispersion lineshape make itundesirable:

It is broader than the absorption lineshape the long dispersive tails may interfere withnearby, low intensity peaks.

It is half the height of the absorption lineshape the SNR is therefore reduced by half.

The positive part of one peak may be cancelled by the negative part of an adjacent one in acomplex spectrum, the result can be very difcult to interpret.

5.2

Setting A( ) = S 0 / 2 R, we obtainS 02 R

= S 0 R

R2 + 2.

Cancelling the factor of S 0 from both sides and inverting the quotient, we obtain

2 R = R2 + 2

R.

Hence,

2 = 2 R2 R2 = R2 = R .

The width of the line is therefore 2 R in rad s1, or R/ in Hz.


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Chapter 5: Fourier transformation and data processing 22

5.3

D( ) can be differentiated using the product rule:d D( )

d=

dd

R2 + 2

= 1 R2 + 2

+ 2 2

( R2 + 2)2

= R2 2 + 2 2( R2 + 2)2

= 2

R2

( R2 + 2)2 .

At the turning pointsd D( )

d= 0,

so, 2 R2

( R2 + 2)2 = 0.

The denominator is always non-zero, so the equation can be solved by setting the numerator tozero:

2 R2 = 0 = R .

Substituting these values into D( ):

D( R) = R2 R2

= 12 R

.

These values are the maximum and minimum heights in the lineshape.

There are two values of at which D( ) is half its maximum positive height. At these frequencies, D( ) = 1/ (4 R). Hence,

R2 + 2

= 14 R

.

Inverting the quotients we obtain, R2 + 2

= 4 R,so,

2 + 4 R + R2 = 0.

This is a quadratic equation in that can be solved by applying the usual formula:

= 12 4 R 16 R2 4 R2 = R(2 3) .

Similarly, D( ) = 1/ (4 R) has two solutions: = R(2 3) .The width, W disp, is the distance between the outer two half-maximum points, as shown in thediagram. Its value is

W disp = R(2 + 3) R(2 3) = 2(2 + 3) R.


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frequency / rad s -1

1/(2R)

1/(4R) W disp

R

R ( 2

+

3 )

R ( 2

-

3 )

R (

- 2

-

3 )

R (

- 2 +

3 )

- R

For comparison, the width of the absorption mode is W abs = 2 R. Therefore, the ratio W disp/ W abs =2 + 3 3.7 . The dispersion lineshape is almost four times wider than the absorption lineshape.


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5.4

S x

S y

S x S y S x S y

real imag

real imagreal imag

y

x

y

x

y

x

= 3 /4

= 2

= 3 /2

(c)

(a) (b)

S x

S y

real imag

y

x = 5 /2

(d)

5.5

A 90( x) pulse rotates the equilibrium magnetization onto the y-axis. The resulting spectrum isphased to absorption, so that magnetization along y can be said to have a phase = 0.A 90( y) pulse rotates the equilibrium magnetization onto the x-axis. This corresponds to a phaseshift of = / 2 with respect to the initial experiment.


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realreal

y

x

90( x )

real

y

x

90( -x )

real

y

x

270( x )

y

x

90( y )

(a) (b)

(a) Applying the pulse about x rotates the magnetization vector onto y. This corresponds to aphase shift of = , therefore the spectrum will exhibit a negative absorption lineshape.(b) A 270 ( x) pulse is equivalent to a 90( x) pulse. The spectrum will be the same as in (a).

5.6

The Larmor frequency of 31P at B0 = 9.4 T is:

02

= B02

= 1.08 108 9.4

2= 1.62 108 Hz or 162 MHz.

The phase correction needed at the edge of the spectrum is given by max

t p, where

max is the

maximum offset. For 31P the maximum offset is 350 ppm, therefore the phase correction is

2 162 350 20 106 = 7.1 radians.This corresponds to 407 , a signicant frequency-dependent phase error.


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5.7

The intensity of the noise in the spectrum depends on both the amplitude of the noise in the time-domain, and the acquisition time. So, recording the time-domain signal long after the NMR signalhas decayed just continues to measure noise and no signal. The resulting spectrum will consequentlyhave a lower SNR than it would for a shorter acquisition time.

A full discussion on how line broadening can be used to improve the SNR is given in section 5.4.3on page 92; the matched lter is discussed in section 5.4.4 on page 94.

5.8

Shortening the acquisition time discards the time-domain data that contains mostly noise and littlesignal. Applying a line broadening weighting function does not discard this section of the time-domain, but reduces its amplitude relative to the earlier part of the FID. Thus, both methods reducethe intensity of the noise in the spectrum.

5.9

Enhancing the resolution of the spectrum by the use of a weighting function that combines a risingexponential and a Gaussian is discussed in section 5.4.5 on page 94.

Zero lling improves the denition of the line in the spectrum by increasing the density of datapoints in the frequency domain. However, it does not improve the fundamental linewidth as no realdata is added to the time-domain.

5.10

Plots of the sine bell weighting functions are given in Fig. 5.21 on page 98.A sine bell that is phase-shifted by 45 initially increases over time, therefore partly cancelling the

decay of the FID; the linewidth of the spectrum will therefore be decreased. The subsequent decayof the sine bell attenuates the noise at the end of the time-domain. The overall effect will be toenhance the resolution, assuming that the original FID has decayed close to zero by the end of theacquisition time.The sine bell with a phase shift of 90 is purely a decaying function, which will broaden the lines injust the same way as a decaying exponential does.


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5.11

The peak due to TMS is likely to be a sharp line. Hence, the corresponding time-domain signaldecays slowly, and is therefore more likely to be truncated. The other lines in the spectrum willusually be broader than TMS, so their time-domain signals decay more rapidly and are less likely tobe truncated.Truncation artefacts (sinc wiggles) can be suppressed by applying a decaying weighting function.This will decrease the resolution, and may reduce the SNR.


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6The quantum mechanics of one spin

6.1

I z = 12 Dirac notation: I z | = 12 |

d Dirac notation: | d Dirac notation: | Q d Dirac notation: | Q|

(a) | = 1(b) | = 0 or | = 0(c) I z | = 12 |(d) | = c | + c | .

6.2

The expectation value of I y is given by:

I y =| I y|

|.

If | is normalized, | = 1, so the expectation value is given by I y = | I y| .

Substituting in | = c | + c | , we obtain I y = c |+ c | I y c | + c |

= c c

| I y

| + c c

| I y

| + c c

| I y

| + c c

| I y

|

= 12 i c c | 12 i c c | + 12 i c c | 12 i c c |

= 12 i c c 12 i c c .


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Chapter 6: The quantum mechanics of one spin 30

To go to the third line, we have used Eq. 6.11 on page 111, I y | = 12 i | I y | = 12 i | ,

and to go to the last line, we have used the fact that| and | are orthonormal (Eq. 6.5 on page 108and Eq. 6.6 on page 108). I y can be interpreted as the average value of the y-component of angular momentum whenmeasured for a large number of spins, each of which has the same wavefunction | .

6.3

The matrix representation of I x is

I x = | I x| | I x| | I x| | I x|

=12 | 12 |12 | 12 |

=0 1212 0

.

To go to the second line, we have used Eq. 6.10 on page 111,

I x| = 12 | I x| = 12 | ,and to go to the last line we have used the fact that| and | are orthonormal (Eq. 6.5 on page 108and Eq. 6.6 on page 108).Similarly,

I y = | I y| | I y| | I y| | I y|

=12 i | 12 i |12 i | 12 i |

=0 12 i12 i 0

.


