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432 Chapter 13
Copyright © 2017 Pearson Education, Inc.
3.
Intense peaks are expected at
m
>
z
=
57 for loss of an ethyl radical (86
-
29) and at
m
>
z
=
71 for loss of
a methyl radical (86
-
15).
CH
3
CH
2
CHCH
2
CH
3
+
•
CH
2
CH
3
•
m
/
z
=
57
CH
3
CH
3
CH
2
CHCH
2
CH
3
m
/
z
=
71
+
CH
3
CH
2
CH
CH
3
+
CH
3
•
CH
3
CH
2
CHCH
2
CH
3
+
•
CH
3
+
+
A secondary carbocation is formed in each case. Because an ethyl radical is more stable than a methyl radical,
the base peak is most likely at
m
>
z
=
57.
4.
Solved in the text.
5.
a.
Dividing 72 by 13 gives 5 with 7 left over, giving a base value of C
5
H
12
. Because the compound
contains only carbons and hydrogens, we know that the base value is also the molecular formula of the
compound.
b.
Dividing 100 by 13 gives 7 with 9 left over, giving a base value of C
7
H
16
. Because the compound
contains one oxygen, an O must be added to the base value and one C and four Hs must be subtracted.
Therefore, the molecular formula is C
6
H
12
O.
c.
Dividing 102 by 13 gives 7 with 11 left over, giving a base value of C
7
H
18
. Because the compound
contains two oxygens, two Os must be added to the base value and two Cs and eight Hs must be
subtracted. Therefore, the molecular formula is C
5
H
10
O
2
.
d.
Dividing 115 by 13 gives 8 with 11 left over, giving a base value of C
8
H
19
. Because the compound
contains one oxygen, an O must be added to the base value and one C and four Hs must be subtracted.
Because the compound contains an N, an N must be added to the base value and one C and two Hs
must be subtracted. Therefore, the molecular formula is C
6
H
13
NO.
6.
a. 1.
15
+
(3
*
14)
+
16
=
73
2.
16
+
(3
*
14)
+
16
=
74
b.
An alkane has an even-mass molecular ion. If a CH
2
group (14) of an alkane is replaced by an NH
group (15) or if a CH
3
group (15) of an alkane is replaced by an NH
2
group (16), the molecular ion will
have an odd mass.
A second nitrogen in the molecular ion will cause it to have an even mass.
Thus, for a molecular ion to have an odd mass, it must have an odd number of nitrogens.
c.
An even-mass molecular ion either has no nitrogens or has an even number of nitrogens.
7.
a.
Dividing 86 by 13 gives 6 with 8 left over, giving a base value of C
6
H
14
.
If the compound contains only carbons and hydrogens, the base value is also the molecular formula of
the compound. Some possible structures are: