Question

$2$-methyl butane on reaction with $B{r}_2$ in the presence of sunlight mainly gives:

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

Answer 1

The correct option is C $2$-Bromo-$2$-methylbutane

This method follows a free radical mechanism and forms alkyl halide.

Here, the reactivity of hydrogen is $3^o>2^o>1^o$. So, $2$-methylbutane on reaction with $B{r}_2$ in the presence of sunlight gives $2$-bromo-$2$-methylbutane.

$C{H}_{3}C{H}_{2}CH\left(C{H}_3\right)C{H}_3\stackrel{B{r}_2}{\to }C{H}_{3}C{H}_{2}C\left(C{H}_3\right)BrC{H}_3$

Hence, option C is correct.