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6.4

Starting with the expression for I y , and substituting in c = r exp(i ) and c = r exp(i ) wend: I y = 12 i c c 12 i c c

= 12 i r r exp(i ) exp(i ) r r exp(i ) exp(i )

= 12 i r r exp i ( ) exp i ( )

= 12i r r exp i ( ) exp i ( ) ,

where to go to the last line we have multiplied top and bottom by i.

Applying the identityexp(i ) exp( i ) 2i sin

to the above expression gives I y = r r sin( ).

The bulk y-magnetization is then given by

M y = I x (1) + I x (2) + . . .

= r (1) r (1) sin(

(1)

(1) ) + r

(2) r

(2) sin(

(2)

(2) ) + . . .

= Nr r sin( ).At equilibrium, the phases are randomly distributed, and so sin( ) is randomly distributedbetween 1. As a result, the equilibrium y-magnetization is zero.

6.5

Starting from Eq. 6.31 on page 120 and premultiplying by |, we obtain:dc (t )

dt | +

dc (t )

dt | =

1

2i c

(t )

| + 1

2i c

(t )

|

|dc (t )

dt | + |dc (t )

dt | = | 12 i c (t ) | + | 12 i c (t ) | .

The derivatives of c and c , and the quantities in square brackets, are numbers, so the aboveexpression can be rearranged to give

dc (t )dt

| +dc (t )

dt | = 12 i c (t ) | + 12 i c (t ) |

dc (t )

dt = 1

2 i c (t ).

To go to the last line, we have used the orthonormality property of |

and | .


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Substituting Eq. 6.58 on page 137 into the left-hand side of Eq. 6.57 on page 137 gives:dc (t )

dt =

ddt

c (0) exp12 i t

= 12 i c (0) exp

12 i t

= 12 i c (t ).

Eq. 6.58 on page 137 is indeed the solution.

6.6

The expectation value of I y is I y = 12 i c c 12 i c c .

Substituting in the expressions for how c and c vary under free evolution (Eq. 6.34 on page 121)gives:

I y (t ) = 12 i c (0) exp 12 i t c (0)exp 12 i t 12 i c (0)exp 12 i t c (0) exp 12 i t = 1

2 i c (0)c (0) exp (i t ) 12 i c (0)c (0) exp (i t )= 1

2 i c (0)c (0) [cos( t ) i sin( t )] 12 i c (0)c (0) [cos( t ) + i sin( t )]= cos( t ) 12 i c (0)c (0) 12 i c (0)c (0) + sin( t ) 12 c (0)c (0) + 12 c (0)c (0)=

cos(

t ) I y (0)+

sin(

t ) I x (0) .To go to the third line, the identities

exp(i ) cos + i sin exp(i ) cos i sin were used, and to go to the last line, the expressions for I x and I y in terms of c and c were used(Eqs 6.12 and 6.13 on p. 111).

This result is summarized in the diagram below. The grey arrow shows the initial position, and theblack arrow shows the position after time t .

(0)

(0)

t

6.7


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The matrix representation of the density operator is given by:

= | | | | | | | |

11 12

21 22.

We can now calculate the 11 element (for clarity, the overbars indicating the ensemble averaginghave been omitted until the last line):

11 = | |= | |= | c | + c |

c |+ c | |

= c | + c | c | + c |= c c .To go to the second line, the denition of was inserted, and on the third line | was expressed asa superposition of | and | .The other elements can be calculated in a similar way to give:

12 = c c 21 = c c 22 = c c .

Hence,

=c c c c c

c

c

c

.


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7Product operators

7.1

exp(i I x) I y exp(i I x) represents a rotation of I y about x through angle . From Fig. 7.4 onpage 148 (a) on p. 148, I y is transformed into I z. Hence,

I y I x

cos I y + sin I z.

This is consistent with the identity on line one of Table 7.1 on page 143. exp(i S y)S z exp(i S y). From (b) of Fig. 7.4 on page 148, S z is transformed into S x by arotation about y:

S z S y

cos S z + sin S x. exp(i I x) I x exp(i I x). Rotating I x about the x-axis has no effect:

I x I x

I x. exp(i I z)( I y) exp(i I z). Fig. 7.4 on page 148 (c) shows the effect of a rotation about z on

I y: the result is a transformation to I x. Hence,

I y I z

cos I y + sin I x. exp(i (/ 2) I y) I x exp(i ( / 2) I y). This represents the rotation of I x about y through angle / 2.From Fig. 7.4 on page 148 (b), I x is transformed to I z. Hence,

I x(/ 2) I y

cos( / 2) I x sin( / 2) I z. exp(i I z)( I z) exp(i I z). Careful inspection of the arguments of the exponentials reveals thatthis represents a z-rotation through angle i.e. the rotation is in a clockwise sense. In thiscase, it does not matter as I z is unaffected by a rotation about the z-axis:

I z I z

I z.


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Chapter 7: Product operators 36

7.2

The 90( x) pulse rotates the equilibrium magnetization (represented by I z) onto the y-axis: I z

(/ 2) I x

cos( / 2) I z sin( / 2) I y= I y.

This transverse term evolves under the offset during the delay to give

I y I z

cos( ) I y + sin( ) I x,where (c) of Fig. 7.4 on page 148 has been used.The 180 ( y) pulse does not affect the I y term, but inverts the I x term:

cos( ) I y + sin( ) I x I y

cos( ) I y + cos sin( ) I x sin sin( ) I z= cos( ) I y sin( ) I x.

Now we consider the evolution during the second delay. Taking each term separately, we obtain

cos( ) I y I z

cos( ) cos( ) I y + sin( ) cos( ) I x,

sin( ) I x I z

cos( ) sin( ) I x sin( ) sin( ) I y.

Combining these terms gives the nal result as

cos 2( ) + sin 2( ) I y = I y,where the terms in I x cancel, and the identity cos2 + sin 2 1 has been used. At the end of thesequence, the magnetization has been refocused onto the y-axis, irrespective of the offset.

7.3

I y (/ 2)

I y I y I y

(/ 2) I y

I yS y

S y

S y.In all three cases, the pulse is applied about the same axis along which the magnetization is aligned,therefore the magnetization is unaffected.In the following cases, we refer to Fig. 7.4 on page 148 to determine how the operator is transformedby the rotation.

I x I y

cos() I x sin() I z= I x.

In this case the magnetization is simply inverted.


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The difference between the next two examples is the sense of the 90 rotation.

I z(/ 2) I y

cos( / 2) I z + sin( / 2) I x= I x.

I z (/ 2) I y

cos(/ 2) I z + sin(/ 2) I x= I x.

The next two are simply inversions:

S zS y

cos S z + sin S x

= S z.

I z I y

cos() I z + sin() I x= I z.

7.4

The 90( x) pulse rotates the equilibrium magnetization I z to I y. Free evolution is a rotation about z, so the state of the system after the delay is

cos( ) I y + sin( ) I x.The 90( y) pulse does not affect the I y term, but rotates I x to I z. The nal result is

cos( ) I y sin( ) I z.The pulse sequence has therefore produced transverse magnetization along y, whose amplitudevaries as cos( ). This becomes zero if cos( ) = 0. Hence, there is a null at = / 2, whichcorresponds to an offset of = / (2 ) in rad s1, or 1/ (4 ) in Hz.There is a maximum in the excitation when cos( ) = 1. This occurs at offsets satisfying = nwhere n = 0, 1, 2, . . . i.e. = (n)/ or n/ (2 ) in Hz.


