Solutions manual for Understanding NMR spectroscopy second edition
James Keeler and Andrew J. Pell University of Cambridge, Department of Chemistry
(a)
(b) Ω1
Ω2
Ω2
Ω1 ω 1 ω 2
Version ersion 2.0
© Jame Jamess Keel Keeler er and and Andr Andrew ew J. Pell Pell
July July 2005 2005 and March March 2010 2010
This solutions manual may be downloaded downloaded and printed for personal use. It may not be copied or distributed, in part or whole, without the permission of the authors.
Preface We 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 adage that ‘practice makes perfect’ certainly applies when it comes to getting to grips with the theory of NMR.
[email protected]) 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 corrections will be maintained maintained on the spectroscopyNOW website http://www.spectroscopynow.com/nmr (follow
the link ‘education’) Cambridge, March 2010
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 field gradient pulses
89
12 Equivalent spins and spin-system analysis
101
13 How the spectrometer works
123
iv
CONTENTS
2 Setting the scene
2.1
We need Eq. 2.1 on page 6: δ(ppm) = 10 6
× υ −υ υ
ref
.
ref
For the first peak δ(ppm)
= 10
6
021 − 500.134 271 × 500.135500 = 1.50 ppm . .134 271
For the second peak the shift is 7.30 ppm . Using Eq. 2.3 on page 8 δ(ppm) = 10 6
× υ −υ υ
ref
,
rx
with υrx = 500.135 271 MHz gives the two shifts as 1.50 ppm and 7.30 ppm i.e. identical values to three significant figures. To all intents and purposes it is perfectly acceptable to use Eq. 2.3 on page 8. The separation of the two peaks can be converted to Hz using Eq. 2.2 on page 7: frequency separation in Hz = ( δ1 − δ2 ) × υref (in MHz). So the separation is (7.30
− 1.50) × 400.130 000 = 2321 Hz .
The conversion to rad s−1 is made using Eq. 2.4 on page 17 ω = 2 π
× υ = 2π × 2321 = 14583 rad s−
1
.
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 first multiplet the 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 α and β , and the line in which the spin states are β and α , lie of top of one another giving a 1:2:1 triplet.
− −
− −
−
Chapter 2: Setting the scene
2
J AB J AB
J AB J AC
J AC
-10
J AC
0
J AB = 10 Hz
10
-10
J AC = 2 Hz
0
J AB = 10 Hz
10
-10
J AC = 12 Hz
0
J AB = 10 Hz
10 J AC = 10 Hz
Introducing a third coupling gives a doublet of doublet of doublets. The line positions are ±1.5, ±3.5, ±6.5, ±8.5 Hz. For clarity, only the spin state of the fourth spin, D, are shown by the grey-headed arrows on the last line of the tree.
J AB J AC J AD
-10
0
J AB = 10 Hz
10
J AC = 2 Hz J AD = 5 Hz
2.3
The frequency, in Hz, is 1/period: υ
=
1 2.5
×
10−9
=
4
108
×
Hz or 400 MHz.
Converting to rad s−1 gives: ω
= 2 πυ =
2.51
(a) 90◦ is one quarter of a rotation so will take
× 1 4
109 rad s−1 .
× 2.5 ×
10−9
(b) As
2π radians is a complete rotation, the fraction of a (3π/2)/(2π) = 3 /4, so the time is 0 .75 2.5 10−9 = 1.875
×
×
(c) 720◦ is two complete rotations, so the time is 2 × 2.5
×
=
6.25
×
10−10
s.
rotation represented by × 10−9 s . 10−9 = 5.0 × 10−9 s .
3π/2
is
Chapter 2: Setting the scene
3
To convert from angular frequency to Hz we need Eq. 2.4 on page 17 υ =
ω
2π
7.85
=
×
104
2π
=
12494 Hz .
The period is 1/frequency: T =
1 υ
1 12 494
=
=
8.00
×
10−5 s .
2.4
(a) & (c)
(b)
y
(d)
y
φ = 135˚
φ = 0 or 2 π
φ = 3π /2 x
x -comp.
y
x
y -comp.
x
time
For φ = 0 or 2π radians, the x-component is a cosine wave, and the y-component is a sine wave. For φ = 3 π/2, the y -component is minus a cosine wave, and the x -component is a sine wave.
2.5
We need the identity sin( A + B)
≡ sin A cos B + cos A sin B.
Using this we find: sin(ωt + π)
= =
sin(ωt ) cos π + cos (ωt ) sin π
− sin(ωt ), where to go to the second line we have used cos π = −1 and sin π = 0 . So the y -component is indeed −r sin(ωt ).
Chapter 2: Setting the scene
4
3 Energy levels and NMR spectra
3.1 ˆ one spin is given by Eq. 3.2 on page 29: The expression for H ˆ one spin H
−γ B I ˆ .
=
0 z
ˆ one spin has on ψ − / : We need to work out the effect that H 1 2
ˆ one spin ψ−1/ H 2
−γ B I ˆ ψ− −γ B − ψ−
=
0
=
0
1/ 2
z
1 2
1/ 2
1 γ B0 ψ 1/2 . 2
=
−
To go to the second line we have used Eq. 3.3 on page 30 i.e. that ψ − / is an eigenfunction of I ˆ z . The wavefunction has been regenerated, multiplied by a constant; ψ− / is therefore an eigenfunction of ˆ one spin with eigenvalue 1 γ B0 . H 2 1
2
1 2
3.2
The Larmor frequency, in Hz, of a nucleus with zero chemical shift is defined by Eq. 3.8 on page 32: υ0
=
−γ B
=
−
0
2π 6.7283
2π 108
−1.01 ×
=
107
×
× 9.4
Hz or
−101 MHz.
To convert to rad s−1 , we need to multiply the frequency in Hz by 2 π: ω0
= 2 πυ0 = 2 π
× −1.01 ×
108
=
−6.32 ×
108 rad s−1 .
In the case of a non-zero chemical shift, the Larmor frequency, in Hz, is: −γ (1 + 10−6 δ) B0 υ0
=
= =
−
2π 6.7283
−1.01 ×
×
107
108
× (1 + 77 × 2π
Hz
.
10−6 )
× 9.4
Chapter 3: Energy levels and NMR spectra
6
This is an identical value to three significant figures. We need to go to considerably more figures to see the difference between these two Larmor frequencies. To seven figures the frequencies are 1.00659 × 108 Hz and 1 .00667 × 108 Hz.
3.3 ˆ one spin act on the wavefunction ψ + / : We let H 1
2
ˆ one spin ψ+1/ H 2
=
ω0 I ˆ z ψ+1/2
=
1 ω ψ 1, 2 0 + /2
where the Hamiltonian has been expressed in angular frequency units. To go to the second line, we have used the fact that ψ + / is an eigenfunction of I ˆ z with eigenvalue + 12 . 1 2
In the same way, ˆ one spin ψ−1/ H 2
=
−
1 ω ψ 1/2 . 2 0
−
ˆ one spin with eigenvalues ± 1 ω0 . Hence, ψ ± / are eigenfunctions of H 2 1
2
3.4
Following the approach in section 3.5 on page 35, we let the Hamiltonian act on the product wavefunction: ˆ two spins, no coupl. ψα,1 ψα,2 H
= = =
υ0,1 I ˆ1 z
+
υ0,2 I ˆ2 z ψα,1 ψα,2
υ0,1 I ˆ1 z ψα,1 ψα,2
ˆ2 z ψα,1 ψα,2 + υ0,2 I
υ0,1 I ˆ1 z ψα,1 ψα,2 + υ0,2 ψα,1 I ˆ2 z ψα,2 .
To go to the third line, we have used the fact that I ˆ1 z acts only on ψ α,1 and not on ψ α,2 . Similarly, I ˆ2 z acts only on ψ α,2 . Using Eq. 3.11 on page 35 i.e. that ψ α,1 and ψ α,2 are eigenfunctions of I ˆ1 z and I ˆ2 z , the terms in the square brackets can be rewritten: ˆ two spins, no coupl. ψα,1 ψα,2 H
= = =
υ0,1 I ˆ1 z ψα,1 ψα,2
+
υ0,2 ψα,1 I ˆ2 z ψα,2
1 υ ψ ψ + 12 υ0,2 ψα,1 ψα,2 2 0,1 α,1 α,2
1 υ + 12 υ0,2 2 0,1
ψα,1 ψα,2 .
ˆ two spins, no coupl. with eigenvalue Hence, ψ α,1 ψα,2 is an eigenfunction of H
1 υ + 12 υ0,2 2 0,1
.
Letting the coupling term act on the product wavefunction: J 12 I ˆ1 z I ˆ2 z ψα,1 ψα,2
ψα,1 ψα,2 is
=
J 12 I ˆ1 z ψα,1 I ˆ2 z ψα,2
=
J 12
=
1 J ψ ψ . 4 12 α,1 α,2
1 ψ 2 α,1
1 ψ 2 α,2
indeed an eigenfunction of the coupling term, with eigenvalue the energy.
1 J 4 12
: this corresponds to
Chapter 3: Energy levels and NMR spectra
7
ˆ two spins, no coupl. and H
the coupling term share the same eigenfunctions (a result of the fact that the two terms commute). Since the Hamiltonian for two coupled spins can be represented as the sum of these two terms, ˆ two spins H
ˆ two spins, no coupl. + = H
2π J 12 I ˆ1 z I ˆ2 z ,
it follows that it must also have the same eigenfunctions. Hence, ψα,1 ψα,2 is an eigenfunction of 1 υ + 12 υ0,2 + 14 J 12 , i.e. the sum of the individual eigenvalues of 2 0,1
ˆ two spins with energy eigenvalue H ˆ two spins, no coupl. and J 12 I ˆ1 z I ˆ2 z . H
3.5
Reproducing Table 3.2 on page 38 for υ 0,1 = −100 Hz, υ 0,2 = −200 Hz and J 12 = 5 Hz: number 1 2 3 4
spin states
eigenfunction
+2
1
αα
ψα,1 ψα,2
1 2 1 +2 1 2
αβ
ψα,1 ψ β,2
βα
ψ β,1 ψα,2
ββ
ψ β,1 ψ β,2
m1
m2
1 1
−
+2 +2
− −
1 2 1 2
−
eigenvalue/Hz 1
1
1
+ 2 υ0,1 + 2 υ0,2 + 4 J 12 = 1
1 υ 2 0,2 1 υ + 12 υ0,2 2 0,1
+ 2 υ0,1
− −
1 υ 2 0,1
−
−
−148.75
1 4 12 1 4 12 1 υ + 14 J 12 = 151 .25 2 0,2
− J = 48.75 − J = −51.25
The set of allowed transitions is: transition spin states αα → αβ 1 → 2 βα → ββ 3 → 4 1 → 3 αα → βα αβ → ββ 2 → 4
frequency/Hz
− E = 197.50 E − E = 202 .50 E − E = 97 .50 E − E = 102 .50 E 2
1
4
3
spin 1 flips spin 2 α β
3
4
1
2
spin 1 spin 2
13 24
80
100
α β flips 12 34
120
140
160
180
200
220
frequency / Hz
Chapter 3: Energy levels and NMR spectra
8
If J 12 = −5 Hz, the table of energies becomes: number 1 2 3 4
spin states
eigenfunction
+2
1
αα
ψα,1 ψα,2
1 2 1 +2 1 2
αβ
ψα,1 ψ β,2
βα
ψ β,1 ψα,2
ββ
ψ β,1 ψ β,2
m1
m2
1 1 2 1 2 1 2
−
+2 +
− −
−
eigenvalue/Hz 1
1
1 υ 2 0,1 1 υ + 2 0,1 1 υ 2 0,1
1 υ 2 0,2 1 υ 2 0,2 1 υ + 2 0,2
spin 1 spin 2
β α flips
+
− −
spin 1 flips spin 2 β α
100
−151.25
−
1 4 12 1 4 12 1 J = 148 .75 4 12
− J = 51.25 − J = −48.75
−
24 13
80
1
+ 2 υ0,1 + 2 υ0,2 + 4 J 12 =
34 12
120
140
160
180
200
220
frequency / Hz
The spectrum in unchanged in appearance. However, the labels of the lines have changed; the spin state of the passive spin for each line of the doublet has swapped over.
3.6
The allowed transitions in which spin two flips are 1–2, 3–4, 5–6 and 7–8. Their frequencies are: transition state of spin one α 1–2 β 3–4 α 5–6 β 7–8
state of spin three α α β β
frequency/Hz
0,2
1 J 2 12 1 J 2 12 1 J + 2 12
0,2
1 J + 12 J 23 = 207 2 12
−υ − −υ + −υ − −υ + 0,2 0,2
− −
1 J = 193 2 23 1 J = 203 2 23 1 J = 197 2 23
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 α state, and two with this spin being in the β state. Changing the sign of J 23 swaps the labels associated with spin three, but leaves those associated with spin one unaffected.
Chapter 3: Energy levels and NMR spectra α α
α β
12
56
spin 1 spin 3
9
β α
β β
α β
α α
34
78
56
12
spin 1 spin 3
−υ0,2
190
195
J 12
β β
β α
78
34
−υ0,2
200
205
210
190
195
200
205
210
frequency / Hz
J 23 J 12 =
10 Hz
J 23 =
J 12 =
4 Hz
10 Hz
J 23 =
-4 Hz
3.7
The six zero-quantum transitions have the following frequencies: transition 2–3 6–7 3–5 4–6 2–5 4–7
initial state
final state
frequency
αβα
βαα
αββ
βαβ
βαα
ααβ
υ0,1
ββα
αββ
υ0,1
αβα
ααβ
υ0,2
ββα
βαβ
υ0,2
−υ −υ
1 J + 2 13 1 0,1 + υ0,2 + 2 J 13 0,1 +
υ0,2
− υ + − υ − − υ + − υ − 0,3 0,3 0,3 0,3
βββ
ββα
αβα
2
ααα
αββ
4
βαα
1
3
−
1 J 2 23 1 J 2 23
− − J
1 J 2 12 1 J + 2 12
1 2 23 1 J 2 23
1 J 2 12 1 J + 2 12
1 J 2 13 1 J 2 13
−
8
6
βαβ
ααβ
5
7
Chapter 3: Energy levels and NMR spectra
10
The six transitions can be divided up into three pairs: • 2–3 and 6–7 in which spins one and two flip, and spin three is passive, • 3–5 and 4–6 in which spins one and three flip, and spin two is passive, • 2–5 and 4–7 in which spins two and three flip, and spin one is passive. Each pair of transitions is centred at the difference in the Larmor frequencies of the two spins which flip, and is split by the difference in the couplings between the two active spins and the passive spin.
4 The vector model
4.1
z
ω eff Ω
θ x
ω 1
The offset of the peak is 5 ppm. This can be converted to Hz using Eq. 2.2 on page 7: Ω
2π
= 10
− 6 ∆δ υ
ref = 10
−6 × 5 × 600 × 106 = 5 × 600 = 3000 Hz or 3 kHz.
From the diagram, tan θ =
ω1 Ω
=
25 3
3
× 10 × 2π = 25 = 8.33, × 10 × 2π 3 3
so θ = 83 ◦ . For a peak at the edge of the spectrum, the tilt angle is within 7 ◦ of that for an on-resonance pulse; the B 1 field is therefore strong enough to give a reasonable approximation to a hard pulse over the full 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 B 1 field giving a poorer approximation to a hard pulse.
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 flip angle β is M 0 sin β. The intensity of the signal will, therefore, vary as sin β, which is a maximum for β = 90 ◦ . (a) If β = 180 ◦ , the magnetization is rotated onto the −z-axis. As sin 180◦ = 0 , the signal intensity is zero. (b) If β = 270 ◦ , the magnetization is rotated onto the y -axis. As sin 270◦ = −1, the signal will have negative intensity of the same magnitude as for β = 90 ◦ .
4.3
From Fig. 4.16 on page 58, the intensity of the signal is proportional to sin β, where the value of the flip angle β is given by Eq. 4.5 on page 58: β
= ω 1 t p .
The pulse lengths of 5 and 10 µ s correspond to flip angles below 90 ◦ . Increasing t p further causes β to increase past 90 ◦ , and so the value of sin β (and hence the signal intensity) decreases. The null at 20.5 µ s corresponds to β = 180 ◦ . From the expression for the flip angle, it follows that π = ω 1 t 180 . Therefore, ω1
=
π
t 180
=
π
20.5
×
10−6
= 1 .5
×
105 rad s−1
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 µs, a complete rotation of 360 ◦ takes 41 .0 µ s. The period of this rotation is thus 41 .0 µ s, so the frequency is 1 4 −6 = 2.4 × 10 Hz . 41.0
×
10
This frequency is ω 1 /2π, the RF field strength in Hz. The length of the 90 ◦ pulse is simply half that of the 180 ◦ pulse: t 90
=
1 2
× 20.5 = 10.25 µ s.
The further null occurs at a pulse length that is twice the value of t 180 . This corresponds to a flip angle of 360 ◦ , for which the magnetization is rotated back onto the z -axis.
Chapter 4: The vector model
13
4.4
-y -y
-y 180˚ pulse
x
φ
-y
x
about y
-y
x
starting position
2π − φ
-y
x
x
x
resolved into x - and y - components
components after 180˚ pulse
final position
The vector has been reflected in the yz-plane, and has a final phase of 2π − φ, measured anti-clockwise from the −y-axis.
4.5
x
Ωτ
-y
e s l u p )
2π−Ωτ
y
( o
0 8 1
0 2π φ
, 3 e s a h p
π /2
τ /2
τ
3τ /2
τ
2τ
φ = 2π − Ωτ
π
π /2
φ = Ωτ
0 τ
time
2τ
The spin echo sequence 90◦ ( x) − τ − 180◦ ( x) − τ − results in the magnetization appearing along the y-axis. In contrast, the 90◦ ( x) − τ − 180◦ ( y) − τ− sequence results in the magnetization appearing along the −y-axis. Shifting the phase of the 180 ◦ pulse by 90 ◦ thus causes the phase of the magnetization to shift by 180◦ . A 180◦ (− x) pulse rotates the magnetization in the opposite sense to a 180◦ ( x) pulse, but the net effect is still to reflect the magnetization vectors in the xz -plane. The sequence 90 ◦ ( x) − τ − 180◦ (− x) − τ− will, therefore, have the same effect as the 90 ◦ ( x) − τ − 180◦ ( x) − τ− sequence i.e. the vector appears on the y -axis at the end of the sequence.
Chapter 4: The vector model
14
4.6
From section 4.11 on page 67, the criterion for the excitation of a peak to at least 90% of its theoretical maximum is for the offset to be less than 1.6 times the RF field strength. The Larmor frequency of 31 P at B 0 = 9 .4 T is: υ0
8
× 9.4 = −1.62 × − γ 2 Bπ = − 1.08 × 210 π 0
=
108
Hz or −162 MHz.
If the transmitter frequency is placed at the centre of the spectrum, the maximum offset is approximately 350 ppm. In Hz, this is an offset of Ω
2π
= 350
104
× 162 = 5.66 ×
Hz or
56.6 kHz.
According to our criterion, the RF field strength must be at least 56 .6/1.6 = 35 .3 kHz, from which the time for a 360◦ pulse is simply 1/(35.3 × 103 ) = 28.28 µs. Thus, the 90◦ pulse length is 14 × 28.28 = 7.07 µ s .
4.7
The flip angle of a pulse is given by Eq. 4.5 on page 58: β = ω 1 t p
So, ω1
=
β
t p
.
For a 90◦ pulse, β = π/ 2, so the B 1 field strength in Hz is: ω1
2π
=
(π/2) 2π t p
=
1 4
× 10 ×
10−6
=
2.5
×
104
Hz or
25 kHz.
The offset of 13 C from 1 H is 300 MHz, which is very much greater than the 13 C nuclei are therefore unaffected by the 1 H pulses.
B1 field
strength. The
Chapter 4: The vector model
15
4.8
From Eq. 4.4 on page 57,
ω21
ωeff =
If we let Ω = κω 1 , ω eff can be written ωeff =
ω21 + κ 2 ω21
+ Ω2 .
= ω 1
√
1 + κ 2 .
(4.1)
If t p is the length of a 90 ◦ pulse, we have ω 1 t p = π/ 2 and so ω1
π
=
2t p
,
and hence substituting this into Eq. 4.1 on page 15 gives ωeff =
π
2t p
√
1 + κ 2 .
Therefore the angle of rotation about the effective field, ω eff t p , is given by =
ωeff tp
=
π
√
1 + κ 2 2t p π 1 + κ 2 . 2
√
× t p
The null condition is when there is a complete rotation about the effective field i.e. ω eff tp = 2 π: 2π =
π
2
√
1 + κ 2 .
Rearranging this gives 4
=
√
i.e.
1 + κ 2
κ =
√
15
or
Ω=
√
15 ω1 ,
which is in agreement with Fig. 4.28 on page 68. The next null appears at ω eff tp = 4 π i.e. two complete rotations; this corresponds to κ =
√
For large offsets, κ 1, so 1 + κ 2 ≈ κ . The general null condition is ωeff tp 1, 2, 3, . . . Combining these two conditions gives 2nπ
for which we find κ = 4 n.
=
π
2
√
1 + κ 2
≈ π2 κ,
=
2nπ,
√
63 .
where
n
=
Chapter 4: The vector model
16
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 which a 90◦ pulse gives significant excitation. Therefore, only peaks with small offsets will be inverted completely. 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 the evolution in the transverse plane is as follows: x
x
x Ωτ
delay τ
-y
90˚(+x )
-y
-y
The x -, y - and z -components after each element of the pulse sequence are: component after first 90◦ ( x) after τ
after second 90◦ ( x)
M 0 sin Ωτ
M 0 sin Ωτ
− M
− M cos Ωτ
0
0
0
− M cos Ωτ
x
0
y z
0
0
0
The final 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 Ωτ. M 0
M x
0
π /2
π
3π /2
2π
-M 0 Ωτ
A null occurs when M x = 0 , i.e. when Ωτ = n π, where n = 0 ,
1, 2, . . .
Chapter 4: The vector model
17
4.11
The initial spin echo sequence refocuses the offset, and aligns the magnetization along the y -axis. If the final pulse is about the y - or −y-axis, then it has no effect on the magnetization as the vector is aligned along the same axis as the B 1 field. The magnetization remains along y. If the final 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 final pulse is about the − x-axis, then the magnetization is rotated from the y-axis to the −z-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-resonance peaks, Ω = 0, so the magnetization does not precess during the delay τ. The final 90◦ (− x) then simply undoes the rotation caused by the first pulse. Overall, the magnetization is returned to its starting position. . During the delay, the magnetization rotates to the x-axis and is therefore not affected by the final 90 ◦ (− x) pulse. The net result is that the magnetization appears along the x-axis. Ωτ = π/ 2
π. During the delay, the magnetization rotates onto the y-axis. The final pulse rotates the magnetization onto the −z-axis. The equilibrium magnetization is inverted: no observable transverse magnetization is produced. Ωτ =
x
Ω =
x
x
delay τ
0
-y
-y
x
x π /2
-y
-y
x
90˚(-x )
x
90˚(-x )
delay τ
-y
-y
x π
Ωτ = π
-y
x
delay τ
Ωτ = π /2
90˚(-x )
-y
-y
The overall effect of the sequence is to produce x-magnetization which varies as M 0 sin(Ωτ).
Chapter 4: The vector model
18
M 0
M x
0
π /2
3π /2
π
2π
-M 0 Ωτ
To suppress a strong solvent peak, it is placed on-resonance. The delay τ is then chosen so that Ωav τ = π/ 2, where Ωav 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 magnetization precesses about the z-axis through an angle of Ωτ. The 90 ◦ ( y) pulse rotates the x-component of the magnetization onto the −z-axis. x
x
x Ωτ
delay τ
-y
90˚(y )
-y
-y
The y -component of the magnetization varies as − M 0 cos Ωτ: M 0
M y
0
π /2
π
3π /2
2π
-M 0 Ωτ
The nulls are located at Ωτ = (2 n + 1)π/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 τ is chosen so that Ωτ = π/2, where Ω is the offset of the solvent. With such a choice, the solvent will not be excited.
Chapter 4: The vector model
19
4.14
Line A is on-resonance, so its magnetization does not precess during the delay τ . The pulse sequence is, 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 first 90◦ ( x) after τ
after second 90◦ ( x)
M 0 sin Ωτ
M 0 sin Ωτ
− M
− M cos Ωτ
0
0
0
− M cos Ωτ
x
0
y z
0
0
0
The final pulse is along the x-axis, so leaves the x-component of the magnetization unchanged. Substituting in the values of Ω and τ we find (note that the offset of 100 Hz has to be converted to rad s−1 ): =
M x
=
M z
3
× 100 × 5 × 10− ) = M sin π = 0 − M cos(2π × 100 × 5 × 10− ) = − M cos π = M . M 0 sin(2π
0
3
0
0
0
The magnetization is therefore returned to the z-axis. The 90◦ pulse rotates the equilibrium magnetization onto the −y-axis. During the delay τ , the vector precesses about z to give the following x - and y-components: M x
= M 0 sin Ωτ
M y
− M cos Ωτ.
=
0
For line A, offset 50 Hz: =
M x
=
M y
3
× 50 × 5 × 10− ) = M sin(π/2) = M − M cos(2π × 50 × 5 × 10− ) = − M cos(π/2) = 0. M 0 sin(2π
0
3
0
0
0
For line B, offset −50 Hz: M x M y
= =
3
× −50 × 5 × 10− ) = M sin(−π/2) = − M − M cos(2π × −50 × 5 × 10− ) = − M cos(−π/2) = 0 . M 0 sin(2π 0
0
3
0
0
The two magnetization vectors rotate at the same rate in the opposite sense. After a delay of τ = 5 ms, they are both aligned along the x -axis, but pointing in opposite directions.
Chapter 4: The vector model
20
5 Fourier transformation and data processing
5.1
One desirable feature of the dispersion lineshape is that it crosses the frequency axis at the frequency of the transition. This allows for a more accurate measurement of the chemical shift than might be possible 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 it undesirable: • It is broader than the absorption lineshape – the long ‘dispersive tails’ may interfere with nearby, 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 a complex spectrum, the result can be very difficult to interpret.
5.2
Setting A (ω) = S 0 /2 R, we obtain
S 0
2 R
=
S 0 R . 2 R + ω2
Cancelling the factor of S 0 from both sides and inverting the quotient, we obtain 2 R
=
R2
+
ω2
R
.
Hence,
The width of the line is therefore
2 R
ω2
=
2 R2
ω
=
± R .
in rad s−1 , or
2
− R
= R
R/π in
2
Hz.
Chapter 5: Fourier transformation and data processing
22
5.3 D(ω) can be differentiated using the product rule:
d D(ω) dω
ω d 2 dω R + ω2 1 2ω2 + ( R2 + ω2 )2 R2 + ω2
=
−
−
=
2
2
2ω2 ( R2 + ω2 )2 +
− R − ω
=
ω2
=
2
− R
( R2 + ω2 )2
.
At the turning points d D(ω) dω
so,
ω2
= 0 ,
2
− R
( R2 + ω2 )2
= 0 .
The denominator is always non-zero, so the equation can be solved by setting the numerator to zero: ω2
2
=
0
ω
=
± R .
− R
Substituting these values into D (ω): D( R) =
±
∓ 2 RR = ∓ 21 R . 2
These values are the maximum and minimum heights in the lineshape. There are two values of ω at which D (ω) is half its maximum positive height. At these frequencies, D(ω) = 1 /(4 R). Hence,
−ω
R2 +
ω2
R2
ω2
=
1 . 4 R
Inverting the quotients we obtain, +
ω
so, ω2
+
=
−4 R,
4 Rω + R2
= 0 .
This is a quadratic equation in ω that can be solved by applying the usual formula: ω
1
= 2
−
4 R
±
√
16 R2
Similarly, D (ω) = −1/(4 R) has two solutions:
ω
=
− 4 R
R(2
±
2
R( 2
=
− ±
√
3) .
√
3) .
The width, W disp , is the distance between the outer two half-maximum points, as shown in the diagram. Its value is √ √ √ W disp
= R (2 +
3) R( 2
− − −
3)
= 2(2 +
3) R.
Chapter 5: Fourier transformation and data processing
23
) 3 √ + 2 -
(
R
1/(2R) W disp R
1/(4R) ) 3 √
R -
-
2 -
(
R ) 3 √
) 3 √ + 2 (
frequency / rad s -1
R
-
2 (
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.
