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(A) The lower stability of ethyl anion $(CH_3CH_2^-)$ compared to methyl anion $(CH_3^-)$ is due to the $+I$-effect of the methyl group,which increase

Vedclass · Accepted Answer

$+I$-effect of the methyl group in ethyl anion; $\sigma \rightarrow p$-orbital conjugation in ethyl radical

Vedclass · Answer

(A) The lower stability of ethyl anion $(CH_3CH_2^-)$ compared to methyl anion $(CH_3^-)$ is due to the $+I$-effect of the methyl group,which increases the electron density on the already negatively charged carbon atom,thereby destabilizing it.The higher stability of the ethyl radical $(CH_3CH_2^\bullet)$ compared to the methyl radical $(CH_3^\bullet)$ is due to $\sigma \rightarrow p$-orbital conjugation,which is known as hyperconjugation. This delocalization of electrons from the $\sigma$-bond of the $C-H$ bond into the half-filled $p$-orbital stabilizes the radical.

The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical,respectively,are due to

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