Question

Which of the following is correct decreasing order of acidic strength for the compounds listed below?
I: p-chlorophenol, II: p-Cresol, III: p-nitrophenol, IV: p-methoxyphenol.

• A. II \(>\) IV \(>\) I \(>\) III
• B. I \(>\) II \(>\) III \(>\) IV
• C. III \(>\) I \(>\) II \(>\) IV
• D. IV \(>\) III \(>\) I \(>\) II

Hint:

Electron withdrawing groups raise acidity of phenol and electron donating groups lower it.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A phenol loses \(\text{H}^+\) to give a phenoxide ion. Anything that stabilises the phenoxide makes the phenol more acidic.

Step 2: Effect of each group:
III: \(-\text{NO}_2\) is strongly electron withdrawing (\(-I\) and \(-R\)), so it stabilises the negative charge most. It is the most acidic.
I: \(-\text{Cl}\) withdraws electrons by \(-I\) effect (this outweighs its weak \(+R\)), so it is more acidic than phenol, but less than nitro.
II: \(-\text{CH}_3\) donates electrons by \(+I\) and hyperconjugation, which destabilises the phenoxide, so it is less acidic than phenol.
IV: \(-\text{OCH}_3\) donates electrons by a strong \(+R\) effect, destabilising the phenoxide even more, so it is the least acidic.

Step 3: Order:
\[ \text{III} > \text{I} > \text{II} > \text{IV} \]
This is option C. The other options put donor-substituted phenols above the acceptor-substituted ones, which is the wrong direction.

Final Answer:
The decreasing order of acidity is III > I > II > IV, option (C). \[ \boxed{\text{III} > \text{I} > \text{II} > \text{IV}} \]