Arrange the following carbanions in the decre | vedclass

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Question

(A) The stability of carbanions is determined by the inductive effect ($-I$ and $+I$ effects) and resonance effect.$1$. $CCl_3^-$: The three chlorine

Vedclass · Accepted Answer

$CCl_3^- > {C_6H_5}CH_2^- > {(CH_3)_2}CH^- > {(CH_3)_3}C^-$

Vedclass · Answer

(A) The stability of carbanions is determined by the inductive effect ($-I$ and $+I$ effects) and resonance effect.$1$. $CCl_3^-$: The three chlorine atoms exert a strong $-I$ effect,which stabilizes the negative charge significantly.$2$. ${C_6H_5}CH_2^-$: The negative charge is stabilized by resonance with the phenyl ring.$3$. ${(CH_3)_2}CH^-$: This is a secondary carbanion,which is destabilized by the $+I$ effect of two methyl groups.$4$. ${(CH_3)_3}C^-$: This is a tertiary carbanion,which is most destabilized by the $+I$ effect of three methyl groups.Therefore,the decreasing order of stability is $CCl_3^- > {C_6H_5}CH_2^- > {(CH_3)_2}CH^- > {(CH_3)_3}C^-$. The correct option is $A$.

Arrange the following carbanions in the decreasing order of their stability: ${(CH_3)_3}C^-$,$CCl_3^-$,${(CH_3)_2}CH^-$,${C_6H_5}CH_2^-$.

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