Question

Write hydrocarbon radicals that can be formed as intermediates during monochlorination of $2$-methylpropane? Which of them is more stable? Give reasons.

Answer 1

Chlorination of alkanes proceeds by a free radical mechanism. $2$-Methylpropane has both primary $\left(1^{\circ }\right)$ and tertiary $\left(3^{\circ }\right)$ carbon atoms. Thus, it will form two free radical intermediates upon chlorination.

Initiation

During monochlorination, the first step is formation of chlorine radical from $C{l}_2$ , in the presence of light.

$Cl-Cl\underrightarrow{hv}2Cl$

Propagation

Chlorine radical then generates the two types of radicals on reaction with $2$-methylpropane, which can be given as




Due to the presence of a $p$-orbital with odd electrons, intermediates like free radicals are stabilized by hyperconjugation via neighbouring $C-H$ bonds. Thus, more the number of alpha hydrogens, greater the stability.

Tertiary free radical have $9$ alpha hydrogens, while primary free radical have $1$ alpha hydrogen. Hence, tertiary free radical will get more stabilized via hyperconjugation than primary free radical.