Chapter 5: Fourier transformation and data processing
24
5.4
y
y φ = 3π /4
(a)
(b)
φ = 3π /2
x
S x
S y
real
x
S x
imag
S y
real
y
(c)
imag
y
(d)
φ = 2π
φ = 5π /2
x
S x
S x
S y
real
x
imag
S y
real
imag
5.5
A 90◦ ( x) pulse rotates the equilibrium magnetization onto the −y-axis. The resulting spectrum is phased to absorption, so that magnetization along − y can be said to have a phase φ = 0 . A 90 ◦ ( y) pulse rotates the equilibrium magnetization onto the x -axis. This corresponds to a phase shift of φ = π/ 2 with respect to the initial experiment.
Chapter 5: Fourier transformation and data processing
y
90˚(x )
25
90˚(y )
y
x
x
real
real
(a) 90˚(-x )
(b)
y
270˚(x )
y
x
x
real
real
(a) Applying the pulse about − x rotates the magnetization vector onto y . This corresponds to a phase shift of φ = π , 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 31 P at B 0 = 9 .4 T is: ω0
2π
8
× 9.4 = −1.62 × − γ 2 Bπ = − 1.08 × 210 π
=
0
108
Hz or
−162 MHz.
The phase correction needed at the edge of the spectrum is given by Ωmax t p , where Ωmax is the maximum offset. For 3 1 P the maximum offset is 350 ppm, therefore the phase correction is 2π
This corresponds to
× 162 × 350 × 20 ×
10−6
=
7.1 radians .
407◦ , a significant frequency-dependent phase error.
Chapter 5: Fourier transformation and data processing
26
5.7
The intensity of the noise in the spectrum depends on both the amplitude of the noise in the timedomain, and the acquisition time. So, recording the time-domain signal long after the NMR signal has decayed just continues to measure noise and no signal. The resulting spectrum will consequently have 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 5.4.3 on page 92; the matched filter 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 little signal. Applying a line broadening weighting function does not discard this section of the timedomain, but reduces its amplitude relative to the earlier part of the FID. Thus, both methods reduce the 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 rising exponential and a Gaussian is discussed in section 5.4.5 on page 94. Zero filling improves the ‘definition’ of the line in the spectrum by increasing the density of data points in the frequency domain. However, it does not improve the fundamental linewidth as no real data 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 decay of the sine bell attenuates the noise at the end of the time-domain. The overall effect will be to enhance the resolution, assuming that the original FID has decayed close to zero by the end of the acquisition time. The sine bell with a phase shift of 90 ◦ is purely a decaying function, which will broaden the lines in just the same way as a decaying exponential does.
Chapter 5: Fourier transformation and data processing
27
5.11
The peak due to TMS is likely to be a sharp line. Hence, the corresponding time-domain signal decays slowly, and is therefore more likely to be truncated. The other lines in the spectrum will usually be broader than TMS, so their time-domain signals decay more rapidly and are less likely to be truncated. Truncation artefacts (‘sinc wiggles’) can be suppressed by applying a decaying weighting function. This will decrease the resolution, and may reduce the SNR.
Chapter 5: Fourier transformation and data processing
28
6 The quantum mechanics of one spin
6.1
1 ψ 2 β
Dirac notation:
ψ β ψα dτ
Dirac notation:
| = − | β β|α
ψ β ψ β dτ
Dirac notation:
β| β
ψ Qˆ ψ dτ
Dirac notation:
ψ|Q ˆ |ψ
I ˆ z ψ β
=
−
I ˆ z β
1 2
(a)
α|α = 1 (b) α| β = 0 or β|α = 0 (c) I ˆ |α = |α (d) |ψ = c |α + c | β. 1 2
z
α
β
6.2
The expectation value of I ˆ y is given by: ˆ y
I = ψψ| I |ψ|ψ . y
If |ψ is normalized, ψ|ψ = 1 , so the expectation value is given by
I = ψ| I ˆ |ψ. y
y
Substituting in |ψ = cα |α + c β | β, we obtain
I y
= = = =
c α α
β
α
β
| + c β| I ˆ c |α + c | β c c α| I ˆ |α + c c α| I ˆ | β + c c β| I ˆ |α + c c β| I ˆ | β i c c α| β − i c c α|α + i c c β| β − i c c β|α i c c − i c c . α α
1 2 1 2
α β
y
α α
β α
y
1 2
1 α β 2 α β
y
β α
1 2
β α
y
β β
1 2
β β
y
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 ˆ y α
I ˆ y β
1 2
| = i | β
1 2
| = − i |α, and to go to the last line, we have used the fact that |α and | β are orthonormal (Eq. 6.5 on page 108 and Eq. 6.6 on page 108).
I can be interpreted as the average value of the
y-component of angular momentum when
y
measured for a large number of spins, each of which has the same wavefunction |ψ.
6.3
The matrix representation of I ˆ x is I x
=
=
=
| | | | | | α I ˆ x α
α| I ˆ | β β| I ˆ | β α|α β|α x
β I ˆ x α
1 2 1 2
β β
0
1 2
1 2
0
x
1 2 1 2
α β
.
To go to the second line, we have used Eq. 6.10 on page 111, I ˆ x α
I ˆ x β
1 2
| = | β
1 2
| = |α, and to go to the last line we have used the fact that |α and | β are orthonormal (Eq. 6.5 on page 108 and Eq. 6.6 on page 108). Similarly, I y
=
=
=
| | | | | | − α I ˆ y α
β I ˆ y α
1 i 2
α β
1 i 2
β β
0
1 i 2
1 i 2
0
α| I ˆ | β β| I ˆ | β − i α|α − i β|α y
y
1 2
1 2
.
Chapter 6: The quantum mechanics of one spin
31
6.4
Starting with the expression for I y , and substituting in find:
I y
=
1 i c c 2 β α
=
1 i 2
=
1 2
=
1 2i α β
−
c α
= r α exp(i φα )
and
c β
= r β exp(i φ β )
we
1 i c α c β 2
r α r β exp( i φ β ) exp(i φα )
− − r r exp(−i φ ) exp(i φ ) i r r exp −i (φ − φ ) − exp i (φ − φ ) r r exp i (φ − φ ) − exp −i (φ − φ ) ,
α β
β
α
β
α
α β
α
β
α
β
α
where to go to the last line we have multiplied top and bottom by i .
β
Applying the identity exp(iθ )
− exp(−iθ ) ≡ 2i sin θ
to the above expression gives
I = r r sin(φ − φ ). α β
y
β
α
The bulk y -magnetization is then given by M y
(1)
=
γ I x
=
(1) γ rα (1) r β
=
α β
+
γ I x
(2)
(1) sin(φ β
+
...
φ(1) α )+
− γ Nr r sin(φ − φ ). β
(2)
(2)
γ rα (2) r β sin(φ β
−φ
(2) α )+
...
α
At equilibrium, the phases φ are randomly distributed, and so sin(φ β − φα ) is randomly distributed between ±1. As a result, the equilibrium y-magnetization is zero.
6.5
Starting from Eq. 6.31 on page 120 and premultiplying by β|, we obtain: dc α (t ) dt dcα (t ) β α dt
|α +
| + β|
|
dc β (t ) dt dc β (t ) dt
1 i Ωc α (t ) 2
| β
=
−
|α +
| β
=
β| −
1 i Ωc β (t ) β 2
|
| |
1 i Ωc α (t ) 2
α
+
β
|
1 i Ωc β (t ) 2
β .
The derivatives of cα and c β , and the quantities in square brackets, are numbers, so the above expression can be rearranged to give dc α (t ) dt
β|α +
dc β (t ) dt
β| β
dc β (t ) dt
1 i Ωc α (t ) 2
=
−
=
1 i Ωc β (t ). 2
β|α +
1 i Ωc β (t ) 2
β| β
To go to the last line, we have used the orthonormality property of |α and | β.
Chapter 6: The quantum mechanics of one spin
32
Substituting Eq. 6.58 on page 137 into the left-hand side of Eq. 6.57 on page 137 gives: dc β (t ) dt
=
d c (0) exp dt β
=
1 i Ωc β (0) exp 12 i Ωt 2
=
1 i Ωc β (t ). 2
1 i Ωt 2
Eq. 6.58 on page 137 is indeed the solution.
6.6
The expectation value of I ˆ y is
I = y
1 i c c 2 β α
−
1 i c α c β . 2
Substituting in the expressions for how c α and c β vary under free evolution (Eq. 6.34 on page 121) gives:
I (t ) y
c β (0) exp
−
1 i 2
=
1 i c (0)c α (0) exp ( i Ωt ) 2 β 1 i c (0)c i α (0) [cos(Ωt ) 2 β
=
1 i Ωt 2
c α (0) exp
=
−
− − 1 i Ωt 2
1 2
β
α
y
c α (0) exp
1 i Ωt 2
1 i c α (0)c β (0) exp (i Ωt ) 2 sin(Ωt )] 12 i c α (0)c β (0) [cos(Ωt ) +
−
− − = cos( Ωt ) i c (0)c (0) − i c (0)c (0) = cos( Ωt ) I (0) + sin(Ωt ) I (0).
1 i 2
1 2
α
β
x
+
sin(Ωt )
c β (0) exp
1 i Ωt 2
i sin(Ωt )]
1 c (0)c β (0) + 12 c β (0)c α (0) 2 α
To go to the third line, the identities
≡ cos θ + i sin θ
exp(iθ )
exp( iθ )
− ≡ cos θ − i sin θ were used, and to go to the last line, the expressions for I and I in terms of c x
y
(Eqs 6.12 and 6.13 on p. 111).
α
and c β were used
This result is summarized in the diagram below. The grey arrow shows the initial position, and the black arrow shows the position after time t .
Ωt
(0) (0)
6.7
Chapter 6: The quantum mechanics of one spin
33
The matrix representation of the density operator is given by: ρ
=
| | ||
| | ≡ | |
α ρˆ α
β ρˆ α
α ρˆ β
ρ11
ρ12
β ρˆ β
ρ21
ρ22
.
We can now calculate the ρ11 element (for clarity, the overbars indicating the ensemble averaging have been omitted until the last line): ρ11
= = = = =
α| ρˆ|α α|ψψ|α α| c |α + c | β c α| + c β| |α c α|α + c α| β c α|α + c β|α
α
α
β
β
c α cα .
α
β
α
β
To go to the second line, the definition of ρˆ was inserted, and on the third line |ψ was expressed as a superposition of |α and | β. The other elements can be calculated in a similar way to give: ρ12
= c α c β
ρ21
Hence, ρ
=
= c β cα
c α cα
c α c β
c β c α
c β c β
ρ22
.
= c β c β .
Chapter 6: The quantum mechanics of one spin
34
7 Product operators
7.1
• exp(−i θ I ˆ x ) I ˆ y exp(i θ I ˆ x ) represents a rotation of I ˆ y about x through angle θ . From Fig. 7.4 on page 148 (a) on p. 148, I ˆ y is transformed into I ˆ z . Hence, I ˆ y
θ I ˆ x
−−→
ˆ y + sin θ I ˆ z . cos θ I
This is consistent with the identity on line one of Table 7.1 on page 143. • exp(−i θ S ˆ y )S ˆ z exp(i θ S ˆ y ). From (b) of Fig. 7.4 on page 148, rotation about y : ˆ z S
ˆ y θ S
−−→
ˆ z is S
transformed into
ˆ x S
by a
ˆ z + sin θ S ˆ x . cos θ S
• exp(−i θ I ˆ x ) I ˆ x exp(i θ I ˆ x ). Rotating I ˆ x about the x-axis has no effect: I ˆ x
θ I ˆ x
−−→ I ˆ . x
• exp(−i θ I ˆ z )(− I ˆ y ) exp(i θ I ˆ z ). Fig. 7.4 on page 148 (c) shows the effect of a rotation about z on − I ˆ y: the result is a transformation to I ˆ x. Hence, θ I ˆ z
− I ˆ −−→ − cos θ I ˆ + sin θ I ˆ . y
y
x
• exp(−i (θ/2) I ˆ y ) I ˆ x exp(i (θ/2) I ˆ y ). This represents the rotation of I ˆ x about y through angle From Fig. 7.4 on page 148 (b), I ˆ x is transformed to − I ˆ z . Hence, I ˆ x
θ/2.
(θ/2) I ˆ y
−−−−−→ cos(θ/2) I ˆ − sin(θ/2) I ˆ . x
z
• exp(i θ I ˆ z )(− I ˆ z ) exp(−i θ I ˆ z ). Careful inspection of the arguments of the exponentials reveals that this represents a z -rotation through angle −θ i.e. the rotation is in a clockwise sense. In this case, it does not matter as − I ˆ z is unaffected by a rotation about the z -axis: θ I ˆ z
− I ˆ −−−−→ − I ˆ . z
z
Chapter 7: Product operators
36
7.2
The 90 ◦ ( x) pulse rotates the equilibrium magnetization (represented by I ˆ z ) onto the −y-axis: (π/2) I ˆ x
I ˆ z
−−−−−→ cos(π/2) I ˆ − sin(π/2) I ˆ ˆ. = − I z
y
y
This transverse term evolves under the offset during the delay τ to give ˆ z Ωτ I
− I ˆ −−−→ − cos(Ωτ) I ˆ + sin(Ωτ) I ˆ , y
y
x
where (c) of Fig. 7.4 on page 148 has been used. The 180 ◦ ( y) pulse does not affect the I ˆ y term, but inverts the I ˆ x term: π I ˆ y
− cos(Ωτ) I ˆ + sin(Ωτ) I ˆ −−→ − cos(Ωτ) I ˆ + cos π sin(Ωτ) I ˆ − sin π sin(Ωτ) I ˆ ˆ − sin(Ωτ) I ˆ . = − cos(Ωτ) I y
x
y
x
y
z
x
Now we consider the evolution during the second delay. Taking each term separately, we obtain ˆ z Ωτ I
− cos(Ωτ) I ˆ −−−→ − cos(Ωτ) cos(Ωτ) I ˆ + sin(Ωτ) cos(Ωτ) I ˆ , − sin(Ωτ) I ˆ −−−→ − cos(Ωτ) sin(Ωτ) I ˆ − sin(Ωτ) sin(Ωτ) I ˆ . y
y
x
ˆ z Ωτ I
x
x
y
Combining these terms gives the final result as
−
cos2 (Ωτ) + sin2 (Ωτ) I ˆ y
− I ˆ , θ ≡ 1 has been used. =
y
where the terms in I ˆ x cancel, and the identity cos2 θ + sin2 At the end of the sequence, the magnetization has been refocused onto the −y-axis, irrespective of the offset.
7.3
I ˆ y
(π/2) I ˆ y
−−−−−→ I ˆ − I ˆ −−−−−−→ I ˆ ˆ −−→ S ˆ . S y
(π/2) I ˆ y
y
y
ˆ y πS
y
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 transformed by the rotation. I ˆ x
π I ˆ y
−−−−→ cos(−π) I ˆ − sin(−π) I ˆ ˆ. = − I x
x
In this case the magnetization is simply inverted.
z
Chapter 7: Product operators
37
The difference between the next two examples is the sense of the 90 ◦ rotation. (π/2) I ˆ y
I ˆ z
−−−−−→ cos(π/2) I ˆ + sin(π/2) I ˆ z
x
ˆ x . = I
I ˆ z
(π/2) I ˆ y
− ˆ + sin(−π/2) I ˆ −−−−−−→ cos(−π/2) I ˆ. = − I z
x
x
The next two are simply inversions:
ˆ z S
ˆ y πS
−−→ cos π S ˆ + sin π S ˆ ˆ. = −S z
x
z
I ˆ z
π I ˆ y
−−−−→ cos(−π) I ˆ + sin(−π) I ˆ ˆ. = − I z
x
z
7.4
The 90 ◦ ( x) pulse rotates the equilibrium magnetization I ˆ z to − I ˆ y . Free evolution is a rotation about z, so the state of the system after the delay τ is
− cos(Ωτ) I ˆ + sin(Ωτ) I ˆ . The 90 ◦ ( y) pulse does not affect the I ˆ term, but rotates I ˆ to − I ˆ . The final result is − cos(Ωτ) I ˆ − sin(Ωτ) I ˆ . y
y
x
x
y
z
z
The pulse sequence has therefore produced transverse magnetization along y, whose amplitude varies as − cos(Ωτ). This becomes zero if cos(Ωτ) = 0. Hence, there is a null at Ωτ = π/2, which corresponds to an offset of Ω = π/ (2τ) in rad s−1 , or 1 /(4τ) in Hz. There is a maximum in the excitation when cos(Ωτ) = ±1. This occurs at offsets satisfying Ωτ = n π where n = 0 , 1, 2, . . . i.e. Ω = ( nπ)/τ or n /(2τ) in Hz.
Chapter 7: Product operators
38
7.5
Figure 7.6 (b) on p. 152 shows that, as a result of evolution of the scalar coupling, the in-phase term − I ˆ1 y is partly transformed into the anti-phase term 2 I ˆ1 x I ˆ2 z ; the angle of rotation is π J 12 τ. This is represented as: ˆ2 z 2π J 12 τ I ˆ1 z I
− I ˆ −−−−−−−−→ − cos(π J τ) I ˆ + sin(π J τ) 2 I ˆ I ˆ Using the same figure, we see that −2 I ˆ I ˆ is partly transformed to − I ˆ : 1 y
12
1 y
1 x 2 z .
12
1 x 2 z
1 y
ˆ2 z 2π J 12 τ I ˆ1 z I
−2 I ˆ I ˆ −−−−−−−−→ − cos(π J
Similarly,
ˆ x S I ˆ2 y
2 I ˆ1 z I ˆ2 y I ˆ2 z
ˆ ˆ
12 τ) 2 I 1 x I 2 z
1 x 2 z
ˆ
− sin(π J
12 τ) I 1 y .
2π J IS (τ/2) I ˆ z ˆ S z
−−−−−−−−−−→ cos(π J τ/2) S ˆ + sin(π J τ/2) 2 I ˆ ˆS . −−−−−−−−→ cos(π J τ) I ˆ − sin(π J τ) 2 I ˆ I ˆ . −−−−−−−−→ cos(π J τ) 2 I ˆ I ˆ − sin(π J τ) I ˆ . −−−−−−−−→ I ˆ . IS
x
IS
z y
ˆ2 z 2π J 12 τ I ˆ1 z I
2 y
12
1 z 2 x
12
ˆ2 z 2π J 12 τ I ˆ1 z I
1 z 2 y
12
12
2 x
ˆ2 z 2π J 12 τ I ˆ1 z I
2 z
In the last example we see that z -magnetization is not affected by evolution under coupling simply because the Hamiltonian for coupling only contains I ˆ z operators.
7.6
The evolution is determined by the Hamiltonian given in Eq. 7.14 on page 150: ˆ two spins H
ˆ1 z + Ω2 I ˆ2 z + = Ω1 I
2π J 12 I ˆ1 z I ˆ2 z .
We will now work out the effect in turn of the three terms in the Hamiltonian. The first is a rotation about z : I ˆ1 y
ˆ1 z Ω1 t I
−−−−→ cos(Ω t ) I ˆ − sin(Ω t ) I ˆ 1
1 y
1 x .
1
The second term, Ω2 I ˆ2 z , does not need to be considered as spin-two operators have no effect on spin-one operators. Finally, we consider the effect of evolution under scalar coupling: ˆ1 y cos( Ω1 t ) I
ˆ2 z 2π J 12 t I ˆ1 z I
− sin(Ω t ) I ˆ −−−−−−−−→ 1
1 x
ˆ1 y cos( π J 12 t ) cos(Ω1 t ) I
−
sin(π J 12 t ) cos(Ω1 t ) 2 I ˆ1 x I ˆ2 z
− −
y-magnetization
ˆ cos(π J t ) sin(Ω t ) I
12 1 1 x
The NMR signal is given by: S (t )
sin(π J 12 t ) sin(Ω1 t ) 2 I ˆ1 y I ˆ2 z .
x-magnetization
=
M x + i M y
=
− cos(π J
12 t ) sin(Ω1 t ) +
i cos(π J 12 t ) cos(Ω1 t )
=
i cos(π J 12 t ) [cos(Ω1 t ) + i sin(Ω1 t )]
=
i cos(π J 12 t ) exp(i Ω1 t )
= =
1 i exp(i π J 12 t ) + exp( i π J 12 t ) exp(i Ω1 t ) 2 1 i exp (i[Ω1 + π J 12 ]t ) + 12 iexp (i[Ω1 π J 12 ]t ) . 2
−
−
Chapter 7: Product operators
39
To go to the fourth line, we have used the identity cos θ + i sin θ ≡ exp(i θ ), and to go to the fifth line, we have used cos θ ≡ 12 [exp(i θ ) + exp(−i θ )]. Finally, to go to the sixth line we have multiplied out the square brackets. Fourier transformation of this signal gives a positive line at Ω1 + π J 12 , and a second positive line at Ω 1 − π J 12 i.e. an in-phase doublet on spin one. The factor of i corresponds to a phase shift of 90◦ , so the imaginary part of the spectrum contains the absorption mode lineshape. 2πJ 12 imaginary
real ω Ω1 πJ 12
-
Ω1+πJ 12
A similar line of argument gives the observable signal arising from 2 I ˆ1 y I ˆ2 z as S (t ) =
1 i exp (i[Ω1 + 2
π J 12 ]t )
−
1 i exp (i[Ω1 2
12 ]t ) .
− π J
The corresponding spectrum is an anti-phase doublet on spin one. Again, the factor of i means that the absorption mode lines will appear in the imaginary part of the spectrum. 2πJ 12 imaginary
real ω Ω1 πJ 12
-
Ω1+πJ 12
Chapter 7: Product operators
40
7.7 I ˆ1 y represents
in-phase magnetization on spin one, aligned along the y -axis. The resulting spectrum will be an in-phase doublet centred on the offset of spin one, both peaks of which are in the absorption mode.
I ˆ2 x represents in-phase magnetization on spin two. However, it is aligned along the x -axis, so has a phase of 3π/2 relative to the y -axis. The spectrum therefore comprises an in-phase doublet that is
centred on the offset of spin two, with both peaks in the dispersion mode. 2 I ˆ1 y I ˆ2 z represents magnetization on spin one that is anti-phase with respect to spin two, and aligned along y . The spectrum is therefore an anti-phase doublet in the absorption mode.
2 I ˆ1 z I ˆ2 x represents anti-phase magnetization on spin two. It is aligned along x , so the lineshape will be
dispersive. Therefore, the spectrum is an anti-phase spin-two doublet with the dispersion lineshape.
I 1y I 2x 2I 1y I 2z
2I 1z I 2x
ω Ω2
Ω1
7.8
In-phase magnetization I ˆ1 x is rotated in the xz -plane towards − I ˆ1 z by the application of the y -pulse of duration t p . I ˆ1 x
ˆ1 y ω1 t p I
−−−−−→ cos(ω t p) I ˆ − sin(ω t p) I ˆ 1
1 x
1
1 z
A 180 ◦ pulse about y applied only to spin two changes the sign of the anti-phase magnetization on spin one. 2 I ˆ1 x I ˆ2 z
π I ˆ2 y
−−−−−→ cos(−π) 2 I ˆ I ˆ ˆ I ˆ = −2 I
1 x 2 z +
sin( π) 2 I ˆ1 x I ˆ2 x
−
1 x 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. ˆ2 z 2π J 12 t I ˆ1 z I
− I ˆ −−−−−−−−→ − cos(π J 1 x
ˆ
12 t ) I 1 x
− sin(π J
ˆ ˆ
12 t ) 2 I 1 y I 2 z
Chapter 7: Product operators
41
Letting each term act sequentially, we obtain 2 I ˆ1 x I ˆ2 z
(π/2) I ˆ1 y
(π/2) I ˆ2 y
−−−−−→ −2 I ˆ I ˆ −−−−−→ −2 I ˆ I ˆ
1 z 2 x .
1 z 2 z
Note that the spin-one operators do not act on spin-two operators and vice versa. The net result is that 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 for the I spin has no effect on the S ˆ x . ˆ x S
ˆ z Ω I t I
−−−→
ˆ z ΩS t S
ˆ x S
−−−−→ cos(Ω
ˆ
S t ) S x +
ˆ y sin(ΩS t ) S
Anti-phase magnetization on spin two evolves under coupling to generate in-phase magnetization on the same spin. ˆ2 z 2π J 12 t I ˆ1 z I
−2 I ˆ I ˆ −−−−−−−−→ − cos(π J
ˆ ˆ
12 t ) 2 I 1 z I 2 y +
1 z 2 y
ˆ2 x sin(π J 12 t ) I
7.9
The Hamiltonian for free evolution is given by Eq. 7.14 on page 150: ˆ two spins H
ˆ1 z + Ω2 I ˆ2 z + = Ω1 I
2π J 12 I ˆ1 z I ˆ2 z .
The spin echo refocuses the evolution due to offset, so we only need to consider the evolution of 2 I ˆ1 x I ˆ2 z under coupling, which gives 2 I ˆ1 x I ˆ2 z
ˆ2 z 2π J 12 τ I ˆ1 z I
ˆ ˆ 12 τ) 2 I 1 x I 2 z +
−−−−−−−−→ cos(π J
ˆ1 y . sin(π J 12 τ) I
The π pulse about the x-axis acts on both spins, leaving I ˆ1 x unaffected, but inverting I ˆ2 z and I ˆ1 y : cos(π J 12 τ) 2 I ˆ1 x I ˆ2 z
ˆ1 y + sin(π J 12 τ) I
π( I ˆ1 x + I ˆ2 x )
−−−−−−−→ − cos(π J
ˆ ˆ
12 τ) 2 I 1 x I 2 z
ˆ
− sin(π J
12 τ) I 1 y .
Finally, evolution under coupling during the second delay gives ˆ2 z 2π J 12 τ I ˆ1 z I
− cos(π J τ) 2 I ˆ I ˆ − sin(π J τ) I ˆ −−−−−−−−→ − cos (π J τ) I ˆ I ˆ − sin(π J τ) cos(π J τ) I ˆ − cos(π J τ) sin(π J τ) I ˆ + sin (π J τ) 2 I ˆ I ˆ ˆ I ˆ − [2 cos(π J τ) sin(π J τ)] I ˆ = − cos (π J τ) − sin (π J τ) 2 I ˆ I ˆ − sin(2π J τ) I ˆ . = − cos(2π J τ) 2 I To go to the last line, we have used the identities cos θ − sin θ ≡ cos 2θ and 2 cos θ sin θ ≡ sin 2θ . 1 x 2 z
12
2
1 x 2 z
12
2
1 y
12
12
12
12
2
12
1 x 2 z
1 y
12
12
1 x 2 z
12
12
12
12
1 y
2
12
1 x 2 z
1 y
1 y
2
2
By a similar method we can show: 2 I ˆ1 y I ˆ2 z
τ π x τ
− − −−−−−→ cos(2π J
ˆ ˆ
12 τ) 2 I 1 y I 2 z
− sin(2π J
ˆ
12 τ) I 1 x .
The effect of the τ − π y − τ spin echo on spin-one and spin-two terms is shown in the table below:
Chapter 7: Product operators
42
final state initial state I ˆ1 x I ˆ1 y
× cos (2π J − I ˆ
12 τ)
× sin (2π J −2 I ˆ I ˆ −2 I ˆ I ˆ
12 τ)
1 x
1 y 2 z
I ˆ1 y
2 I ˆ1 x I ˆ2 z 2 I ˆ1 y I ˆ2 z
1 x 2 z
2 I ˆ1 x I ˆ2 z
I ˆ1 y
−2 I ˆ I ˆ − I ˆ
I ˆ1 x
1 y 2 z
I ˆ2 x
−2 I ˆ I ˆ −2 I ˆ I ˆ
2 x
I ˆ2 y
1 z 2 y
I ˆ2 y
2 I ˆ1 z I ˆ2 x 2 I ˆ1 z I ˆ2 y
1 z 2 x
2 I ˆ1 z I ˆ2 x
I ˆ2 y
−2 I ˆ I ˆ
I ˆ2 x
1 z 2 y
The results for the in- and anti-phase operators on spin two can be obtained from those for spin one simply by swapping the labels 1 and 2. Likewise for the τ − π x − τ spin echo: final state initial state I ˆ2 x
× cos (2π J
12 τ)
12 τ)
I ˆ2 x
I ˆ2 y
2 I ˆ1 z I ˆ2 x 2 I ˆ1 z I ˆ2 y
× sin (2π J I ˆ1 z I ˆ2 y
2 I ˆ1 z I ˆ2 x
− I ˆ
2 y
−2 I ˆ I ˆ
− I ˆ − I ˆ
1 z 2 x
2 y
2 I ˆ1 z I ˆ2 y
2 x
7.10
A spin echo in a homonuclear two-spin system is equivalent to: (a) evolution of the coupling for time 2 τ, (b) a 180◦ ( x) pulse. Applying this to the first example, we obtain I ˆ2 y
τ π x τ
− − −−−−− → − cos(2π J
ˆ 12 τ) I 2 y +
sin(2π J 12 τ) 2 I ˆ1 z I ˆ2 x .
