Which one of the carbanions is the least stab | vedclass

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Question

(C) The stability of carbanions is determined by the electron-withdrawing or electron-donating nature of the groups attached to the negatively charged

Vedclass · Accepted Answer

$^{\ominus}CH_2-CH_3$

Vedclass · Answer

(C) The stability of carbanions is determined by the electron-withdrawing or electron-donating nature of the groups attached to the negatively charged carbon.Electron-withdrawing groups ($-NO_2$,$-CHO$) stabilize the carbanion by dispersing the negative charge.Electron-donating groups $(-CH_3)$ destabilize the carbanion by increasing the electron density on the negatively charged carbon.Comparing the given options:$1$. $^{\ominus}CH_2-NO_2$: $-NO_2$ is a strong electron-withdrawing group ($-I$ and $-M$ effect).$2$. $^{\ominus}CH_2-CHO$: $-CHO$ is an electron-withdrawing group ($-I$ and $-M$ effect).$3$. $^{\ominus}CH_2-CH_3$: $-CH_3$ is an electron-donating group ($+I$ effect).$4$. $^{\ominus}CH_3$: This is a methyl carbanion with no additional alkyl groups.Between $^{\ominus}CH_2-CH_3$ and $^{\ominus}CH_3$,the ethyl carbanion $(^{\ominus}CH_2-CH_3)$ is less stable due to the $+I$ effect of the additional methyl group,which increases electron density on the anionic center.

Which one of the carbanions is the least stable?

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Which of the following are secondary radicals?
$(1) \ CH_3 - \dot{C}H - C_2H_5$
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$(3) \ CH_2 = CH - \dot{C}H_2$
$(4) \ (CH_3)_2 \dot{C}H$

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