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Assignment #3
Simulating NMR Spectra
March 9, 2003
This assignment is intended to introduce you to NMR spectral simulation of
solution-phase spectra. You will be using MESTRE, a software package available free
from the Internet, which can be used on a P.C. We will run through the basic steps of
simulating a simple AB spectrum.
To illustrate how the Hamiltonian can be factorized, the theory of the ABX case
is then explained.
The ABX spectrum is one of the simplest cases.
You are then
required to answer some questions for which you are expected to use the program.
Getting started
To start change disk to A:\ drive and click on the Mestrec 2.1 with the icon that
looks like a scroll, shown below.
Figure 1
The MESTRE Icon
Now click on simulation at the top of the screen, then select new. At this stage the
simulation editor should be activated as seen in figure 2.
Now enter the numbers as shown in figure 2.
You are simulating an AB
spectrum, so, the number of spins is 2, and there is only one subspectrum. Make sure that
the units are set to Hz and the spectrometer frequency is at 300 MHz. Now enter the
number of spins, 1, for A and 1 for B and chemical shifts (in Hz from an arbitrary
2
reference) of each type of nucleus, say A=300 Hz and B= 600 Hz, in the table, along
with the coupling constant, JAB=10 Hz . Make sure that the frequency range is set from 0
to 3000 Hz, the threshold is 10-5, the number of points is 64000, and the linewidth is 0.5
Hz (these settings are found at the bottom of the window). After you have entered all the
parameters, click on O.K.
Figure 2 Simulation editor.
At this stage you should see an AB spectrum in the main working environment as
shown in figure 3. Notice that the units on the axis are seconds. To change this to Hz
select Hz in the box at the bottom center of the window. To change the spectral
parameters just click on simulation and select edit, make your changes and click on O.K.
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Figure 3
The AB spectrum in the main program window.
To modify the size of the spectrum, make use of the tool bar at the top of the
window, shown below.
The + and - buttons increase and decrease the intensity of the spectrum. The looking
glass button is used to expand a region. Just click on the button and box the area of
interest. The full button restores the complete spectrum.
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Now investigate the effect of the ratio of the difference in the chemical shift to the
coupling constant on the line positions and intensities of the AB spectrum. Try to
reproduce the spectra given in figures 6 and 7. You might want to use the interactive
simulation mode. To enter this mode, click on simulation and select interactive. A small
window opens, as in figure 4, which allows you to selectively change one parameter at a
time, by either entering a specific value or clicking on the increase or decrease buttons
(make sure that the increment is set to an appropriate value).
Figure 4
Interactive simulation window
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The AB spectrum
The Hamiltonian of a two-spin system is:
ˆ   Iˆ A  Iˆ A  J Iˆ A Iˆ B 
h 1 H
A z
B z
AB z z

where Iˆ A Iˆ B  Iˆ A Iˆ B


J AB ˆ A ˆ B ˆ A ˆ B
I I  I I
2

are known as the flip-flop terms. Choosing the product basis
and the matrix form of the Hamiltonian is:


 

 
1 ˆ
h H






v A  v B 
2
J
 AB
4


0
0
v A  v B 
0
J AB
2
J AB
2
v A  v B  J AB


2
4
0
0
0
0
2

J AB
4



0



0


0


v  v B  J AB 
 A

2
4 
Diagonalizing the center 2x2 block gives eigenvalues:

with
eigenvectors
1
2
2
ν A  νB 2  J AB
cos   sin  

J AB
4
and
 sin    cos 
where
tan 2  J AB /(v B  v A ) .(See figure 5 )
Transition intensities can be obtained from the expression 4 *   i
 Iˆ
x
k

2
,
where the factor of 4 normalizes the intensities so that the total is 4 (implying that in
first-order spectra each of the 4 lines has unit intensity).
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Question 1
i)
As you reach the extreme where the difference in the chemical shift is very
small compared to the coupling constant, what type of spin system does it
become?
ii)
Why do the outer lines disappear when vA = vB?
iii)
Draw the energy level diagram for this case and state the eigenfunctions.
With these eigenfunctions calculate the relative transition intensities.
iv)
What happens to the spectrum when the coupling constant is negative?
Figure 5
The Energy Level Diagram of an AB spin system.
+