For complete transformation to 2 I ˆ1 z I ˆ2 x , we need sin(2π J 12 τ) when 2 π J 12 τ = π/ 2, i.e. τ = 1 /(4 J 12 ). I ˆ1 x
τ π x τ
− − −−−−−→ cos(2π J
ˆ
12 τ) I 1 x +
=
1 and cos(2π J 12 τ)
=
0.
These occur
sin(2π J 12 τ)2 I ˆ1 y I ˆ2 z .
√
Setting 2π J 12 τ = π/4 gives cos(2π J 12 τ) = sin(2π J 12 τ) = 1/ 2. The required delay is therefore τ 1/(8 J 12 ). To achieve conversion to − I ˆ1 x , we need cos(2π J 12 τ) = −1 and sin(2π J 12 τ) = 0 i.e. τ = 1 /(2 J 12 ). 2 I ˆ1 z I ˆ2 x
τ π x τ
− − −−−−− → − cos(2π J
ˆ ˆ
12 τ)2 I 1 z I 2 x
− sin(2π J
ˆ
12 τ) I 2 y .
=
Chapter 7: Product operators
43
Setting the delay to τ = 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 τ
τ
S
The 180 ◦ ( x) pulse is applied to only the S spin, so the evolution of the offset of the S spin will be refocused. We need to consider the evolution of the coupling. Starting with S ˆ x , the state of the system after the first delay is ˆ x + sin(π J 12 τ) 2 I ˆ z ˆ cos(π J 12 τ) S S y .
The 180◦ (x) pulse is applied only to the S spin, and so does not affect I ˆ z or S ˆ x . However, the term in ˆ y changes sign to give: S ˆ x cos(π J 12 τ) S
ˆ ˆ
− sin(π J
12 τ) 2 I z S y .
Evolution of the coupling during the second delay gives
ˆ x + [sin(π J 12 τ) cos(π J 12 τ) cos2 (π J 12 τ) + sin2 (π J 12 τ) S
where the anti-phase terms cancel, and the identity cos 2 of the coupling has therefore been refocused.
− cos(π J τ) sin(π J τ)] 2 I ˆ ˆS = S ˆ , θ + sin θ ≡ 1 has been used. The evolution 12
12
z y
x
2
Repeating the calculation for the anti-phase term, we see that 2 I ˆ z ˆS x is unaffected by the spin echo sequence. Again, the coupling is refocused. Both operators are unchanged, which is the same effect that a 180◦ ( x) pulse to the S spin would have: ˆ x S 2 I ˆ z ˆ S x
ˆ x πS
−−→ −−→ ˆ x πS
ˆ x S 2 I ˆ z ˆ S x .
Likewise, the operators I ˆ x and 2 I ˆ x ˆS 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 for the duration of the spin echo (time 2 τ).
7.12
The pulse sequence for the INEPT experiment is reproduced below from Fig. 7.15 on page 168: y τ 1
τ 1
τ 2
τ 2
I
S
A
B
C
Chapter 7: Product operators
44
At the end of period A it was shown in section 7.10.2 on page 168 that the state of the spin system is ˆ y k I cos(2π J IS τ1 ) I
ˆ ˆ
− k sin(2π J
IS τ1 ) I x S z .
I
The purpose of the two 90◦ pulses in period B is to transfer the anti-phase magnetization (the second term) from the I spin to the S spin. This requires the pulse acting on the I spin to cause the transformation I ˆ x → I ˆ z , which requires a rotation about the y -axis. ˆ z to k I I ˆ y . At the end of the If the initial 90 ◦ pulse is about the −x-axis, it rotates the equilibrium k I I spin echo in period A, the system is in the following state: ˆ ˆ ˆ IS τ1 ) I y + k I sin(2π J IS τ1 ) 2 I x S z .
−k cos(2π J I
As before, the I ˆ y term is not affected by the 90 ◦ ( y) pulse on the I spin, and can be discarded. The two pulses affect the ant-phase term as follows: k I sin(2π J IS τ1 ) 2 I ˆ x ˆ S z
(π/2) I ˆ y
ˆ ˆ
−−−−−→ −k sin(2π J
IS τ1 ) 2 I z S z
I
ˆ x (π/2)S
−−−−−→
k I sin(2π J IS τ1 ) 2 I ˆ z ˆ S y .
This term evolves under coupling during the spin echo in C to give: k I cos(2π J IS τ2 ) sin(2π J IS τ1 ) 2 I ˆ z ˆ S y
ˆ
− k sin(2π J
IS τ2 ) sin(2π J IS τ1 ) S x ,
I
the observable term of which is the one in S ˆ x . The 90 ◦ ( x) pulse acting on the S spin during B also rotates equilibrium k S S ˆ z to −k I ˆS y , which evolves during the spin echo in C to give: ˆ
IS τ2 ) S y +
−k cos(2π J S
k S sin(2π J IS τ2 ) 2 I ˆ z ˆ S x .
This also has an observable term in S ˆ y . Hence, the two observable terms are combined to give:
−k cos(2π J S
ˆ
IS τ2 ) S y
ˆ
− k sin(2π J
IS τ2 ) sin(2π J IS τ1 ) S x .
I
The first term is unaffected by changing the phase of the I spin 90◦ pulse from x to − x, whereas the second term changes sign.
7.13
• By definition, I ˆ+ has coherence order +1. • I ˆ z is unaffected by a z -rotation, so has coherence order zero. • I ˆ− has coherence order −1, again by definition. • Using the definitions of I ˆ1+ and I ˆ1− (Eq. 7.28 on page 174) as applied to spin one:
we can write I ˆ1 x as:
I ˆ1+ I ˆ1−
≡ ≡
I ˆ1 x
1 2
≡
I ˆ1 x I ˆ1 x
+
−
ˆ1 y i I ˆ1 y , i I
I ˆ1+ + I ˆ1− .
Therefore, I ˆ1 x is an equal mixture of coherence orders
+1
and −1.
Chapter 7: Product operators
45
• Similarly, I ˆ2 y can be written as I ˆ2 y
≡
1 2i
Hence, 2 I ˆ1 z I ˆ2 y can be written as 2 I ˆ1 z I ˆ2 y
I ˆ2+
− I ˆ − 2
.
I ˆ2+
− I ˆ − , which is an equal mixture of coherence orders +1 and −1, found by summing the coherence ≡ 2 ×
1 ˆ I 2i 1 z
2
orders of spins one and two (spin one has coherence order zero). • Since both I ˆ1 z and I ˆ2 z have coherence order zero, so does 2 I ˆ1 z I ˆ2 z . •
2 I ˆ1+ I ˆ2− has two is 1.
coherence order zero since the coherence order of spin one is
+1
and that of spin
−
•
2 I ˆ1 x I ˆ2 y can be written as: 2 I ˆ1 x I ˆ2 y
2
≡ ≡
×
1 2i
2 I ˆ1 x I ˆ2 y is
1 2
I ˆ1+ + I ˆ1−
I ˆ1+ I ˆ2+
× 1 2i
I ˆ2+
− I ˆ − 2
− I ˆ − I ˆ − − I ˆ 1
ˆ 1+ I 2−
2
therefore an equal mixture of coherence orders ence, and coherence order 0, zero-quantum coherence.
+2
ˆ1 + I
− I ˆ2+ .
and −2, double-quantum coher-
7.14
Using the definitions of I ˆi± given by Eq. 7.28 on page 174, we can write 2 I ˆ1 x I ˆ2 y as: 2 I ˆ1 x I ˆ2 y
≡ ≡
− × − − − − − − − − 2
×
1 2i
1 2
I ˆ1+ + I ˆ1
I ˆ1+ I ˆ2+
I ˆ2+
1 2i
I ˆ1 I ˆ2
I ˆ1 I ˆ2+
1
+ 2i
I ˆ2
I ˆ1+ I ˆ2
.
zero-quantum part
double-quantum part
The other relationships in the table can be verified in the same way.
7.15
The first 90 ◦ ( x) pulse rotates the equilibrium I ˆ1 z to − I ˆ1 y . During the spin echo sequence, the offset is refocused, but the coupling evolves throughout. The state of the spin system at the end of the spin echo is ˆ1 y cos(2π J 12 τ) I
The final pulse acts to give
ˆ ˆ
− sin(2π J
12 τ) 2 I 1 x I 2 z .
ˆ1 z + sin(2π J 12 τ) 2 I ˆ1 x I ˆ2 y . cos(2π J 12 τ) I ˆ and ZQ ˆ given in the last table of section 7.12.1 on page 174, we see Using the definitions of DQ y y that we can rewrite the second term as 1 sin(2π J 12 τ) 2
DQ − ZQ ˆ
y
ˆ
which is a mixture of double- and zero-quantum coherence.
y
,
Chapter 7: Product operators
46
The amplitude of this multiple quantum term is a maximum when sin(2π J 12 τ) when τ = 1 /(4 J 12 ).
1,
=
which occurs
Starting with equilibrium magnetization on spin two, I ˆ2 z , the terms present after the final pulse are ˆ2 z + sin(2π J 12 τ) 2 I ˆ1 y I ˆ2 x ; cos(2π J 12 τ) I
we have taken the terms from the previous calculation and swapped the labels 1 and 2. Again, from ˆ and ZQ ˆ in section 7.12.1 on page 174, we can write the multiple quantum the definitions of DQ y y term as 1 ˆ + ZQ ˆ sin(2π J 12 τ) DQ y y . 2
Therefore, adding this term to the one originating from I ˆ1 z , we obtain; 1 sin(2π J 12 τ) 2
DQ − ZQ ˆ
ˆ
y
1
y + 2 sin(2π J 12 τ)
DQ ˆ
y +
ˆ ZQ y
= sin(2π J 12 τ)
ˆ , DQ y
which is pure double-quantum coherence. It is a rather unusual feature of this sequence that, in a two-spin system, it generates pure double-quantum coherence.
7.16
ˆ is equal to 2 I ˆ1 x I ˆ2 x + 2 I ˆ1 y I ˆ2 y . Zero-quantum coherence between spins From the table on p. 176, ZQ x one and two does not evolve under the coupling between these two spins, so we need only consider the evolution under offset. Considering first the 2 I ˆ1 x I ˆ2 x term: 2 I ˆ1 x I ˆ2 x
ˆ1 z +Ω2 t I ˆ2 z Ω1 t I
−−−−−−−−−→
ˆ1 x 2 cos(Ω1 t ) I
ˆ1 y + sin(Ω1 t ) I
ˆ2 x + sin(Ω2 t ) I ˆ2 y . cos(Ω2 t ) I
− sin(Ω t ) I ˆ
ˆ2 y cos( Ω2 t ) I
We will now look at the 2 I ˆ1 y I ˆ2 y term; 2 I ˆ1 y I ˆ2 y
ˆ1 z +Ω2 t I ˆ2 z Ω1 t I
−−−−−−−−−−→
ˆ1 y 2 cos(Ω1 t ) I
1
1 x
Collecting these terms together, we obtain:
− sin(Ω t ) I ˆ 2
[cos(Ω1 t ) cos(Ω2 t ) + sin(Ω1 t ) sin(Ω2 t )] (2 I ˆ1 x I ˆ2 x + 2 I ˆ1 y I ˆ2 y ) + [sin(Ω1 t ) cos(Ω2 t )
− cos(Ω t ) sin(Ω t ) ](2 I ˆ I ˆ − 2 I ˆ I ˆ 1
1 x 2 y ).
1 y 2 x
2
Using the identities: cos( A B)
− sin( A − B)
=
cos A cos B + sin A sin B
=
sin A cos B
− cos A sin B,
ˆ and ZQ ˆ : and the definitions of ZQ x y ˆ x ZQ
we obtain
≡ (2 I ˆ I ˆ
1 x 2 x +
ˆ y ZQ
2 I ˆ1 y I ˆ2 y )
≡ (2 I ˆ I ˆ − 2 I ˆ I ˆ
ˆ − Ω ]t ) ZQ
cos ([Ω1
2
x +
1 y 2 x
1 x 2 y ),
ˆ . − Ω ]t ) ZQ
sin ([Ω1
2
y
2 x
.
8 Two-dimensional NMR
8.1
In each example, the preparation period is highlighted with a grey box, and the mixing period with a grey box with a dashed border. COSY t 1
DQF COSY t 1
t 2
t 2
DQ spectroscopy τ
TOCSY
t 1
τ
t 1
t 2
τ
mix
t 2
HSQC y τ1
τ1
τ2
t 2
τ2
I
t 1
S
HMQC τ
HMBC τ
t 2 I
I
S
t 2
τ
t 1
t 1
S
HETCOR t 1
τ
1
τ
1
τ
2
τ
2
I
t 2 S
Chapter 8: Two-dimensional NMR
48
8.2
ω 2
t 1
4
5 1
2
6 3
1 4 2 5 3 6 t 1
ω 2
1, 2 and 3 are cross-sections of the damped cosine wave, whose amplitude provides the modulation in t 1 . The period is the same for each wave, and the amplitude increases as we approach the centre of the peak in ω 2 . 4, 5 and 6 are cross-sections through the ω2 dimension. The amplitude and sign of the peak is modulated by a damped cosine wave in t 1 .
8.3
The COSY pulse sequence is given in Fig. 8.8 on page 192. t 1
t 2
Starting with equilibrium magnetization on spin two, the state of the system at t 2 = 0 can be determined from terms [1]–[4] on p. 191 by swapping the spin labels 1 and 2. The result is:
− cos (π J t )cos(Ω t ) I ˆ − sin(π J t )cos(Ω t ) 2 I ˆ I ˆ 12 1
2 1
12 1
2 1
2 z
1 y 2 x
ˆ2 x + cos (π J 12 t 1 )sin(Ω2 t 1 ) I
ˆ ˆ
12 t 1 )sin(Ω2 t 1 ) 2 I 1 y I 2 z .
− sin(π J
[1] [2] [3] [4]
The observable terms are [3] and [4]. The operator in term [3] is I ˆ2 x , which will give rise to a doublet on spin two in the ω2 dimension. It is modulated in t 1 by sin(Ω2 t 1 ) i.e. at the offset of spin two. Thus, [3] produces a diagonal-peak multiplet. The operator in term [4] is 2 I ˆ1 y I ˆ2 z ; this gives rise to an anti-phase doublet centred at the offset of spin one in the ω 2 dimension. It is also modulated in t 1 by sin(Ω2 t 1 ). Therefore, it produces a cross-peak multiplet.
Chapter 8: Two-dimensional NMR
49
It was shown in section 7.5.2 on page 154 that the evolution of 2 I ˆ1 y I ˆ2 z during t 2 gives rise to the following time domain signal: 1 i exp(i[Ω1 + 2
π J 12 ]t 2 )
−
1 i exp(i[Ω1 2
12 ]t 2 ).
− π J
Imposing an exponential decay on this signal and Fourier transforming, we obtain the following spectrum 1 i [ A2 (Ω1 + 2
π J 12 ) + i D2 (Ω1
+ π J 12 )]
−
1 i [ A2 (Ω1 2
−
1 [ A2 (Ω1 2
12 ) +
− π J
i D2 (Ω1
12 )] .
− π J
To ensure that the absorption mode lineshape appears in the real part of the spectrum, we multiply the expression above by a −90◦ phase correction factor i.e. by exp(−i π/2). Noting that exp(−i π/2) ≡ −i, we obtain: 1 [ A2 (Ω1 + 2
π J 12 ) + i D2 (Ω1
+ π J 12 )]
12 ) +
i D2 (Ω1
− π J
12 )] .
− π J
Clearly this is an anti-phase doublet on spin one. The t 1 modulation of term [4] has the form − sin(π J 12 t 1 )sin(Ω2 t 1 ). Applying the identity sin A sin B
≡
1 [cos( A 2
− B) − cos( A + B)] ,
gives 1 [cos(Ω2 + 2
π J 12 )t 1
− cos(Ω − π J
12 )t 1 ] .
2
Imposing an exponential decay and taking the cosine Fourier transform yields the spectrum 1 [ A1 (Ω2 + 2
π J 12 )
− A (Ω − π J 1
12 )] .
2
This is clearly an anti-phase doublet on spin two. Multiplying the ω 1 and ω 2 spectra together, and taking the real part, gives the following four lines which form the cross-peak multiplet. Note that they form an anti-phase square array. 1
+ 4 A1 (Ω2 +
−
1 A (Ω2 4 1
−
1 A (Ω2 + 4 1 1 π J 12 ) A2 (Ω1 + π J 12 ) + 4 A1 (Ω2
π J 12 ) A2 (Ω1
+ π J 12 )
−
π J 12 ) A2 (Ω1
− π J
12
12 )
− π J ) A (Ω − π J 2
12 ).
1
The operator in the diagonal peak term [3] is I ˆ2 x . Evolution of this operator during t 2 gives the following time domain signal: 1 exp(i[Ω2 + 2
π J 12 ]t 2 ) + 12 exp(i[Ω2
12 ]t 2 ).
− π J
Imposing an exponential decay to this, and Fourier transforming gives the spectrum 1 [ A2 (Ω2 + 2
1
π J 12 ) + i D2 (Ω2
+ π J 12 )] + 2 [ A2 (Ω2
12 ) +
− π J
i D2 (Ω2
12 )] .
− π J
This is an in-phase doublet on spin two. The t 1 modulation is: cos(π J 12 t 1 ) sin(Ω2 t 1 )
≡
1 [sin(Ω2 + 2
π J 12 )t 1 + sin(Ω2
12 )t 1 ] ,
− π J
where we have used the identity sin A sin B
≡
1 [sin( A + B) + 2
sin( A B)] .
−
Assuming an exponential decay and applying a sine Fourier transform gives the spectrum: 1 [ A1 (Ω2 + 2
This is an in-phase doublet on spin two.
π J 12 ) + A1 (Ω2
12 )] .
− π J
Chapter 8: Two-dimensional NMR
50
Multiplying together the ω1 and ω2 parts of the spectrum and taking the real part yields the following four components of the diagonal-peak multiplet. Note that they all have the same sign. 1
+ 4 A1 (Ω2 + +
π J 12 ) A2 (Ω2
1 A (Ω2 4 1
1
+ π J 12 ) + 4 A1 (Ω2 + π J 12 ) A2 (Ω2
12 ) A2 (Ω2 +
− π J
π J 12 ) + 14 A1 (Ω2
− π J
12
12 )
− π J ) A (Ω − π J 2
12 ).
2
8.4
The DQF COSY pulse sequence is given in Fig. 8.15 on page 200. t 1
t 2
Starting with equilibrium magnetization on spin two, I ˆ2 z , the state of the spin system after the second pulse is exactly the same as for the COSY experiment at t 2 = 0 as calculated in Exercise 8.3. Of the four terms present, the only one that contains double-quantum coherence is [2]: ˆ ˆ
12 t 1 )cos(Ω2 t 1 ) 2 I 1 y I 2 x .
− sin(π J
In section 7.12.1 on page 174, it was shown that 2 I ˆ1 y I ˆ2 x is a mixture of double- and zero-quantum ˆ , and the zero-quantum operator ZQ ˆ , are defined coherence. The double-quantum operator DQ y y as: ˆ ≡ 2 I ˆ1 x I ˆ2 y + 2 I ˆ1 y I ˆ2 x ˆ ≡ 2 I ˆ1 y I ˆ2 x − 2 I ˆ1 x I ˆ2 y . DQ ZQ y y Hence, ˆ + ZQ ˆ 2 I ˆ1 y I ˆ2 x = 12 DQ y y .
The double-quantum part that is retained is therefore:
−
1 sin (π J 12 t 1 )cos(Ω2 t 1 ) 2
ˆ = − 1 sin (π J 12 t 1 )cos(Ω2 t 1 ) DQ y 2
The third 90◦ pulse acts to give:
−
1 sin (π J 12 t 1 )cos(Ω2 t 1 ) 2
2 I ˆ1 x I ˆ2 y + 2 I ˆ1 y I ˆ2 x .
2 I ˆ1 x I ˆ2 z + 2 I ˆ1 z I ˆ2 x .
2 I ˆ1 x I ˆ2 z and 2 I ˆ1 z I ˆ2 x represent anti-phase magnetization on spins one and two, respectively. Both modulated in t 1 at Ω2 , so the first term therefore gives the cross-peak multiplet, and the second
diagonal-peak multiplet. Expanding the t 1 modulation, we obtain
−
1 sin (π J 12 t 1 )cos(Ω2 t 1 ) 2
≡ −
1 [sin(Ω2 + 4
π J 12 )t 1
are the
− sin(Ω − π J 2
12 )t 1 ] ,
which is an anti-phase doublet on spin two. Hence, both the cross- and diagonal-peak multiplets are anti-phase in both dimensions. Furthermore, both terms have the same t 1 modulation, and both appear along the x-axis at the start of acquisition, so the spectrum can be phased so that all the peaks appear in the double absorption mode.
8.5
The pulse sequence is given in Fig. 8.18 on page 204.
Chapter 8: Two-dimensional NMR
51
τ
τ
t 2
t 1
The first 90◦ pulse rotates equilibrium I ˆ1 z to − I ˆ1 y , which then evolves under coupling during the spin echo (the offset is refocused) to give ˆ1 y cos(2π J 12 τ) I
ˆ ˆ
− sin(2π J
12 τ) 2 I 1 x I 2 z .
This is rotated by the second 90 ◦ pulse to give ˆ1 z + sin(2π J 12 τ) 2 I ˆ1 x I ˆ2 y . cos(2π J 12 τ) I
We select just zero-quantum coherence at this point. From the table on p. 176, the zero-quantum ˆ , so at the start of t 1 we have: part of 2 I ˆ1 x I ˆ2 y is − 12 ZQ y
−
1 sin(2π J 12 τ) 2
ˆ . ZQ y
This evolves during t 1 according to the rules in section 7.12.3 on page 176:
− where
1 sin(2π J 12 τ) 2
ˆ1 z +Ω2 t 1 I ˆ2 z Ω1 t 1 I
ˆ −−−−−−−−−−−→ ZQ y
−
1 + 2
ˆ ≡ 2 I ˆ1 x I ˆ2 x + 2 I ˆ1 y I ˆ2 y ZQ x
ˆ − Ω ]t ) sin(2π J τ) ZQ ˆ , sin ([Ω − Ω ]t ) sin(2π J τ) ZQ
1 cos 2
([Ω1
2
1
12
y
1
2
1
12
x
ˆ ≡ 2 I ˆ1 y I ˆ2 x − 2 I ˆ1 x I ˆ2 y . ZQ y
Note that the zero-quantum coherence between spins one and two does not evolve due to the coupling between these two spins. The final pulse rotates the zero-quantum terms to give
− Ω ]t ) 2 I ˆ I ˆ − 2 I ˆ I ˆ τ)sin ([Ω − Ω ]t ) 2 I ˆ I ˆ + 2 I ˆ I ˆ
−
1 sin(2π J 12 τ)cos ([Ω1 2
2
1
+
1 sin(2π J 12 2
2
1
1
the observable terms of which are: 1 sin(2π J 12 τ)cos ([Ω1 2
− Ω ]t ) 2
1
1 z 2 x
1 x 2 z
1 x 2 x
1 z 2 z
2 I ˆ1 x I ˆ2 z
− 2 I ˆ I ˆ
1 z 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 ω 2 the anti-phase doublet on spin two, which arises from the 2 I ˆ1 z I ˆ2 x term, appears with the opposite sign. • The frequency of the peaks in ω 1 is ( Ω1 − Ω2 ) i.e. the zero-quantum frequency. The information that can be gained from this spectrum is the same as for the double-quantum spectrum.
8.6
From section 8.7 on page 209, the terms present after the first spin echo are cos(2π J IS τ1 ) I ˆ y
− sin(2π J
ˆ ˆ
IS τ1 )2 I x S z .
Chapter 8: Two-dimensional NMR
52
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 spin pulse to rotate I ˆ x to I ˆ z , which is only possible if the pulse is about y . Applying the I spin pulse about − y gives: ˆ ˆ
− sin(2π J
IS τ1 ) 2 I x S z
( π/2) I ˆ y
− −−−−−− → − sin(2π J
ˆ ˆ
IS τ1 ) 2 I z S z
ˆ x (π/2)S
−−−−−→ sin(2π J
ˆ ˆ
IS τ1 ) 2 I z S y .
The 2 I ˆ z ˆS y term, present at the start of t 1 , simply changes sign when the I spin pulse is changed in phase from + y to − y.
8.7
The pulse sequence is given in Fig. 8.22 on page 210(a). y τ1
t 2
τ1
I
y t 1
S
A
C
B
D
The state of the spin system after the spin echo ( A) is, from section 8.7 on page 209: ˆ y cos(2π J IS τ1 ) I
ˆ ˆ
− sin(2π J
IS τ1 ) 2 I x S z .
The pulses during period B have the following effect on the anti-phase term: ˆ ˆ IS τ1 )2 I x S z
− sin(2π J
ˆ y ) (π/2)( I ˆ y +S
−−−−−−−−→ sin(2π J
ˆ ˆ
IS τ1 ) 2 I z S x .
Period C is a spin echo, during which the coupling is refocused, but the offset of the S spin evolves for time t 1 . At the end of this period, the terms are:
− cos(Ω
ˆ ˆ
− sin(Ω
S t 1 ) sin(2π J IS τ1 ) 2 I z S x
ˆ ˆ
S t 1 ) sin(2π J IS τ1 ) 2 I z S y .
The final two pulses (period D ) produce the following state at t 2 = 0 : cos(ΩS t 1 ) sin(2π J IS τ1 ) 2 I ˆ y ˆ S x + sin(ΩS t 1 ) sin(2π J IS τ1 ) 2 I ˆ y ˆ S z .
The observable signal is due to the 2 I ˆ y ˆS z term, and is now modulated in t 1 according to sin(ΩS t 1 ). So, shifting the phase of the first 90◦ pulse to the S spin from x to y does indeed alter the modulation in 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 first 90 ◦ S spin pulse is about − x.
Chapter 8: Two-dimensional NMR
53
F τ
t 2
τ
I
-x t 1
S
A
C
B
D
E
As argued in section 8.8 on page 212, the offset of the I spin is refocused over the whole of period F. The first pulse creates − I ˆ y , which evolves during period A under coupling to give ˆ
IS τ) I y +
− cos(π J
sin(π J IS τ) 2 I ˆ x ˆ S z .
Taking just the second term (the first does not produce any useful peaks), and applying to it the first S spin pulse (with phase − x) gives: sin(π J IS τ) 2 I ˆ x ˆ S y ,
which is of opposite sign to the corresponding term in section 8.8 on page 212. This sign change propagates throughout the rest of the calculation so that the observable term sin2 (π J IS τ) cos(ΩS t 1 ) I ˆ y ,
also has the opposite sign. The same result is produced on changing the phase of the second 90 ◦ S spin pulse to − x. I spins that are not coupled to S spins do not give rise to anti-phase magnetization, and so are not affected by the S spin pulses. This I spin magnetization is therefore unaffected by altering the phase of the first S spin pulse. So, recording two spectra, the first with the first S spin pulse about x, and the 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 2
ˆ
− sin (π J
IS τ) cos(ΩS t 1 ) I y .
The amplitude of the signal is given by sin 2 (π J IS τ), which has a maximum value of 1 . This occurs when the argument of the sine is an odd multiple of π/2 i.e. when π J IS τ = nπ/2, n = 1, 3, 5, . . . Hence, τ = n /(2 J IS ), n = 1 , 3, 5, . . . sin2 (π J IS τ) = 0 when π J IS τ = nπ/2, n = 0, 2, 4, . . . i.e. is an even multiple of π/2. Hence the amplitude will be zero when τ = n /(2 J IS ), n = 0 , 2, 4, . . .
8.10
The HSQC pulse sequence, without decoupling during acquisition, is shown in Fig. 8.22 on page 210 (b) on p. 210. y τ1
t 2
τ1
I
t 1
S
A
B
C
D
Chapter 8: Two-dimensional NMR
54
At the start of acquisition, the observable terms are:
− cos(2π J
ˆ ˆ
IS τ2 ) sin(2π J IS τ1 ) cos(ΩS t 1 ) 2 I y S z
ˆ x . + sin(2π J IS τ2 ) sin(2π J IS τ1 ) cos(ΩS t 1 ) I
The modifications for detecting long-range correlation are essentially the same as those discussed for the HMQC experiment in section 8.8 on page 212. They are: • Increase the length of the delay long-range coupling constants.
