The correct stability order for the following | vedclass

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(C) The stability of carbanions is increased by electron-withdrawing groups $(EWG)$ through $-I$ and $-M$ effects.$1$. In $(IV)$,there is no substitue

Vedclass · Accepted Answer

$I > III > II > IV$

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(C) The stability of carbanions is increased by electron-withdrawing groups $(EWG)$ through $-I$ and $-M$ effects.$1$. In $(IV)$,there is no substituent,only the phenyl ring providing resonance stabilization.$2$. In $(I)$,$(II)$,and $(III)$,the $-CN$ group is present,which is a strong electron-withdrawing group ($-I$ and $-M$ effects).$3$. In $(III)$ $(p-CN)$,the $-CN$ group exerts both $-I$ and $-M$ effects,providing maximum stabilization via resonance.$4$. In $(I)$ $(o-CN)$,the $-CN$ group exerts both $-I$ and $-M$ effects,but it is closer to the negative charge,making it highly stable due to the strong $-I$ effect.$5$. In $(II)$ $(m-CN)$,the $-CN$ group exerts only the $-I$ effect because the resonance effect $(-M)$ does not operate at the meta position.Comparing the effects:- $(III)$ $(p-CN)$ and $(I)$ $(o-CN)$ are more stable than $(II)$ $(m-CN)$ due to the additional $-M$ effect.- Between $(I)$ and $(III)$,the ortho-effect and proximity of the $-CN$ group make $(I)$ the most stable.- Thus,the order is $(I) > (III) > (II) > (IV)$.

The correct stability order for the following species is:
$(I)$ $o-CN-C_6H_4-CH_2^-$
$(II)$ $m-CN-C_6H_4-CH_2^-$
$(III)$ $p-CN-C_6H_4-CH_2^-$
$(IV)$ $C_6H_5-CH_2^-$
(Note: The image shows the structures of these carbanions.)

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The correct stability order for the following species is:$(I)$ $o-CN-C_6H_4-CH_2^-$$(II)$ $m-CN-C_6H_4-CH_2^-$$(III)$ $p-CN-C_6H_4-CH_2^-$$(IV)$ $C_6H_5-CH_2^-$(Note: The image shows the structures of these carbanions.)