Name _________________________
Exam 4, 2010:
CHEM/BCMB 4190/6190/8189
1
Exam 4:
CHEM/BCMB 4190/6190/8189 (118 points)
Thursday, 9 December, 2010
1
). Consider the HETCOR pulse sequence, and
the energy diagram for the simple
1
H
13
C spin
system (i.e. CHCl
3
).
Just
before
the second
1
H 90
°
pulse,
for a particular t
1
value, the M
H
C
α
and
M
H
C
β
vectors happen to be positioned
along the –y and y axes, respectively.
a.
Define the populations for the individual states N
1
N
4
and the population differences for the
A
1
, A
2
, X
1
and X
2
transitions
both
before
the beginning of the pulse sequence and
after
the
second
1
H 90
°
pulse (assuming the particular value of t
1
mentioned above, and no relaxation
losses during t
1
).
Assume that N
4
=N at equilibrium, and
Δ
H is the difference in the number of
spins in
α
and
β
states for
1
H, and
Δ
X is the difference in the number of spins in
α
and
β
states
for
13
C.
Also, define the relationship between
Δ
H and
Δ
X.
(
8 points
)
After the second
1
H pulse, the M
H
C
β
vector will be inverted from its equilibrium position (
z), whereas the M
H
C
α
vector will be in the same position as it was at equilibrium (z).
Thus, the N
2
and N
4
populations are inverted after the second
1
H pulse.
Also, the ratio of
Δ
H and
X is the ratio of their gyromagnetic ratios, which is ~4:
equilibrium (beginning
after second
of pulse sequence)
1
H pulse
N
4
= N
N
4
= N +
H
N
3
= N +
X
N
3
= N +
X
N
2
= N +
H
N
2
= N
N
1
= N +
H +
X
N
1
= N +
H +
X
A
1
= N
2
– N
4
=
H
A
1
= N
2
– N
4
= 
H
A
2
= N
1
– N
3
=
H
A
2
= N
1
– N
3
=
H
X
1
= N
3
– N
4
=
X
X
1
= N
3
– N
4
=
X –
H = 3
X
X
2
= N
1
– N
2
=
X
X
2
= N
1
– N
2
=
X +
H = 5
X
13
C
1
H
N
2
N
1
N
3
N
4
X
1
αβ
βα
αα
ββ
X
2
A
1
A
2
13
C
1
H
90x
90x
90x
t
1
t
2
M
H
C
β
M
H
C
α
y
z
x
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Exam 4, 2010:
CHEM/BCMB 4190/6190/8189
2
b.
Draw the vectors M
C
H
α
and M
C
H
β
that are present just before the
13
C pulse.
Make sure to label
each with the correct name (M
C
H
α
or M
C
H
β
) and with the proper magnitude (expressed in terms of
Δ
X).
(
4 points
)
c.
In the modified HETCOR experiment (right) an
additional pulse (
13
C 180
°
x) and an additional delay (
Δ
)
have been added in order to remove the
1
J
CH
splitting in
the t
1
dimension (decouple).
For the simple
1
H
13
C spin
system (i.e. CHCl
3
), at point “a” in the sequence, the M
H
C
α
and M
H
C
β
vectors are positioned as shown below (in the
transverse plane). For an arbitrary value of t
1
/2, the vectors
have moved into the positions shown in “b”.
Show the results at points “c” and “d”.
Then explain why decoupling is achieved.
(
8 points
)
Decoupling is achieved because the
1
J
CH
coupling is refocused during the t
1
/2
13
C 180x
t
1
/2 period.
At “d”, the M
H
C
α
and M
H
C
β
vectors are refocused, and the net precession to
this point for each vector is the same.
d.
The delay
Δ
has to be set appropriately in order for the pulse sequence to operate properly.
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