-sin cos
cos sin 

n.b. The product basis functions  and are mixed to form the eigenbasis.
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Figure 6
Stack Plot of an AX spin system changing into an AB spin system. The
ratio of the chemical shift difference with respect to the coupling constant
ranges from 10 to 2.
/J = 10
/J = 3
/J = 2
600
500
400
300
Frequency (Hz)
200
100
0
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Figure 7
Stack Plot of an AB spin system changing into an A2 spin system. The
ratio of the chemical shift difference with respect to the coupling constant
ranges from 0.5 to 0.
/J = 0.5
/J = 0.1
/J = 0
600
500
400
300
Frequency (Hz)
200
100
0
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The ABX Spectrum
For larger spin systems it is possible to get analytical solutions if the Hamiltonian
can be factorized, e.g. by making certain assumptions. This happens if the Hamiltonian
can be transformed into a block diagonal form where largest block is of order 2. This is
possible for ABX, ABX2 to ABXn, and ABn systems.
There are three ways in which the Hamiltonian structure can be block
diagonalized.
The m=±1 selection rule provides blocking based on the total z-
component of angular momentum. In other words energy levels with different Fˆ z   Iˆ z
j
values do no mix.
By ignoring the flip-flop terms, known as the first order
approximation, off-diagonal terms can be removed, often simplifying the structure of the
Hamiltonian. This will be illustrated with an ABX spin system. Finally, when several
spins are magnetically equivalent, couplings between then do not manifest themselves in
the spectrum, and thus may be left out of the Hamiltonian. The eigenbasis of the
equivalent nuclei then have symmetry properties which can be exploited to block
diagonalize the Hamiltonian. This will be illustrated with an AB2 spectrum.
Take the general Hamiltonian for a three-spin-1/2 system, for instance an ABC
system:
ˆ   Iˆ A   Iˆ A   Iˆ A  J Iˆ A Iˆ B  J Iˆ A Iˆ B  J Iˆ A Iˆ B
h 1 H
A z
B z
C z
AB z z
AC z z
BC z z






J AB ˆ A ˆ B ˆ A ˆ B J AC ˆ A ˆ C ˆ A ˆ C J BC ˆ B ˆ C ˆ B ˆ C
I I  I I 
I I  I I 
I I  I I
2
2
2

Using a spin-1/2 product basis the Hamiltonian can be represented in a block diagonal
form when choosing the basis function in the appropriate order.
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
 

 

 
1 ˆ
h H  

 












H
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
G
J BC / 2 J AC / 2
J BC / 2
F
J AB / 2
J AC / 2 J AB / 2
D
0
0
0
0
0
0
0
0
E
J AB / 2 J AC / 2
C
J BC / 2
J AB / 2
0
0
0
0
0
0
0
0
J AC / 2 J BC / 2
0
0
B
0
 
0 
0 

0 
0 

0 
0 

0 

A 
where:
A  1 / 2 v A  v B  v C   1 / 4( J AB  J AC  J BC )
C  1 / 2 v A  v B  v C   1 / 4( J AB  J AC  J BC )
B  1 / 2 v A  v B  v C   1 / 4( J AB  J AC  J BC )
D  1 / 2 v A  v B  v C   1 / 4( J AB  J AC  J BC )
E  1 / 2  v A  v B  v C   1 / 4( J AB  J AC  J BC ) F  1 / 2  v A  v B  v C   1 / 4( J AB  J AC  J BC )
G  1 / 2  v A  v B  v C   1 / 4( J AB  J AC  J BC ) H  1 / 2  v A  v B  v C   1 / 4( J AB  J AC  J BC )
This matrix contains 1x1 blocks corresponding to the basis states with mT=3/2 and –3/2,
and 3x3 blocks corresponding to those with mT=-1/2 and 1/2. There are no analytical
solutions for matrices of this size. For an ABX system the situation is better. If we
factorize according to mX, giving rise to a matrix of the form:
H
0

0

0
1 ˆ
h H
0

0
0

 0
0
G
0
0
0
0
0
0
0
0
F
0
0
0
0
0
0
E
0
0
0
0
0
0
J AB 2
0
J AB 2
D
0
0
0
J AB 2 0
0
0
J AB 2
C
0
0
0
0
0
B
0
0
0
0
0
0
0 
0

