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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
attached to a carbon that is not attached to a proton. Because we know that the compound has only four
carbons and contains a bromine, the compound must be
2-bromo-2-methylpropanoic acid
.
Br
OH
O
26.
The quintet
( 2.2 ppm)
that integrates to 2 protons indicates that the protons that give this signal have
four identical neighboring protons. A carbon cannot be bonded to four protons and still be able to bond to
anything else. Therefore, the two protons that give the quintet must be bonded to a carbon that is attached
to two methylene groups in the same environment.
The triplet
( 3.8 ppm)
that integrates to 4 protons must be the signal for the four protons of the two
methylene groups. The two methylene groups must be on either side of a carbon that is bonded to two
protons (that is, the protons that give the quintet).
C
H
C
H
H HH
H
C
Two bonds are left unaccounted for, so this is where the two chlorines shown in the molecular formula go.
Therefore, the compound is
1,3-dichloropropane
. The highly deshielded nature of the signal at 3.8 ppm
for the protons bonded to the carbons that are attached to chlorines is further evidence that the chlorines are
attached to these carbons.
ClCH
2
CH
2
CH
2
Cl
27.
The IR spectrum shows a strong and sharp absorption at
1720 cm
-
1
,
indicating a carbonyl
(C
“
O)
group. The two absorptions at 2720 and
2820 cm
-
1
are characteristic of an aldehyde; they are due to the
C
¬
H
stretch of the bond between the carbonyl carbon and the aldehydic hydrogen. Because the reactant
has three carbons, the aldehyde that produces the IR spectrum must also have three carbons. Therefore, the
product of the reaction is
propanal
.
CH
3
CH
2
C
O
H
Thus, the reaction that occurred was the conversion of 1-propyne to propanal. This reaction can occur by
hydroboration–oxidation of the alkyne (Section 7.8 in the text).
1. R
2
BH, THF
2. H
2
O
2
, HO
−
, H
2
O
C H
3
C CH
CH
3
CH
2
C
O
H