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7.5

Figure 7.6 (b) on p. 152 shows that, as a result of evolution of the scalar coupling, the in-phaseterm I 1 y is partly transformed into the anti-phase term 2 I 1 x I 2 z; the angle of rotation is J 12 . Thisis represented as:

I 1 y2 J 12 I 1 z I 2 z

cos( J 12 ) I 1 y + sin( J 12 ) 2 I 1 x I 2 z.Using the same gure, we see that 2 I 1 x I 2 z is partly transformed to I 1 y:

2 I 1 x I 2 z2 J 12 I 1 z I 2 z

cos( J 12 ) 2 I 1 x I 2 z sin( J 12 ) I 1 y.Similarly,

S x2 J IS (/ 2) I z S z

cos( J IS / 2) S x + sin( J IS / 2) 2 I z S y. I 2 y

2 J 12 I 1 z I 2 z

cos( J 12 ) I 2 y sin( J 12 ) 2 I 1 z I 2 x.2 I 1 z I 2 y

2 J 12 I 1 z I 2 z

cos( J 12 ) 2 I 1 z I 2 y sin( J 12 ) I 2 x. I 2 z

2 J 12 I 1 z I 2 z

I 2 z.In the last example we see that z-magnetization is not affected by evolution under coupling simplybecause the Hamiltonian for coupling only contains I z operators.

7.6

The evolution is determined by the Hamiltonian given in Eq. 7.14 on page 150: H two spins = 1 I 1 z + 2 I 2 z + 2 J 12 I 1 z I 2 z.

We will now work out the effect in turn of the three terms in the Hamiltonian. The rst is a rotationabout z:

I 1 y 1 t I 1 z

cos( 1 t ) I 1 y sin( 1 t ) I 1 x.The second term, 2 I 2 z, does not need to be considered as spin-two operators have no effect onspin-one operators. Finally, we consider the effect of evolution under scalar coupling:

cos(

1 t ) I 1 y sin(

1 t )

I 1 x

2 J 12 t I 1 z I 2 z

cos( J 12 t ) cos( 1 t ) I 1 y

y-magnetizationsin( J 12 t ) cos( 1 t ) 2 I 1 x I 2 z

cos( J 12 t ) sin( 1 t ) I 1 x

x-magnetizationsin( J 12 t ) sin( 1 t ) 2 I 1 y I 2 z.

The NMR signal is given by:S (t ) = M x + i M y

= cos( J 12 t ) sin( 1 t ) + i cos( J 12 t ) cos( 1 t )= i cos( J 12 t ) [cos( 1 t ) + i sin( 1 t )]= i cos( J 12 t ) exp(i 1 t )= 1

2 i exp(i J 12 t ) + exp( i J 12 t ) exp(i 1 t )= 1

2 i exp (i[ 1 + J 12 ]t ) + 12 iexp (i[ 1 J 12 ]t ) .


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To go to the fourth line, we have used the identity cos + i sin exp(i ), and to go to the fth line,we have used cos 12 [exp(i ) + exp( i )]. Finally, to go to the sixth line we have multiplied outthe square brackets. Fourier transformation of this signal gives a positive line at 1 + J 12 , and asecond positive line at 1 J 12 i.e. an in-phase doublet on spin one. The factor of i corresponds toa phase shift of 90, so the imaginary part of the spectrum contains the absorption mode lineshape.

real

imaginary

1- J 12

2 J 12

1+ J 12

A similar line of argument gives the observable signal arising from 2 I 1 y I 2 z as

S (t ) = 12 i exp (i[ 1 + J 12 ]t ) 12 i exp (i[ 1 J 12 ]t ) .The corresponding spectrum is an anti-phase doublet on spin one. Again, the factor of i means thatthe absorption mode lines will appear in the imaginary part of the spectrum.

real

imaginary

1- J 12

2 J 12

1+ J 12


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7.7

I 1 y represents in-phase magnetization on spin one, aligned along the y-axis. The resulting spectrumwill be an in-phase doublet centred on the offset of spin one, both peaks of which are in theabsorption mode. I 2 x represents in-phase magnetization on spin two. However, it is aligned along the x-axis, so has aphase of 3/ 2 relative to the y-axis. The spectrum therefore comprises an in-phase doublet that iscentred on the offset of spin two, with both peaks in the dispersion mode.2 I 1 y I 2 z represents magnetization on spin one that is anti-phase with respect to spin two, and alignedalong y. The spectrum is therefore an anti-phase doublet in the absorption mode.

2 I 1 z I 2 x represents anti-phase magnetization on spin two. It is aligned along x, so the lineshape will bedispersive. Therefore, the spectrum is an anti-phase spin-two doublet with the dispersion lineshape.

1

I 1 y

2 I 1 y I 2 z

2

I 2 x

2 I 1 z I 2 x

7.8

In-phase magnetization I 1 x is rotated in the xz-plane towards I 1 z by the application of the y-pulseof duration t p.

I 1 x1 t p I 1 y

cos( 1 t p) I 1 x sin( 1 t p) I 1 z

A 180 pulse about y applied only to spin two changes the sign of the anti-phase magnetization onspin one.

2 I 1 x I 2 z I 2 y

cos() 2 I 1 x I 2 z + sin() 2 I 1 x I 2 x= 2 I 1 x I 2 z

In-phase magnetization on spin one is allowed to evolve under coupling for time t , thus generating

anti-phase magnetization on the same spin.

I 1 x2 J 12 t I 1 z I 2 z

cos( J 12 t ) I 1 x sin( J 12 t ) 2 I 1 y I 2 z


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Letting each term act sequentially, we obtain

2 I 1 x I 2 z(/ 2) I 1 y

2 I 1 z I 2 z(/ 2) I 2 y

2 I 1 z I 2 x.Note that the spin-one operators do not act on spin-two operators and vice versa. The net result isthat the non-selective 90( y) pulse has caused a coherence transfer from spin one to spin two.

Transverse, in-phase magnetization on the S spin evolves under offset for time t . The offset term forthe I spin has no effect on the S x.

S x I t I z

S x S t S z

cos( S t ) S x + sin( S t ) S y

Anti-phase magnetization on spin two evolves under coupling to generate in-phase magnetizationon the same spin.

2 I 1 z I 2 y2 J 12 t I 1 z I 2 z

cos( J 12 t ) 2 I 1 z I 2 y + sin( J 12 t ) I 2 x

7.9

The Hamiltonian for free evolution is given by Eq. 7.14 on page 150:

H two spins

= 1

I 1 z

+ 2

I 2 z

+ 2 J 12

I 1 z

I 2 z

.

The spin echo refocuses the evolution due to offset, so we only need to consider the evolution of 2 I 1 x I 2 z under coupling, which gives

2 I 1 x I 2 z2 J 12 I 1 z I 2 z

cos( J 12 ) 2 I 1 x I 2 z + sin( J 12 ) I 1 y.The pulse about the x-axis acts on both spins, leaving I 1 x unaffected, but inverting I 2 z and I 1 y:

cos( J 12 ) 2 I 1 x I 2 z + sin( J 12 ) I 1 y( I 1 x+ I 2 x)

cos( J 12 ) 2 I 1 x I 2 z sin( J 12 ) I 1 y.Finally, evolution under coupling during the second delay gives

cos( J 12 ) 2 I 1 x I 2 z sin( J 12 ) I 1 y2 J 12 I 1 z I 2 z

cos 2( J 12 ) I 1 x I 2 z sin( J 12 ) cos( J 12 ) I 1 y cos( J 12 ) sin( J 12 ) I 1 y + sin 2( J 12 ) 2 I 1 x I 2 z= cos 2( J 12 ) sin 2( J 12 ) 2 I 1 x I 2 z [2 cos( J 12 ) sin( J 12 )] I 1 y= cos(2 J 12 ) 2 I 1 x I 2 z sin(2 J 12 ) I 1 y.

