Question

2-Methylbutane on reacting with bromine in presence of sunlight gives mainly:

• A. 1-Bromo-2-methylbutane
• B. 2-Bromo-2-methylbutane
• C. 2-Bromo-3-methylbutane
• D. 1-Bromo-3-methylbutane

Answer 1

The correct option is B 2-Bromo-2-methylbutane

$\begin{array}{c}C{H}_{3}CH\left(C{H}_3\right)C{H}_{2}C{H}_3+B{r}_2\stackrel{sunlight}{\to }\end{array}{H}_{3}CC\left(Br\right)\left(C{H}_3\right)C{H}_{2}C{H}_3$
The reaction proceeds via the free radical mechanism.
Since the radical formed in this case is tertiary, as it is more stable than primary and secondary, 2-Bromo-2-methylbutane is formed.
Option B is correct.