τ1 so
that sin(2π J IS τ1 ) is significant for typical values of the
• Acquire immediately after the final transfer pulses D, thus avoiding loss of signal due to relaxation during the final 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 .
Chapter 8: Two-dimensional NMR
55
8.11
The diagonal peak is A1→1 cos(π J IS t 1 ) cos(Ω1 t 1 ) I ˆ1 y .
It was shown in section 7.5.1 on page 153 that evolution of I ˆ1 y during t 2 gives the following timedomain signal: 1 i exp(i[Ω1 + 2
π J 12 ]t 2 ) + 12 i exp(i[Ω1
12 ]t 2 ).
− π J
Imposing an exponential decay and Fourier transforming yields the following spectrum: 1 i [ A2 (Ω1 + 2
1
π J 12 ) + i D2 (Ω1
+ π J 12 )] + 2 i [ A2 (Ω1
12 ) +
− π J
i D2 (Ω1
12 )] .
− π J
Applying a −90◦ phase correction and taking the real part, we obtain an in-phase doublet on spin one: 1 A (Ω1 + 2 2
The modulation with respect to identity
t 1
is
π J 12 ) + 12 A2 (Ω1
12 ).
− π J
A1→1 cos(π J IS t 1 ) cos(Ω1 t 1 ),
cos A cos B
≡
to give
1 [cos( A + B) + 2
which can be expanded using the
cos( A B)] ,
−
1 A 2 1
→1 [cos(Ω1 + π J IS )t 2 + cos(Ω1 − π J IS )t 2 ] . Imposing an exponential decay, and then taking the cosine Fourier transform gives: 1 A 2 1
→1 [ A1 (Ω1 + π J 12 ) + A1 (Ω1 − π J 12 )] ,
which is an in-phase doublet in ω 1 . Multiplying the spectra in the ω 1 and ω 2 dimensions together gives the following four peaks for the diagonal-peak multiplet: 1 A 4 1 1 + 4 A1
→1 A1 (Ω1 + π J 12 ) A2 (Ω1 + π J 12 ) + 14 A1→1 A1 (Ω1 + π J 12 ) A2 (Ω1 − π J 12 ) →1 A1 (Ω1 − π J 12 ) A2 (Ω1 + π J 12 ) + 14 A1→1 A1 (Ω1 − π J 12 ) A2 (Ω1 − π J 12 ).
+
All the peaks are positive and in the absorption mode. The cross peak term A1→2 cos(π J IS t 1 ) cos(Ω1 t 1 ) I ˆ2 y
has the same modulation in t 1 as the diagonal peak, and in t 2 the operator is I ˆ2 y , rather than I ˆ1 y , so in ω 2 the doublet appears at Ω2 . We can simply write down the four peaks which contribute to the cross-peak multiplet as: 1
π J 12 ) A2 (Ω2 + π J 12 ) + 14 A1→2 A1 (Ω1 + π J 12 ) A2 (Ω2 − π J 12 ) → 1 1 + A1→2 A1 (Ω1 − π J 12 ) A2 (Ω2 + π J 12 ) + A1→2 A1 (Ω1 − π J 12 ) A2 (Ω2 − π J 12 ). 4 4 + 4 A1 2 A1 (Ω1 +
Again, these are in the absorption mode, and are all positive.
8.12
The phase-twist lineshape is S (ω1 , ω2 )
= [ A1 (Ω A ) A2 (Ω B )
D (Ω ) D (Ω )] +i [ A1 (Ω A ) D2 (Ω B ) + D1 (Ω A ) A2 (Ω B )] .
−
1 A 2 B
The plot shows the imaginary part.
real
imaginary
Chapter 8: Two-dimensional NMR
56
8.13
The observable signal, acquired with broadband decoupling, is ˆ x . sin(2π J IS τ2 ) cos(ΩS t 1 ) sin(2π J IS τ1 ) I
(a) Applying the SHR method to the HSQC sequence requires the acquisition of two time-domain signals: one with cos(ΩS t 1 ) modulation in t 1 , the second with sin(ΩS t 1 ) modulation in t 1 . It was shown in section 8.12.1 on page 227 that the modulation can be changed from cosine to sine by shifting the phase of the first 90◦ S spin pulse by 90 ◦ . (b) For TPPI, each time t 1 is incremented, the phase of the first 90◦ pulse on the S spin must be incremented by 90 ◦ .
Chapter 8: Two-dimensional NMR
57
8.14
In order to obtain a sine modulated data set from cos ([Ω1
we need to set 2φ = −π/2 i.e. cosine using the identity hence cos ([Ω1 +
Ω2 ]t 1
+ Ω2 ]t 1 +
2φ),
−π/4. To show this explicitly, we expand the argument of the cos( A − B) ≡ cos A cos B + sin A sin B,
φ
=
− π/2) ≡ cos ([Ω + Ω ]t )cos(π/2) + sin ([Ω ≡ sin ([Ω + Ω ]t ), 1
2
1
1
2
1
1 + Ω2 ]t 1 )sin(π/2)
where we have used cos (π/2) = 0 and sin (π/2) = 1. So, shifting the phase by − π/4 alters the modulation 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◦ .
Chapter 8: Two-dimensional NMR
58
9 Relaxation and the NOE
9.1
The equilibrium populations of the α and β levels are given by Eq. 9.6 on page 259: n0α
1 N exp( 2
=
− E /k B T )
n β0
=
1 γ B0 2
E β
= + 2 γ B0 .
α
1 N exp( 2
− E /k B T ), β
where E α
=
−
Evaluating the energies yields: E α
=
E β
=
34
1 2
1
108
− × 1.055 × 10− × 2.675 × − J. +1.326 × 10 25
× 9.4 = −1.326 ×
10−25 J,
Hence, at 298 K, the populations are: n0α
= =
n β0
= =
1 2
13
25
23
× 10 × exp(1.326 × 10− /(1.381 × 10− × 298)) 5.00016 × 10 , × 10 × exp(−1.326 × 10− /(1.381 × 10− × 298)) 4.99984 × 10 . 12
1 2
13
25
23
12
On account of the very small energy gap, these populations are very similar, although as expected n0α > n β0 . The energy of the system is given by E
=
nα E α
+ n β E β
=
1 γ B0 2
n β
Initially, n α = n β , so E initial = 0 . At equilibrium, E equ. = 1.326 × 10−25 × (4.99984 −17 J. = −4.243 × 10
− n
α
×
.
1012
− 5.00016 ×
The total change in energy is therefore ∆ E = E equ.
− E initial = − 4.243 ×
10−17 J .
1012 )
Chapter 9: Relaxation and the NOE
60
The thermal energy of N molecules is of the order Nk B T = 10 13
10−23
× 1.381 ×
× 298 =
4.115
10−8 J ,
×
which is nine orders of magnitude greater than the value of ∆ E calculated above. This reinforces the point that the energy of interaction between the spins and the magnetic field is minuscule compared to the thermal energy.
9.2
The reduced spectral density function is given by Eq. 9.4 on page 257 j(ω)
=
2τ c
.
1 + ω2 τ2c
For a fixed frequency ω , the maximum value of j(ω) occurs at a value of τ c given by d j (ω) dτc
= 0 .
Using the product rule, we obtain: d j(ω) dτc
=
=
2 1 + ω2 τ2c
−
2 + 2ω2 τ2c
4ω2 τ2c
1 + ω2 τ2c
1 + ω2 τ2c
=
−
2 1
2 2 c
− 4ω τ
ω2 τ2c
1 + ω2 τ2c
2
2
2
.
The denominator is always non-zero, so the above expression can be solved by setting the numerator to zero:
−
2 1
ω2 τ2c
τc
=
0
=
1 ω
Since the rate constant for longitudinal relaxation depends on this rate constant has its maximum value when τ c = 1 /ω0 .
j(ω0 ),
the above result indicates that
Chapter 9: Relaxation and the NOE
61
9.3
At equilibrium, the lower state ( α) must have a greater population than the upper state ( β), as predicted by the Boltzmann distribution (assuming that the gyromagnetic ratio is positive). Suppose we start with equal populations of the α and β states. The only way in which the population of the α state can increase relative to that of the β state is for the rate of transitions from β to α to exceed the rate from α to β . As the populations are equal, this implies that the rate constant for the transition from β to α must be greater than that for the transition from α to β .
9.4
In the inversion–recovery experiment, the peak height S (τ) is given by
S (τ) = S (0) 2 exp( R z τ)
−
−1
,
where S (0) is the peak height at time zero. Rearranging this, we get:
S (τ) + S (0) ln 2S (0)
=
− R τ, z
from which we can see that a plot of ln[(S (τ) + S (0))/(2S (0))] against τ will be a straight line of gradient − R z = −1/T 1 . τ
/s
0.0
S (τ)
ln[(S (τ) + S (0))/(2S (0))]
0.1
0.5
0.9 62.6
1.3 93.4
2.1 118.9
2.9 126.4
−129.7 −93.4 7.6 −0.151 −0.754 −1.353 −1.968 −2.554 −3.179 −4.370 0.000 τ /
0.0
1.7 109.5
0.5
1.0
1.5
2.0
s 2.5
3.0
0
] ) 0 (
-1
2 / ) ) 0 (
-2
S
S
+ ) τ (
S
( [ n l
-3 -4 -5
The gradient is −1.508 s−1 , so R z = 1 .508 s−1 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 τ null / ln 2. The values of T 1 are therefore:
Chapter 9: Relaxation and the NOE
62
/ s 0.5 0.6 0.8 0.72 0.87 1.15 T 1 / s
τnull
The fact that the solvent was still inverted after a delay of 1.5 s shows that it has a T 1 value that is greater 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 z0 exp( R z t ) + M z0 .
Setting M z (0) = 0 and t = τ , we obtain
M z (τ)
− 0
= M z
1
exp( R z τ) .
1
0
z
M / ) τ (
z
M
0
0
τ
The peak height S (τ) is proportional to the z -magnetization present just before the 90◦ pulse. Thus, S (τ) can be written as S (τ)
= c
− 1
exp( R z τ) .
−
Letting τ → ∞, S ∞ = c; this will be the height of the peak in a simple Substituting this into the above equation gives
−
S (τ) = S ∞ 1
Rearranging this yields: S (τ) S (τ) S ∞
ln
−
S ∞ S (τ)
S ∞ S (τ)
∞ − S
S ∞
−
−
S ∞ 1
=
1
=
experiment.
exp( R z τ) .
=
=
90◦ –acquire
exp( R z τ)
−
− exp(− R τ) z
exp( R z τ)
−
− R τ, z
where we have taken the natural logarithm to go to the last line. Hence, a plot of ln[(S ∞ − S (τ))/S ∞ ] against τ gives a straight line of gradient − R z .
Chapter 9: Relaxation and the NOE
63
9.7
Assuming that the rate is proportional to the deviation from the equilibrium population, we can write the rate of change of the population of level 1 (using the labelling in Fig. 9.17 on page 268) as dn1 dt
− − − − − − − −
=
(2,α) 1
−W
n01
n1
(1,α)
W 1
n01
n1
n01
W 2 n1
loss from level 1
(2,α) +W 1
n02
n2
(1,α)
+W 1
n03
n3
+W 2
n04 .
n4
gain from level 2
gain from level 3
gain from level 4
Similarly, the rates of change of the populations of the other levels are: dn2 dt
=
− − − − − − − − − − − − − − − − − −
(2,α) 1
−W
n02
n2
n02
W 0 n2
− − − − −
(1,β)
W 1
n2
0 2
− n
loss from level 2
(2,α) +W 1
n01 +W 0
n1
n03
n3
(1,β)
+W 1
n04 .
n4
dn3 dt
=
gain from level 1 (1,α) 1
−W
gain from level 3
n03
n3
gain from level 4
n03
W 0 n3
(2,β)
W 1
n03
n3
loss from level 3
(1,α) +W 1
n01 +W 0
n1
n02
n2
(2,β)
+W 1
n04 .
n4
dn4 dt
=
gain from level 1
−W
(1,β)
n04
n4
2
gain from level 2
W 1
gain from level 4
n04
n4
(2,β)
W 1
n04
n4
loss from level 4
+W 2
(1,β) n01 +W 1
n1
n02
n2
(2,β)
+W 1
n03 .
n3
gain from level 1
gain from level 2
gain from level 3
9.8
(a) The expression for b is (from section 9.6.3 on page 271) b =
2 µ0 γ H
Hence, b 2 =
4πr 3 1.675
=
×
4π
×
1010
10−7
8 2
× (2.675 × 10 ) × 1.055 × 4π × (1.8 × 10− )
10−34
10 3
= 1 .294
×
105
s−1 .
s−2 .
(b) The expressions for the transition rate constants are given in section 9.6.3 on page 271: (1)
W 1 W 2
3
2
= 10 b
3
2
(2)
j(ω0,1 )
W 1
j(ω0,1 + ω0,2 )
W 0
= 40 b
3
2
= 40 b 1
2
= 20 b
j(ω0,2 )
j(ω0,1
0,2 ).
− ω
Chapter 9: Relaxation and the NOE
64
In the fast motion limit, j(ω) following numerical values: (1)
= 20 b
(2)
= 20 b
W 1 W 1
W 2 W 0
2τc for
=
all frequencies ω, so the rate constants have the
3
2
τc
=
3 20
10
12
=
0.0503 s−1 ,
3
2
τc
=
3 20
10
12
=
0.0503 s−1 ,
= 5b
3 2
τc
=
3 5
1
τc
=
1 10
2
= 10 b
× 1.675 × 10 × 1.675 × 10 × 1.675 × 10 × 1.675 × 10
10
× 20 × 10− × 20 × 10− × 20 × 10− × 20 × 10−
12
10
0.201 s−1 ,
=
12
0.0335 s−1 .
=
From Eq. 9.19 on page 271: R z(1)
=
R z(2)
=
σ12
=
(c) Substituting
(1)
× 0.0503) + 0.201 + 0.0335 = 0.335 s− 2W + W + W = (2 × 0.0503) + 0.201 + 0.0335 = 0.335 s− W − W = 0 .201 − 0.0335 = 0.168 s− . 2W 1
+
W 2
(2) 1
+ W 0 = (2
2
2
0
1
,
1
,
1
0
j(ω) = 2 τc for all values of ω in Eq. 9.20 on page 272, we obtain: R z(1)
Similarly, R z(2) =
=
b2
=
b2 τc
=
1.675
=
0.335 s−1 .
0.335 s−1
3 3 j (ω0,1 ) + 10 j (ω0,1 + 20
×
1010
, and σ 12 =
× 20 ×
0.168 s−1
ω0,2 ) +
1 j (ω0,1 20
0,2 )
− ω
10−12
.
(1) (d) The value of R xy can be calculated from the expression in section 9.8.5 on page 296: (1) R xy
=
b2
=
b2 τc
=
1.675
=
0.335 s−1 .
1 3 3 3 j (0) + 20 j (ω0,2 ) + 40 j (ω0,1 ) + 20 j (ω0,1 + 10
1010
×
To go to the second line, we set
× 20 ×
ω0,2 ) +
1 j (ω0,1 40
0,2 )
− ω
10−12
(2)
j(ω) = 2 τc . Similarly, R xy
=
0.335 s−1
.
(e) As expected in the fast motion limit, the rate constants for the self-relaxation of both longitudinal and transverse magnetization have the same value. The rate constant for the cross-relaxation of longitudinal magnetization has half the value of the self-relaxation rate constant and is positive, again as expected. (f) The Larmor frequency is: ω0
= 2 π
× 500 ×
106
= 3 .140
×
109 rad s−1 .
From the expression for the reduced spectral density, j(ω)
we can calculate the values of j(ω0 ), 2τ c
j(ω0 )
=
j(2ω0 )
=
9.20
j(0)
=
1.00
1
2τ c 1 + ω2 τ2c
,
j(2ω0 ) and j(0): 12
× 500 × 10− = 1 + (3.140 × 10 × 500 × 2
2 2 + ω0 τc
× ×
=
10−11 s, 10−9 s.
9
= 2 .88 10−12 )2
×
10−10 s,
Chapter 9: Relaxation and the NOE
65
The values of R z(1) , R z(2) and σ12 can be calculated by substituting Eq. 9.20 on page 272, giving R z(1) = R z(2) = 2.025 s−1 , and σ12 = (1) (2) −1 . from section 9.8.5 on page 296, R xy = R xy = 3.41 s
= ω0 into − 1 −0.375 s . Similarly,
ω0,1
=
ω0,2
(g) As ω0 τc = 1 .6, we are now outside the fast motion limit, and beyond the zero-crossing point where σ12 = 0. As a result, σ12 is negative and the rate constant for transverse relaxation exceeds that for longitudinal relaxation. We are not very far beyond ω0 τc = 1 , so the rate of longitudinal relaxation is significantly faster than for τ c = 20 ps.
9.9
For a 13 C–1 H pair, the value of b is: b
=
µ0 γ C γ H
4πr 3
4π
= =
10−7
×
1.427
105
×
107
× 6.728 ×
8
× 2.675 × 10 × 1.055 × 4π × (1.1 × 10− )
s−1 .
10−34
10 3
Hence, b 2 = 2 .035 × 1010 s−2 . In the fast motion limit ( τc = 20 ps), the values of the rate constants can be calculated from those in the previous question by multiplying by the ratio of the b 2 values. Note that we can only do this because j(ω) is independent of τ c in this limit. So, R z(1)
C– H 13
1
=
b2C–H b2H–H
2.035 1.675
=
H– H R z(1)
× ×
1
1010 1010
1
× 0.335
0.407 s−1 .
=
(1) (2) −1 . All these values are greater Similarly, R z(2) = 0.407 s−1 , σ 12 = 0.204 s−1 , and R xy = R xy = 0.407 s than for the 1 H–1 H pair due to the smaller separation between the 13 C and 1 H. γ C is a quarter the value of γ H , so for the same distance we would expect the relaxation to be sixteen times slower. However, the rate constant goes as 1 /r 6 , which changes by a factor of 19.2 on going from r = 1 .8 Å to r = 1 .1 Å.
9.10
The necessary equations are given in section 9.10.2 on page 305. At B 0 = 4 .7 T, c2 is given by: c2
= = =
2
− ⊥ × × γ B0 σ
σ
6.728
107
1.00
×
109
s−2
4.7
× 100 ×
10−6
,
2
where we have used the gyromagnetic ratio of 13 C. In the fast motion limit, of ω , so the rate constants are: R z
=
1 c2 15 j (ω0 )
=
2 τc c2 15
=
1.00
=
0.00267 s−1 ,
×
109
2 15
× × 20 ×
10−12
j(ω) = 2 τc for all values
Chapter 9: Relaxation and the NOE R xy
66
=
c2
=
7 τc c2 45
=
1.00
=
0.00311 s−1 .
2 1 j (0) + 30 j (ω0 ) 45
109
×
7 45
10−12
× × 20 ×
At B 0 = 11 .74 T, the rate constants are greater by a factor of (11 .74/4, 7)2 : c2
=
R z
=
R xy
=
11.742 4.72 11.742 4.72 11.742 4.72
109
× 1.00 ×
=
6.24
s−2
109
×
× 0.00267 =
0.0167 s−1 ,
× 0.00311 =
0.0194 s−1 .
,
The values of the CSA relaxation rate constants at B 0 = 11 .74 T are an order of magnitude smaller than those for dipolar relaxation of 13 C due to an attached 1 H. However, as the CSA contribution goes as B 20 it will become more significant at higher fields.
9.11
The formulae are as for the previous question. For B 0 = 4 .7 T, c2
2
− ⊥ × ×
= =
γ B0 σ
σ
2.675
108
1.581
=
108
×
4.7
s−2
× 10 ×
.
10−6
2
Hence, R z
R xy
At
=
1 c2 15 j (ω0 )
=
2 τc c2 15
=
1.581
=
0.00042 s−1 ,
=
c2
=
7 τc c2 45
=
1.581
=
0.00049 s−1 .
B0 = 11.74 T, the values are greater 0.00263 s−1 , and R xy = 0.00307 s−1 .
×
108
2 15
× × 20 ×
10−12
2 1 j (0) + 30 j (ω0 ) 45
×
108
7 45
× × 20 ×
10−12
by a factor of (11.74/4.7)2 :
c2
=
9.864
×
108
s−1 ,
R z
=
At B 0 = 23 .5 T, c2 = 3.953 × 109 s−1 , R z = 0.01054 s−1 , and R xy = 0.01230 s−1 . Even at a field of B0 = 23.5 T, the rate constants are still an order of magnitude smaller than the dipolar relaxation rate constants at B 0 = 11.74 T.
Chapter 9: Relaxation and the NOE
67
9.12
We are going to apply the initial rate limit, in which we assume that, on the right hand side of Eq. 9.21 on page 274, d I 1 z (1) 0 0 = − R z I 1 z − I 1 z − σ12 I 2 z − I 2 z , dt I 1 z and I 2 z have their initial values:
d I 1 z
dt
(1) z
=
− R − R
init
(1) z
=
0 1 z
I 1 z (0)
0 2 z
− I − σ I (0) − I I − I − σ 0 − I
12
2 z
0 1 z
0 1 z
d I 1 z (t )
=
I 1 z (t )
=
0 σ12 I 2 z t + const.
σ12 I 20 z .
=
12
0 2 z
Integrating this, we obtain:
σ12 I 20 z dt
We know that at time t = 0 , I 1 z (0) = I 10 z , so the constant of integration is I 10 z . At t = τ : I 1 z (τ)
0
0
= σ 12 I 2 z τ + I 1 z .
Now we will look at the z-magnetization on spin two in the initial rate limit. Starting from d I 2 z dt
(2) z
=
− R
I 2 z
0 2 z
− I − σ
12
I 1 z
0 1 z
− I
,
we obtain:
d I 2 z
dt
=
init = =
(2) z
− R − R
(2) z
− − − − − − I 20 z
I 2 z (0)
0 I 20 z
σ12 I 1 z (0)
0 σ12 I 1 z
I 10 z
I 10 z
0 . R z(2) I 2 z
Integrating this, and noting that I 2 z (0) = 0 , we get, at t = τ : I 20 z (τ)
(2) 0 I 2 z τ.
= R z
The height of the peak due to spin one is proportional to I 1 z , and the height of that due to spin two is proportional to I 2 z . Furthermore, both spins are of the same type, so I 1 z0 = I 20 z . The peak heights for the irradiated, reference and difference spectra are: spectrum
S 1 (τ)
S 2 (τ)
irradiated: (a)
c (σ12 τ + 1)
cR z τ
reference: (b)
c
NOE difference: (a) − (b)
c σ12 τ
(2)
c
(2)
c R z τ
−1
Chapter 9: Relaxation and the NOE
68
Note that |σ12 | τ 1 and R z(2) τ 1 in the initial rate limit. Ω1
Ω2
(a)
(b)
(c) = (a) - (b)
The NOE enhancement is given by: η
=
= =
peak height in irradiated spectrum − peak height in reference spectrum peak height in reference spectrum c (σ12 τ + 1) c σ12 τ.
−c
9.13
The NOE difference spectrum is convenient as it only shows the target resonance, and the resonances which are receiving an NOE enhancement.
9.14
(a) The observation that the NOE enhancement depends only upon the cross-relaxation rate constant is a property of the initial rate limit i.e. the assumption that the target peak is still fully inverted after the delay τ . We are effectively ignoring self relaxation during this delay. (b) At longer times, the inverted spin begins to relax back to equilibrium. This reduces the zmagnetization on that spin and so slows the growth of the NOE: hence the dependence on the self-relaxation rate constant of that spin. The spin receiving the enhancement can also relax, resulting in the NOE enhancement being lost: hence the dependence on its self relaxation rate constant. (c) Spin one is held saturated throughout the experiment, so its relaxation is of no importance. Cross relaxation gives the rate of transfer of magnetization from spin one to spin two, while self relaxation of spin two leads to a loss of this transferred magnetization. Therefore, there is competition between these two processes, which is reflected in the observation that the enhancement depends upon the ratio of the rate constants for cross and self relaxation.
Chapter 9: Relaxation and the NOE
69
9.15
In the initial rate limit, the enhancement in a transient NOE experiment depends only upon the cross-relaxation rate constant for the transfer of magnetization between the inverted spin and the spin receiving the enhancement. In this example, σAB and σBC will be approximately equal, so when HB is inverted, the enhancement of H A and HC will be the same. On inverting HA , the enhancement at HB still depends only on σAB , so will be the same as for H A and HC when HB is irradiated. HC is too far from H A to receive an enhancement. In a steady state experiment, the enhancement depends upon the ratio of the cross-relaxation rate constant to the self-relaxation rate constant of the spin receiving the enhancement. R zA and R zC are equal to each other, so saturation of H B will give equal enhancements on H A and HC . Irradiation of HA gives a smaller enhancement on H B as the self relaxation of this spin is faster than for HA or HC . This is because HB has two nearby protons which relax it, whereas H A and HC only have one nearby proton.
Chapter 9: Relaxation and the NOE
70
9.16
The NOESY pulse sequence is given in Fig. 9.24 on page 281. t 1
t 2
τ
We will start with equilibrium magnetization on spin one, and assume that spins one and two are not coupled. If the phase of the first 90◦ pulse is − x, it rotates equilibrium I ˆ1 z to I ˆ1 y . This evolves under the offset during t 1 to give: I ˆ1 y
ˆ1 z +Ω2 t 1 I ˆ2 z Ω1 t 1 I
−−−−−−−−−−−→ cos(Ω t ) I ˆ − sin(Ω t ) I ˆ 1 1
1 y
1 1
1 x .
The second 90 ◦ pulse acts on the above terms to give: ( ) − sin(Ω t ) I ˆ −−−−−−−−−−→ cos(Ω t ) I ˆ − sin(Ω t ) I ˆ (π/2) I ˆ1 x + I ˆ2 x
cos(Ω1 t 1 ) I ˆ1 y
1 x
1 1
1 1
1 z
1 1
1 x .
There are also similar terms due to spin two. We select only longitudinal terms after this pulse, so at τ = 0 , the z-magnetization on each spin is: I 1 z
and
0
= cos(Ω1 t 1 ) I 1 z
I 2 z
0
= cos(Ω2 t 1 ) I 2 z .
The Solomon equations are (from Eq. 9.26 on page 282): d I 1 z (t ) dt d I 2 z (t ) dt where we have assumed that I 10 z initial conditions:
I 1 z (t )
=
− R
=
−σ I
z
0
0
= I 2 z = I z
1 z (t )
−
0 z
0 z
− I
0 z
2 z (t )
− I
R z I 2 z (t )
− I
− I − σ I
0 z
,
. Using the initial rate approximation with the following and
I 1 z (0) = cos(Ω1 t 1 ) I z0
I 2 z (0)
0
= cos(Ω2 t 1 ) I z ,
we obtain:
d d d I 1 z (t ) t
dt
init
0 z
− R [cos(Ω t ) − 1] I − σ [cos(Ω t ) − 1] I
=
−σ [cos(Ω t ) − 1] I − R [cos(Ω t ) − 1] I .
init
I 2 z (t )
0 z
=
1 1
z
1 1
0 z
2 1
z
0 z
2 1
Integrating these, and using the initial conditions to determine the values of the constants of integration, we obtain: I 1 z (τ) I z0 I 2 z (τ) I z0
=
cos(Ω1 t 1 ) (1 R z τ)
diagonal peak
=
cos(Ω2 t 1 )στ + ( R z
+
σ) τ,
− − − −
cos(Ω2 t 1 ) (1 R z τ)
diagonal peak
cross peak
axial peak
cos(Ω1 t 1 )στ + ( R z
cross peak
+
σ) τ .
axial peak
Comparing these with Eq. 9.28 on page 282 and Eq. 9.29 on page 282, we see that the terms which give the diagonal and cross peaks have changed sign, while the axial peak terms have not. The axial peaks can therefore be suppressed by difference spectroscopy: we record two spectra with the phase
Chapter 9: Relaxation and the NOE
71
of the first pulse set to + x and − x in turn, then we subtract one spectrum from the other. The cross and diagonal peaks reinforce, and the axial peaks cancel.
9.17
The Larmor frequency in rad s−1 is ω0
= 2 π
106
× 500 ×
= 3 .142
×
109 rad s−1 .
