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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
Now that the three fragments have been identified, we know that the compound is
ethylmethylamine
.
N
H
The second spectrum shows that a broad singlet (2.8 ppm) must be due to hydrogens that are attached to
nitrogens. Because the signal integrates to 2 protons, we know that the compound is a primary amine.
The triplet (0.8 ppm) that integrates to 3 protons is due to a methyl group that is adjacent to a methylene
group. The triplet (2.7 ppm) that integrates to 2 protons must also be adjacent to a methylene group. The
multiplet (1.5 ppm) that integrates to 2 protons is the methylene group that splits both the methyl and
methylene groups. (The two triplets and multiplet are characteristic of a propyl group.)
Therefore, the compound is
propylamine
.
N
H
H
42.
The relatively weak absorption in the IR spectrum at
1650 cm
-
1
tells us that it is probably due to a
carbon–carbon double bond. This is reinforced by the presence of absorptions at
3080 cm
-
1
,
indicating
C
¬
H
bond stretches of hydrogens attached to
sp
2
carbons.
The shape of the two absorptions at
3300 cm
-
1
suggests the presence of an
NH
2
group of a primary
amine. (Compare these to the shape of the
N
¬
H
stretches of an
NH
2
group of an amide in Problem 40.)
The three signals in the NMR spectrum between 5.0 and 6.0 ppm that integrate as a group to 3 protons
indicate that there are three vinylic protons. Therefore, we know that the alkene is monosubstituted.
The two remaining signals in the NMR spectrum are a doublet (3.3 ppm) and a singlet (1.3 ppm) that each
integrate to 2 protons. Because splitting is not typically seen with protons attached to nitrogens, we can
identify the singlet at 1.3 ppm as due to the two amine protons. The doublet must be due to a methylene
group that is attached to an
sp
2
carbon and split by a vinylic proton that is attached to the same carbon. The
compound, therefore, is
allylamine
.
CH
2
NH
2
H
H H
CC
Now we can understand why the signal at 5.9 ppm is a multiplet. This vinylic proton is split by the
methylene group and two unique vinylic protons. The signals for the other two vinylic protons are doublets
because each is split by the single proton attached to the adjacent
sp
2
carbon. Notice that the higher-
frequency doublet has the larger
J
value. This is the signal for the proton that is trans to its coupled proton.
43.
The molecular formula tells us that the compound does not have any degrees of unsaturation. Therefore,
the oxygen must be the oxygen of either an ether or an alcohol. Because there are no signals that integrate
to one proton, we can conclude that the compound is an ether.