432 / 912
424 Chapter 12
Copyright © 2017 Pearson Education, Inc.
43.
a.
HO
HBr
peroxide
HO
Br
NaH
_
O
Br
O
b.
HBr
peroxide
Br
CH
3
O
_
OCH
3
c.
HBr
peroxide
OH
H
2
SO
4
Br
d.
PdL
2
(CH
3
CH
2
)
3
N
Br
Br
Br
Br
2
HO
−
H
2
Pd/C
44.
2-Methylbutane forms two primary alkyl halides, one secondary alkyl halide, and one tertiary alkyl halide.
First, we need to calculate how much alkyl halide is formed from each primary hydrogen available, from
each secondary hydrogen available, and from each tertiary hydrogen available.
a primary alkyl halide
ClCH
2
CHCH
2
CH
3
CH
3
Substitution of any one of six hydrogens leads to this product.
percentage of this product that is formed
=
36
,
percentage formed per hydrogen available
=
36
>
6
=
6
,
CH
3
CHCH
2
CH
2
Cl
CH
3
Substitution of any one of three hydrogens leads to this product.
percentage of this product that is formed
=
18
,
percentage formed per hydrogen available
=
18
>
3
=
6
,
a secondary alkyl halide
CH
3
CHCHCH
3
CH
3
Cl
Substitution of any one of two hydrogens leads to this product.
percentage of this product that is formed
=
28
,
percentage formed per hydrogen available
=
28
>
2
=
14
,
a tertiary alkyl halide
CH
3
CCH
2
CH
3
Cl
CH
3
Substitution of the one tertiary hydrogen leads to this product.
percentage of this product that is formed
=
18
,
percentage formed per hydrogen available
=
18
>
1
=
18
,
From the above calculations, we see that at 300
°
C, the relative rates of removal of a hydrogen atom from a
tertiary, secondary, and primary carbocation are:
18 : 14 : 6
=
3.0 : 2.3 : 1