Arrange the following carbanions in decreasin | vedclass

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(C) The stability of carbanions is increased by electron-withdrawing groups ($-I$ or $-M$ effect) and decreased by electron-donating groups ($+I$ or $

Vedclass · Accepted Answer

$Q > R > S > P$

Vedclass · Answer

(C) The stability of carbanions is increased by electron-withdrawing groups ($-I$ or $-M$ effect) and decreased by electron-donating groups ($+I$ or $+M$ effect). $1.$ In $(Q)$,the $-CHO$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which stabilizes the negative charge significantly. $2.$ In $(R)$,the $-Cl$ group has a $-I$ effect (electron-withdrawing) but a $+M$ effect (electron-donating). The $-I$ effect dominates,providing some stabilization. $3.$ In $(P)$,the $-OCH_3$ group has a strong $+M$ effect (electron-donating),which destabilizes the carbanion. $4.$ In $(S)$,the $-CH_3$ group has a $+I$ effect (electron-donating) and hyperconjugation,which also destabilizes the carbanion. Comparing the effects: $(Q)$ is most stable due to strong $-M$ effect. $(R)$ is next due to $-I$ effect. $(P)$ is less stable than $(S)$ because the $+M$ effect of $-OCH_3$ is stronger than the $+I$ effect of $-CH_3$ in destabilizing the carbanion. Therefore,the order of stability is $Q > R > S > P$.

Arrange the following carbanions in decreasing order of stability $:$

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