The value of ω 0 τc is 0.03, which is much less than 1 . Therefore, we are working in the fast motion limit, where j(ω) = 2 τc for all frequencies. The rate constant for longitudinal relaxation is given by Eq. 9.32 on page 294: 2
where R z obtain:
=
1/T 1 .
2
loc j(ω0 ),
= γ B
R z
Substituting in this, and using the fast motion limit expression for 1 T 1
=
B2loc
= =
j(ω0 ),
we
2γ 2 τc B2loc 1 2T 1 γ 2 τc 1 8 2
12
× 1 × (2.675 × 10 ) × 10 × 10− = 6.99 × 10− T . This corresponds to a root mean square field of 8 .4 × 10− T, which is 10 − 2
2
7
4
The local fields are indeed very weak.
4
times smaller than B 0 .
9.18
τ
τ
Any effects of inhomogeneous broadening are refocused by the spin echo, so the amplitude of the transverse magnetization present at the start of acquisition depends only upon R xy and the time 2 τ. The peak height is therefore given by: S (τ) S (τ) S 0
= =
S 0 exp( 2 R xy τ)
−
exp( 2 R xy τ).
−
Taking logarithms of both sides gives us ln
S (τ) S 0
=
−2 R
xy τ,
so a plot of ln( S (τ)/S 0 ) against τ is a straight line of gradient −2 R xy .
Chapter 9: Relaxation and the NOE τ/
72
s
0
0.1
0.2
S (τ)
65
39.4
23.9
ln(S (τ)/S 0 )
0
0.3 14.5
0.4 8.8
0.5 5.34
0.1
0.2
0.3
0.4
-0.5 -1.0 -1.5
/ ) τ (
-2.0
[ n l
-2.5
S S
-3.0 -3.5 -4.0
The gradient is −5.00 s−1 , giving R xy =
2.50 s−1
s
0.5
0.0
] ) 0 (
0.7 1.96
−0.501 −1.001 −1.500 −2.000 −2.499 −2.999 −3.501 τ /
0.0
0.6 3.24
, or T 2 = 0 .4 s.
0.6
0.7
10 Advanced topics in two-dimensional NMR
10.1
J 12 = 2 Hz
(a)
J 23 = 6 Hz
J 12 = 6 Hz
(b)
spin 3
α α
β β
spin 3
α
spin 1
α β
α β
spin 1
α
J 12
J 23 = 6 Hz
α/β
β
β/α
J 23
J 23
β J 12
I 2y
2I 1z I 2y 2I 2y I 3z 4I 1z I 2y I 3z
-10
-5
0
Ω2 /2π
5
10
-10
-5
0
5
10 Hz
Ω2 /2π
(a) Assuming that the offset of spin two is 0 Hz, the line positions are −4, −2, 2 and 4 Hz. (b) Assuming that the offset of spin two is 0 Hz, the line positions are −6, 0 , 0 and 6 Hz; we have a doublet of doublets, with the central two lines falling on top of one another, giving a triplet.
Chapter 10: Advanced topics in two-dimensional NMR
74
10.2
We do not need to consider the 1–3 coupling as this does not affect the evolution of a spin-two operator. First, let us consider the evolution due to the 1–2 coupling: I ˆ2 y
ˆ2 z 2π J 12 t I ˆ1 z I
ˆ
−−−−−−−−→ cos (π J
12 t ) I 2 y
− sin(π J
ˆ ˆ
12 t ) 2 I 1 z I 2 x .
We will now consider the effect of the 2–3 coupling separately on each of the terms on the right. For the term in I ˆ2 y the evolution is straightforward: ˆ2 y cos (π J 12 t ) I
ˆ3 z 2π J 23 t I ˆ2 z I
−−−−−−−−→ cos (π J t ) cos(π J t ) I ˆ − sin(π J t ) cos(π J t ) 2 I ˆ I ˆ . t ) 2 I ˆ I ˆ term, the factor − sin (π J t ) 2 I ˆ is unaffected by the evolution of the 2–3 23
12
For the − sin(π J 12 1 z 2 x coupling: writing this factor as A we have ˆ2 x A I
12
ˆ3 z 2π J 23 t I ˆ2 z I
−−−−−−−−→ A cos (π J
2 y
23
12
2 x 3 z
1 z
ˆ
ˆ ˆ
23 t ) I 2 x + A sin(π J 23 t ) 2 I 2 y I 3 z .
Reinserting the factor A gives ˆ ˆ
− cos(π J
23 t ) sin(π J 12 ) 2 I 1 z I 2 x
ˆ ˆ ˆ
− sin(π J
23 t ) sin(π J 12 t ) 4 I 1 z I 2 y I 3 z .
The overall result of the evolution of I ˆ2 y under coupling is summarized in the table:
term I ˆ2 y
dependence dependence axis description on J 12 on J 23 cos (π J 12 t )
−2 I ˆ I ˆ sin (π J −2 I ˆ I ˆ cos (π J −4 I ˆ I ˆ I ˆ sin (π J
y in-phase
cos (π J 23 t )
1 z 2 x
12 t )
cos (π J 23 t )
2 x 3 z
12 t )
sin (π J 23 t )
1 z 2 y 3 z
12 t )
sin (π J 23 t )
− x − x − y
anti-phase with respect to J 12 anti-phase with respect to J 23 doubly anti-phase with respect to J 12 and J 23
As expected, going anti-phase with respect to the coupling between spins i and j introduces a factor sin(π J i j t ), whereas remaining in-phase with respect to this coupling introduces a factor cos (π J i j t ). The in-phase term is along y , singly anti-phase terms are along − x, and the doubly anti-phase term is along − y i.e. they follow around in the usual sequence x → y → − x → − y. The corresponding tree diagram is I 2y J 12 I 2y J 23 I 2y
-2I 1z I 2x J 23
-2I 2x I 3z -2I 1z I 2x -4I 1z I 2y I 3z
Chapter 10: Advanced topics in two-dimensional NMR
75
10.3
The trick to getting the signs right is just to think about the usual way in which y evolves into − x and then into − y: 2I 2y I 3z
J 23 2I 2y I 3z
-I 2x
J 12 2I 2y I 3z
J 12
-4I 1z I 2x I 3z -I 2x
-2I 1z I 2y
The term 4 I ˆ1 z I ˆ2 x I ˆ3 z arises from splitting first to the left, giving the coefficient cos (π J 23 t ), and second to the right, giving the coefficient sin (π J 12 t ). Note also that there is a minus sign introduced. So the overall factor multiplying 4 I ˆ1 z I ˆ2 x I ˆ3 z is − cos(π J 23 t ) sin(π J 12 t ).
10.4
Term [1] is ˆ1 x . cos (π J 13 t 1 )cos(π J 12 t 1 )sin(Ω1 t 1 ) I
First, let us consider the modulation in t 1 . We use the identity sin A cos B
≡
1 [sin ( A + B) + 2
sin ( A B)]
−
to combine the terms cos (π J 12 t 1 )sin(Ω1 t 1 ) to give 1 cos (π J 13 t 1 )[sin(Ω1 t 1 + 2
π J 12 t 1 ) + sin(Ω1 t 1
12 t 1 )] .
− π J
Next we multiply out the square brace: 1 cos (π J 13 t 1 )sin(Ω1 t 1 + 2
π J 12 t 1 ) + 12 cos (π J 13 t 1 )sin(Ω1 t 1
12 t 1 ).
− π J
(10.1)
Now we combine the two terms cos (π J 13 t 1 )sin(Ω1 t 1 + π J 12 t 1 ) to give 1 sin (Ω1 t 1 + 2
π J 12 t 1
+
π J 13 t 1 ) + 12 sin (Ω1 t 1 + π J 12 t 1
13 t 1 ).
− π J
Doing the same for the two terms cos (π J 13 t 1 )sin(Ω1 t 1 − π J 12 t 1 ) gives 1 sin (Ω1 t 1 2
12 t 1 +
− π J
π J 13 t 1 ) + 12 sin (Ω1 t 1
13 t 1 ).
− π J t − π J 12 1
So overall Eq. 10.1 on page 75 expands to four terms 1 4
sin (Ω1 t 1 + π J 12 t 1 + π J 13 t 1 ) + sin(Ω1 t 1 + π J 12 t 1
+ sin (Ω1 t 1
12 t 1 + π J 13 t 1 ) + sin(Ω1 t 1
− π J
− π J
13 t 1 )
− π J t − π J
12 1
13 t 1 ) .
Therefore, what we have in ω 1 is a completely in-phase doublet of doublets on spin one. In ω2 the operator is I ˆ1 x , which also gives rise to an in-phase doublet of doublets on spin one. ‘Multiplying’ these two multiplets together in the manner of Fig. 10.6 on page 327 gives rise to a
Chapter 10: Advanced topics in two-dimensional NMR
76
two-dimensional multiplet consisting of sixteen lines, all with the same sign; this is in contrast to the cross-peak multiplet, which consists of four anti-phase square arrays. Note, too, that the magnetization which gives rise to the diagonal peak is along x in t 2 and is sine modulated in t 1 . This is the complete opposite of the cross peak, which is along y in t 2 and cosine modulated. Thus, as in the COSY of the two-spin system, the diagonal and cross peaks are 90 ◦ out of phase with one another in both dimensions. The reason why the splittings due to J 12 and J 13 are in-phase in the ω1 dimension is that the modulation with respect to these couplings takes the form of a cosine: cos (π J i j t 1 ).
Chapter 10: Advanced topics in two-dimensional NMR
10.5
(a)
(b)
(c)
(d)
(e)
(f)
For (f) there are only two anti-phase square arrays.
77
Chapter 10: Advanced topics in two-dimensional NMR
78
(b)
(a)
Ω1
Ω2
(c)
(d)
ω 1 ω 2
In each case, the region plotted is ±10 Hz from the centre of the cross-peak multiplet; for clarity, only one anti-phase square array is shown. The linewidth is 0.5 Hz in each dimension.
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79
10.6
(a)
(b)
(c)
(d)
(e)
(f)
ω 1
ω 2
In each case ± 10 Hz is plotted from the centre of the cross-peak multiplet. Note that in (c), where J 12 = J 23 , the column of peaks down the centre of the cross peak no longer cancel one another out, as four of the peaks are missing from the reduced multiplet. In the series (a) to (c), J 23 is increasing, thus increasing the ω2 separation of the two anti-phase square arrays. In the series (d) to (f), J 13 is decreasing, thus decreasing the ω 1 separation of the two anti-phase square arrays.
10.7
It is not usually possible to measure a value for the active coupling constant since this appears as an anti-phase splitting. If the positive and negative peaks overlap significantly, the separation between the maxima and minima of the anti-phase peaks is no longer equal to the value of the active coupling constant. See section 10.3.3 on page 334 for a description of how, under some circumstances, the values of passive couplings may be determined from reduced multiplets.
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80
10.8 I ˆ1α I ˆ2− I ˆ3α :
observable magnetization corresponding to the line of the spin-two multiplet which is associated with spin one and spin three both being in the α state.
I ˆ1α I ˆ2− I ˆ3− :
double-quantum coherence, with p = −2, between spins two and three. More specifically, this operator is associated with one of the lines of the double-quantum ‘doublet’ – see section 3.7.3 on page 42. This term is not observable.
I ˆ1 β I ˆ2 β I ˆ3 β :
the population of the βββ energy level. This term is not observable.
I ˆ1α I ˆ2 β I ˆ3+ :
single-quantum coherence, with p = +1, corresponding to the line of the spin-three multiplet which is associated with spin one being in the α state and spin two being in the β state. Although it is single quantum, this term is not observable as only coherence order −1 is observable.
As described in section 10.4.2 on page 338, free evolution simply gives a phase factor, with the frequency depending on the offset of the spin in question and on the spin states of the passive spins. If the passive spin is in the α state, a term −π J is contributed to the frequency, whereas if it is in the β state, a term +π J is contributed. The overall sense of the phase factor depends on whether the operator is I ˆ+ or I ˆ− . I ˆ1α I ˆ2− I ˆ3α
ˆ ˆ − I ˆ3α
−→ exp (i[Ω − π J − π J 2
I ˆ1α I ˆ2 β I ˆ3+
23 ]t 1 ) I 1α I 2
12
−→ exp(−i[Ω − π J
13 +
3
ˆ1α I ˆ2 β I ˆ3+ π J 23 ]t 1 ) I
10.9
During t 1 the term I ˆ1+ I ˆ2 β I ˆ3α acquires a phase factor: exp ( i[Ω1 + π J 12
−
ˆ
ˆ ˆ
13 ]t 1 ) I 1+ I 2 β I 3α .
− π J
The small flip angle pulse causes the following transfers to observable operators on spin two (the coefficients come from Eq. 10.7 on page 340) I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α
−→ + −→ − −→ + −→ −
1 iθ 2
+ 2 iθ
ˆ1α I ˆ2− I ˆ3α (1) I
1 iθ 2
+ 2 iθ
1
ˆ1 β I ˆ2− I ˆ3α (1) I
1 iθ 2
+ 2 iθ
1 2 θ 4
ˆ1α I ˆ2− I ˆ3 β I
1 iθ 2
+
1 iθ 2
1 2 θ 4
ˆ1 β I ˆ2− I ˆ3 β . I
1
1
For a small flip angle, we discard the third and fourth terms as these go as θ 4 . This leaves I ˆ1+ I ˆ2 β I ˆ3α
−→ −
1 2 ˆ ˆ θ I 1α I 2 4
− I ˆ3α
I ˆ1+ I ˆ2 β I ˆ3α
−→ +
1 2ˆ ˆ θ I 1 β I 2 4
− I ˆ3α .
(10.2)
These two transfers can be found in the table on p. 342. Next we consider the behaviour of the term I ˆ1− I ˆ2 β I ˆ3α . For this term the sense of the phase modulation is opposite to that of I ˆ1+ I ˆ2 β I ˆ3α : exp (+i[Ω1 + π J 12
ˆ − I ˆ2 β I ˆ3α .
13 ]t 1 ) I 1
− π J
Chapter 10: Advanced topics in two-dimensional NMR
81
For this term, the transfer I ˆ1− −→ I ˆ1α has associated with it a factor of ( − 12 iθ ), which is the opposite sign to that for the transfer I ˆ1+ −→ I ˆ1α . So, the cross-peak components arising from I ˆ1+ I ˆ2 β I ˆ3α and I ˆ1− I ˆ2 β I ˆ3α have opposite signs.
10.10
Starting with I ˆ1+ I ˆ2 β I ˆ3α the first small flip angle pulse creates four possible population terms, which are the ones of interest in ZCOSY, in which spin one is in the α state: I ˆ1+ I ˆ2 β I ˆ3α
−→ −→ −→ −→
I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α
+
1 iθ 2 1
+ 2 iθ 1
+ 2 iθ 1
+ 2 iθ
ˆ1α I ˆ2 β I ˆ3α (1) (1) I ˆ1α I ˆ2α I ˆ3α (1) I
1 2 θ 4
(1)
1 2 θ 4
1 2 θ 4
ˆ1α I ˆ2 β I ˆ3 β I
1 2 θ 4
ˆ1α I ˆ2α I ˆ3 β . I
Of these four terms, only the first will be significant for the case of a small flip angle. There are four additional transfers from I ˆ1+ I ˆ2 β I ˆ3α to operators in which spin one is in the β state, but as before only one of these is significant in the small flip angle case: I ˆ1+ I ˆ2 β I ˆ3α
ˆ1 β I ˆ2 β I ˆ3α (1) (1) I
and
−
−→ −
1 iθ 2
So, we have just two population terms at this stage:
+
1 iθ 2
ˆ1α I ˆ2 β I ˆ3α I
1 iθ 2
ˆ1 β I ˆ2 β I ˆ3α . I
From all that we have done so far we can see that, for small flip angles, the significant contributions that these terms will make to the 1–2 cross peak arise from the transfer I ˆ2 β −→ I ˆ2− , with both of the other operators remaining the same:
−
ˆ1α I ˆ2 β I ˆ3α + 2 iθ I 1
1 iθ 2
−→ −→
ˆ1 β I ˆ2 β I ˆ3α I
− 1
+ 2 iθ 1
+ 2 iθ
ˆ1α I ˆ2 + 2 iθ I 1
1 iθ 2
− I ˆ3α
ˆ1 β I ˆ2− I ˆ3α . I
So the overall transfers from I ˆ1+ I ˆ2 β I ˆ3α caused by the two small flip angles pulses are I ˆ1+ I ˆ2 β I ˆ3α I ˆ1+ I ˆ2 β I ˆ3α
−→ − −→ +
1 2ˆ ˆ θ I 1α I 2 4 1 2ˆ ˆ θ I 1 β I 2 4
− I ˆ3α − I ˆ3α .
These are exactly the same as found for small flip angle COSY in the previous exercise (see Eq. 10.2 on page 80 on the previous page).
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82
10.11
From section 10.8.1 on page 354, we found that at the end of the constant time T the the following operators are present: ˆ cos cos (Ω1 t 1 )cos(π J 12 12 T ) I 1 y
− cos cos (Ω t )sin(π J 1 1
12 12
− sin(Ω t )cos(π J T ) I ˆ T ) 2 I ˆ I ˆ − sin(Ω t )sin(π J T ) 2 I ˆ I ˆ 1 x 2 z
1 1
12 12
1 1
12 12
1 x
1 y 2 z .
The third of these is rotated by the second 90◦ pulse to a mixture of double- and zero-quantum coherence:
− cos cos (Ω t )sin(π J 1 1
ˆ ˆ
12 12 T ) 2 I 1 x I 2 z
(π/2)( I ˆ1 x + I ˆ2 x )
−−−−−−−−−−→ + cos cos (Ω t )sin(π J 1 1
ˆ ˆ
12 12 T ) 2 I 1 x I 2 y .
Following section 7.12.1 on page 174, the pure double quantum part of 2 I ˆ1 x I ˆ2 y is so the double quantum term between the final two pulses is 1 ˆ ˆ cos cos (Ω1 t 1 )sin(π J 12 12 T ) (2 I 1 x I 2 y + 2
,
1 (2 I ˆ1 x I ˆ2 y + 2 I ˆ1 y I ˆ2 x ) 2
2 I ˆ1 y I ˆ2 x ).
The final 90◦ pulse makes this observable: 1 ˆ ˆ cos cos (Ω1 t 1 )sin(π J 12 12 T ) (2 I 1 x I 2 y + 2
2 I ˆ1 y I ˆ2 x )
(π/2)( I ˆ1 x + I ˆ2 x )
−−−−−−−−−−→
1 ˆ ˆ cos cos (Ω1 t 1 )sin(π J 12 12 T ) (2 I 1 x I 2 z + 2
2 I ˆ1 z I ˆ2 x ).
ˆ ˆ The term cos cos (Ω1 t 1 )sin(π J 12 12 T ) 2 I 1 x I 2 z gives rise to a diagonal peak centred at {Ω1 , Ω1 }, as it is modulated modulated in t 1 at Ω1 and appears on spin one in t 2 . There is a single modulatin modulatingg frequency frequency of Ω1 in ω 1 i.e. no splitting due to couplings, as expected. In ω 2 the multiplet is in anti-phase. ˆ ˆ cos (Ω1 t 1 )sin(π J 12 The term cos 12 T ) 2 I 1 z I 2 x gives rise to a cross peak centred at {Ω1 , Ω2 }, as it is modulated in t 1 at Ω1 and appears on spin two in t 2 . Like Like the diago diagona nall peak peak,, it is in anti-p anti-pha hase se in ω2 . Furthermore, note that the terms which give rise to both the diagonal and cross peak appear along x, so they will have the same lineshape in ω2 : this contrasts with the simple constant time COSY experiment. both the diagonal and cross peaks goes as sin sin (π J 12 The intensity of both 12 T ): again, this contrasts with the simple constant time COSY, where the two kinds of peaks have a different dependence on T . The advant advantage age of double double-qu -quant antum um filtrati filtration on is that it results results in both diagon diagonal al and cross cross peaks peaks having having the same lineshape in ω 2 , as well as in ω 1 .
Chapter 10: Advanced topics in two-dimensional NMR
83
10.12
Following the same kind of analysis as in section 10.8.1 on page 354, we first let − I ˆ1 y evolve under the coupling for time T : a ‘tree’ is perhaps useful here:
-I 1y J 12 2I 1x I 2z
-I 1y J 13
-I 1y
J 13 2I 1x I 3z
2I 1x I 2z
4I 1y I 2z I 3z
Using this, we can simply read off the four terms which arise as a result of the evolution of the coupling ˆ 13 13 T )cos(π J 12 12 T ) I 1 y +
ˆ ˆ sin(π J 13 13 T )cos(π J 12 12 T ) 2 I 1 x I 3 z ˆ ˆ ˆ ˆ ˆ + cos cos (π J 13 13 T )sin(π J 12 12 T ) 2 I 1 x I 2 z + sin(π J 13 13 T )sin(π J 12 12 T ) 4 I 1 y I 2 z I 3 z .
− cos cos (π J
The 180◦ pulse in the constant time period simply flips the sign of any y or z operators: ˆ cos (π J 13 + cos 13 T )cos(π J 12 12 T ) I 1 y
− cos cos (π J
13 13 T )sin(π J 12 12
− sin(π J T )cos(π J T ) 2 I ˆ I ˆ T ) 2 I ˆ I ˆ − sin(π J T )sin(π J T ) 4 I ˆ I ˆ I ˆ 13 13
1 x 2 z
1 x 3 z
12 12
13 13
12 12
1 y 2 z 3 z .
Now we have to let each of these terms evolve under the offset of spin one for time t 1 . The result will be all of the above terms, multiplied by cos cos (Ω1 t 1 ):
cos cos (Ω1 t 1 )
− cos(π J
+
ˆ cos cos (π J 13 13 T )cos(π J 12 12 T ) I 1 y ˆ ˆ
13 13 T )sin(π J 12 12 T ) 2 I 1 x I 2 z
ˆ ˆ
− sin sin (π J
13 13 T )cos(π J 12 12 T ) 2 I 1 x I 3 z
13 13 T )sin(π J 12 12 T ) 4 I 1 y I 2 z I 3 z
sin (Ω1 t 1 ): and a related set of terms multiplied by sin
−
sin sin (Ω1 t 1 )
ˆ ˆ ˆ ,
− sin sin (π J
ˆ cos cos (π J 13 13 T )cos(π J 12 12 T ) I 1 x
ˆ ˆ
− sin(π J
ˆ ˆ 13 13 T )sin(π J 12 12 T ) 2 I 1 y I 2 z +
− cos cos (π J
13 13 T )cos(π J 12 12 T ) 2 I 1 y I 3 z
ˆ ˆ ˆ sin(π J 13 13 T )sin(π J 12 12 T ) 4 I 1 x I 2 z I 3 z .
After the final 90 ◦ pulse the first set of terms become
cos cos (Ω1 t 1 )
+
ˆ ˆ ˆ cos cos (π J 13 sin (π J 13 13 T )cos(π J 12 12 T ) I 1 z + sin 13 T )cos(π J 12 12 T ) 2 I 1 x I 3 y
ˆ ˆ + cos(π J 13 13 T )sin(π J 12 12 T ) 2 I 1 x I 2 y
ˆ ˆ ˆ
− sin sin (π J
13 13 T )sin(π J 12 12 T ) 4 I 1 z I 2 y I 3 y
none of which are observable. sin (Ω1 t 1 ), become The second set of terms, those multiplied by sin
−
sin sin (Ω1 t 1 )
ˆ cos cos (π J 13 13 T )cos(π J 12 12 T ) I 1 x
ˆ ˆ + cos cos (π J 13 13 T )sin(π J 12 12 T ) 2 I 1 z I 2 y +
+
,
ˆ ˆ sin(π J 13 13 T )cos(π J 12 12 T ) 2 I 1 z I 3 y
ˆ ˆ ˆ sin(π J 13 13 T )sin(π J 12 12 T ) 4 I 1 x I 2 y I 3 y .
The term in I ˆ1 x is the diagonal peak: in ω2 it will appear as the in-phase doublet of doublets of spin one, and as the t 1 modulation is simply sin(Ω1 t 1 ), there will be a single frequency in ω 1 i.e. no splitting due to couplings, as expected.
Chapter 10: Advanced topics in two-dimensional NMR
84
The term in 2 I ˆ1 z I ˆ2 y is the 1–2 cross peak: in ω 2 it will appear as the doublet of doublets of spin two, anti-phase with respect to the 1–2 coupling, but in-phase with respect to the 2–3 coupling. In ω1 there is a single modulating frequency of Ω1 , just as for the diagonal peak. The cross- and diagonal-peak terms have the same modulation in t 1 , and so will have the same lineshape is this dimension. However, in t 2 the magnetization which gives rise to the diagonal peak appears along x, whereas that which gives rise to the cross peak appears along y. So, So, as for the the two-spin case, in ω 2 the cross and diagonal peaks will have different lineshapes. The intensities of the two type of peaks show a different dependence on the couplings: spin-one diagonal peak :
cos(π J 13 13 T )cos(π J 12 12 T )
1–2 cross peak :
cos ( π J 13 13 T )sin(π J 12 12 T ).
As for the two-spin case, the cross-peak goes as sin(π J active active T ), whereas the diagonal peak goes as cos cos (π J active dependencee active T ): here the active coupling is J 12 12 . The two kinds of peaks have a common dependenc on the passive coupling J 13 cos (π J 13 13 , going as the cosine: cos 13 T ). In words, to give rise to the 1–2 cross peak, the magnetization needs to be anti-phase with respect to the 1–2 coupling, and in-phase with respect to the 1–3 coupling, hence the sine dependence on J 12 12 and the cosine dependence on J 13 13 . For the cross peak to have a maximum intensity π J 12 12 T must be an odd multiple of π/2, whereas π J 13 π/ 2. It might be difficult to satisfy this requirement exactly. must be an even multiple of π/ 13 T must This analysis reveals the main problem with constant time experiments, which is the complex dependence of the cross-peak intensity on the couplings in the system, and the value of the fixed delay T .
10.13
Following section 8.7 on page 209, we found for a two-spin system the following S spin operator after the first S spin 90◦ pulse: ˆ ˆ
− sin sin (2π J
IS τ1 ) 2 I z S y .
We need to adapt this for the more complex spin system we are dealing with here. Firstly, the S spin has to become the spin S 1 , and the coupling becomes that between I and and S 1 , J IS IS : 1
ˆ ˆ
− sin sin (2π J
IS 1 τ1 ) 2 I z S 1 y .
τ 1 = 1 /(4 J IS ), then the sine term goes to 1 and so we just have −2 I ˆ z ˆ If τ S 1 y at the start of t t 1 . Just as before, we now allow the homonuclear coupling, which in this case is between S 1 and S 2 , to evolve for the whole time T , giving 1
ˆ ˆ 12 12 T ) 2 I z S 1 y +
− cos cos (π J
ˆ ˆ ˆ sin(π J 12 12 T ) 4 I z S 1 x S 2 z ,
where J 12 12 is the coupling between the two S spins. Note the generation of anti-phase magnetization with respect respect to this coupling. coupling. We also need to take account account of the S spin 180◦ pulse which inverts the operators S ˆ1 y and S ˆ1 z to give ˆ ˆ cos (π J 12 + cos 12 T ) 2 I z S 1 y
− sin(π J
ˆ ˆ ˆ
12 12 T ) 4 I z S 1 x S 2 z .
S 1 has an effect, giving We now allow the S spin offset terms to act for time t 1 ; only the offset of S
ˆ ˆ ˆ ˆ ˆ cos cos (ΩS 1 t 1 )cos(π J 12 cos cos (ΩS 1 t 1 )sin(π J 12 12 T ) 2 I z S 1 y 12 T ) 4 I z S 1 x S 2 z ˆ ˆ ˆ ˆ ˆ sin sin (ΩS 1 t 1 )cos(π J 12 sin(ΩS 1 t 1 )sin(π J 12 12 T ) 2 I z S 1 x 12 T ) 4 I z S 1 y S 2 z .
− −
−
Finally, we need to take account of the I spin spin 180◦ pulse, which inverts all of the terms, as they all contain I ˆ z : ˆ ˆ 12 T ) 2 I z S 1 y + S 1 t 1 )cos(π J 12
ˆ ˆ ˆ cos(ΩS 1 t 1 )sin(π J 12 12 T ) 4 I z S 1 x S 2 z ˆ ˆ ˆ ˆ ˆ + sin(ΩS 1 t 1 )cos(π J 12 12 T ) 2 I z S 1 x + sin(ΩS 1 t 1 )sin(π J 12 12 T ) 4 I z S 1 y S 2 z .