To go to the last line, we have used the identities cos 2 sin 2 cos 2 and 2 cos sin sin 2 .By a similar method we can show:

2 I 1 y I 2 z x cos(2 J 12 ) 2 I 1 y I 2 z sin(2 J 12 ) I 1 x.

The effect of the y spin echo on spin-one and spin-two terms is shown in the table below:


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nal stateinitial state cos (2 J 12 ) sin (2 J 12 )

I 1 x I 1 x 2 I 1 y I 2 z I 1 y I 1 y 2 I 1 x I 2 z

2 I 1 x I 2 z 2 I 1 x I 2 z I 1 y2 I 1 y I 2 z 2 I 1 y I 2 z I 1 x

I 2 x I 2 x 2 I 1 z I 2 y I 2 y I 2 y 2 I 1 z I 2 x

2 I 1 z I 2 x 2 I 1 z I 2 x I 2 y2 I 1 z I 2 y 2 I 1 z I 2 y I 2 x

The results for the in- and anti-phase operators on spin two can be obtained from those for spinone simply by swapping the labels 1 and 2.Likewise for the x spin echo:

nal stateinitial state cos (2 J 12 ) sin (2 J 12 )

I 2 x I 2 x I 1 z I 2 y I 2 y I 2 y 2 I 1 z I 2 x

2 I 1 z I 2 x 2 I 1 z I 2 x I 2 y2 I 1 z I 2 y 2 I 1 z I 2 y I 2 x

7.10

A spin echo in a homonuclear two-spin system is equivalent to:

(a) evolution of the coupling for time 2 ,

(b) a 180 ( x) pulse.

Applying this to the rst example, we obtain

I 2 y x cos(2 J 12 ) I 2 y + sin(2 J 12 ) 2 I 1 z I 2 x.

For complete transformation to 2 I 1 z I 2 x, we need sin(2 J 12 ) = 1 and cos(2 J 12 ) = 0. These occurwhen 2 J 12 = / 2, i.e. = 1/ (4 J 12 ).

I 1 x x cos(2 J 12 ) I 1 x + sin(2 J 12 )2 I 1 y I 2 z.

Setting 2 J 12 = / 4 gives cos(2 J 12 ) = sin(2 J 12 ) = 1/ 2. The required delay is therefore =1/ (8 J 12 ).To achieve conversion to I 1 x, we need cos(2 J 12 ) = 1 and sin(2 J 12 ) = 0 i.e. = 1/ (2 J 12 ).

2 I 1 z I 2 x x cos(2 J 12 )2 I 1 z I 2 x sin(2 J 12 ) I 2 y.


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Setting the delay to = 1/ (4 J 12 ) gives complete conversion to in-phase magnetization.

7.11

The pulse sequence is given in Fig. 7.14 on page 164:

I

S

The 180 ( x) pulse is applied to only the S spin, so the evolution of the offset of the S spin will berefocused. We need to consider the evolution of the coupling. Starting with S x, the state of thesystem after the rst delay is

cos( J 12 ) S x + sin( J 12 ) 2 I z S y.

The 180 (x) pulse is applied only to the S spin, and so does not affect I z or S x. However, the term inS y changes sign to give:

cos( J 12 ) S x sin( J 12 ) 2 I z S y.Evolution of the coupling during the second delay gives

cos 2( J 12 ) + sin 2( J 12 ) S x + [sin( J 12 ) cos( J 12 ) cos( J 12 ) sin( J 12 )] 2 I z S y = S x,

where the anti-phase terms cancel, and the identity cos 2 + sin2

1 has been used. The evolutionof the coupling has therefore been refocused.Repeating the calculation for the anti-phase term, we see that 2 I z S x is unaffected by the spin echosequence. Again, the coupling is refocused.Both operators are unchanged, which is the same effect that a 180( x) pulse to the S spin wouldhave:

S x S x

S x2 I z S x

S x

2 I z S x.

Likewise, the operators I x and 2 I x S z will have their evolution under coupling refocused. However,as the 180 ( x) pulse is not applied to the I spin, the offset will not be refocused, but will evolve forthe duration of the spin echo (time 2).

7.12

The pulse sequence for the INEPT experiment is reproduced below from Fig. 7.15 on page 168:

1 1 2 2

y

A CB

I

S


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At the end of period A it was shown in section 7.10.2 on page 168 that the state of the spin systemis

k I cos(2 J IS 1) I y k I sin(2 J IS 1) I x S z.The purpose of the two 90 pulses in period B is to transfer the anti-phase magnetization (thesecond term) from the I spin to the S spin. This requires the pulse acting on the I spin to cause thetransformation I x I z, which requires a rotation about the y-axis.If the initial 90 pulse is about the x-axis, it rotates the equilibrium k I I z to k I I y. At the end of thespin echo in period A, the system is in the following state:

k I cos(2 J IS 1) I y + k I sin(2 J IS 1) 2 I x S z.As before, the I y term is not affected by the 90( y) pulse on the I spin, and can be discarded. Thetwo pulses affect the ant-phase term as follows:

k I sin(2 J IS 1) 2 I x S z(/ 2) I y

k I sin(2 J IS 1) 2 I z S z(/ 2)S x

k I sin(2 J IS 1) 2 I z S y.This term evolves under coupling during the spin echo in C to give:

k I cos(2 J IS 2) sin(2 J IS 1) 2 I z S y k I sin(2 J IS 2 ) sin(2 J IS 1) S x,the observable term of which is the one in S x.The 90( x) pulse acting on the S spin during B also rotates equilibrium k S S z to k I S y, which evolvesduring the spin echo in C to give:

k S cos(2 J IS 2) S y

+k S sin(2 J IS 2) 2

I z

S x.

This also has an observable term in S y. Hence, the two observable terms are combined to give:

k S cos(2 J IS 2) S y k I sin(2 J IS 2) sin(2 J IS 1) S x.The rst term is unaffected by changing the phase of the I spin 90 pulse from x to x, whereas thesecond term changes sign.

7.13

By denition, I + has coherence order + 1.

I z is unaffected by a z-rotation, so has coherence order zero.

I has coherence order 1, again by denition. Using the denitions of I 1+ and I 1 (Eq. 7.28 on page 174) as applied to spin one:

I 1+ I 1 x + i I 1 y I 1 I 1 x i I 1 y,

we can write I 1 x as: I 1 x

12

I 1+ + I 1

.

Therefore, I 1 x is an equal mixture of coherence orders + 1 and 1.


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Similarly, I 2 y can be written as I 2 y 12i I 2+ I 2 .

Hence, 2 I 1 z I 2 y can be written as

2 I 1 z I 2 y 2 12i I 1 z I 2+ I 2 ,which is an equal mixture of coherence orders + 1 and 1, found by summing the coherenceorders of spins one and two (spin one has coherence order zero).

Since both I 1 z and I 2 z have coherence order zero, so does 2 I 1 z I 2 z.

2 I 1+ I 2 has coherence order zero since the coherence order of spin one is + 1 and that of spin

two is 1. 2 I 1 x I 2 y can be written as:2 I 1 x I 2 y 2 12 I 1+ + I 1 12i I 2+ I 2

12i I 1+ I 2+ I 1 I 2 I 1+ I 2 + I 1 I 2+ .2 I 1 x I 2 y is therefore an equal mixture of coherence orders + 2 and 2, double-quantum coher-ence, and coherence order 0, zero-quantum coherence.