0
0

0
0

A
This matrix contains four 1x1 blocks as well as two 2x2 blocks, which can be solved
exactly, just like an AB spectrum. Each block may be diagonalized separately, giving
two AB “subspectra” (figure 8), while X transitions occur between the two groups of
energy levels (see figure 9).
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Figure 8
The AB-transitions of the ABX Spectrum



mixing




mixing

Question 2
i)
How many lines do you expect to see in the AB region of the ABX spectrum?
ii)
Pick out the AB subspectra in figure 10.
iii)
How many lines are there in the X region?
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Question 3
i)
Reproduce the spectra in figures 10 and 11 using the parameters:
A = 150 Hz, B = 200 Hz, X=3000 Hz ;
JAB = 50 Hz,
ii)
JAX=150 Hz, JBX = 10 Hz.
Starting with the spectrum you just simulated vary the ratio JAB to (vA-vB) from 0
to 100.
a)
Do any lines disappear in the AB spectrum, in the extreme of the vA = vB?
b)
Do any AB lines disappear when vA = vB and JAX=JBX?
c)
What happens to the X region? Explain?
Figure 9
The X transitions of the ABX spectrum



mixing




mixing

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Figure 10
The AB region of an ABX spectrum
0
100
200
300
Frequency (Hz)
Figure 11
The X region of an ABX spectrum
3100
3000
Frequency (Hz)
2900
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The AB2 Spectrum
Magnetic equivalence can be exploited to simplify (block diagonalize) the
Hamiltonian. Take for instance the AB2 spin system. Instead of using the regular
product basis we can exploit the symmetry properties of the B2 basis. This basis has
three symmetric,  ,
1
2
   
and  , and one antisymmetric,
1
2
    ,
functions. Transitions between the symmetric and antisymmetric states are symmetry
forbidden.
Therefore if you construct a product basis between the above B2 basis
functions and a spin-1/2 basis for A (i.e.  and  ), such as
1S 
 2S


1
2
    
 3S 

 4S 

 5S 
 6S 
1
2
 7A 
 8A 
1
2
1
2
    
    
    

(n.b. the superscript S and A indicate whether the function is symmetric or
antisymmetric).
In the regular product basis there are two 3x3 blocks.
In this
representation these 3x3 blocks are reduced to a 1x1 and a 2x2 block since certain cross
terms are removed. The Hamiltonian takes on the following form:
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vA
 2  v B  J AB


0


0



0

ˆ 
h 1 H

0



0


0



0

0
vA
0
0
0
0
0
J AB
0
0
0
0
0
0
0
0
vA
 v B  J AB
2
J AB
J AB
0
0
 vA
0
0
2
J AB
2
v
 A  v B  J AB
2
0
0
0
0
2
2
0
0
0
0
v
 A  v B  J AB
2
0
0
0
0
0
0
v
 A
2
0
0
0
0
0
0

0 

0 


0 


0 


0 

0 


0 

vA 
2 
This matrix is block diagonal with only two 2x2 blocks, and therefore can be solved
analytically. The corresponding energy level diagram is seen below in figure 11.
Figure 11
The energy level diagram of an AB2 3-spin-1/2 system

21/2
mixing

mixing


21/2
21/2
21/2
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Question 4
i)
How many lines do you expect in an AB2 spectrum?
ii)
Which transition/s is/are purely due to A only?
iii)
Why are there no transitions between the energy levels on either side of the
vertical line in the diagram (figure 11)?
iv)
Simulate a series of spectra changing the ratio of the chemical shift difference to
the coupling constant from 0 to 100.
a) Which transitions disappear when the ratio is very large?
b) Why do you get only one line when the ratio is zero?
Question 5
i)
Sketch an energy level diagram for an A2B2 four-spin-1/2 system using symmetrized
wavefunctions for both groups (A2 and B2). Note that symmetrization of A2 is entirely
independent of symmetrization of B2 – so how many groups of wavefunctions with
different symmetry are there?
ii)
What is the maximum number of transitions you expect to observe? Compare this with
the number expected for a general four-spin (ABCD) system.
iii)
Simulate a series of spectra varying the ratio of the chemical shift difference with the
coupling constant from 0 to 100, and observe what happens.
iv)
Which lines disappear when the ratio is large, e.g. 100? (Hint: draw the energy level
diagram when the ratio is large.)