Question

Arrange the following carbanions in the decreasing order of stability:
I. $p$-$\mathrm{Br{-}C_6H_4{-}CH_2^-}$ 
II. $\mathrm{C_6H_5{-}CH_2^-}$ 
III. $p$-$\mathrm{CH_3O{-}C_6H_4{-}CH_2^-}$ 
IV. $p$-$\mathrm{CHO{-}C_6H_4{-}CH_2^-}$ 
V. $p$-$\mathrm{CH_3{-}C_6H_4{-}CH_2^-}$ 
Choose the correct answer from the options given below: 
 

• A. IV $>$ I $>$ II $>$ V $>$ III
• B. I $>$ IV $>$ II $>$ V $>$ III
• C. I $>$ II $>$ IV $>$ V $>$ III
• D. IV $>$ II $>$ I $>$ III $>$ V

Hint:

Electron withdrawing groups stabilize carbanions, while electron donating groups destabilize them.

Answer 1

The Correct Option is A

Step 1: Stability principle for carbanions.
Carbanions are stabilized by electron withdrawing groups and destabilized by electron donating groups. 
Step 2: Analysing substituent effects. 
(IV) $-CHO$: Strong $-M$ and $-I$ effect, maximum stabilization. 
(I) $-Br$: $-I$ effect dominates, stabilizing the carbanion. 
(II) H: No substituent effect, moderate stability. 
(V) $-CH_3$: $+I$ effect, destabilizes carbanion. 
(III) $-OCH_3$: Strong $+M$ effect, maximum destabilization. 
Step 3: Ordering stability. 
\[ \text{IV}>\text{I}>\text{II}>\text{V}>\text{III} \] Step 4: Final conclusion. 
The correct decreasing order of stability is IV $>$ I $>$ II $>$ V $>$ III, corresponding to option (1).