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Chapter 14 477
Copyright © 2017 Pearson Education, Inc.
73.
Using the formula given on page 621 of the text:
∆
E
=
h
n
=
h
g
2
p
B
0
Given: Planck
,
s constant
=
h
=
6.626
*
10
-
34
Js
1 cal
=
4.184 J
300 MHz
=
7.046 T
1
page 622 of the text
2
g
for
1
H
=
2.675
*
10
8
T
-
1
s
-
1
∆
E
=
h
g
2
p
B
0
=
6.626
*
10
-
34
Js
*
2.675
*
10
8
T
-
1
s
-
1
2
1
3.1416
2
*
1.4092 T
*
1 cal
4.184 J
=
9.50
*
10
-
26
cal
74.
All four spectra show a singlet at
2.0
ppm, suggesting that they are all esters with a methyl group attached
to the carbonyl group. Now the problem becomes determining the nature of the group attached to the oxygen.
C
H
3
C
O
O R
a.
The highest frequency signal in the first spectrum is a triplet that integrates to 2 protons, indicating
that a
CH
2
group is attached to the oxygen and is bonded to another
CH
2
group. The lowest frequency
signal is a triplet that integrates to 3 protons, indicating a methyl group that is attached to a
CH
2
group.
The presence of two multiplets that each integrate to 2 protons confirms the structure.
C
H
3
C O
O
CH
2
CH
2
CH
2
CH
3
b.
The highest frequency signal in the second spectrum is a multiplet that integrates to 1 proton, indicat-
ing that the carbon attached to the oxygen is attached to one proton and two nonequivalent carbons
bonded to hydrogens. The lowest frequency signal is a triplet that integrates to 3 protons, indicating a
methyl group attached to a
CH
2
group. The doublet at
1.2
ppm that integrates to 3 protons is due to a
methyl group attached to a carbon bonded to one hydrogen.
C
CHCH CH
3 2
O
O
CH
3
H
3
C
c.
The highest frequency signal is a doublet that integrates to 2 protons, indicating that the methylene
group that is attached to the oxygen is attached to a carbon bonded to one hydrogen. The lowest fre-
quency signal that integrates to 6 protons indicates two equivalent methyl groups that are attached to a
carbon bonded to one hydrogen.
C
O
O
CH
2
CHCH
3
CH
3
CH
3