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462 Chapter 14
Copyright © 2017 Pearson Education, Inc.
12.
Recall that the highest frequency signal is the one that is farthest to the left on the spectrum. The proton(s)
that is underlined in the answer gives the higher frequency signal.
a.
CH
3
CHCHBr
Br Br
b.
CH
3
CHOCH
3
CH
3
c.
CH
3
CH
2
CHCH
3
Cl
d.
CH
3
CH CH
2
CH
3
O
CH
3
_
e.
CH
3
CH
2
CH CH
2
f.
CH
3
OCH
2
CH
2
CH
3
13.
a.
CH
3
CH
2
CH
2
Cl
b.
CH
3
CH
2
CHCH
3
Cl
c.
CH
3 2
CH H
O
_
14.
a.
O
C
CH
a b
c
3
CH
2
H
b.
CH
3
CH
2
CHCH
3
OCH
3
a c e b
d
c.
ClCH
2
CH
2
CH
2
Cl
b
b a
d.
CH
3
CH
2
CHCH
2
CH
3
a b d
OCH
3
b a
c
e.
CH
3
CH
2
CH
2
a b c
OCH
3
d
O
C
f.
CH
3
CH
2
CH
2
OCHCH
3
a c d e
b
CH
3
b
g.
CH
3
CH
2
CH
2
CH
3
a b d
c
O
C
h.
CH
3
CHCH
2
OCH
3
a b d c
CH
3
a
i.
CH
3
CHCHCH
3
a c d b
CH
3
Cl
a
15.
From the direction of the electron flow around the benzene ring pictured in Figure 14.6 on page 631 of the
text, you can see that the magnetic field induced in the region of the hydrogens that protrude out from the
compound in this problem is in the same direction as the applied magnetic field. However, the magnetic
field induced in the region of the hydrogens that protrude into the center of the compound is in the opposite
direction of the applied magnetic field.
Therefore, the signal at 9.25 ppm is for the hydrogens that protrude out because they need a higher
frequency to come into resonance due to the fact that they sense a larger effective magnetic field since the
induced and applied magnetic fields are in the same direction. The signal at
-
2.88
ppm is for the hydro-
gens that protrude inward because a smaller frequency is necessary since the induced magnetic field and
the applied magnetic field are in opposite directions.
16.
Each of the compounds would show two signals, but the ratio of the integrals for the two signals will be
different for each of the compounds. The ratio of the integrals for the signals given by the first compound
will be 2:9 (or 1:4.5), the ratio of the integrals for the signals given by the second compound will be 1:3,
and the ratio of the integrals for the signals given by the third compound will be 1:2.