Question:
The correct decreasing order of acidic strength is
The correct decreasing order of acidic strength is
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In aromatic compounds, electron-donating groups (like \( \text{CH}_3 \)) decrease acidity, while electron-withdrawing groups increase acidity.
Updated On: Apr 28, 2026
- \( \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
- \( \text{C}_2\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_6\text{H}_5\text{OH} \)
- \( m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
- \( \text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \)
- \( \text{C}_6\text{H}_5\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} \)
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The Correct Option is D
Solution and Explanation
Step 1: Understand the acidic strength concept.
The presence of electron-withdrawing groups increases acidity, while electron-donating groups decrease acidity.
Step 2: Identify the effect of substituents.
- \( \text{C}_6\text{H}_5\text{OH} \) (phenol) has no substituent and is the most acidic.
- \( m\text{CH}_3\text{C}_6\text{H}_4\text{OH} \) (meta-hydroxy-toluene) has an \( \text{CH}_3 \) group at the meta position, which is slightly electron-donating but not enough to reduce acidity significantly.
- \( p\text{CH}_3\text{C}_6\text{H}_4\text{OH} \) (para-hydroxy-toluene) has the \( \text{CH}_3 \) group in the para position, where it is electron-donating through both inductive and hyperconjugative effects, making it less acidic than the meta derivative.
- \( \text{C}_2\text{H}_5\text{OH} \) (ethanol) has an \( \text{C}_2\text{H}_5 \) group, which is electron-donating and makes it the least acidic.
Step 3: Compare the acidic strengths.
The acidic strength decreases in the order: \[ \text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \]
Step 4: Final conclusion.
Therefore, the correct order is: \[ \boxed{\text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH}} \] Thus, the correct option is: \[ \boxed{(4)} \]
The presence of electron-withdrawing groups increases acidity, while electron-donating groups decrease acidity.
Step 2: Identify the effect of substituents.
- \( \text{C}_6\text{H}_5\text{OH} \) (phenol) has no substituent and is the most acidic.
- \( m\text{CH}_3\text{C}_6\text{H}_4\text{OH} \) (meta-hydroxy-toluene) has an \( \text{CH}_3 \) group at the meta position, which is slightly electron-donating but not enough to reduce acidity significantly.
- \( p\text{CH}_3\text{C}_6\text{H}_4\text{OH} \) (para-hydroxy-toluene) has the \( \text{CH}_3 \) group in the para position, where it is electron-donating through both inductive and hyperconjugative effects, making it less acidic than the meta derivative.
- \( \text{C}_2\text{H}_5\text{OH} \) (ethanol) has an \( \text{C}_2\text{H}_5 \) group, which is electron-donating and makes it the least acidic.
Step 3: Compare the acidic strengths.
The acidic strength decreases in the order: \[ \text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH} \]
Step 4: Final conclusion.
Therefore, the correct order is: \[ \boxed{\text{C}_6\text{H}_5\text{OH} > m\text{CH}_3\text{C}_6\text{H}_4\text{OH} > p\text{CH}_3\text{C}_6\text{H}_4\text{OH} > \text{C}_2\text{H}_5\text{OH}} \] Thus, the correct option is: \[ \boxed{(4)} \]
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