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Chapter 13 431
Copyright © 2017 Pearson Education, Inc.
Solutions to Problems
1.
Only positively charged fragments are accelerated through the analyzer tube.
+
CH CH CH
3 2 2
CH
3
CH
2
CH
3
[
]
+
.
+
CH CH CH
2
2
2.
The peak at
m
>
z
=
57 is more intense for 2,2-dimethylpropane than for isopentane or pentane. The peak
at
m
>
z
=
57 is due to loss of a methyl radical: loss of a methyl radical from 2,2-dimethylpropane forms
a tertiary carbocation, whereas loss of a methyl radical from isopentane or pentane forms a less stable
secondary and primary carbocation, respectively.
CH
3
CH
3
CCH
3
CH
3
+
•
+
CH
3
•
+
CH
3
•
+
CH
3
•
2,2-dimethylpropane
CH
3
CHCH
2
CH
3
+
•
2-methylbutane
CH
3
CH
3
a secondary carbocation
m
/
z
=
57
3
CHCH
2
CH
+
CH
3
CH
2
CH
2
CH
2
CH
3
+
•
pentane
CH
3
CH
2
CH
2
CH
2
a primary carbocation
m
/
z
=
57
+
CH
3
CH
3
CCH
3
+
a tertiary carbocation
m
/
z
=
57
Notice that the mass spectrum of isopentane can be distinguished from those of the other isomers by the
peak at
m
>
z
=
43. The peak at
m
>
z
=
43 is intense for isopentane because such a peak is due to loss of
an ethyl radical, which forms a secondary carbocation. Pentane gives a less intense peak at
m
>
z
=
43
because loss of an ethyl radical from pentane forms a primary carbocation. 2,2-Dimethylpropane does not
show a peak at
m
>
z
=
43 because it does not have an ethyl group.
CH
3
CHCH
2
CH
3
+
•
CH
3
CH
2
•
m
/
z
=
43
CH
3
CH
3
CH
2
CH
2
CH
2
CH
3
+
•
CH
3
CH
2
CH
2
m
/
z
=
43
+
CH
3
CH
CH
3
+
+
+
CH
3
CH
2
•
a secondary carbocation
a primary carbocation