7.14

Using the denitions of I i given by Eq. 7.28 on page 174, we can write 2 I 1 x I 2 y as:

2 I 1 x I 2 y 2 12 I 1+ + I 1 12i I 2+ I 2 12i I 1+ I 2+ I 1 I 2

double-quantum part+ 1

2i I 1

I 2+ I 1+ I 2

zero-quantum part.

The other relationships in the table can be veried in the same way.

7.15

The rst 90( x) pulse rotates the equilibrium I 1 z to I 1 y. During the spin echo sequence, the offset isrefocused, but the coupling evolves throughout. The state of the spin system at the end of the spinecho is

cos(2 J 12 ) I 1 y sin(2 J 12 ) 2 I 1 x I 2 z.The nal pulse acts to give

cos(2 J 12 ) I 1 z + sin(2 J 12 ) 2 I 1 x I 2 y.

Using the denitions of DQ y and ZQ y given in the last table of section 7.12.1 on page 174, we seethat we can rewrite the second term as

12 sin(2 J 12 ) DQ y ZQ y ,

which is a mixture of double- and zero-quantum coherence.


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The amplitude of this multiple quantum term is a maximum when sin(2 J 12 ) = 1, which occurswhen = 1/ (4 J 12 ).Starting with equilibrium magnetization on spin two, I 2 z, the terms present after the nal pulse are

cos(2 J 12 ) I 2 z + sin(2 J 12 ) 2 I 1 y I 2 x;

we have taken the terms from the previous calculation and swapped the labels 1 and 2. Again, fromthe denitions of DQ y and ZQ y in section 7.12.1 on page 174, we can write the multiple quantumterm as

12 sin(2 J 12 ) DQ y + ZQ y .

Therefore, adding this term to the one originating from I 1 z, we obtain;

12 sin(2 J 12 ) DQ y ZQ y + 12 sin(2 J 12 ) DQ y + ZQ y = sin(2 J 12 ) DQ y,

which is pure double-quantum coherence. It is a rather unusual feature of this sequence that, in atwo-spin system, it generates pure double-quantum coherence.

7.16

From the table on p. 176, ZQ x is equal to 2 I 1 x I 2 x + 2 I 1 y I 2 y . Zero-quantum coherence between spinsone and two does not evolve under the coupling between these two spins, so we need only considerthe evolution under offset. Considering rst the 2 I 1 x I 2 x term:

2 I 1 x I 2 x 1 t I 1 z+ 2 t I 2 z

2 cos( 1 t ) I 1 x + sin( 1 t ) I 1 y cos( 2 t ) I 2 x + sin( 2 t ) I 2 y .We will now look at the 2 I 1 y I 2 y term;

2 I 1 y I 2 y 1 t I 1 z+ 2 t I 2 z

2 cos( 1 t ) I 1 y sin( 1 t ) I 1 x cos( 2 t ) I 2 y sin( 2 t ) I 2 x .Collecting these terms together, we obtain:

[cos( 1 t ) cos( 2 t ) + sin( 1 t ) sin( 2 t )] (2 I 1 x I 2 x + 2 I 1 y I 2 y)+ [sin( 1 t ) cos( 2 t ) cos( 1 t ) sin( 2 t )](2 I 1 y I 2 x 2 I 1 x I 2 y).

Using the identities:

cos( A B) = cos A cos B + sin A sin Bsin( A B) = sin A cos B cos A sin B,

and the denitions of ZQ x and ZQ y:

ZQ x (2 I 1 x I 2 x + 2 I 1 y I 2 y) ZQ y (2 I 1 y I 2 x 2 I 1 x I 2 y),we obtain

cos ([ 1 2]t ) ZQ x + sin ([ 1 2 ]t ) ZQ y.


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8Two-dimensional NMR

8.1

In each example, the preparation period is highlighted with a grey box, and the mixing period witha grey box with a dashed border.

t 2t 1 t 2t 1

t 2t 1

1 1

y 2 2 t 2

t 1

I

S

t 2

t 1

I

S

t 2

t 1

I

S

t 2t 1 mix

1 1 2 2

t 2

t 1 I

S

COSY

HSQC

HMQC HMBC

HETCOR

DQF COSY

DQ spectroscopy TOCSY


52/128

Chapter 8: Two-dimensional NMR 48

8.2

2

1

2

3

4

5

6

t 1

2

t 14

5

612

3

1, 2 and 3 are cross-sections of the damped cosine wave, whose amplitude provides the modulationin t 1. The period is the same for each wave, and the amplitude increases as we approach the centre

of the peak in 2.4, 5 and 6 are cross-sections through the 2 dimension. The amplitude and sign of the peak ismodulated by a damped cosine wave in t 1.

8.3

The COSY pulse sequence is given in Fig. 8.8 on page 192.

t 2t 1

Starting with equilibrium magnetization on spin two, the state of the system at t 2 = 0 can bedetermined from terms [1][4] on p. 191 by swapping the spin labels 1 and 2. The result is:

cos ( J 12 t 1 )cos( 2 t 1) I 2 z [1]sin( J 12 t 1)cos( 2 t 1 ) 2 I 1 y I 2 x [2]+ cos ( J 12 t 1 )sin( 2 t 1 ) I 2 x [3]

sin( J 12 t 1)sin( 2 t 1) 2 I 1 y I 2 z. [4]The observable terms are [3] and [4]. The operator in term [3] is I 2 x, which will give rise to adoublet on spin two in the 2 dimension. It is modulated in t 1 by sin( 2 t 1) i.e. at the offset of spintwo. Thus, [3] produces a diagonal-peak multiplet.

The operator in term [4] is 2 I 1 y I 2 z; this gives rise to an anti-phase doublet centred at the offset of spinone in the 2 dimension. It is also modulated in t 1 by sin( 2 t 1 ). Therefore, it produces a cross-peakmultiplet.


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It was shown in section 7.5.2 on page 154 that the evolution of 2 I 1 y I 2 z during t 2 gives rise to thefollowing time domain signal:

12 i exp(i[ 1 + J 12 ]t 2) 12 i exp(i[ 1 J 12 ]t 2 ).

Imposing an exponential decay on this signal and Fourier transforming, we obtain the followingspectrum

12 i [ A2 ( 1 + J 12 ) + i D2( 1 + J 12 )] 12 i [ A2( 1 J 12 ) + i D2 ( 1 J 12 )] .

To ensure that the absorption mode lineshape appears in the real part of the spectrum, we multiplythe expression above by a 90 phase correction factor i.e. by exp(i / 2). Noting that exp(i / 2) i, we obtain:

12 [ A2 (

1

+ J 12 )

+i D2(

1

+ J 12 )]

12 [ A2(

1 J 12 )

+i D2 (

1 J 12 )] .Clearly this is an anti-phase doublet on spin one.

The t 1 modulation of term [4] has the form sin( J 12 t 1)sin( 2 t 1). Applying the identitysin A sin B 12 [cos( A B) cos( A + B)] ,

gives12 [cos( 2 + J 12 )t 1 cos( 2 J 12 )t 1] .

Imposing an exponential decay and taking the cosine Fourier transform yields the spectrum12 [ A1( 2 + J 12 ) A1( 2 J 12 )] .

This is clearly an anti-phase doublet on spin two.Multiplying the 1 and 2 spectra together, and taking the real part, gives the following four lineswhich form the cross-peak multiplet. Note that they form an anti-phase square array.

+ 14 A1( 2 + J 12 ) A2 ( 1 + J 12 ) 14 A1( 2 + J 12 ) A2( 1 J 12 )

14 A1( 2 J 12 ) A2 ( 1 + J 12 ) + 14 A1( 2 J 12 ) A2( 1 J 12 ).The operator in the diagonal peak term [3] is I 2 x. Evolution of this operator during t 2 gives thefollowing time domain signal:

12 exp(i[ 2 + J 12 ]t 2) +

12 exp(i[ 2 J 12 ]t 2 ).

