The decreasing order of stability of the foll | vedclass

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Question

(A) The stability of carbanions is determined by the electron-withdrawing or electron-donating nature of the substituents attached to the benzene ring

Vedclass · Accepted Answer

$Q > R > S > P$

Vedclass · Answer

(A) The stability of carbanions is determined by the electron-withdrawing or electron-donating nature of the substituents attached to the benzene ring.Electron-withdrawing groups $(EWG)$ increase the stability of the carbanion by dispersing the negative charge,while electron-donating groups $(EDG)$ decrease the stability by intensifying the negative charge.$(Q)$ contains the $-CHO$ group,which is a strong electron-withdrawing group ($-I$ and $-M$ effect),making it the most stable.$(R)$ contains the $-Cl$ group,which has a $-I$ effect (electron-withdrawing) but a $+M$ effect (electron-donating). The $-I$ effect dominates,providing moderate stability.$(S)$ contains the $-CH_3$ group,which is an electron-donating group ($+I$ and hyperconjugation),decreasing stability.$(P)$ contains the $-OCH_3$ group,which is a strong electron-donating group ($+M$ effect),making it the least stable.Thus,the order of stability is $Q > R > S > P$.

The decreasing order of stability of the following anions is:
$(P)$ $p-OCH_3-C_6H_4-CH_2^-$
$(Q)$ $p-CHO-C_6H_4-CH_2^-$
$(R)$ $p-Cl-C_6H_4-CH_2^-$
$(S)$ $p-CH_3-C_6H_4-CH_2^-$

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The decreasing order of stability of the following anions is:$(P)$ $p-OCH_3-C_6H_4-CH_2^-$$(Q)$ $p-CHO-C_6H_4-CH_2^-$$(R)$ $p-Cl-C_6H_4-CH_2^-$$(S)$ $p-CH_3-C_6H_4-CH_2^-$