Arrange the following carbanions in the decreasing order of stability.
I., p—Br—C₆H₄—C̈(⊖)H₂; II., C₆H₅—C̈(⊖)H₂; III., p—CH₃O—C₆H₄—C̈(⊖)H₂; IV., p—CHO—C₆H₄—C̈(⊖)H₂; V., p—CH₃—C₆H₄—C̈(⊖)H₂
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

A negative charge is happy when the para group pulls electrons away and unhappy when it pushes them in.

1. The rule. In each ion the negative charge on CH₂⁻ spreads into the ring and reaches the para position. A group there that withdraws electrons stabilises the carbanion, and a group that donates electrons destabilises it.

   Ion	Para group	Electronic effect	Effect on the carbanion
   IV	-CHO	-M and -I, strong withdrawal	most stable
   I	-Br	-I is stronger than its weak +M, net withdrawal	stabilised, but less than IV
   II	-H	reference	benzyl carbanion only
   V	-CH₃	+I and hyperconjugation, weak donation	destabilised
   III	-OCH₃	+M is much stronger than -I, strong donation	least stable

2. Ends of the order. The C=O of -CHO can take the negative charge onto oxygen by resonance, so IV comes first. The lone pair on the oxygen of -OCH₃ pushes electron density into the ring by resonance. That piles more charge onto an already negative system, so III comes last.
3. So. The decreasing order of stability is IV > I > II > V > III, which is option (d).

The idea to keep. Carbanions are the mirror image of carbocations: electron-withdrawing groups (-M, then -I) stabilise them, and electron-donating groups (+M, then +I) destabilise them. Resonance effects beat inductive effects.

Answer from the NTA final answer key. Solution by PrepWiser.