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

To arrange the given carbanions in the order of stability, we need to understand the factors that influence the stability of carbanions: 

1. Resonance Stabilization: Carbanions are more stable if the negative charge can be delocalized through resonance.
2. Inductive Effect: Electron-withdrawing groups increase stability by dispersing the negative charge, whereas electron-donating groups decrease stability.
3. Substituent Effects: Specific groups attached to the phenyl ring can either stabilize or destabilize the carbanion based on their electronic nature.

Let's analyze each carbanion:

1. IV. $p$-$\mathrm{CHO{-}C_6H_4{-}CH_2^-}$: The $p$-CHO group is an electron-withdrawing group through resonance and inductive effects. It stabilizes the carbanion by delocalizing the negative charge.
2. I. $p$-$\mathrm{Br{-}C_6H_4{-}CH_2^-}$: Bromine is slightly electron-withdrawing due to its inductive effect, providing moderate stability to the carbanion.
3. II. $\mathrm{C_6H_5{-}CH_2^-}$: The benzyl carbanion is stabilized by resonance with the benzene ring but lacks additional substituents that further stabilize or destabilize it.
4. V. $p$-$\mathrm{CH_3{-}C_6H_4{-}CH_2^-}$: The $p$-CH₃ group is electron-donating, destabilizing the carbanion slightly due to a positive inductive effect (+I effect).
5. III. $p$-$\mathrm{CH_3O{-}C_6H_4{-}CH_2^-}$: The $p$-OCH₃ group is strongly electron-donating through resonance (+R effect), which significantly destabilizes the carbanion.

Based on this analysis, the stability order of the carbanions is:

\(IV > I > II > V > III\)

Therefore, the correct answer is:

IV $>$ I $>$ II $>$ V $>$ III