Th amount of energy needed to form the radicals $C H_{3}, 1^{o}, 2^{o}$ and $3^{o}$ decreases in the order

3^{o} > 2^{o} > 1^{o} > C H_{3}

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C H_{3} > 1^{o} > 2^{o} > 3^{o}

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1^{o} > 2^{o} > 3^{o} > C H_{3}

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C H_{3} > 3^{o} > 2^{o} > 1^{o}

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Solution

The correct option is B $C H_{3} > 1^{o} > 2^{o} > 3^{o}$
We know that the stability of radicals in the following order due to the $+ I$ effect of the alkyl group(s) in more branched radicals: $3^{o} > 2^{o} > 1^{o} > C H_{3}$ .
Since the energy of an unstable species is more and the energy of a stable species is less, the methyl radical will be at the highest energy level and so on. Hence , the energy required to form the least stable methyl radical is the greatest and so on.
so b is the answer

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