Question:
Which of the following compounds is most acidic?
Which of the following compounds is most acidic?
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Electron-withdrawing groups, especially when located at the para position, increase the acidity of phenolic compounds.
Updated On: Apr 6, 2025
- \( \text{C}_6\text{H}_5\text{OH} \)
- \( \text{o}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \)
- \( \text{m}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \)
- \( \text{p}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \)
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The Correct Option is D
Solution and Explanation
The presence of electron-withdrawing groups (such as nitro groups) in aromatic compounds increases the acidity of hydroxyl groups attached to the ring. Among the given options, the nitro group (\( \text{NO}_2 \)) at the para position exerts a strong electron-withdrawing effect, making the hydroxyl group more acidic.
- In \( \text{p}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \), the nitro group at the para position is most effective in withdrawing electron density, thus making this compound the most acidic.
- \( \text{m}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \) also has a nitro group, but its effect is weaker because it is in the meta position.
- \( \text{o}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \) has the nitro group in the ortho position, but the effect is still less than the para position due to steric hindrance.
- \( \text{C}_6\text{H}_5\text{OH} \) has no electron-withdrawing group and, therefore, is the least acidic.
Thus, the most acidic compound is \( \text{p}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \).
- In \( \text{p}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \), the nitro group at the para position is most effective in withdrawing electron density, thus making this compound the most acidic.
- \( \text{m}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \) also has a nitro group, but its effect is weaker because it is in the meta position.
- \( \text{o}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \) has the nitro group in the ortho position, but the effect is still less than the para position due to steric hindrance.
- \( \text{C}_6\text{H}_5\text{OH} \) has no electron-withdrawing group and, therefore, is the least acidic.
Thus, the most acidic compound is \( \text{p}-\text{O}_2\text{N}-\text{C}_6\text{H}_4\text{OH} \).
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