Imposing an exponential decay to this, and Fourier transforming gives the spectrum12 [ A2 ( 2 + J 12 ) + i D2( 2 + J 12 )] +

12 [ A2( 2 J 12 ) + i D2 ( 2 J 12 )] .

This is an in-phase doublet on spin two.The t 1 modulation is:

cos( J 12 t 1) sin( 2 t 1) 12 [sin( 2 + J 12 )t 1 + sin( 2 J 12 )t 1] ,where we have used the identity

sin A sin B 12 [sin( A + B) + sin( A B)] .Assuming an exponential decay and applying a sine Fourier transform gives the spectrum:

12 [ A1( 2 + J 12 ) + A1( 2 J 12 )] .This is an in-phase doublet on spin two.


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Multiplying together the 1 and 2 parts of the spectrum and taking the real part yields the followingfour components of the diagonal-peak multiplet. Note that they all have the same sign.

+ 14 A1( 2 + J 12 ) A2 ( 2 + J 12 ) +

14 A1( 2 + J 12 ) A2( 2 J 12 )

+ 14 A1( 2 J 12 ) A2 ( 2 + J 12 ) + 14 A1( 2 J 12 ) A2( 2 J 12 ).

8.4

The DQF COSY pulse sequence is given in Fig. 8.15 on page 200.

t 2t 1

Starting with equilibrium magnetization on spin two, I 2 z, the state of the spin system after the secondpulse is exactly the same as for the COSY experiment at t 2 = 0 as calculated in Exercise 8.3. Of thefour terms present, the only one that contains double-quantum coherence is [2]:

sin( J 12 t 1)cos( 2 t 1 ) 2 I 1 y I 2 x.In section 7.12.1 on page 174, it was shown that 2 I 1 y I 2 x is a mixture of double- and zero-quantumcoherence. The double-quantum operator DQ y, and the zero-quantum operator ZQ y, are denedas:

DQ y 2

I 1 x

I 2 y

+2

I 1 y

I 2 x

ZQ y 2

I 1 y

I 2 x 2

I 1 x

I 2 y.

Hence,2 I 1 y I 2 x = 12 DQ y + ZQ y .

The double-quantum part that is retained is therefore:

12 sin ( J 12 t 1)cos( 2 t 1) DQ y = 12 sin ( J 12 t 1)cos( 2 t 1) 2 I 1 x I 2 y + 2 I 1 y I 2 x .The third 90 pulse acts to give:

12 sin ( J 12 t 1)cos( 2 t 1 ) 2 I 1 x I 2 z + 2 I 1 z I 2 x .2 I 1 x I 2 z and 2 I 1 z I 2 x represent anti-phase magnetization on spins one and two, respectively. Both are

modulated in t 1 at 2, so the rst term therefore gives the cross-peak multiplet, and the second thediagonal-peak multiplet.Expanding the t 1 modulation, we obtain

12 sin ( J 12 t 1)cos( 2 t 1) 14 [sin( 2 + J 12 )t 1 sin( 2 J 12 )t 1] ,which is an anti-phase doublet on spin two. Hence, both the cross- and diagonal-peak multipletsare anti-phase in both dimensions. Furthermore, both terms have the same t 1 modulation, and bothappear along the x-axis at the start of acquisition, so the spectrum can be phased so that all thepeaks appear in the double absorption mode.

8.5

The pulse sequence is given in Fig. 8.18 on page 204.


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t 2t 1

The rst 90 pulse rotates equilibrium I 1 z to I 1 y, which then evolves under coupling during the spinecho (the offset is refocused) to givecos(2 J 12 ) I 1 y sin(2 J 12 ) 2 I 1 x I 2 z.

This is rotated by the second 90 pulse to give

cos(2 J 12 ) I 1 z + sin(2 J 12 ) 2 I 1 x I 2 y.

We select just zero-quantum coherence at this point. From the table on p. 176, the zero-quantumpart of 2 I 1 x I 2 y is 12 ZQ y, so at the start of t 1 we have:

12 sin(2 J 12 ) ZQ y.This evolves during t 1 according to the rules in section 7.12.3 on page 176:

12 sin(2 J 12 ) ZQ y 1 t 1 I 1 z+ 2 t 1 I 2 z

12 cos ([ 1 2]t 1) sin(2 J 12 ) ZQ y+ 1

2 sin ([ 1 2]t 1) sin(2 J 12 ) ZQ x,where

ZQ x 2 I 1 x I 2 x + 2 I 1 y I 2 y ZQ y 2 I 1 y I 2 x 2 I 1 x I 2 y.Note that the zero-quantum coherence between spins one and two does not evolve due to thecoupling between these two spins.The nal pulse rotates the zero-quantum terms to give

12 sin(2 J 12 )cos ([ 1 2]t 1) 2 I 1 z I 2 x 2 I 1 x I 2 z+ 1

2 sin(2 J 12 )sin ([ 1 2]t 1) 2 I 1 x I 2 x + 2 I 1 z I 2 z ,the observable terms of which are:

12 sin(2 J 12 )cos ([ 1 2]t 1) 2 I 1 x I 2 z 2 I 1 z I 2 x .

The spectrum has the same form as the double-quantum spectrum shown in Fig. 8.19 on page 205

with the following differences: In 2 the anti-phase doublet on spin two, which arises from the 2 I 1 z I 2 x term, appears with the

opposite sign.

The frequency of the peaks in 1 is ( 1 2) i.e. the zero-quantum frequency.The information that can be gained from this spectrum is the same as for the double-quantumspectrum.

8.6

From section 8.7 on page 209, the terms present after the rst spin echo arecos(2 J IS 1 ) I y sin(2 J IS 1)2 I x S z.


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The subsequent 90 pulses are required to transfer the anti-phase magnetization (the second term)to the S spin, so that it can evolve under the offset of the S spin during t 1. This requires the I spinpulse to rotate I x to I z, which is only possible if the pulse is about y.Applying the I spin pulse about y gives:

sin(2 J IS 1) 2 I x S z(/ 2) I y sin(2 J IS 1 ) 2 I z S z

(/ 2)S x

sin(2 J IS 1) 2 I z S y.The 2 I z S y term, present at the start of t 1, simply changes sign when the I spin pulse is changed inphase from + y to y.

8.7

The pulse sequence is given in Fig. 8.22 on page 210(a).

1 1

y

y

A CB D

t 2

t 1

I

S

The state of the spin system after the spin echo (A) is, from section 8.7 on page 209:

cos(2 J IS 1) I y sin(2 J IS 1) 2 I x S z.The pulses during period B have the following effect on the anti-phase term:

sin(2 J IS 1)2 I x S z(/ 2)( I y+ S y)

sin(2 J IS 1) 2 I z S x.Period C is a spin echo, during which the coupling is refocused, but the offset of the S spin evolvesfor time t 1. At the end of this period, the terms are:

cos( S t 1 ) sin(2 J IS 1) 2 I z S x sin( S t 1) sin(2 J IS 1) 2 I z S y.

The nal two pulses (period D) produce the following state at t 2 =

0:cos( S t 1) sin(2 J IS 1) 2 I y S x + sin( S t 1) sin(2 J IS 1 ) 2 I y S z.

The observable signal is due to the 2 I y S z term, and is now modulated in t 1 according to sin( S t 1).So, shifting the phase of the rst 90 pulse to the S spin from x to y does indeed alter the modulationin t 1 from cosine to sine.

