What is the major product obtained in the bromination of #(CH_3)_(2)CHCH_2CH_3#?
1 Answer
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:
B:
C:
D:
So the major product is C, 2-bromo-2-methylbutane.
