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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

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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 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

Similar Questions

Choose the correct statement regarding the formation of carbocations $A$ and $B$ given:
$CH_3-CH_2-CH=CH_2 + HBr \to CH_3-CH_2-CH_2-CH_2^+ + Br^-$ $(A)$
$CH_3-CH_2-CH=CH_2 + HBr \to CH_3-CH_2-CH^+-CH_3 + Br^-$ $(B)$

Which of the following is most stable?

What is the minimum energy required for the homolytic bond fission of the $C-H$ bond in the following molecule?

The correct stability order for the following carbocations is:

Stability order of the following carbocations:
$(i) CH_3 - C^{+}(OCH_3) - CH_3$
$(ii) CH_3 - C^{+}(CH_3) - CH_3$
$(iii) CH_3 - NH - C^{+}H_2$
$(iv) CH_3 - C^{+}H_2$

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