8.8

The pulse sequence is given in Fig. 8.24 on page 213 (a) on p. 213. We will now modify it so that

the rst 90 S spin pulse is about x.


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A

EC

F

B D

t 2

t 1

I

S

-x

As argued in section 8.8 on page 212, the offset of the I spin is refocused over the whole of periodF. The rst pulse creates I y, which evolves during period A under coupling to give

cos( J IS ) I y + sin( J IS ) 2 I x S z.Taking just the second term (the rst does not produce any useful peaks), and applying to it the rstS spin pulse (with phase x) gives: sin( J IS ) 2 I x S y,which is of opposite sign to the corresponding term in section 8.8 on page 212. This sign changepropagates throughout the rest of the calculation so that the observable term

sin 2( J IS ) cos( S t 1) I y,

also has the opposite sign. The same result is produced on changing the phase of the second 90 Sspin pulse to x.I spins that are not coupled to S spins do not give rise to anti-phase magnetization, and so are notaffected by the S spin pulses. This I spin magnetization is therefore unaffected by altering the phaseof the rst S spin pulse. So, recording two spectra, the rst with the rst S spin pulse about x, andthe second with it about

x, and then subtracting one from the other will retain the wanted signal

and eliminate the unwanted signal.

8.9

It was shown in section 8.8 on page 212 that the observable term at the start of acquisition is

sin 2( J IS ) cos( S t 1) I y.The amplitude of the signal is given by sin 2 ( J IS ), which has a maximum value of 1. This occurswhen the argument of the sine is an odd multiple of / 2 i.e. when J IS = n/ 2, n = 1, 3, 5, . . .Hence, = n/ (2 J IS ), n = 1, 3, 5, . . .

sin2( J IS ) = 0 when J IS = n/ 2, n = 0, 2, 4, . . . i.e. is an even multiple of / 2. Hence theamplitude will be zero when = n/ (2 J IS ), n = 0, 2, 4, . . .

8.10

The HSQC pulse sequence, without decoupling during acquisition, is shown in Fig. 8.22 onpage 210 (b) on p. 210.

1 1

y

A CB D

t 2

t 1

I

S


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At the start of acquisition, the observable terms are:

cos(2 J IS 2) sin(2 J IS 1 ) cos( S t 1) 2 I y S z+ sin(2 J IS 2) sin(2 J IS 1) cos( S t 1) I x.

The modications for detecting long-range correlation are essentially the same as those discussedfor the HMQC experiment in section 8.8 on page 212. They are:

Increase the length of the delay 1 so that sin(2 J IS 1) is signicant for typical values of thelong-range coupling constants.

Acquire immediately after the nal transfer pulses D, thus avoiding loss of signal due torelaxation during the nal spin echo E, as in sequence (b) of Fig. 8.22 on page 210.

Acquire without broadband decoupling, as the wanted term is anti-phase with respect to J IS .


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8.11

The diagonal peak is A11 cos( J IS t 1) cos(

1 t 1 ) I 1 y.

It was shown in section 7.5.1 on page 153 that evolution of I 1 y during t 2 gives the following time-domain signal:

12 i exp(i[ 1 + J 12 ]t 2) +

12 i exp(i[ 1 J 12 ]t 2 ).

Imposing an exponential decay and Fourier transforming yields the following spectrum:12 i [ A2 ( 1 + J 12 ) + i D2( 1 + J 12 )] +

12 i [ A2( 1 J 12 ) + i D2 ( 1 J 12 )] .

Applying a 90 phase correction and taking the real part, we obtain an in-phase doublet on spinone:12 A2( 1 + J 12 ) +

12 A2 ( 1 J 12 ).

The modulation with respect to t 1 is A11 cos( J IS t 1) cos( 1 t 1), which can be expanded using theidentitycos A cos B 12 [cos( A + B) + cos( A B)] ,

to give12 A11 [cos(

1 + J IS )t 2 + cos( 1 J IS )t 2] .Imposing an exponential decay, and then taking the cosine Fourier transform gives:

12 A11 [ A1(

1 + J 12 ) + A1( 1 J 12 )] ,which is an in-phase doublet in

1.

Multiplying the spectra in the 1 and 2 dimensions together gives the following four peaks for thediagonal-peak multiplet:

+ 14 A11 A1(

1 + J 12 ) A2 ( 1 + J 12 ) + 14 A11 A1( 1 + J 12 ) A2 ( 1 J 12 )

+ 14 A11 A1(

1 J 12 ) A2 ( 1 + J 12 ) + 14 A11 A1( 1 J 12 ) A2 ( 1 J 12 ).All the peaks are positive and in the absorption mode.

The cross peak term A12 cos( J IS t 1) cos(

1 t 1) I 2 yhas the same modulation in t 1 as the diagonal peak, and in t 2 the operator is I 2 y, rather than I 1 y, soin 2 the doublet appears at 2. We can simply write down the four peaks which contribute to the

cross-peak multiplet as:+ 1

4 A12 A1( 1 + J 12 ) A2 ( 2 + J 12 ) + 14 A12 A1(

1 + J 12 ) A2 ( 2 J 12 )+ 1

4 A12 A1( 1 J 12 ) A2 ( 2 + J 12 ) + 14 A12 A1( 1 J 12 ) A2 ( 2 J 12 ).

Again, these are in the absorption mode, and are all positive.

8.12

The phase-twist lineshape is

S ( 1 , 2 ) = [ A1 ( A) A2( B)

D1( A) D2 ( B)]

real+ i [ A1( A) D2( B) + D1( A) A2( B)]

imaginary.

The plot shows the imaginary part.


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8.13

The observable signal, acquired with broadband decoupling, is

sin(2 J IS 2) cos( S t 1) sin(2 J IS 1) I x.

(a) Applying the SHR method to the HSQC sequence requires the acquisition of two time-domainsignals: one with cos( S t 1 ) modulation in t 1 , the second with sin( S t 1 ) modulation in t 1 . Itwas shown in section 8.12.1 on page 227 that the modulation can be changed from cosine tosine by shifting the phase of the rst 90 S spin pulse by 90.

(b) For TPPI, each time t 1 is incremented, the phase of the rst 90 pulse on the S spin must beincremented by 90.


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8.14

In order to obtain a sine modulated data set from

cos ([ 1 + 2]t 1 + 2),

we need to set 2 = / 2 i.e. = / 4. To show this explicitly, we expand the argument of thecosine using the identitycos( A B) cos A cos B + sin A sin B,

hence

cos ([ 1 + 2]t 1 / 2) cos ([ 1 + 2]t 1 )cos( / 2) + sin ([ 1 + 2]t 1)sin( / 2) sin ([ 1 + 2]t 1),where we have used cos ( / 2) = 0 and sin (/ 2) = 1. So, shifting the phase by / 4 alters themodulation from cosine to sine. Thus, to implement TPPI, each time we increment t 1 the phases of the pulses preceding t 1 are incremented by 45.


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63/128

9Relaxation and the NOE

9.1

The equilibrium populations of the and levels are given by Eq. 9.6 on page 259:

n0 = 12 N exp( E / k BT ) n

0

= 12 N exp( E / k BT ),

where E = 12 B0 E = + 12 B0 .

Evaluating the energies yields: E = 12 1.055 1034 2.675 108 9.4 = 1.326 1025 J, E = + 1.326 1025 J.

Hence, at 298 K, the populations are:

n0 = 1

2 1013 exp(1 .326 1025 / (1 .381 1023 298))= 5.00016 1012 ,

n0 = 1

2 1013

exp( 1.326 1025 / (1 .381 10

23

298))= 4.99984 1012 .

