Chemistry Reference
In-Depth Information
to the hydroxyl group, and the remaining 1H signal is due to the -OH. Compound
B
must therefore be (CH
3
)
2
CHCH=O.
12.47
The formula C
5
H
12
O = C
5
H
11
OH tells us that the alcohols are saturated. The peak at
m/z
= 59 cannot be caused by a four-carbon fragment (C
4
= 4 x 12 = 48; this would
require 11 hydrogens, too many for four carbons). Thus the peak must contain one
oxygen, leaving 59 - 16 = 43 for carbon and hydrogen. A satisfactory composition for
the peak at 59 is C
3
H
7
O
+
(or C
3
H
6
OH
+
). Similarly, the
m/z
= 45 peak corresponds to
C
2
H
5
O
+
(or C
2
H
4
OH
+
). Fragmentation of alcohols often occurs between the hydroxyl-
bearing carbon and an attached hydrogen or carbon (eq. 12.8). Possible structures
for the first alcohol are
H
CH
3
CH
3
CH
2
C
OH
H
3
CCOH
CH
2
CH
3
or
CH
2
CH
3
3-pentanol
2-methyl-2-butanol
Possible structures of its isomer are
H
H
or
H
3
CC
OH
CH
3
C
OH
CH
2
CH
2
CH
3
CH(CH
3
)
2
2-pentanol
3-methyl-2-butanol
The correct structure for the first isomer could be deduced by NMR spectroscopy as
follows:
δ
4, 1 H, quintet
H
H
3
CCH
2
C
CH
2
CH
3
OH
δ
1.3, 4 H, quintet or multiple t
δ
1, 3 H, tripl et
three
13
C peaks
δ
1.3, 2 H, quartet
CH
3
CH
3
C
CH
2
CH
3
OH
δ
1, 3 H, triplet
δ
1, 6 H, singlet
four
13
C peaks
The correct structure for the second isomer also could be deduced by NMR
spectroscopy. However, both proton spectra are likely to be quite complex because
of similar chemical shifts and a great deal of spin-spin coupling. The
13
C spectra are
simpler and diagnostic:
CH
3
C HCH
2
CH
2
CH
3
OH
CH
3
CHC H(CH
3
)
2
OH
six types of protons
five types of protons
five
13
C peaks
four
13
C peaks
