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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
9.
From the reaction conditions provided, we know that the product is a monochlorinated toluene.
Cl
CH
3
CH
3
Cl
2
, AlCl
3
The singlet (2.3 ppm) that integrates to 3 protons is due to the methyl group.
The signals in the 7–8 ppm region that integrate to 4 protons are due to the protons of a disubstituted
benzene ring. Because both signals are doublets, we know that each proton is coupled to one adjacent
proton. Thus, the compound has a 1,4-substituted benzene ring.
Therefore, the compound is
4-chloromethylbenzene
.
10.
The strong and broad absorption in the IR spectrum at
3400 cm
-
1
indicates a hydrogen-bonded
O
¬
H
group. The absorption bands between 2800 and
3000 cm
-
1
indicate hydrogens bonded to
sp
3
carbons.
Only one signal in the
1
H NMR
spectrum integrates to 1 proton, so it must be due to the hydrogen of the
OH group. The singlet that integrates to 3 protons can be attributed to a methyl group that is attached to a
carbon that is not attached to any hydrogens.
Because the other two signals show splitting, we know that they represent coupled protons (that is, protons
on adjacent carbons). The quartet and triplet combination indicates an ethyl group. Because the quartet and
triplet integrate to 6 and 4 protons, respectively, the compound must have two ethyl groups.
The identified fragments of the molecule are:
C
H
d
H
d
2
H
a
C
H
a
H
a
C
H
b
H
b
H
b
O H
c
When these fragments are subtracted from the molecular formula, only one carbon remains. Therefore, this
carbon must connect the four identified fragments. The compound is
3-methyl-3-pentanol
.
OH
11.
The
1
H NMR
spectrum contains only one signal, so only one type of hydrogen is present in the molecule.
Because the compound has 4 carbons and 9 identical hydrogens, the compound must be
tert
-butyl
bromide
.
CH
3
Br
C
CH
3
CH
3