− cos cos (Ω
Chapter 10: Advanced topics in two-dimensional NMR
85
Note that we do not need to worry about the evolution of the heteronuclear coupling as this is refocused by the 180◦ pulses in periods A and B. Next comes the 90◦ pulses to the I and and S spins: these have the following effect on the operators (the trigonometric terms have been left out): 2 I ˆ z ˆ S 1 y
−→ −→ −2 I ˆ ˆS
y 1 z
4 I ˆ z ˆ S 1 x ˆ S 2 z
−→ −→ 4 I ˆ ˆS
ˆ
y 1 x S 2 y
2 I ˆ z ˆ S 1 x
−→ −→ −2 I ˆ ˆS
y 1 x
4 I ˆ z ˆ S 1 y ˆ S 2 z
−→ −→ 4 I ˆ ˆS
ˆ
y 1 z S 2 y .
Of these terms, only the first becomes becomes observable observable on the I spin. spin. We can see that the feature of this term is that it has remained in-phase with respect to the S 1 –S 2 coupling, and is cosine modulated in t 1 . After these two 90 ◦ pulses the observable term on the I spin spin is ˆ ˆ cos(ΩS 1 t 1 )cos(π J 12 12 T ) 2 I y S 1 z .
After the following spin echo, assuming τ 1 = 1 /(4 J IS ), this term simply becomes in-phase along − x: 1
− cos cos (Ω
ˆ
S 1 t 1 )cos(π J 12 12 T ) I x .
We then observe this term with broadband S spin decoupling, giving a single peak at {ΩS , Ω I }. As a result of using the constant time procedure, there is no splitting in ω 1 due to the coupling between the S spins. cos (π J 12 The intensity of the peak depends on cos 12 T ); for a maximum, π J 12 12 T must be a multiple of π, i.e. π J 12 n = 1 , 2, . . .. This condition corresponds to the magnetization being 12 T = n π or T = n / J 12 12 in-phase with respect to the coupling between the S spins at the end of the constant time T . In the case that the S spins are 13 C in a globally labelled sample, the couplings we need to worry about are the one-bond 13 C–13 C couplings, simply because these are the largest. Such couplings do not vary very much with structure, so it should be possible to find a value of T which is a reasonable T which compromise for all the carbons in the system. If there are further S spins coupled to S 1 , then we can see that the intensity of the cross peak will go cos (π J 12 as cos 12 T )cos(π J 13 13 T ) . . .. Again, if the couplings do not cover too wide a range, we can find a T which value of T which will give reasonable intensity for all cross peaks. 1
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86
10.14
We just use the idea that the selective 180 ◦ pulse ‘drags’ the curly line (the coherence) from the energy level shared by the pulse and the coherence, to the energy level at the ‘other end’ of the 180 ◦ pulse. 4 ββ
4 ββ 3 βα
2 αβ
4 ββ 3 βα
2 αβ 1 αα
3 βα
2 αβ 1 αα
180˚ to 1 3
1 αα
180˚ to 1 2
-
-
4 ββ
4 ββ 3 βα
2 αβ
4 ββ 3 βα
2 αβ 1 αα
3 βα
2 αβ 1 αα
180˚ to 1 3
1 αα
180˚ to 3 4
-
result
-
The same idea is used below. Note that the selective 180 ◦ pulse and the coherence must share an energy level for anything to happen. (a) 4 ββ
4 ββ 3 βα
2 αβ
4 ββ 3 βα
2 αβ 1 αα
3 βα
2 αβ 1 αα
180˚ to 1-2
1 αα
180˚ to 2-4
(b) 4 ββ
4 ββ 3 βα
2 αβ
3 βα
2 αβ 1 αα 180˚ to 3-4
4 ββ 3 βα
2 αβ 1 αα 180˚ to 2-4
1 αα result
Transfer (a) can also be achieved by pulses to 3–4 and then 1–3; similarly, transfer (b) can also be achieved by pulses to 1–2 and 1–3.
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87
10.15
After the 90◦ (y) pulse to the I spin and the first 90◦ pulse to the S spin, and assuming that τ = 1 /(4 J IS ), we have already worked out that the state of the system is − 2 I ˆ z ˆ S y (see section 8.7 on page 209). It is then just a question of following the evolution of this term under the influence of the S spin offset and the I–S coupling. The final stage is to use the trigonometric identities (given in the appendix). For example, the term ˆ x is multiplied by the trigonometric term cos (ΩS t 1 )sin(π J IS t 1 ). Applying the identity S cos A sin B
≡
1 [sin ( A + B) 2
− sin ( A − B)]
gives cos(ΩS t 1 )sin(π J IS t 1 )
≡
1 [sin (ΩS t 1 + 2
π J IS t 1 )
− sin(Ω t − π J S 1
IS t 1 )] .
This is indeed 12 ( s+ − s − ), as stated. We now follow through the fate of the term S ˆ x for the rest of the sequence (Fig. 10.33 on page 363). The I spin 90◦ pulse at the start of period A has no effect, and there then follows a spin echo of total duration 1/(2 J IS ) during which the in-phase term is completely transferred to anti-phase, giving 2 I ˆ z ˆS y . We need to take account of the two 180 ◦ pulses which invert both I ˆ z and S ˆ y , leaving the term overall unaffected. The 90 ◦ (y) pulse to the I spin transforms this term to 2 I ˆ x ˆS y ; this brings us to the end of period A . The 90◦ pulse to the S spin which starts period B rotates the operator to 2 I ˆ x ˆS z , and this anti-phase term evolves completely into in-phase during the subsequent spin echo, giving I ˆ y . We need to take account of the two 180◦ pulses in the spin echo, which invert this term to give − I ˆ y . This term is unaffected by the final 90◦ pulse to the S spin, so the observable term arising from S ˆ x is
−
− s −) I ˆ .
1 (s 2 +
y
This term can be found on the first line of the table on p. 364.
10.16
The combinations S 3 and S 4 are: S 3
1
= 2
(a) (d) +
S 4
where (a) and (d) are given in the table on p. 364:
expt (a) (b) (c) (d)
I ˆ x
φ I
φS
y
y
( c+
− y
y
+
y
− y − y
− y
1
= 2
(a) − (d)
,
observable operator at t 2 = 0 2 I ˆ x ˆS z 2 I ˆ y ˆS z I ˆ y
− − c−) (− s + s− ) (c − c−) (s + s−) (−c − c− ) ( s − s − ) (−c + c− ) ( s + s − ) (−c − c− ) (− s + s − ) (−c + c− ) (− s − s − ) (−c − c− ) ( s − s − ) (c − c− ) (− s − s − )
Forming these combinations we have
+ +
+
+
+
+
+
+ +
+
+ +
+ +
Chapter 10: Advanced topics in two-dimensional NMR
S 3
= =
88
(a) (d) − − − − − (a) − (d) − − − 1 2
+
ˆ c ) I
( c+
x
+ (c+
c ) 2 I ˆ ˆ S
x z
E
S 4
= =
F
1 2
ˆ y + ( s+ ( s+ + s ) I
+ s
) 2 I ˆ y ˆ S z .
G
H
As before, we have a clean separation of x- and y-magnetization. If the two combinations are processed separately, and a 90◦ phase correction applied to one combination in both dimensions, we will have two spectra in which all peaks are in the absorption mode. Term E is in-phase in ω2 and also in-phase in ω1 , so all four peaks of the multiplet have the same sign, which is this case is negative. The multiplet is the same as from term A given in Eq. 10.12 on page 365. Term F is anti-phase in each dimension, so gives rise to an anti-phase square array. Note, however, that the overall sign is opposite to that of term B given in Eq. 10.12 on page 365. Term G is in-phase in ω2 and anti-phase in ω1 , and is again opposite in overall sign to term C in Eq. 10.13 on page 365. Finally, term H is anti-phase in ω2 , but in-phase in ω 1 : it is identical to term D in Eq. 10.13 on page 365. The multiplets from the four terms, along with the way they combine to give S 3 and S 4 , are shown in the diagram below, which should be compared to Fig. 10.34 on page 365. E
F
+
G
S 3 + S 4
S 4
S 3
=
H
+
S 3
- S 4
=
We see from the digram that by combining the spectra S 3 and S 4 , either as ( S 3 + S 4 ) or ( S 3 − S 4 ), we are left with just one line of the multiplet, either top left, or bottom right.
10.17
Aside from the extra complication of the pulse sequence and data processing, probably the only significant difficulty is that the peak does not appear at {ΩS , Ω I }, but offset from this by 12 J IS in each dimension. Account needs to be take of this when comparing TROSY type spectra with other spectra.
11 Coherence selection: phase cycling and field gradient pulses
11.1
φ I ˆiz
I ˆi−
≡ I ˆ − i I ˆ −−→ cos φ I ˆ + sin φ I ˆ − i cos φ I ˆ − sin φ I ˆ ˆ − i I ˆ + i sin φ I ˆ − i I ˆ = cos φ I ˆ − i I ˆ = (cos φ + i sin φ) I ix
iy
ix
iy
ix
iy
ˆi = exp (iφ) I
−.
iy
ix
ix
ix
iy
iy
Assigning coherence orders I ˆ1+ I ˆ2− :
2 I ˆ1+ I ˆ2+ I ˆ3 z :
p1 I ˆ1 x I ˆ2 y
2 I ˆ1 z I ˆ2 y
≡ 2 ×
2 I ˆ1 x I ˆ2 x + 2 I ˆ1 y I ˆ2 y
≡ ≡ ≡
p1
= 1
p2
= 1
p2
= 1
≡
1 2
≡
1 2i
1 ˆ I 2i 1 z
−1
=
1 2
p p
=
I ˆ2+
p
=
− I ˆ
2
:
− I ˆ − : 2
p1
2 12 12 I ˆ1+ + I ˆ1− I ˆ2+ + I ˆ2−
I ˆ1+ I ˆ2+ + I ˆ1+ I ˆ2− + I ˆ1− I ˆ2+
+
= 0
p2
−1
=
or
ˆ1 + I
p1
=
=
2
± 1 ± 1 p2
=
±1
− I ˆ −
− I ˆ2−
−1
=
0
=
p1 + p 2 + p 3
2 12i 2i1 I ˆ1+
I ˆ1+ I ˆ2− + I ˆ1− I ˆ2+
hence p1 = 1
p1 + p 2
I ˆ1+ + I ˆ1− :
I ˆ2+
=
= 0
p3
−
p
1
− I ˆ
=
p
I ˆ2+
± 1
− I ˆ − 2
ˆ ˆ ˆ ˆ ˆ 1+ I 2+ + I 1+ I 2− + I 1− I 2+
p2
= 1
p =
0
− I ˆ − I ˆ − 1
2
Chapter 11: Coherence selection: phase cycling and field gradient pulses
90
Heteronuclear spin system I ˆ x
1 2
≡
ˆ y S
≡
2 I ˆ x ˆ S z
≡ 2 ×
2 I ˆ x ˆ S y
≡ 2
1 1 2 2i
1 2
I ˆ+ + I ˆ− :
ˆ+ S
1 2i
p I =
− S ˆ − :
I ˆ+ + I ˆ−
p I =
ˆ+ S
=
pS
ˆ z : I ˆ+ + I ˆ− S
− S ˆ − :
± 1 ± 1
± 1
p I =
=
0
pS
=
pS
± 1
± 1
Following section 11.1.2 on page 384, free evolution results in these operators acquiring a phase exp
−
i
( p1 + p2 +...)
Ω
t ,
where Ω( p + p +...) = p1 Ω1 + p 2 Ω2 + . . .. The table gives this phase term for each operator: 1
2
operator
p1 or p I
I ˆ1+
+1
p2 or pS
+1
I ˆ+ ˆ S −
+1
I ˆ1− I ˆ2− I ˆ3−
−1
−
exp (i
−Ω
−1
2
+1
Ω1 + Ω2
−1 −1
Ω I
− Ω
−1
phase term exp ( i
Ω1
I ˆ2− I ˆ1+ I ˆ2+
( p1 + p2 +...)
Ω
p3
S
Ω1 + Ω2 + Ω3
Ω1 t )
Ω2 t )
exp ( i [Ω1 +
− Ω ]t ) exp (−i [Ω − Ω ]t ) 2
I
S
exp (i [Ω1 +
Ω2 + Ω3 ]t ).
11.2
(a) TQF COSY
(b) zero-quantum spectroscopy
(c) ZCOSY θ θ
t 2
t 1
τ
+3 +2 +1 0 -1 -2 -3
τ
t 1
t 2
t 1 +2 +1 0 -1 -2
+2 +1 0 -1 -2
y τ
(d) HSQC
τ
τ
I
S
+1
p I 0
-1
+1 p S 0 -1
t 1
τ
t 2
t 2
Chapter 11: Coherence selection: phase cycling and field gradient pulses
91
Note that in HSQC, sequence (d), we have pS = ±1 and p I = 0 during t 1 i.e. S spin single-quantum coherence, and that during t 2 we have p I = −1 and pS = 0, as these are the coherence orders for observable signals on the I spin. HMQC τ
τ
t 2
I
t 1
S
p I
+1 0 -1
p S
+1 0 -1
(a) As described in section 11.3 on page 389, the P-type spectrum has the same sign of p in t 1 and t 2 : this is the solid line in the CTP. The resulting spectrum will be phase modulated in t 1 , and so is frequency discriminated. (b) The N-type spectrum has the opposite sign of p in t 1 and t 2 : this is the dashed line in the CTP; like the P-type spectrum, the N-type spectrum is frequency discriminated. (c) To be able to give absorption mode lineshapes we need to retain symmetrical pathways in t 1 i.e. pS = ±1. Thus we need to select both the solid and dashed CTP. The resulting spectrum is not frequency discriminated, but discrimination can be achieved using the SHR or TPPI methods (section 8.12 on page 226).
11.3
By inspecting Fig. 11.5 on page 392 we can determine the form of the signal from detectors A and B using simple trigonometry. For example in (b) it is clear that the component along A is − sin (Ωt ) whereas that along B is cos (Ωt ). The table gives these components and the required combinations for all four cases: A
B
(a)
cos (Ωt )
(b)
− sin(Ωt ) cos (Ωt ) − cos(Ωt ) − sin(Ωt ) sin(Ωt ) − cos(Ωt )
(c) (d)
sin (Ωt )
combination result A + i B
cos(Ωt ) + i sin (Ωt )
− i A − A − i B − B + i A
cos(Ωt )
B
= exp (i Ωt )
− i [− sin(Ωt )] = exp (i Ωt ) −[− cos (Ωt )] − i [− sin (Ωt )] = exp (i Ωt ) −[− cos (Ωt )] + i sin(Ωt ) = exp (i Ωt )
Each combination gives modulation of the form exp (i Ωt ), which will all give the same lineshape on Fourier transformation. Following the approach of Fig. 11.6 on page 394, for the case where the pulse goes [ x, y , − x, − y] and the receiver phase goes [ −180◦ , −270◦ , 0 ◦ , −90◦ ] we have
Chapter 11: Coherence selection: phase cycling and field gradient pulses
90˚(x )
90˚(y )
y
y
90˚(-x )
rx phase
y
x
-180˚
90˚(-y )
y
x
92
x
-270˚
x
0˚
-90˚
A 90◦ (x) pulse places the magnetization along − y and then precession through an angle Ωt rotates the vector towards + x. Similarly, a 90 ◦ (y) pulse places the magnetization along + x, and then precession rotates the vector towards + y. The receiver phase is measured clockwise from 3 o’clock, and is indicated by the bullet • . We see that in each diagram there is a constant angle between the position of the magnetization and the receiver phase. As a result, each combination of pulse and receiver phase will give the same lineshape, and so all four spectra will add up.
11.4
For ∆ p = −1 the phase shift experienced by the pathway when the pulse is shifted in phase by ∆ φ is −∆ p × ∆φ = −(−1)∆φ = ∆φ. Similarly for ∆ p = 0 the phase shift is −0 × ∆φ = 0 , and for ∆ p = 5 the phase shift is −5 × ∆φ = −5∆φ. The table gives the phase shifts for each of these three pathways: pulse phase
∆ p =
−1
∆ p = 0
step
∆φ
∆φ
0
1
0◦ 90◦
0◦ 90◦
0◦ 0◦
180◦ 270◦
180◦ 270◦
0◦ 0◦
2 3 4
∆ p = 5
− 5 ∆φ 0◦
equiv(−5 ∆φ) 0◦
−450◦ −900◦ −1350◦
These phases can be represented in the manner of Fig. 11.8 on page 397:
270◦ 180◦ 90◦
Chapter 11: Coherence selection: phase cycling and field gradient pulses
93
signal phases for ∆p = -1, receiver phases to select ∆p = -3 (1)
(2)
(3)
(4)
signal phases for ∆p = 0, receiver phases to select ∆p = -3 (1)
(2)
(3)
(4)
For ∆ p = −1, steps (1) and (3) have the signal and receiver in alignment, whereas in steps (2) and (4) the signal and the receiver are opposed. As a result steps (1) and (3) will cancel steps (2) and (4). For ∆ p = 0, steps (1) and (3) will cancel as the signal and the receiver are aligned in one and opposed in the other. Similarly, steps (2) and (4) will cancel as in step (2) the signal is 90 ◦ ahead of the receiver, whereas in step (4) it is 90 ◦ behind i.e. there is an overall shift of 180 ◦ . For ∆ p = 5 the signal phase shifts are exactly the same as those for ∆ p = −3, so both pathways are selected. This is of course exactly what is expected for a four-step cycle since −3 + 2 × 4 = +5 i.e. ∆ p = −3 and ∆ p = 5 are separated by a multiple of four.
11.5
The second pulse has ∆ p = −2, so if the pulse phase goes [0◦ , 90◦ , 180◦ , 270◦ ] the receiver phase shifts must be [0◦ , 180◦ , 0◦ , 180◦ ]. The first pulse has ∆ p = +1, so if the pulse phase goes [0◦ , 90 ◦ , 180◦ , 270◦ ] the receiver phase shifts must be [0 ◦ , 270◦ , 180◦ , 90 ◦ ]. In the first four steps, ∆φ2 therefore goes [0◦ , 90◦ , 180◦ , 270◦ ], ∆φ1 remains fixed, and the receiver goes [0◦ , 180◦ , 0 ◦ , 180◦ ]. In the second group of four steps, ∆φ2 does the same, but ∆φ1 is now 90◦ , and this results in an extra 270◦ which must be added to the receiver phase shifts from the first group of four. The required receiver phase shifts are therefore [0◦ +270◦ , 180◦ +270◦ , 0 ◦ +270◦ , 180◦ +270◦ ] ≡ [270◦ , 90 ◦ , 270◦ , 90 ◦ ]. In the third group of four steps ∆φ1 is 180◦ , and this results in an extra 180◦ which must be added to the receiver phase shifts from the first group of four. Finally, for the fourth group of four steps ◦ ◦ ∆φ1 is 270 , and 90 must be added to the receiver phase shifts. The complete sixteen-step cycle is therefore
Chapter 11: Coherence selection: phase cycling and field gradient pulses
step
∆φ1
∆φ2
φrx
step
∆φ1
∆φ3
φrx
1
0◦ 0◦
0◦ 90◦
0◦
9 10
0◦ 90◦
180◦ 0◦
0◦ 0◦
180◦ 270◦
180◦ 0◦
180◦ 180◦
12
180◦ 270◦
180◦ 0◦
90◦ 90◦
0◦ 90◦
180◦ 270◦
180◦ 180◦
14
0◦ 90◦
90◦ 270◦
90◦ 90◦
180◦ 270◦
90◦ 270◦
270◦ 270◦
90◦
16
270◦ 270◦
180◦ 270◦
90◦ 270◦
2 3 4 5 6 7 8
Selection of ∆ p
94
11 13 15
−1 and then ∆ p = +3
=
The first pulse has ∆ p = − 1, so if the pulse phase goes [0 ◦ , 90◦ , 180◦ , 270◦ ] the receiver phase shifts must be [0◦ , 90◦ , 180◦ , 270◦ ]. The second pulse has ∆ p = +3, so if the pulse phase goes [0◦ , 90 ◦ , 180◦ , 270◦ ] the receiver phase shifts must be [0 ◦ , 90◦ , 180◦ , 270◦ ]. For these four-step cycles the receiver phases needed to select ∆ p = −1 and +3 are, of course, the same. The sixteen-step cycle is: step
∆φ1
∆φ2
φrx
step
∆φ1
∆φ3
φrx
1
0◦ 90◦
0◦ 0◦
0◦ 90◦
9
0◦ 90◦
180◦ 180◦
180◦ 270◦
180◦ 270◦
0◦ 0◦
180◦ 270◦
11
180◦ 270◦
180◦ 180◦
0◦ 90◦
0◦ 90◦
90◦ 90◦
90◦ 180◦
13
0◦ 90◦
270◦ 270◦
270◦ 0◦
180◦ 270◦
90◦ 90◦
270◦ 0◦
15
180◦ 270◦
270◦ 270◦
90◦ 180◦
2 3 4 5 6 7 8
10 12 14 16
11.6
For ∆ p = −2 the phase shift experienced by the pathway when the pulse is shifted in phase by ∆ φ is −∆ p × ∆φ = −(−2)∆φ = 2 ∆φ. So, as the pulse goes [0 ◦, 120◦, 240◦] the pathway experiences phase shifts of [0◦ , 240◦ , 480◦ ] which are equivalent to [0 ◦ , 240◦ , 120◦ ]. So, to select ∆ p = −2, we would use the cycle: pulse: [0◦ , 120◦ , 240◦ ] receiver: [0◦ , 240◦ , 120◦ ]. On modern spectrometers, the receiver phase can be shifted by arbitrary amounts, not just multiples of 90◦ . The selectivity of this three-step sequence can be represented in the manner described on p. 398:
−5 (−4) (−3) −2 (−1)
(0)
1
(2)
(3)
4
Chapter 11: Coherence selection: phase cycling and field gradient pulses
95
Here the boldface numbers are the values of ∆ p which are selected, and the numbers in brackets are the values which are rejected; these selected values are separated by three, as we are dealing with a three-step cycle. The CTP for N-type COSY is: t 2
t 1
+1 0 1
-
The second pulse has ∆ p = −2, so we can use the three-step cycle described above to select this. As the first pulse can only generate p = ±1, this three step cycle is sufficient to select the overall pathway we require. To be specific, p = −1 present during t 1 would only lead to observable coherence via the pathway ∆ p = 0 on the second pulse, which is blocked by this three-step cycle. For P-type COSY (Fig. 11.4 (b) on p. 390), ∆ p = 0 on the second pulse. This is selected by the following three-step cycle of the second pulse: pulse: [0◦ , 120◦ , 240◦ ] receiver: [0◦ , 0 ◦ , 0 ◦ ]. Such a cycle would be sufficient to select the wanted pathway as it would reject the ∆ p = −2 pathway on the second pulse.
11.7 φ1
φ2 φ3 t 1
φrx t 2
+3 +2 +1 0 1 2 3
-
Grouping together the first two pulses means that they are required to achieve the transformation ∆ p = ±3. Concentrating for the moment on the pathways with ∆ p = −3, shifting the phase of the first two pulses by ∆ φ will result in a phase shift of −∆ p × ∆φ = −(−3)∆φ = 3 ∆φ. If the pulse goes through the phases [0 ◦ , 60◦ , 120◦ , 180◦ , 240◦ , 300◦ ] then the phase acquired by the pathway with ∆ p = −3 is [0◦ , 180◦ , 360◦ , 540◦ , 720◦ , 900◦ ]. Reducing these to the range 0 ◦ to 360◦ gives [0◦ , 180◦ , 0 ◦ , 180◦ , 0 ◦ , 180◦ ]. So the phase cycle needed is φ1 and φ 2 : [0◦ , 60 ◦ , 120◦ , 180◦ , 240◦ , 300◦ ] receiver: [0◦ , 180◦ , 0 ◦ , 180◦ , 0 ◦ , 180◦ ]. This six-step phase cycle also selects ∆ p = +3. Since p = ±3 has been selected prior to the last pulse, and as the first pulse can only generate p = ±1, no further phase cycling is needed (with the possible exception of axial peak suppression, see section 11.7 on page 403). Other pathways selected by this six-step cycle include ∆ p = +3 + 6 = +9 and ∆ p = − 3 − 6 = −9. These involve such high orders of coherence that we can safely ignore them. The final pulse has ∆ p = −4 and ∆ p = +2; as these are separated by 6, they will both be selected by a six-step cycle. The phase experienced by the pathway with ∆ p = − 4 will be 4 ∆φ so as the pulse goes [0◦ , 60◦ , 120◦ , 180◦ , 240◦ , 300◦ ] then the phase acquired by the pathway will be [0◦ , 240◦ , 480◦ , 720◦ , 960◦ , 1200◦ ]. Reducing these to the range 0 ◦ to 360◦ gives the following cycle:
Chapter 11: Coherence selection: phase cycling and field gradient pulses φ3 :
[0◦ , 60 ◦ , 120◦ , 180◦ , 240◦ , 300◦ ]
96
receiver: [0◦ , 240◦ , 120◦ , 0 ◦ , 240◦ , 120◦ ].
11.8 φ1
φ2 t 1
φ3
φrx t 2
τ
+2 +1 0 1 2
-
The first two pulses achieve the transformation ∆ p = 0 , so a four-step cycle will be: φ1 and φ 2 : [0◦ , 90 ◦ , 180◦ , 270◦ ] receiver: [0◦ , 0 ◦ , 0 ◦ , 0 ◦ ]. Axial peak suppression (section 11.7 on page 403) involves shifting the phase of the first pulse [0◦ , 180◦ ] and similarly for the receiver. Combining these two cycles gives eight steps: step φ1 φ2 φrx
1 0◦ 0◦ 0◦
2
3
4
5
6
90◦ 90◦
180◦ 180◦
270◦ 270◦
180◦ 0◦
270◦ 90◦
0◦
0◦
0◦
180◦
180◦
7 0◦ 180◦ 180◦
8 90◦ 270◦ 180◦
The CTP for N-type NOESY is φ1
φ2 t 1
φ3
φrx t 2
τ
+2 +1 0 1 2
-
We need to select ∆ p = −1 on the last pulse. A suitable four-step cycle is [0 ◦ , 90◦ , 180◦ , 270◦ ] for φ 3 and [0◦ , 90◦ , 180◦ , 270◦ ] for the receiver. We also need to select ∆ p = +1 on the first pulse. The four-step cycle [0 ◦ , 90 ◦ , 180◦ , 270◦ ] for φ1 and [0◦ , 270◦ , 180◦ , 90 ◦ ] for the receiver achieves this selection. The complete sixteen-step cycle is step
∆φ1
∆φ3
φrx
step
∆φ1
∆φ3
φrx
1
0◦ 90◦
0◦ 0◦
0◦
9 10
180◦ 180◦
180◦ 90◦
180◦ 270◦
0◦ 0◦
270◦ 180◦
0◦ 90◦
12
180◦ 180◦
0◦
0◦ 90◦
90◦ 90◦
90◦ 90◦
180◦ 270◦
0◦
14
0◦ 90◦
270◦ 270◦
180◦ 270◦
90◦ 90◦
270◦ 180◦
15
180◦ 270◦
270◦ 270◦
2 3 4 5 6 7 8
11 13
16
270◦ 270◦ 180◦ 90◦ 0◦
Chapter 11: Coherence selection: phase cycling and field gradient pulses
97
It is not necessary to add explicit axial peak suppression to this cycle as we are selecting ∆ p = +1 on the first pulse, and so all of the peaks we see in the spectrum must derive from the first pulse.
11.9
RF G 2
G 1
G
τ
τ
1
2
+2 +1 0
The spatially dependent phase is given by Eq. 11.8 on page 412: φ( z) = p
− × γ G z t .
Hence the phases due to the two gradient pulses are φ1
−(1) × γ G zτ
=
1
1
and
φ2
−(2) × γ G zτ .
=
2
2
The refocusing condition is that the total phase, φ 1 + φ2 , is zero: φ1
−γ G zτ − 2γ G zτ
+ φ2 =
1
1
2
2 = 0 .
The factors of z and γ cancel to give, after some rearrangement:
− 12
G 2 τ2 G 1 τ1
=
(a) If the gradients have the same length, then G2 /G1 = − 12 , i.e. the second gradient needs to be half the strength of the first, and applied in the opposite sense. (b) If the gradients have the same absolute strength, they still have to be applied in the opposite sense i.e. G1 = − G2 . Inserting this gives the refocusing condition as (G2 τ2 )/(−G2 τ1 ) = − 12 , which means that τ2 = 12 τ1 .
