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

2-methylpropane gives two types of radicals.

\[\begin{array}{cc}
\phantom{...}\ce{CH3}\phantom{..................}\ce{CH3}\phantom{................}\ce{CH3}\phantom{..}\\
\phantom{..}|\phantom{.....................}|\phantom{....................}|\phantom{...}\\
\ce{\underset{(2-methylpropane)}{CH3 - CH - CH3} -> CH3 - \underset{(I)}{\underset{\bullet}{C}} - CH3 and CH3 - \underset{(II)}{\underset{\bullet}{CH}} - CH2}
\end{array}\]

Radical (I) is more stable because it is 3° and stabilised by nine hyperconjugative structures (as it has 9 CC-hydrogens).

Radical (II) is less stable because it is 1° and stabilised by only one hyperconjugative structure (as it only 1 a-hydrogens).