Name _________________________
Exam 3, 2009:
CHEM/BCMB 4190/6190/8189
1
Exam 3:
CHEM/BCMB 4190/6190/8189 (114 points)
Thursday, 29 October, 2009
1
). For a simple two spin (
1
H-
13
C) coupled system (i.e.
13
CHCl
3
), the population diagram is
shown (below, right). Here “H
1
” and “H
2
” represent the
1
H transitions, and “C
1
” and “C
2
”
represent the
13
C transitions. We’ll consider the
effect of the pulse sequence shown (below, left)
assuming the Larmor frequency is equal to the
reference frequency (
±
L
=
±
ref
). Assume
²
=1/(4J) and
ignore the effects of magnetic field inhomogeneity.
a. What is the name of the pulse sequence shown. If it is an acronym, state what the acronym
stands for?
(
2 points
)
INEPT (Insensitive Nuclei Enhanced by Polarization Transfer)
b.
Complete the vector diagrams below for points ‘c’, ‘d’, ‘e’, and ‘f’ in the pulse sequence.
Be
sure to label the vectors (M
H
C
³
, M
H
C
´
), to include arrows indicating the direction of precession for
the vectors in the rotating frame, and to indicate the angle (
µ
)
between the vectors at each point.
In this example, the precession frequency of M
H
C
´
is slightly faster than the Larmor frequency.
(
8 points
)
e
²
a
b
c
d
f
g
90x
180x
90y
180x
90x
13
C
²
1
H
g
¶
h
1
H
13
C
N
2
N
1
N
3
N
4
C
1
³´
´³
³³
´´
C
2
H
1
H
2
x
µ
= 0°
M
H
C
´
M
H
C
´
M
H
C
³
transverse
plane
y
z
x
90
°
x
y
x
²
M
H
C
³
M
H
C
´
x
µ
= 90°
y
a
b
c
180
°
x
1
H
M
H
C
´
M
H
C
³
µ
= 90°
y
180
°
x
13
C
²
M
H
C
³
M
H
C
´
µ
= 180°
x
y
d
e
f
x
M
H
C
³
M
H
C
´
µ
= 90°
y
M
H
C
³

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Name _________________________
Exam 3, 2009:
CHEM/BCMB 4190/6190/8189
2
c. Which of the hydrogen transitions, H
2
(N
1
-N
3
) or H
1
(N
2
-N
4
) corresponds to M
H
C
±
?
(
2 points
)
H
2
d. Draw the vector diagram showing the M
H
C
±
and M
H
C
²
vectors at point ‘g’ in the pulse
sequence. Make sure to label each vector (M
H
C
±
, M
H
C
²
).
(
4 points
).
e. Ignoring the effects of relaxation and magnetic field inhomogeneity, write down the
populations of the states N
1
, N
2
, N
3
and N
4
at each point to ‘g/g
³
’ in the sequence. The values at
equilibrium (at point ‘a’) are given below (note: point ‘d’ is following the
1
H 180
°
x pulse, and
‘e/f’ is, subsequently, following the
13
C 180
°
x pulse).
(
8 points
)
‘a’
‘b/c’
‘d’
N
4
= N
N
4
= N +
´
H/2
N
4
= N +
´
H/2
N
3
= N +
´
C
N
3
= N +
´
C +
´
H/2
N
3
= N +
´
C +
´
H/2
N
2
= N +
´
H
N
2
= N +
´
H/2
N
2
= N +
´
H/2
N
1
= N +
´
C +
´
H
N
1
= N +
´
C +
´
H/2
N
1
= N +
´
C +
´
H/2
‘e/f’
‘g/g
³
’
N
4
= N +
´
H/2
N
4
= N +
´
H
N
3
= N +
´
C +
´
H/2
N
3
= N +
´
C
N
2
= N +
´
H/2
N
2
= N
N
1
= N +
´
C +
´
H/2
N
1
= N +
´
C +
´
H
M
H
C
²
M
H
C
±
y
z
x
g