11.10
I
S
G +1 I 0 1
p
-
+1 S 0 1
p
-
G 1 τ
1
G 2 τ
2
Chapter 11: Coherence selection: phase cycling and field gradient pulses
98
In the heteronuclear case we use Eq. 11.9 on page 414 to find the spatially dependent phase: φ( z)
During the first gradient p I = 0 and spatially dependent phases are φ1
pS
− ( p γ + p γ ) G z t . = −1, whereas during the second p = −1 and p =
= γ S G1 zτ1
I I
S S
I
and
φ2
= γ I G2 zτ2 .
The refocusing condition is φ 1 + φ2 = 0 , which in this case rearranges to G1 τ1 G2 τ2
− γ γ .
=
I
S
If I is 1 H and S is 1 5 N, then γ I /γ S = 10 /(−1) so the refocusing condition becomes G1 τ1 G2 τ2
= 10 .
If the gradients have the same duration, τ 1 = τ 2 then
G1
.
= 10G 2
Note that ratio of the gyromagnetic ratios of 1 H and 15 N is in fact 9 .86 : −1.
S = 0
. So the
Chapter 11: Coherence selection: phase cycling and field gradient pulses
99
11.11
(a) P-type DQF COSY t 2
t 1 τ
G
1
τ
2
(c) N-type COSY
(b) N-type TQF COSY t 2
t 1 τ
τ
3
G
G 1 G 2 G 3
1
τ
τ1
τ
2
3
G
G 1 G 2 G 3
+3 +2 +1 0 -1 -2 -3
+2 +1 0 -1 -2
τ
G 1 G 2
(e) N-type HSQC y
t 2
t 1
τ2
+1 0 -1
(d) double-quantum spectroscopy τ
t 2
t 1
τ
τ
τ
τ
t 2
I τ
G +2 +1 0 -1 -2
1
τ
1
τ2
τ3
t 1
S
G 1 G 1
G 2
τ1
G 3 G p I
+1 0 -1
p S
+1 0 -1
G 1
τ
2
G 2
τ
3
G 3
(a) P-type DQF COSY We have chosen p = 2 in the interval between the last two pulses, but it would have been just as acceptable to choose p = −2. The pathway will give a P-type spectrum as p = −1 is present during t 1 . The refocusing condition is G1 τ1
− 2 G τ
2 2 + G3 τ3 = 0 .
If the gradients are all the same length, then one choice is for the strengths to be in the ratio G1 : G 2 : G 3
= 1
: 1 : 1.
(b) N-type TQF COSY We have chosen p = 3 in the interval between the last two pulses, but it would have been just as acceptable to choose p = − 3. The pathway will give an N-type spectrum as p = +1 is present during t 1 . The refocusing condition is
−G τ − 3 G τ
2 2 + G3 τ3 = 0 .
1 1
If the gradients are all the same length, then one choice is for the strengths to be in the ratio G1 : G 2 : G 3
= 1
: 1 : 4.
(c) N-type COSY The pathway will give a N-type spectrum as p = +1 is present during t 1 . The refocusing condition is 1 1 + G 2 τ2 = 0 .
−G τ
If the gradients are all the same length, then the strengths must be in the ratio G1 : G 2
= 1
:1
Chapter 11: Coherence selection: phase cycling and field gradient pulses
100
(d) Double-quantum spectroscopy The two gradients G 1 serve to ‘clean up’ the 180 ◦ pulse in the spin echo (see section 11.12.3 on page 418). Double-quantum coherence is dephased by G2 and then rephased by G 3 ; to control phase errors due to the underlying evolution of the offsets, both gradients are placed within spin echoes (see section 11.12.5 on page 419). We will need to record separate P- and N-type spectra, and then recombine them in order to obtain an absorption mode spectrum (see section 11.12.2 on page 417); the N-type pathway is given by the solid line, and the P-type by the dashed line. The refocusing condition for the N-type pathway is 2 G 2 τ2
+ G3 τ3 = 0 .
If the gradients are the same length, then the strengths must be in the ratio = 1
G2 : G 3
:
−2.
The refocusing condition for the P-type pathway is 2 2 + G3 τ3 = 0 .
−2 G τ
If the gradients are the same length, then the strengths must be in the ratio G2 : G 3
= 1
: 2.
(e) N-type HSQC G1 is a ‘purge’ gradient (see section 11.12.6 on page 420). S spin magnetization is dephased by G 2 and rephased after transfer to I by G 3 . The refocusing condition is 3 3 = 0 .
−γ G τ + γ G τ S
I
2 2
If the gradients are both the same length, then the strengths must be in the ratio G2 : G 3
= γ I
: γ S .
For the case where the I spin is 1 H and the S spin is 13 C, γ I : condition is G2 : G 3
= 4
: 1.
γ S
=
4,
and so the refocusing
12 Equivalent spins and spin-system analysis
12.1
In the following ‘Ha ≡ H b ’ is a short hand for saying that the environments, environments, and hence shifts, of H H a and Hb are the same. Likewise Ha –Hb is a shorthand for the coupling between H a and H b . In this notation, H a –Hb ≡ H c –Hd means that the H a –Hb coupling is equivalent to the H c –Hd coupling. (i) All the Fe equivalent due to the four-fold axis along the Br–F a bond.
(a)
(ii) Each Fe is two bonds from each Fa and the angle between the two bonds is 90◦ in each case. Thus all the F e –Fa couplings are the same.
Fa Fe
Fe
Br Fe
(iii) Fe are all magnetically equivalent.
Fe
(iv) AX4 spin system. (i) Mirror planes planes (grey) (grey) show that F 1 ≡ F 2 and H 1 ≡ H2 .
(b) F1 H1
Br
H2
Br F2
(ii)
H1 –F1 is
a three-bond coupling whereas they are not the same.
H2 –F1 is
over four bonds: thus
(iii) Likewise H1 –F2 H 2 –F2 . (iv)
H1 and H 2 chemically equivalent; F 1 and F 2 chemically equivalent.
(v) AA XX . (i) Reflection Reflection in plane of ring means means H 3 ≡ H 4 and H 5 ≡ H 6 . (c)
H4
H3 H 6
H5
(ii) Vertical plane (grey) means H 1 ≡ H2 , H 3 ≡ H 5 and H 4 ≡ H 6 . (iii) Thus H 1 and H2 are equivalent, as are H 3 , H4 , H 5 and H6 . (iv) Plainly H 1 –H4 H 1 –H5 , and H 1 –H3 H 1 –H6 .
H1
H2
(v) Chemical, but not magnetic equivalence. (vi) AA XX X X .
Chapter 12: Equivalent spins and spin-system analysis
(i) Reflection in plane indicated means H 1 ≡ H2 and F 1 ≡ F 2 .
(d) F1
H1
F2
H2
H6
(e)
(ii)
H1 –F1
H4
H2
(f)
(ii) Ignoring Ignoring couplings couplings over more than four bonds means that H 1 , H 2 and H 3 are isolated from H 4 , H 5 and H 6 . (iii)
H3
H6
H1
H4
H2 Cl
.
(ii) Ignoring Ignoring couplings couplings over more than four bonds means that H 1 and H 2 are isolated from H 3 , H4 , H 5 and H6 . (iii)
H1
≡ H so these form an A system. 2
(iv)
H4 –H6
H 4
2
–H3 : not magnetically equivalent. (v) Remaining Remaining spins spins form an AA XX system.
(i) All three rotamers about the C–CH 3 bond are likely to be equally populated since three identical identical groups (in fact protons) are attached to one of the carbons.
H COOH
H H
H 2 H 3
(i) Due to the mirror plane plane H 1 ≡ H 2 , H 4 ≡ H 5 , and H 3 ≡ H 6 .
Cl
H3
H1
(iv) Thus we have two isolated (and identical) AMX systems.
H5
H
–F2 , since one is cis and one is trans.
(i) Rotation Rotation about about C 2 axis out of the plane of the paper (indicated by the grey dot) means that H 1 ≡ H 4 , H 2 ≡ H 5 , and H 3 ≡ H 6 .
Cl H1
Cl
H 1
(iii) Chemical, but not magnetic equivalence. (iv) AA XX .
H5
(g)
102
NH2
(ii) This is analogous to the case of chloroethane described in the text i.e. the couplings between the CH proton and each of the CH 3 protons are the same when averaged over the rotamers. (iii) Thus the three methyl protons are magnetically equivalent. (iv) AX3 .
Chapter 12: Equivalent spins and spin-system analysis (h)
COOH H2 H1
103
COOH
H3
Cl
NH2
H1
COOH
H2
H3
Cl
NH2
H1
NH2
H3 R2
Cl R1
H2
R3
(i) There is no reason to assume that the three rotamers are equally populated. (ii)
H2
H 3
.
(iii) AMX spin spin system (H2 and H 3 are pro-chiral).
12.2
(a) First compute evolution under the offset ˆ x x S
S z ΩS t ˆ
−−−−→
ˆ x x cos (ΩS t ) S
ˆ y . + sin (ΩS t ) S
We are only interested in the subsequent evolution of the first term as this is the one with cos (ΩS t ) modulation. This term evolves under the I 1 –S coupling to give ˆ x x cos (ΩS t ) S
ˆ1 z ˆ 2π Jt I S z
−−−−−−−→
cos (π Jt ) cos (ΩS t ) ˆS x x
+
sin (π Jt ) cos (ΩS t ) 2 I ˆ1 z ˆ S y .
Again it is only the evolution of the first term we need to follow, and under the influence of the I 2 –S coupling this is ˆ x x cos (π Jt ) cos (ΩS t ) S
ˆ2 z ˆ 2π Jt I S z
−−−−−−−→
ˆ x x + sin (π Jt ) cos (π Jt ) cos (ΩS t ) 2 I ˆ2 z ˆ cos (π Jt ) cos (π Jt ) cos (ΩS t ) S S y .
The first term on the right is the one required. (b) Working with the trigonometric factor and applying the various identities we have cos2 (π Jt ) cos (ΩS t )
≡ ≡ ≡ ≡ ≡
1 [1 + cos (2π Jt )]cos (ΩS t ) 2 1 cos (ΩS t ) + 12 [cos (2π Jt ) cos (ΩS t )] 2 1 cos (ΩS t ) + 14 [cos (2π Jt + ΩS t ) + cos (2π Jt 2 1 cos (ΩS t ) + 14 cos (ΩS t + 2π Jt ) + 14 cos (ΩS t 2 1 [2 [ 2 cos cos (ΩS t ) + cos (ΩS t + 2π Jt ) + cos (ΩS t 4
− Ω t )] − 2π Jt ) − 2π Jt )] . S
(c) Assuming that x-magnetization is detected, the above expression gives the form of the freeinduction signal. Cosine Fourier transformation of this cosine modulated signal gives a peak corresponding to each of these terms. There will be a peak at ΩS of relative intensity 2, and peaks at ( ΩS ± 2π J ) with relative intensity 1 i.e. a 1:2:1 triplet. rad s−1 , which is 2 J Hz. The spacing between the outer lines is therefore 4 π J rad Hz. (d) First compute evolution under the offset 2 I ˆ1 z ˆ S x x
S z ΩS t ˆ
−−−−→
cos (ΩS t ) 2 I ˆ1 z ˆ S x x + sin (ΩS t ) 2 I ˆ1 z ˆ S y .
Chapter 12: Equivalent spins and spin-system analysis
104
We are only interested in the subsequent evolution of the second term as this is the one with sin (ΩS t ) modulation. This term evolves under the I 1 –S coupling to give sin (ΩS t ) 2 I ˆ1 z ˆ S y
ˆ1 z ˆ 2π Jt I S z
cos (π Jt ) sin (ΩS t ) 2 I ˆ1 z ˆ S y
− sin (π Jt ) sin (Ω
−−−−−−−→
ˆ
S t ) S x .
Ultimately, we are only interested in the term in S ˆ x , so we only need to consider the evolution of the second term on the right under the I 1 –S coupling
− sin (π Jt ) sin (Ω
ˆ
S t ) S x
ˆ2 z ˆ 2π Jt I S z
−−−−−−→ − cos (π Jt ) sin (π Jt ) sin (Ω
ˆ
S t ) S x
ˆ ˆ
− sin (π Jt ) sin (π Jt ) sin (Ω
S t ) 2 I 2 z S y .
Now applying the trigonometric identities gives 1 2 1 4
− cos (π Jt ) sin (π Jt ) sin (Ω
S t )
≡ − sin (2π Jt ) sin (Ω t ) ≡ − [cos (2π Jt − Ω t ) − cos (2π Jt + Ω t )] ≡ [cos (Ω t + 2π Jt ) − cos (Ω t − 2π Jt )] . S
S
1 4
S
S
S
In addition we have used cos (−θ ) ≡ cos (θ ) to go to the last line. Cosine Fourier transformation of this cosine modulated signal gives a peak corresponding to each term. There will be a peak at (ΩS − 2π J ) of relative intensity −1, and a peak at ( ΩS + 2π J ) of relative intensity + 1 i.e. a −1:0:+1 triplet. (e) First compute evolution under the offset S z ΩS t ˆ
4 I ˆ1 z I ˆ2 z ˆ S x
−−−−→
cos (ΩS t ) 4 I ˆ1 z I ˆ2 z ˆ S x + sin (ΩS t ) 4 I ˆ1 z I ˆ2 z ˆ S y .
(12.1)
The first term evolves under the I 1 –S coupling to give cos (ΩS t ) 4 I ˆ1 z I ˆ2 z ˆ S x
ˆ1 z ˆ 2π Jt I S z
−−−−−−−→ cos (π Jt ) cos(Ω
ˆ ˆ ˆ
S t ) 4 I 1 z I 2 z S x +
sin (π Jt ) cos(ΩS t ) 2 I ˆ2 z ˆ S y .
Evolution of the second term on the right under the I 2 –S coupling will result in the required term S ˆ x : sin (π Jt ) cos (ΩS t ) 2 I ˆ2 z ˆ S y
ˆ2 z ˆ 2π Jt I S z
−−−−−−−→
− sin (π Jt ) sin (π Jt ) cos (Ω
S t )
ˆ x . cos (π Jt ) sin (π Jt ) cos (ΩS t ) 2 I ˆ2 z ˆ S y sin (π Jt ) sin (π Jt ) cos (ΩS t ) S
−
≡ ≡ ≡ ≡ ≡
− − − −
1 [1 cos (2π Jt )]cos (ΩS t ) 2 1 cos (ΩS t ) + 12 cos (2π Jt ) cos (ΩS t ) 2 1 cos (ΩS t ) + 14 [cos (2π Jt + ΩS t ) + cos (2π Jt 2 1 cos (ΩS t ) + 14 [cos (2π Jt + ΩS t ) + cos (ΩS t 2
1 [ 4
−
−2cos (Ω
S t ) +
cos (2π Jt + ΩS t ) + cos (ΩS
− Ω t )] − 2π Jt )] t − 2π Jt )] . S
There will be peaks at ( ΩS ± 2π J ) of relative intensity +1, and a peak at ΩS of relative intensity −2 i.e. a + 1:−2:+1 triplet. The second term in Eq. 12.1 does not lead to a term in S ˆ x .
Chapter 12: Equivalent spins and spin-system analysis
105
12.3
We need to be intelligent about the way we approach this or we will waste a lot of time computing terms which ultimately do not end up as S ˆ x . The evolution under the offset, the I 1 –S , the I 2 –S and I 3 –S couplings needs to be considered. If we start with S ˆ x then from the evolution under the offset and the three couplings we need to pick up the term in which this operator does not change, which means that each will have a cosine factor cos (ΩS t ) for the offset term, and cos (π Jt ) for each coupling term. Thus, without any calculation we deduce that the wanted term is ˆ x . cos (π Jt ) cos (π Jt ) cos (π Jt ) cos (ΩS t ) S
Repeatedly applying an appropriate selection of the given identities gives 1 cos (ΩS t 8
− 3π Jt ) +
3 cos (ΩS t 8
3 8
1 8
− π Jt ) + cos (Ω t + π Jt ) + cos (Ω t + 3π Jt ) This is a 1:3:3:1 quartet with peaks at ( Ω − 3π J ), ( Ω − π J ), ( Ω + π J ) and ( Ω + 3π J ), respectively. S
S
S
S
S
S
If we start with 2 I ˆ1 z ˆS x , then in order to generate a term in S ˆ x , we need first to allow this anti-phase term to go become in phase with respect to the I 1 –S coupling, generating S ˆ y . This term then needs to remain in phase under the evolution of the I 2 –S and I 3 –S couplings. Finally, the S ˆ y term needs to evolve under the offset to give the wanted S ˆ x term. In summary 2 I ˆ1 z ˆ S x
− coupling ˆ − coupling ˆ − coupling ˆ offset ˆ −−−−−−−−−−→ S −−−−−−−−−−→ S −−−−−−−−−−→ S −−−−−→ −S I 1 S
I 2 S
I 3 S
y
S
y
y
x
The first arrow is associated with a factor sin (π Jt ), and the second and third arrows both generate factors of cos (π Jt ). The final arrow generates a factor sin (ΩS t ). Overall the S ˆ x term is of the form ˆ
− sin (Ω
S t ) cos (π Jt ) cos (π Jt ) sin (π Jt ) S x .
Repeatedly applying an appropriate selection of the given identities gives 1 8
1 8
1 8
1 8
− cos (Ω t − 3π Jt ) − cos (Ω t − π Jt ) + cos (Ω t + π Jt ) + cos (Ω t + 3π Jt ). This is a − 1:−1:+1:+1 quartet with peaks at (Ω − 3 π J ), (Ω − π J ), (Ω + π J ) and (Ω S
S
S
S
respectively.
S
S
,
S + 3 π J )
S
If we start with 4 I ˆ1 z I ˆ2 z ˆS x , then in order to generate a term in S ˆ x we need first to allow this doubly anti-phase term to become in phase with respect to the I 1 –S coupling, and then in phase with respect to the I 2 –S coupling. This term then needs to remain in phase under the evolution of the I 3 –S coupling, and as it is already along x no further evolution is needed under the offset. In summary 4 I ˆ1 z I ˆ2 z ˆ S x
− coupling ˆ ˆ − coupling ˆ − coupling ˆ offset ˆ −−−−−−−−−−→ 2 I S − −−−−−−−−−→ −S −−−−−−−−−−→ −S −−−−−→ −S I 1 S
I 2 S
2 z y
I 3 S
S
x
x
x
A factor of sin (π Jt ) is associated with the first two arrows, and a factor of cos (π Jt ) is associated with the third; the factor arising from the final arrow is cos (ΩS t ). Overall the S ˆ x term is of the form ˆ
− cos (Ω
S t ) cos (π Jt ) sin (π Jt ) sin (π Jt ) S x .
Repeatedly applying an appropriate selection of the given identities gives +
1 cos (ΩS t 8
− 3π Jt ) −
1 cos (ΩS t 8
− π Jt ) −
1 cos (ΩS t + 8
π Jt ) + 18 cos (ΩS t + 3π Jt ).
Chapter 12: Equivalent spins and spin-system analysis
This is a +1:−1:−1:+1 quartet with peaks at respectively.
(ΩS
106
− 3 π J ), (Ω − π J ), (Ω
S +
S
and
π J )
(ΩS
,
+ 3 π J )
If we start with 8 I ˆ1 z I ˆ2 z I ˆ3 z ˆS x , then in order to generate a term in S ˆ x we need first to allow this triply anti-phase term to become in phase with respect to the I 1 –S , I 2 –S and I 3 –S couplings, thus generating the term −S ˆ y . This then needs to evolve under the offset to give S ˆ x . 8 I ˆ1 z I ˆ2 z I ˆ3 z ˆ S x
− coupling ˆ ˆ ˆ − coupling − coupling ˆ offset ˆ −−−−−−−−−−→ −−−−−−−−− → −S −−−−−→ S 4 I I S −−−−−−−−−→ −2 I ˆ ˆ S − I 1 S
I 2 S
I 3 S
2 z 3 z y
S
3 z x
y
x
A factor of sin (π Jt ) is associated with the first three arrows, and a factor of sin (ΩS t ) arises from the final arrow. Overall the S ˆ x term is of the form ˆ x . sin (ΩS t ) sin (π Jt ) sin (π Jt ) sin (π Jt ) S
Repeatedly applying an appropriate selection of the given identities gives 1 8
3 8
3 8
1 8
− cos (Ω t − 3π Jt ) + cos (Ω t − π Jt ) − cos (Ω t + π Jt ) + cos (Ω t + 3π Jt ). This is a − 1:+3:−3:+1 quartet with peaks at (Ω − 3 π J ), (Ω − π J ), (Ω + π J ) and (Ω S
S
S
S
respectively.
S
S
,
S + 3 π J )
S
12.4
(a) There is only one coupling present in an I S spin system; the coupling Hamiltonian therefore contains the single term 2 π J I ˆ1 z ˆS z . The operator S ˆ y evolves into a term that is anti phase with respect to this coupling: ˆ y S
2π J I ˆ1 z ˆ S z t
−−−−−−→
ˆ y cos (π Jt ) S
− sin (π Jt ) 2 I ˆ ˆS . 1 z x
The I 2 S spin system contains an additional coupling between the S spin and a second I spin. We can therefore take the result for the I S system above and calculate its evolution under this second coupling term 2 π J I ˆ2 z ˆS z : ˆ y cos (π Jt ) S
2π J I ˆ2 z ˆ S z t
2
− sin (π Jt ) 2 I ˆ ˆS −−−−−−→ cos (π Jt ) S ˆ − cos (π Jt ) sin (π Jt ) 2 I ˆ ˆS − sin (π Jt ) cos (π Jt ) 2 I ˆ ˆS − sin (π Jt ) 4 I ˆ I ˆ ˆS . 1 z x
y
2 z x
2
1 z x
1 z 2 z y
To calulate the evolution of S ˆ y under the influence of the couplings in the I 3 S spin system we take the result obtained for the I 2 S system, and let the operators evolve under the term in the Hamiltonian that is due to the coupling between the S spin and the third I spin i.e. 2 π J I ˆ3 z ˆS z . The result is ˆ y cos3 (π Jt ) S
ˆ y cos2 (π Jt ) S
2
− cos
(π Jt ) sin(π Jt ) 2 I ˆ3 z ˆ S x
− cos (π Jt ) sin (π Jt ) 2 I ˆ ˆS − cos (π Jt ) sin (π Jt ) 2 I ˆ ˆS − cos (π Jt ) sin (π Jt ) 4 I ˆ I ˆ ˆS −−−−−−→ − sin (π Jt ) cos (π Jt ) 2 I ˆ ˆS − sin (π Jt ) cos (π Jt ) 2 I ˆ ˆS − sin (π Jt ) cos (π Jt ) 4 I ˆ I ˆ ˆS − sin (π Jt ) 4 I ˆ I ˆ ˆS − sin (π Jt ) cos (π Jt ) 4 I ˆ I ˆ ˆS + sin (π Jt ) 8 I ˆ I ˆ I ˆ ˆS . S z t 2 z x 2π J I ˆ3 z ˆ
2
2
1 z x
2
1 z 2 z y
2
2
2 z x 1 z x
1 z 2 z y
2
2 z 3 z y 1 z 3 z y
3
1 z 2 z 3 z x
Chapter 12: Equivalent spins and spin-system analysis
107
(b) If we allow the triply anti-phase operator to evolve under the influence of the three couplings of the I 3 S spin system, we obtain: 8 I ˆ1 z I ˆ2 z I ˆ3 z ˆ S y
2π J I ˆ1 z ˆ S z t
−−−−−−→ cos (π Jt ) 8 I ˆ I ˆ I ˆ ˆS − sin (π Jt ) 4 I ˆ I ˆ ˆS −−−−−−→ cos (π Jt ) 8 I ˆ I ˆ I ˆ ˆS − cos (π Jt ) sin (π Jt ) 4 I ˆ I ˆ ˆS − sin (π Jt ) cos (π Jt ) 4 I ˆ I ˆ ˆS − sin (π Jt ) 2 I ˆ ˆS −−−−−−→ cos (π Jt ) 8 I ˆ I ˆ I ˆ ˆS − cos (π Jt ) sin (π Jt ) 4 I ˆ I ˆ ˆS − cos (π Jt ) sin (π Jt ) 4 I ˆ I ˆ ˆS − cos (π Jt ) sin (π Jt ) 2 I ˆ ˆS − sin (π Jt ) cos (π Jt ) 4 I ˆ I ˆ ˆS − sin (π Jt ) cos (π Jt ) 2 I ˆ ˆS − sin (π Jt ) cos (π Jt ) 2 I ˆ ˆS + sin (π Jt ) S ˆ . 1 z 2 z 3 z y
2π J I ˆ2 z ˆ S z t
2
2 z 3 z x
1 z 2 z 3 z y
1 z 3 z x
2
2 z 3 z x
2π J I ˆ3 z ˆ S z t
3
2
1 z 2 z 3 z y
2
3 z y
1 z 2 z x
2
1 z 3 z x
2
2 z 3 z x
2
1 z y
2
2 z y
3
3 z y
x
The times at which only in-phase operator S ˆ x is generated are such that sin (π Jt ) = ±1, i.e. when t = 1 /(2 J ), 3 /(2 J ), 5 /(2 J ), and so on.
cos (π Jt )
=
0
and
12.5
The function cos2 (θ ) sin(θ ) can be rewritten using the identity expressed entirely in terms of sin(θ ): cos2 (θ ) sin(θ )
sin2 (θ )
+ cos
2
(θ )
≡
1 so
that it is
3
≡ sin(θ ) − sin (θ ).
Differentiating this with respect to θ gives: d dθ
sin(θ )
3
− sin (θ )
2
= cos(θ )
− 3sin (θ ) cos(θ ) = cos(θ ) 1 − 3 sin (θ ) .
2
The turning points are found by setting the derivative to zero so either cos(θ ) = 0 or 1 − 3 sin2 (θ ) = 0 . The first equality gives θ = π/ 2, which is plainly not a maximum since cos 2 (θ ) sin(θ ) = 0 for θ = π/ 2. The second equality can be solved to give 1
2
− 3 sin (θ ) = 0 sin(θ ) =
1 . 3
Therefore the greatest amount of the singly anti-phase term is generated when τ is given by θ = sin
− 1 3
1
rad s−1 .
= 0 .6155
Given that θ = 2 π J τ it therefore follows that
τ = 0 .0980/ J .
The amount of the doubly anti-phase term that is present goes as cos(2π J τ)sin2 (2π J τ). The delay for which this is a maximum can be found by maximizing the function cos(θ ) sin2 (θ ). Rewriting sin 2 (θ ) as 1 − cos2 (θ ) expresses everything in terms of cos(θ ). Differentiating with respect to θ gives: d dθ
cos( θ )
3
2
− cos (θ ) = − sin(θ ) + 3 sin(θ ) cos (θ ).
Chapter 12: Equivalent spins and spin-system analysis
108
Setting the derivative to zero means that either sin(θ ) = 0 or 3 cos2 (θ ) − 1 = 0 . Plainly sin(θ ) = 0 is not a maximum. Solving the other inequality gives cos(θ )
Hence θ = 0 .9553 rad, and so
τ
=
1 . 3
.
= 0 .1520/ J
12.6
The intensity of the peak due to an I n S spin system in the APT spectrum is given by cos n (2π J τ), and so setting τ = 1 /(2 J ) ensures that peaks from the I S and I 3 S spin systems are negative, whilst those from the S and I 2 S systems are positive. If the chosen compromise value of J is 130 Hz, the value of τ is 1 = 3.85 ms . τ= 2
× 130
The acetylene group forms an IS spin system with a coupling constant of 250 Hz. The resulting peak would be expected to have a negative intensity. However, using the compromise value of τ that was calculated above, we find that the actual intensity is: cos(2π × 250 × 3.58 × 10−3 ) = 0 .97. This peak therefore has the opposite sign to that expected.
12.7
We will begin by calculating the form of the COSY spectrum of the I 2 S spin system. Starting with I ˆ1 z , the first 90 ◦ pulse generates − I ˆ1 y . This term then evolves under the offset of I 1 , and the coupling between I 1 and S . So, at the end of t 1 , we have: ˆ1 z t 1 Ω I I
− I ˆ −−−−−→ − cos(Ω t ) I ˆ + sin(Ω t ) I ˆ −−−−−−−→ − cos(Ω t ) cos(π Jt ) I ˆ + cos(Ω t ) sin(π Jt )2 I ˆ 1 y
I 1
1 y
I 1
1 x
2π J I ˆ1 z ˆ S z t 1
I 1
+
1
1 y
I 1
ˆ
1 x S z
1
sin(Ω I t 1 ) cos(π Jt 1 ) I ˆ1 x + sin(Ω I t 1 ) sin(π Jt 1 )2 I ˆ1 y ˆ S z .
