Bromination of isobutane gives 99% *
Answers
Answer:
Explanation:
The major product obtained in the bromination of (CH₃)₂CHCH₂CH₃ is (CH₃)₂CBrCH₂CH₃.
The free-radical bromination of this alkane produces four products:
1-bromo-3-methylbutane (A)
2-bromo-3-methylbutane (B)
2-bromo-2-methylbutane (C)
1-bromo-2-methylbutane(D)
But bromine is highly regioselective in where it attacks.
The relative rates of attack at the different types of H atom are 3°:2°:1° = 1640:82:1.
To calculate the relative amounts of each product, we multiply the numbers of each type of atom by the relative reactivity. Then
A: (1°): 3 × 1 = 3
B: (2°): 2 × 82 = 164
C: (3°): 1 × 1640 = 1640
D: (1°): 6 × 1 = 6
The percentages of each isomer are
A:
3
1813
×
100
%
=
0.1
%
B:
164
1813
×
100
%
=
9.0
%
C:
1640
1813
×
100
%
=
90.5
%
D:
6
1813
×
100
%
=
0.3
%
So the major product is C, 2-bromo-2-methylbutaneThe major product obtained in the bromination of (CH₃)₂CHCH₂CH₃ is (CH₃)₂CBrCH₂CH₃.
The free-radical bromination of this alkane produces four products:
1-bromo-3-methylbutane (A)
2-bromo-3-methylbutane (B)
2-bromo-2-methylbutane (C)
1-bromo-2-methylbutane(D)
But bromine is highly regioselective in where it attacks.
The relative rates of attack at the different types of H atom are 3°:2°:1° = 1640:82:1.
To calculate the relative amounts of each product, we multiply the numbers of each type of atom by the relative reactivity. Then
A: (1°): 3 × 1 = 3
B: (2°): 2 × 82 = 164
C: (3°): 1 × 1640 = 1640
D: (1°): 6 × 1 = 6
The percentages of each isomer are
A:
3
1813
×
100
%
=
0.1
%
B:
164
1813
×
100
%
=
9.0
%
C:
1640
1813
×
100
%
=
90.5
%
D:
6
1813
×
100
%
=
0.3
%
So the major product is C, 2-bromo-2-methylbutane