Question

Which one of the following substituents at para-position is most effective in stabilizing the phenoxide ion ?

• A. $-CH_3$
• B. $-OCH_3$
• C. $-COCH_3$
• D. $-CH_2OH$

Answer 1

The Correct Option is C

To determine which substituent at the para-position is most effective in stabilizing the phenoxide ion, we need to consider electron-withdrawing and electron-donating properties of each substituent. The stability of the phenoxide ion can be significantly affected by the nature of these substituents.

Phenoxide ion (\( \text{C}_6\text{H}_5\text{O}^-\)) stability is influenced by the resonance and inductive effects of substituents. A substituent that can either withdraw electrons through resonance or push electrons away through inductive effects is likely to stabilize the negative charge on the oxygen atom in the phenoxide ion.

1. -CH_3: This is a simple alkyl group and is an electron-donating group. Being an electron-donating group, it tends to destabilize the phenoxide ion by increasing electron density on an already negatively charged oxygen atom.
2. -OCH_3: The methoxy group is an electron-donating group through resonance (\( +M \) effect), contributing electron density to the phenoxide ion. Although it can donate electrons via resonance, this generally results in reduced stability of the negatively charged phenoxide ion compared to electron-withdrawing groups.
3. -COCH_3: The acetyl group (\( -COCH_3 \)) is an electron-withdrawing group through resonance (\( -M \) effect) as well as through the inductive effect (\( -I \)). It stabilizes the phenoxide ion by withdrawing electron density, thereby delocalizing the negative charge more effectively over the aromatic system.
4. -CH_2OH: The hydroxymethyl group can act as an electron-donating group through resonance and inductive effect. It is not as effective in stabilizing the negative charge as electron-withdrawing groups like \( -COCH_3 \).

Based on these analyses, the \( -COCH_3 \) substituent is the most effective in stabilizing the phenoxide ion due to its strong electron-withdrawing properties, which help to delocalize and stabilize the negative charge on the oxygen atom.

Conclusion: The correct choice is -COCH_3 as it effectively stabilizes the phenoxide ion by pulling electron density via the \( -M \) and \( -I \) effects, thus minimizing the electron density on the phenoxide oxygen.