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Spectroscopy Problems
Copyright © 2017 Pearson Education, Inc.
3.
First, we need to determine the most abundant cationic fragments for each compound.
The possible fragments for
pentane
are:
CH
3
CH
2
CH
2
CH
2
CH
3
1
•
+
CH
3
m/z
= 57
+
+
CH
2
CH
2
CH
2
CH
3
CH
3
CH
2
CH
2
CH
2
CH
3
2
•
+
CH
2
CH
3
m/z
= 43
+
+
CH
2
CH
2
CH
3
CH
3
CH
2
CH
2
CH
2
CH
3
3
•
+
CH
2
CH
2
CH
3
m/z
= 29
+
+
CH
2
CH
3
CH
3
CH
2
CH
2
CH
2
CH
3
4
•
+
CH
2
CH
2
CH
2
CH
3
m/z
= 15
+
+
CH
3
The most abundant fragments result from bond cleavages that produce the most stable cations and radicals.
Fragments from
2
and
3
are the most abundant because, in each case, a primary carbocation and a primary
radical are formed.
2
is expected to give the base peak (the most stable fragment). The cation formed in
2
(
m
>
z
=
43)
is more
stable than the cation formed in
3
(
m
>
z
=
29),
because the former is more stabilized by inductive electron
donation from the alkyl group.
Fragments from
1
and
4
are expected to be less abundant. They each form one primary species, but the
second species is a methyl fragment (either a radical or a carbocation), which is less stable than the second
species formed in
2
and
3
.
Four sets of fragments are shown for
isopentane
. Fragmentations that result in a primary fragment and
a methyl fragment have been excluded because they would be less abundant than those shown here.
CH
3
CH CH
2
CH
3
CH
3
CH
3
3
•
+
m/z
= 43
+
CH
3
CH
CH
2
CH
3
CH
3
CHCH
2
CH
3
CH
3
1
•
+
CH
3
m/z
= 57
+
CH
3
CHCH
2
CH
3
+
+
CH
3
CH CH
2
CH
3
CH
3
CH
3
CH
3
CH
4
•
+
m/z
= 29
+
+
CH
2
CH
3
CH
3
CHCH
2
CH
3
CH
3
2
•
+
CH
3
CHCH
2
CH
3
m/z
= 15
+
+
CH
3