The second 90 ◦ pulse then acts to give two observable terms, which arise from the third and fourth terms above. These are: sin(Ω I t 1 ) cos(π Jt 1 ) I ˆ1 x
− sin(Ω t ) sin(π Jt )2 I ˆ ˆS I 1
1
1 z y
: diagonal-peak multiplet : cross-peak multiplet.
The t 1 modulation of the diagonal peak can be expressed as sin(Ω I t 1 ) cos(π Jt 1 )
≡
1 sin(Ω I + 2
π J )t 1 + 12 sin(Ω I
− π J )t , 1
which is an in-phase doublet centred at the offset of the I spins. In addition, the operator I ˆ1 x will evolve during t 2 to give an in-phase doublet on the same spin. The t 1 modulation of the cross-peak multiplet is
− sin(Ω t ) sin(π Jt ) ≡ I 1
1
1 cos(Ω I + 2
π J )t 1
−
1 cos(Ω I 2
− π J )t , 1
Chapter 12: Equivalent spins and spin-system analysis
109
which represents a doublet on the I spin that is anti phase with respect to the I 1 –S coupling. During t 1 , the operator 2 I ˆ1 z ˆ S y will evolve to give a − 1:0:+1 “triplet” which is centred on the S spin (see Fig. 12.6 on page 452), and which is anti phase with respect to one of the I –S couplings. Both the diagonal- and cross-peak multiplets are illustrated below, where as usual we have formed the two-dimensional multiplet by “multiplying” together the two one-dimensional multiplets along each axis. 2πJ
2πJ
ΩI
ΩS
ΩI
2πJ
ω 1
ΩI
2πJ
ω 1 ω 2
ω 2
Diagonal-peak multiplet
Cross-peak multiplet
If we repeat the calculation for the I 3 S spin system, we find that we obtain the same mathematical forms for the diagonal- and cross-peak multiplets. The reason for this is that − I ˆ1 y can only evolve under the I 1 –S coupling and not under any of the other couplings between the I spins and the S spin. Therefore, both terms have the same form of t 1 modulation as for the I 2 S spin system, i.e. either an in-phase or anti-phase doublet. The structure of the diagonal-peak multiplet in ω 2 is also the same as for the I 2 S spin system since I ˆ1 x evolves to give an in-phase doublet. However the t 2 modulation of the cross-peak multiplet is rather different, as a result of the additional coupling between the S spin and the third I spin; it now takes the form of a −1:−1:+1:+1 “quartet” that is anti phase with respect to one coupling (see Fig. 12.6 on page 452). Both multiplets are illustrated below. 2πJ
2πJ
ΩI
ΩS
ΩI
ω 1
2πJ
ΩI
ω 1 ω 2
Diagonal-peak multiplet
ω 2
Cross-peak multiplet
2πJ
Chapter 12: Equivalent spins and spin-system analysis
110
We now return to the I 2 S spin system and repeat the calculation, this time starting with ˆ y which evolves under the S spin offset and both I –S couplings: 90◦ pulse gives us −S S z t 1 ΩS ˆ
−S ˆ −−−−−→ − cos(Ω −−−−−−−→ − cos(Ω
ˆ
S t 1 )S y +
y
2π J I ˆ1 z ˆ S z t 1
ˆ x sin(ΩS t 1 )S
ˆ S t 1 ) cos(π Jt 1 )S y +
+
ˆ x sin(ΩS t 1 ) cos(π Jt 1 )S
2π J I ˆ2 z ˆ S z t 1
ˆ z . The first S
+
cos(ΩS t 1 ) sin(π Jt 1 )2 I ˆ1 z ˆ S x sin(ΩS t 1 ) sin(π Jt 1 )2 I ˆ1 z ˆ S y
2 ˆ S t 1 )cos (π Jt 1 )S y +
−−−−−−−→ − cos(Ω
cos(ΩS t 1 ) cos(π Jt 1 ) sin(π Jt 1 )2 I ˆ2 z ˆ S x
+
cos(ΩS t 1 ) sin(π Jt 1 ) cos(π Jt 1 )2 I ˆ1 z ˆ S x + cos(ΩS t 1 )sin2 (π Jt 1 )4 I ˆ1 z I ˆ2 z ˆ S y
+
ˆ x sin(ΩS t 1 )cos2 (π Jt 1 )S
+
sin(ΩS t 1 ) sin(π Jt 1 ) cos(π Jt 1 )2 I ˆ1 z ˆ S y
+
sin(ΩS t 1 ) cos(π Jt 1 ) sin(π Jt 1 )2 I ˆ2 z ˆ S y 2 ˆ ˆ ˆ S t 1 )sin (π Jt 1 )4 I 1 z I 2 z S x .
− sin(Ω
Only three of the terms (the fifth, sixth, and seventh) are rendered observable by the final pulse. The result is therefore ˆ x sin(ΩS t 1 )cos2 (π Jt 1 )S
− sin(Ω − sin(Ω
ˆ ˆ
S t 1 ) cos(π Jt 1 ) sin(π Jt 1 )2 I 2 y S z
ˆ ˆ
S t 1 ) cos(π Jt 1 ) sin(π Jt 1 )2 I 1 y S z
: diagonal-peak multiplet : cross-peak multiplet : cross-peak multiplet.
The t 1 modulation of the diagonal-peak term can be expanded to give sin(ΩS t 1 )cos2 (π Jt 1 )
≡
1 sin(ΩS t 1 + 4
π Jt 1 ) +
1 1 + sin(ΩS t 1 2 4
− π Jt ), 1
which is an in-phase 1:2:1 triplet centred on ΩS . During t 2 , S ˆ x will evolve to also give an in-phase triplet. The cross-peak term has the following t 1 modulation:
− sin(Ω
S t 1 ) cos(π Jt 1 ) sin(π Jt 1 )
≡
1 cos(ΩS t 1 + 4
2π Jt 1 )
−
1 cos(ΩS t 1 4
− 2π Jt ). 1
This is a −1:0:+1 “triplet” that is centred on ΩS in which one of the splittings is in anti phase. The evolution of both 2 I ˆ1 y ˆS z and 2 I ˆ1 y ˆS z gives exactly the same multiplet structure in ω2 , namely a doublet that is anti phase with respect to the relevant I –S coupling. Both multiplets are shown below. 2πJ
2πJ
ΩS
ΩI
2πJ
2πJ
ΩS
ω 1
ΩS
ω 1 ω 2
Diagonal-peak multiplet
ω 2
Cross-peak multiplet
Chapter 12: Equivalent spins and spin-system analysis
111
It now remains to compute the spectrum of the I 3 S spin system starting out from S ˆ z . The calculation results in the following four observable operators: ˆ x sin(ΩS t 1 )cos3 (π Jt 1 )S S t 1 )cos
2
(π Jt 1 ) sin(π Jt 1 )2 I ˆ3 y ˆ S z
S t 1 )cos
2
(π Jt 1 ) sin(π Jt 1 )2 I ˆ2 y ˆ S z
S t 1 )cos
2
(π Jt 1 ) sin(π Jt 1 )2 I ˆ1 y ˆ S z
− sin(Ω − sin(Ω − sin(Ω
: : : :
diagonal-peak multiplet cross-peak multiplet cross-peak multiplet cross-peak multiplet.
The modulation in t 1 of the diagonal-peak multiplet is sin(ΩS t 1 )cos3 (π Jt 1 )
≡
1 sin(ΩS t 1 + 3π Jt 1 ) + 38 sin(ΩS t 1 + 8
π Jt 1 ) + 38 sin(ΩS t 1
− π Jt ) + 1
1 sin(ΩS t 1 8
− 3π Jt ), 1
which is an in-phase quartet on the S spin. Likewise, S ˆ x will evolve during t 2 to give an in-phase quartet. All three cross-peak multiplets lie on top of each other. The t 1 modulation can be expanded to give
− sin(Ω
S t 1 )cos
2
(π Jt 1 ) sin(π Jt 1 )
≡
1 cos(ΩS t 1 +3π Jt 1 )+ 18 cos(ΩS t 1 +π Jt 1 ) 8
−
1 cos(ΩS t 1 8
−π Jt )− 1
1 cos(ΩS t 1 8
−3π Jt ), 1
which is a − 1:−1:+1:+1 S spin “quartet” that is anti-phase with respect to one of the splittings. During t 2 , the anti-phase operators will evolve to give an anti-phase doublet on the I spin. Both multiplets are shown below. 2πJ
2πJ
ΩS
ΩI
ΩS
ω 1
2πJ
ΩS
ω 1 ω 2
Diagonal-peak multiplet
ω 2
Cross-peak multiplet
2πJ
Chapter 12: Equivalent spins and spin-system analysis
112
12.8
18.7 Hz 9.18 Hz
31.1 Hz 40.8 Hz D
D J AB
J AB
average = −1 ⁄ 2 Σ
frequency / Hz
(a) D
−
= 31 .1
− 9.18 = 21.92
1 1 Σ = (40.8 + 2 2
9.18)
= 24 .99
| J | = 18.7 − 9.18 = 9.52
−
1 1 Σ = (31.1 + 2 2
=
(υA
2 2 B ) + J AB
B)
=
− 9.61
=
− υ
22.022
average: D = 22.02 Hz
= 24 .9
average: Σ = −49.8 Hz
average: | J AB | = 9 .61 Hz
2
19.81 Hz
taking positive root
=
−49.8 ( −49.8 + 19.81) = = −15.00 Hz ( −49.8 − 19.81) = = −34.81 Hz. = −34.81 Hz | J | = 9 .7 Hz 1 2
υA
1 2
υB
−15.00 Hz υ − υ ) = 19.81 Hz and J
=
18.7)
AB
(υA + υB )
υA
− 18.7 = 22.1
D2
− υ
In summary
= 40 .8
| J | = 40.8 − 31.1 = 9.7
AB
(υA
D
B
(b) We note that (υA B means that ξ must be in the range 0 to
AB
AB =
.
1 π 2
9.7 Hz, so both quantities
tan ξ
= =
intensity of inner lines intensity of inner lines
J AB (υA υB ) 9.61
−
=
19.81 0.4851 rad
hence ξ
=
0.4517 rad
sin ξ
=
0.4365
sin ξ )
=
0.718
− sin ξ )
=
0.282
1 (1 + 2 1 (1 2
are positive which
or 25 .9◦
Chapter 12: Equivalent spins and spin-system analysis
113
The spectrum therefore looks like:
0
10
20
30
40 frequency / Hz
50
(c) If the field is doubled, the offsets are doubled so that υ A = 2 × −15.00 = −30.00 Hz and υ B 2 × −34.81 = −69.62 Hz, but the coupling remains the same. Σ
D
=
υA + υB
=
−30.00 − 69.62 = −99.62 Hz (υ − υ ) + J (−30.00 − (−69.62)) + 9.61
= =
tan ξ
B
2
2 AB
2
(υA
− υ
ξ
=
sin ξ
=
0.236
sin ξ )
=
0.618
− sin ξ )
=
0.382
−
line expression
−−
= 0 .2426
frequency / Hz
intensity
1
1 ( + D 2
AB )
1 ( +40.77 + 2
99.62
2
1 2
AB )
1 ( +40.77 + 2
99.62 + 9.61)
3
1 2
AB )
4
1 2
AB )
− Σ − J ( + D − Σ + J ( − D − Σ − J ( − D − Σ + J
30
Hz
B)
9.61 30.00 ( 69.62) 0.2380
1 (1 + 2 1 (1 2
2 = 40 .77
J AB
= =
20
A
40
− 9.61) = 65.4
0.618
= 75 .0
0.382
1 ( 2
0.382
1 2
−40.77 + 99.62 − 9.61) = 24.6 ( −40.77 + 99.62 + 9.61) = 34 .2
0.618
50
60
70 frequency / Hz
80
=
Chapter 12: Equivalent spins and spin-system analysis
114
As expected, there is less roofing at this higher field since the separation of the offsets is greater.
12.9
(a) See spreadsheet. Note that the horizontal scales on the graphs will need to be adjusted by hand in order to make them suitable for the parameters specified. (b) For the given parameters we find υ0,A+ υ0,B+
−14.00 Hz −25.50 Hz
D+
12.54 Hz
Σ+
−39.50 Hz
ξ +
0.410
υ0,A− υ0,B−
D− Σ
−
−26.00 Hz −28.50 Hz 5.59 Hz
−54.50 Hz 1.107
ξ −
The frequencies and intensities are
AB part, X up
AB part, X down
frequency / Hz intensity
frequency / Hz intensity
11.0
0.301
22.0
0.053
16.0
0.699
27.0
0.947
23.5
0.699
27.5
0.947
28.5
0.301
32.5
0.053
X part frequency / Hz intensity 990.9
0.029
992.5
0.250
996.5
0.221
1003.5
0.221
1007.5
0.250
1009.1
0.029
The corresponding spectra are shown over.
Chapter 12: Equivalent spins and spin-system analysis υ0,A+
115 υ0,B+
υ0,B−
υ0,A−
X up AB subspectrum
X down AB subspectrum
AB part 10
20
30
40
frequency / Hz
X part of ABX
980
985
990
995
1000
1005
1010
1015
frequency / Hz
1020
Chapter 12: Equivalent spins and spin-system analysis
116
12.10 c d e
g
AB part
X part b
j
h
a
k
l
m
f i
0
10
20
30
40
50 980
n
990
1000
1010
1020
frequency / Hz
line a b c d e f g h
frequency / Hz 12.3 18.1 19.3 21.2 25.1 28.2 32.4 39.4
line i j k l m n
frequency / Hz 988.4 991.5 997.3 1002.7 1008.5 1011.6
(a) By looking at the pattern of intensities and checking for the repeated spacing corresponding to J AB we deduce that one subspectrum consists of lines b, e, g, h and other consists of lines a, c, d, f . (b) For the b, e, g, h subspectrum we determine the following (average) values: J AB
= 7 .00
Hz
−57.5 Hz
Σ =
D
= 14 .3
Hz
From these we can determine that
−22.5 Hz
=
υA
υB
−35.0 Hz
=
To check these we use them to predict the form of the spectrum tan ξ =
7.00 22.5 + 35.0
−
= 0 .560
sin ξ = 0 .489
The intensities of the inner lines are 12 (1 + sin ξ ) = 0 .74, and the intensity of the outer lines are 1 (1 − sin ξ ) = 0 .25. The frequencies of the four lines are 2 line expression
frequency / Hz
intensity
1
1 ( + D 2
AB )
1 ( +14.3 + 2
57.5
2
1 2
AB )
1 ( +14.3 + 2
57.5 + 7.00)
3
1 2
AB )
4
1 2
AB )
− Σ − J ( + D − Σ + J ( − D − Σ − J ( − D − Σ + J
− 7.00) = 32.4
0.74
= 39 .4
0.25
1 ( 2
0.25
1 2
−14.3 + 57.5 − 7.00) = 18.1 ( −14.3 + 57.5 + 7.00) = 25 .1
0.75
Chapter 12: Equivalent spins and spin-system analysis
117
Lines 1, 2, 3, 4 in the table correspond to lines g, h, b, e in the spectrum: everything matches up. For the a, c, d, f subspectrum we determine the following (average) values: J AB
= 7 .00
Hz
−40.45 Hz
Σ =
D
= 8 .90
Hz
From these we can determine that
−17.48 Hz
=
υA
−22.97 Hz
=
υB
To check these we use them to predict the form of the spectrum tan ξ =
7.00 17.48 + 22.97
−
= 1 .273
sin ξ = 0 .7865
The intensities of the inner lines are 12 (1 + sin ξ ) = 0 .89, and the intensity of the outer lines are 1 (1 − sin ξ ) = 0 .11. The frequencies of the four lines are 2 line expression
frequency / Hz
intensity
1
1 ( + D 2
AB )
1 ( +8.90 + 2
40.45
2
1 2
AB )
1 ( +8.90 + 2
40.45 + 7.00)
3
1 2
AB )
4
1 2
AB )
− Σ − J ( + D − Σ + J ( − D − Σ − J ( − D − Σ + J
− 7.00) = 21.2
0.89
= 28 .2
0.11
1 ( 2
0.89
1 2
−8.90 + 40.45 − 7.00) = 12.3 ( −8.90 + 40.45 + 7.00) = 19 .3
0.11
Lines 1, 2, 3, 4 in the table correspond to lines d, f, a, c in the spectrum: everything matches up. (c) The two possibilities are Hz υ1 υ2 υ3 υ4
−22.5 −35.0 −17.5 −23.0
first possibility second possibility υ0,A+
υ0,A+
υ0,B+
υ0,B+
υ0,A−
υ0,B−
υ0,B−
υ0,A−
Starting with the first possibility we have 1
−22.5 −17.5
υ0,A+
= υ 0,A + 2 J AX =
υ0,A−
1 J = 2 AX
= υ 0,A
−
hence υ 0,A = −20.0 Hz and J AX = −5.0 Hz. 1
υ0,B+
= υ 0,B + 2 J BX =
υ0,B−
= υ 0,B
hence υ 0,B = −29.0 Hz and J BX = −12 Hz.
−
1 J 2 BX
−35.0 = −23.0
Chapter 12: Equivalent spins and spin-system analysis
118
The second possibility gives an alternative set of parameters υ0,A+
= υ 0,A +
υ0,A−
= υ 0,A
1 J = 2 AX 1 J = 2 AX
−22.5 −23.0
−
hence υ 0,A = −22.75 Hz and J AX = 0 .5 Hz. υ0,B+
= υ 0,B +
υ0,B−
= υ 0,B
1 J = 2 BX 1 J = 2 BX
−35.0 −17.5
−
hence υ 0,B = −26.25 Hz and J BX = −17.5 Hz. (d) The line frequencies are intensities are given by Table 12.5 on page 474. Using the first set of parameters we have
D+
=
D−
υ0,A+
0,B+
υ0,A
0,B
=
− υ − − υ
tan ξ +
=
tan ξ −
2
2 + J AB =
2
−
− −
2
υ0,A+
− υ
=
−
2
7.0 = 0 .56 22.5 + 35.0 7.0 = 1 .27 17.5 + 23.0
−
0,B+
J AB υ0,A− υ0,B−
=
2
− (−35.0)) + 7.0 = 14.33 Hz 17.5 − (−23.0)) + 7.0 = 8 .90 Hz
(
+ J AB =
J AB
2
( 22.5
=
−
2
ξ +
= 0 .5105
ξ −
= 0 .905
In the calculation of ξ ± there are no complications since both J AB and (υ0,A± −υ0,B± ) are positive, placing the angle in the top-right quadrant. The intensity of both the X2 and X3 transitions are 1 cos 2 4
1 ξ 2 +
−
−
1 ξ 2
= 0 .25
2
× cos
(0.5
× 0.5105 − 0.5 × 0.905) = 0.24
The intensity of both the X5 and X6 transitions are 1 sin 2 4
1 ξ 2 +
−
−
1 ξ 2
= 0 .25
× sin
2
(0.5
× 0.5105 − 0.5 × 0.905) = 0.0096
We assume υ 0,X = −1000 Hz, and so can determine the line positions as transition frequency
Hz
1 J 2 AX
1 2 BX
X1
υ0,X
X2
υ0,X +
1 2
X3
υ0,X +
1 2
X4
υ0,X + 12 J AX
X5
υ0,X + 12 ( D+
X6
υ0,X + 12 ( + D+ + D− ) 1011.6
−
− J 1008.5 ( + D − D− ) 1002.7 ( − D + D− ) 997.3
intensity line 0.25
m
+
0.24
l
+
0.24
k
1
991.5
0.25
j
− − D−)
988.4
0.0096
i
0.0096
n
+ 2 J BX
The line positions and intensities are a good match to the X part of the spectrum: the assignment is listed in the right-most column of the table.
Chapter 12: Equivalent spins and spin-system analysis
119
(e) Using the second set of parameters we have
D+
=
D−
=
− υ − − υ
υ0,A+
0,B+
υ0,A
0,B
2
+ J AB =
2
−
2
2
− −
2
( 22.5 (
+ J AB =
2
− (−35.0)) + 7.0 = 14.33 Hz 23.0 − (−17.5)) + 7.0 = 8 .90 Hz 2
2
7.0 J AB = = 0 .56 υ0,A+ υ0,B+ 22.5 + 35.0 7.0 J AB = = 1.27 tan ξ − = υ0,A− υ0,B− 23.0 + 17.5
tan ξ +
=
−
−
−
−
ξ +
= 0 .5105
ξ −
−
= 2 .24
In the calculation of ξ − J AB is positive but (υ0,A− − υ0,B− ) is negative, placing the angle in the top-left quadrant (see Fig. 12.19 on page 470), so the angle must be between π/ 2 and π . The intensity of both the X2 and X3 transitions are 1 cos 2 4
1 ξ 2 +
−
−
1 ξ 2
= 0 .25
2
× cos
(0.5
× 0.5105 − 0.5 × 2.24) = 0 .105
The intensity of both the X5 and X6 transitions are 1 sin 2 4
1 ξ 2 +
−
1 ξ 2
−
= 0 .25
2
× sin
(0.5
× 0.5105 − 0.5 × 2.24) = 0.145
We assume υ 0,X = −1000 Hz, and so can determine the line positions as transition frequency
Hz
1 J 2 AX
1 2 BX
X1
υ0,X
X2
υ0,X
+
1 2
X3
υ0,X
+
1 2
X4
υ0,X
+ 2 J AX + 2 J BX
X5
υ0,X
+ 2 (
X6
υ0,X
+ 2 ( + D+ + D
−
1 1 1
intensity
− J 1008.5 ( + D − D− ) 1002.7 ( − D + D− ) 997.3
0.25
+
0.105
+
0.105
1
991.5
0.25
− D − D− )
988.4
0.145
+
− ) 1011.6 0.145
The frequencies are exactly the same, but the pattern of intensities is quite different, and certainly does not match the experimental spectrum. We therefore reject the second set of parameters. In conclusion, therefore, the determined parameters are υ0,A
−20.0 Hz
=
υ0,B
−29.0 Hz
=
J AX
−5.0 Hz
=
J BX
−12 Hz
=
| J | = 7.00 Hz AB
We do not know the sign of the A–B coupling, but we do know that the A–X and B–X couplings have the same sign (either both positive or both negative, we do not know which).
Chapter 12: Equivalent spins and spin-system analysis
120
12.11
(a) The effective Larmor frequencies in the two sub-spectra are 1
υ0,B+
= υ 0,B + 2 J BX = υ 0,B +
− 75
υ0,B−
= υ 0,B
=
−
= υ 0,A + 2 J AX = υ 0,A +
υ0,A−
= υ 0,A
−
1
75
υ0,A+
1 J = υ 0,A 2 AX
−
7.5
1 J = υ 0,B 2 BX
− 7.5
If υ 0,A + 75 ≈ υ 0,B + 7.5, then the X up subspectrum shows strong coupling. Rearranging this condition gives υ 0,A − υ0,B ≈ −67.5 Hz. Using this condition, in the X down sub-spectrum we have υ0,A−
− υ
0,B
−
− 75 − υ − 7.5 υ − υ −67.5
υ0,A
=
0,A
0,B
67.5 Hz, from
=
0,B
above
−67.5 − 67.5 −135 Hz.
=
What this means is that when there is very strong coupling in the X up subspectrum, the two effective shifts in the X down subspectrum are still separated by −135 Hz, and so will not be very strongly coupled at all. Put another way ∆+ = υ 0,A+
0,B+
=
υ0,A
0,B
∆
0,B
=
υ0,A
0,B
− υ − − υ
− = υ 0,A
−
− υ + − υ −
1 ( J AX 2 1 ( J AX 2
BX )
− J − J
BX )
Thus the differences in the two effective Larmor frequencies in the sub spectra, ∆+ and ∆− , themselves differ by ∆+ − ∆− = ( J AX − J BX ). Since J AX is quite large, when we compare the difference in the effective Larmor frequencies in one sub spectrum with that in the other, the change can be substantial. (b) For there to be ‘infinite’ strong coupling in the X up subspectrum we need υ0,A+ υ0,A + 75 υ0,A
−
− υ
0,B+
υ0,B + 7.5
− υ
0,B +
67.5
=
0
=
0
=
0
−67.5 Hz. Given that, at 500 MHz, 1 ppm ≡ 500 Hz, if follows that δ − δ = −0.135 ppm. υ0,A
− υ
0,B
=
A
B
For the X down sub spectrum the calculation is the same apart from a minus sign in front of the coupling terms so the strong coupling condition is δ A − δB = +0.135 ppm.
(c) The degree of strong coupling, and hence the spectrum, is invariant to the sign of the A–B coupling. However, if the sign of the B–X coupling is reversed, the effective Larmor frequencies are changed. Assuming that J BX = −15 Hz, the condition for infinite strong coupling in the X up subspectrum is υ0,A+ υ0,A + 75 υ0,A
−
− υ − 7.5
0,B+
υ0,B
− υ
0,B +
υ0,A
82.5
− υ
0,B
=
0
=
0
=
0
=
−82.5 Hz.
The shift difference is thus −0.165 ppm, a different value from that given above.
Chapter 12: Equivalent spins and spin-system analysis
121
(d) In an AB spectrum the degree of strong coupling depends on the difference in the Larmor frequencies of the two spins υ0,A − υ0,B . If the magnitude of this difference increases, the degree of strong coupling decreases. As the applied magnetic field is increased, the Larmor frequencies will scale directly and as a result |υ0,A − υ0,B | will increase and the degree of strong coupling will decrease. In the ABX spectrum the degree of strong coupling, for example in the X up subspectrum, depends on ∆+ = υ 0,A+
0,B+ = υ 0,A
− υ
− υ
1 0,B + 2 ( J AX
BX ) .
− J
Let us suppose that J AX and J BX and both positive, and that J AX >> J BX . Imagine that the chemical shifts of A and B are such that υ0,A − υ0,B is positive. As the applied field is increased this difference increases, thus ∆+ increases and the degree of strong coupling decreases.
Now imagine that the chemical shifts of A and B are such that υ0,A − υ0,B is negative. Let us imagine starting at very low field so that υ0,A − υ0,B is smaller in magnitude than 1 + 2 ( J AX − J BX ). As the applied field is increased υ0,A − υ0,B becomes more negative and, as a result, ∆+ decreases. Consequently, increasing the field strength at first increases the degree of strong coupling. Eventually ∆+ goes to zero, which is infinite strong coupling. A further increase in the applied field then makes ∆+ more and more negative, thus decreasing the strength of the coupling. The diagram below may help.
y c n e u q e r f
1 / 2(J – AX
J BX)
strong coupling decreases
0 applied field
strong coupling increases
∆+
(υ0,A – υ0,B)
The point here is that infinite strong coupling in one of the AB sub spectra occurs when the difference in Larmor frequencies between A and B is a specific value, determined by the values of J AX and J AX . As the applied field is increased we may move away from this condition, thus decreasing the degree of strong coupling, or move towards it, thus increasing the degree of strong coupling.
Chapter 12: Equivalent spins and spin-system analysis
122
13 How the spectrometer works
13.1
The magnetic field strength can be computed from the given Larmor frequency, f 0 , and gyromagnetic ratio using 2 π f 0 = γ B0 . Hence B0
=
2π f 0 γ
=
6
2π
× 180 × 10 1.08 × 10
= 10 .47
8
T.
A homogeneity of one part in 108 means that the magnetic field varies by 1.047 × 10−7 T. This translates to a variation in frequency, ∆ f , of γ ∆ B 1.08 × 108 × 1.047 × 10−7 ∆ f = = = 1 .8 Hz. 2π
∆ B =
10−8
× 10 .47 =
2π
This is significantly less than the expected linewidth of 25 Hz, so the magnet is useable. The calculation is much simpler if we realise that a homogeneity of 1 part in Larmor frequency will vary by 10 −8 times its nominal value i.e. −8 × 180 × 106 = 1 .8 Hz. ∆ f = 10
108
means that the
13.2
For a 180◦ (or π ) pulse, π = ω 1 t 180 , so ω1
=
π
t 180
=
π
24.8
×
10−6
= 1 .27
×
105 rad s−1 .
Therefore ω 1 /(2π) = 20 .2 kHz. The same result can be found more simply by noting that a 360 ◦ pulse takes 2 × 24.8 = 49.6 µs; this is the period of the rotation about the RF field, so the frequency is just the reciprocal of this: ω1 /(2π) = 1 /(49.6 × 10−6 ) = 20 .2 kHz. Using Eq. 13.2 on page 488 we have attenuation = 20 log
2 20.2
=
− 20.1 dB .
