Arrange the radicals (a) CH3^., (b) CH3CH2^., (c) (CH3)2CH^. and (d) CH2=CH-CH2^. in decreasing order of stability.

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Arrange the radicals (a) $CH X_{3} X^{∙}$ , (b) $CH X_{3} CH X_{2} X^{∙}$ , (c) $(CH X_{3}) X_{2} CH X^{∙}$ and (d) $CH X_{2} = CH - CH X_{2} X^{∙}$ in decreasing order of stability.

A $d > a > b > c$
B $d > c > b > a$
C $c > d > b > a$
D $a > b > c > d$

Answer: B

The allyl radical is delocalised over two carbons by resonance, so it is more stable than any simple alkyl radical. Among alkyl radicals the stability follows the number of hyperconjugating alkyl groups: secondary > primary > methyl. The order is therefore d > c > b > a.

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Arrange the radicals (a) CHX3​X∙, (b) CHX3​CHX2​X∙, (c) (CHX3​)X2​CHX∙ and (d) CHX2​=CH−CHX2​X∙ in decreasing order of stability.

Answer: B

More questions on Free radicals

Arrange the radicals (a) $CH X_{3} X^{∙}$ , (b) $CH X_{3} CH X_{2} X^{∙}$ , (c) $(CH X_{3}) X_{2} CH X^{∙}$ and (d) $CH X_{2} = CH - CH X_{2} X^{∙}$ in decreasing order of stability.