Chemistry Reference
In-Depth Information
12.17
Using the approach described in Problem 12.16, the molecular formula of the
compound is C
3
H
3
Cl. The molecular formula indicates that the compound has two
π
bonds or one
π
bond and one ring. Possible structures are:
H
CC
HCCl
H
2
CCCHCl
C
H
3
C
CCl
Cl
Cl
12.18
The molecular ion peak (C
7
H
14
O
+
) will appear at
m/z
= 114 (14 mass units, or one -
CH
2
- group less than for 4-octanone). Since the ketone is symmetrical, it should
fragment to yield a C
3
H
7
CO
+
peak (
m/z
= 71) and a C
3
H
7
+
peak (
m/z
= 43). Only one
set of daughter ions (instead of the two seen in Figure 12.10) will be observed.
ADDITIONAL PROBLEMS
12.19
Possible solutions are:
a.
b.
Cl
c.
CH
3
H
3
CC
CH
3
H
3
C
CH
3
Cl
H
3
C
CH
3
CH
3
d.
e.
CH
3
f.
CH
3
OCH
3
H
3
CC CH
3
CH
3
12.20
a.
There are four different types of protons:
b.
There are five different types of protons:
c
a
b
d
e
H
H
3
CCH
CH
3
N
CH
2
CH
3
a