On account of the very small energy gap, these populations are very similar, although as expectedn0 > n0 .

The energy of the system is given by

E = n E + n E = 1

2 B0 n n .Initially, n = n , so E initial = 0. At equilibrium,

E equ. = 1.326 1025 (4 .99984 1012 5.00016 1012 )= 4.243 1017 J.

The total change in energy is therefore E = E equ. E initial = 4.243 1017 J .


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Chapter 9: Relaxation and the NOE 60

The thermal energy of N molecules is of the order

Nk BT = 1013 1.381 1023 298 = 4.115 108 J ,which is nine orders of magnitude greater than the value of E calculated above. This reinforces thepoint that the energy of interaction between the spins and the magnetic eld is minuscule comparedto the thermal energy.

9.2

The reduced spectral density function is given by Eq. 9.4 on page 257 j( ) =

2 c1 + 2 2c

.

For a xed frequency , the maximum value of j( ) occurs at a value of c given by

d j( )d c

= 0.

Using the product rule, we obtain:

dd c

j( ) = 21 + 2 2c

4 2 2c

1+

2

2c

2

=2 + 2 2 2c 4 2 2c

1 + 2 2c2

=2 1 2 2c

1 + 2 2c2 .

The denominator is always non-zero, so the above expression can be solved by setting the numeratorto zero:

2 1 2 2c = 0

c = 1

Since the rate constant for longitudinal relaxation depends on j( 0), the above result indicates thatthis rate constant has its maximum value when c = 1/ 0 .


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9.3

At equilibrium, the lower state ( ) must have a greater population than the upper state ( ), aspredicted by the Boltzmann distribution (assuming that the gyromagnetic ratio is positive). Supposewe start with equal populations of the and states. The only way in which the population of the state can increase relative to that of the state is for the rate of transitions from to toexceed the rate from to . As the populations are equal, this implies that the rate constant for thetransition from to must be greater than that for the transition from to .

9.4

In the inversionrecovery experiment, the peak height S ( ) is given by

S ( ) = S (0) 2 exp( R z ) 1 ,where S (0) is the peak height at time zero. Rearranging this, we get:

lnS ( ) + S (0)

2S (0) = R z,

from which we can see that a plot of ln[(S ( ) + S (0)) / (2S (0))] against will be a straight line of gradient R z = 1/ T 1.

/ s 0.0 0.1 0.5 0.9 1.3 1.7 2.1 2.9S ( ) 129 .7 93 .4 7.6 62.6 93.4 109.5 118.9 126.4

ln[( S ( ) + S (0)) / (2S (0))] 0 .000 0.151 0.754 1.353 1.968 2.554 3.179 4.370

0.0 0.5 1.0 1.5 2.0 2.5 3.0

-5

-4

-3

-2

-1

0

/ s

l n [ ( S ( ) +

S ( 0 ) ) / 2 S ( 0 ) ]

The gradient is 1.508 s1, so R z = 1.508 s1 and T 1 = 0.663 s .

9.5

In section 9.5.2 on page 266, it was shown that an estimate for T 1 is given by null/ ln 2 . The valuesof T 1 are therefore:


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null / s 0.5 0.6 0.8T 1 / s 0.72 0.87 1.15

The fact that the solvent was still inverted after a delay of 1.5 s shows that it has a T 1 value that isgreater than 1.5/ ln 2 = 2.16 s i.e. the solvent relaxes at a slower rate than the other spins.

9.6

The z-magnetization relaxes according to Eq. 9.15 on page 264:

M z(t ) = M z(0) M 0 z exp( R zt ) + M 0 z .Setting M z(0) = 0 and t = , we obtain

M z( ) = M 0 z 1 exp( R z ) .

00

1

M z (

) / M

z 0

The peak height S ( ) is proportional to the z-magnetization present just before the 90 pulse. Thus,S ( ) can be written as

S ( ) = c 1 exp( R z ) .Letting , S = c; this will be the height of the peak in a simple 90acquire experiment.Substituting this into the above equation gives

S ( ) = S 1 exp( R z ) .Rearranging this yields:

S ( ) = S 1 exp( R z )S ( )S

= 1 exp( R z )S S ( )

S = exp( R z )

lnS S ( )

S = R z,

where we have taken the natural logarithm to go to the last line. Hence, a plot of ln[(S

S ( ))/ S

]against gives a straight line of gradient R z.


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9.7

Assuming that the rate is proportional to the deviation from the equilibrium population, we canwrite the rate of change of the population of level 1 (using the labelling in Fig. 9.17 on page 268) as

dn1dt

= W (2 , )1 n1 n01 W

(1 , )1 n1 n01 W 2 n1 n01

loss from level 1+ W (2 , )1 n2 n02

gain from level 2+ W (1 , )1 n3 n03

gain from level 3+ W 2 n4 n04

gain from level 4.

Similarly, the rates of change of the populations of the other levels are:dn2dt

= W (2 , )1 n2 n02 W 0 n2 n02 W

(1 ,)1 n2 n02

loss from level 2+ W (2 , )1 n1 n01

gain from level 1+ W 0 n3 n03

gain from level 3+ W (1 ,)1 n4 n04

gain from level 4.

dn3dt

= W (1 , )1 n3 n03 W 0 n3 n03 W

(2 ,)1 n3 n03

loss from level 3+ W (1 , )1

n1

n0

1

gain from level 1+ W 0 n2

n0

2

gain from level 2+ W (2 ,)

1n4

n0

4

gain from level 4.

dn4dt

= W 2 n4 n04 W (1 ,)1 n4 n04 W

(2 ,)1 n4 n04

loss from level 4+ W 2 n1 n01



gain from level 3.

9.8

(a) The expression for b is (from section 9.6.3 on page 271)

b = 0 2H

4r 3 =

4 107 (2 .675 108)2 1.055 10344 (1 .8 1010 )3

= 1.294 105 s1 .

Hence, b2 = 1.675 1010 s2 .(b) The expressions for the transition rate constants are given in section 9.6.3 on page 271:

W (1)1 = 340 b

2 j( 0,1) W (2)1

= 340 b

2 j( 0,2)

W 2 = 310 b2 j( 0,1 + 0,2) W 0 = 120 b

2 j( 0,1

0,2).


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In the fast motion limit, j( ) = 2 c for all frequencies , so the rate constants have thefollowing numerical values:

W (1)1 = 320 b

2 c = 320 1.675 1010 20 1012 = 0.0503 s1 ,W (2)1 =

320 b

2 c = 320 1.675 1010 20 1012 = 0.0503 s1 ,W 2 = 35 b

2 c = 35 1.675 1010 20 1012 = 0.201 s1 ,W 0 = 110 b

2 c = 110 1.675 1010 20 1012 = 0.0335 s1 .From Eq. 9.19 on page 271:

R(1) z = 2W (1)1

+ W 2 + W 0 = (2 0.0503) + 0.201 + 0.0335 = 0.335 s1 , R

(2) z = 2W

(2)1 + W 2 + W 0 = (2 0.0503) + 0.201 + 0.0335 = 0.335 s1 ,

12 = W 2 W 0 = 0.201 0.0335 = 0.168 s1 .(c) Substituting j( ) = 2 c for all values of in Eq. 9.20 on page 272, we obtain:

R(1) z = b2 3

20 j( 0,1) + 310 j( 0,1 + 0,2 ) +

120 j( 0,1 0,2)

= b2 c= 1.675 1010 20 1012= 0.335 s1 .

Similarly, R(2) z = 0.335 s1 , and 12 = 0.168 s1 .

(d) The value of R(1) xy can be calculated from the expr

solution manual for understanding nmr spectroscopy

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