Question

What is the major product obtained in the bromination of `(CH_3)_(2)CHCH_2CH_3`?

Answer 1

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-methylbutane.