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

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

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Due to hyperconjugation and 9 \alpha hydrogen tertiary 3^{\circ} free radical is more stable and sit stabilized; whereas, 1^{\circ} free radical has 1 \alpha hydrogen and one hyper conjugative structure and hence is less stable.

Chlorination of alkanes proceeds by a free radical mechanism. 2-Methylpropane has both primary (1?) and tertiary (3?) carbon atoms. Thus, it will form two free radical intermediates upon chlorination.

Initiation

During monochlorination, the first step is the formation of chlorine radicals from Cl2, in the presence of light.

Propagation

Chlorine radical then generates the two types of radicals on reaction with 2-methylpropane.

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

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

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