midterm-2003-answers - Name Midterm Exam CHEM/BCMB...

Name _________________________
Midterm Exam: CHEM/BCMB 4190/6190/8189
1
Midterm Exam: CHEM/BCMB 4190/6190/8189 (155 pts) Tuesday, 7 October, 2003
1). In the example (right), the effect of a 90 ° ( π /2) pulse
applied along the “x” axis (90 ° x) is shown. For a-f below,
show the effect of the indicated pulses on the indicated
magnetization vectors by drawing the resulting vectors on
the coordinate axes. (12 pts)
a. d.
b. e.
c. f.
2). A diagram of the Zeeman levels/states for 13 C is shown to the right.
Draw the Zeeman diagram for 10 B ( I = 3). (4 pts)
90 ° x
M
0
y
z
x
y
z
x
0
E
m
+1/2
-1/2
E =
+1/2 γ hB
0
E =
-1/2 γ hB
0
E = 0 γ hB
0
= 0
E = -1 γ hB
0
E = -2 γ hB
0
E = -3 γ hB
0
E = +1 γ hB
0
E = +2 γ hB
0
E = +3 γ hB
0
E
m
+1
-1
+3
0
-2
+2
-3
90 ° -x
y
z
x
y
z
x
90 ° -y
y
z
x
y
z
x
180 ° x
y
z
x
y
z
x
270 ° y
y
z
x
y
z
x
360 ° -y
y
z
x
y
z
x
90 ° -y
y
z
x
y
z
x

Name _________________________
Midterm Exam: CHEM/BCMB 4190/6190/8189
2
3). If we directly compare 10 B spins and 13 C spins;
a. What is the ratio of the bulk/macroscopic magnetization ( M
0 10B
/ M
0 13C
) produced for
equal numbers of nuclei? (6 pts)
b. How does the sensitivity of the NMR signal compare for 10 B and 13 C spins? (6 pts)
a.
M
0 10B
M
0 13C
=
N γ
10B
2
h 2 B
0
I
10B
(I
10B
+1)
3k
B
T
N γ
13C
2
h 2 B
0
I
13C
(I
13C
+1)
3k
B
T
=
γ
10B
2
× 3 3 + 1
( )
γ
13C
2
×
1
2
1
2
+ 1
⎛
⎝
⎜
⎞
⎠
⎟
=
(2.8747 × 10 7 rad/Ts) 2 × 12
(6.7283 × 10 7 rad/Ts) 2 ×
3
4
= 2.9207
So, the bulk magnetization is about 2.9 times as large for 10 B as for 13 C.
b.
Sensitivity is proportional to the electromotive force ( ε ) induced in the receiver coil by the bulk
magnetic moment. The magnitude of ε is proportional to the rate of change in the magnetic
moment (
ε ∝ dM/dt = γ M
0
B
).
Remember, M
0
=
N γ
2
h
2
B
0
I(I +1)
3k
B
T
(“h” here is “h-bar”, Planck’s constant divided by 2 π ),
so ε ∝ γ M
0
B =
N γ 3 h 2 B
0
2
I(I +1)
3k
B
T
. So, whereas the magnitude of the bulk magnetization was
dependent on γ 2 , the sensitivity is dependent on γ 3 .
Thus,
ε
10B
ε
13C
=
N γ
10B
3
h 2 B
0
2
I
10B
(I
10B
+1)
3k
B
T
N γ
13C
3
h 2 B
0
2
I
13C
(I
13C
+1)
3k
B
T
=
γ
10B
3
I
10B
(I
10B
+ 1)
γ
13C
3
I
13C
(I
13C
+ 1)
=
(2.8747 × 10 7 rad/Ts) 3 × 12
(6.7283 × 10 7 rad/Ts) 3 ×
3
4
= 1.248
So, 10 B nuclei are about 1.25 times more sensitive than 13 C nuclei.

Name _________________________
Midterm Exam: CHEM/BCMB 4190/6190/8189
3
4). The frequency difference ( ∆ν ) between two 1 H signals in a 1 H spectrum is 4320 Hz when the
spectrum is acquired with a magnetic field strength of 9.4 T (corresponding to a 1 H
resonance/observe frequency of 400 MHz).
a. What would be the frequency difference between the two 1 H signals if the spectrum
was acquired with a 18.79 T magnet? (5 pts)
b. What is the chemical shift difference ( ∆δ ) between the two signals? (5 pts)
a. We know that 18.79 T ≡ 800 MHz,
∆ δ
400
=
∆ ν
400
400 MHz
× 10 6 and ∆ δ
800
=
∆ ν
800
800 MHz
× 10 6
. We
also know that, by definition, ∆δ
400
= ∆δ
800
(chemical shift is independent of B
0
). So, the ratio of
the frequencies is equal to the ratio of the field strengths:
∆ ν
400
∆ ν
800
=
400 MHz
800 MHz
, therefore , ∆ ν
800
= 4320 Hz ×
800 MHz
400 MHz
= 8640 Hz
b. ∆ δ =
∆ ν
observe frequency
× 10
6
=
4320 Hz
400 MHz
× 10
6
=
4320 Hz
400 Hz
= 10.8 "ppm"
5). a. For the hydrocarbon below, the T
1
values for the 13 C nuclei (in seconds, shown below
the respective carbon atoms) increase towards the ends of the hydrocarbon chain.