Question

The percentage of isomers formed during monobromination of $2,3-dimethyl$ butane at room temperature relative to $1,2,3$ H atoms to chlorination is $(1.0:3.8:5.0)$

• A. 81%
• B. 56%
• C. 15%
• D. 22%

Answer 1

Answer: (D)

22%

(A) is obtained from the reaction of $12$ equivalent $1$ H atoms.

(B) is obtained from the reaction of two equivalent $3$ H atoms.

Percentage of (A) $=\frac{12\times 100}{33}=54.5$

Percentage of (B) $=\frac{10\times 100}{22}=45.5$

$12\times 1=12.0$
$2\times 5=10.0$
Total $=22.0$