Question

When phenol is treated with NaOH, CO_2 (heat), and then acidified, what results?

• A. Benzoic acid
• B. Salicylic acid
• C. Benzene
• D. Anisole

Hint:

The Kolbe-Schmitt reaction involves phenol, NaOH, CO\(_2\), and acidification to form salicylic acid.

Answer 1

The Correct Option is B

Step 1: Understanding the reaction.
This reaction is known as the Kolbe-Schmitt reaction. In this reaction, phenol reacts with sodium hydroxide (NaOH) and carbon dioxide (CO\(_2\)) at high temperature to form the sodium salt of salicylic acid. Upon acidification, the sodium salt is converted into salicylic acid.
Step 2: Conclusion.
The correct product of this reaction is salicylic acid, which corresponds to option (B).

Answer 2

Step 1: This reaction is called the Kolbe-Schmitt reaction.

Step 2: Phenol reacts with NaOH and CO₂ under heat to form the sodium salt of salicylic acid.

Step 3: Acidifying that salt converts it into salicylic acid.

Step 4: So the final product is salicylic acid — option (B).

Answer 3

Elimination approach: Benzoic acid would need oxidation of a side chain, which isn't happening here. Plain benzene would mean losing the hydroxyl group entirely, which this reaction doesn't do. Anisole would require methylation, but there's no methylating agent present. What actually happens is the phenoxide ion attacking carbon dioxide next to the oxygen, the Kolbe-Schmitt carboxylation, and acidifying the resulting salt frees salicylic acid, option (B).