JEE Chemistry 2025 Question with Solution

Given: Two statements about the reactivity of vanillin are to be checked.

Find: Which statement is correct.

Vanillin is 4-hydroxy-3-methoxybenzaldehyde. It contains a phenolic $-OH$ group and an aldehyde $-CHO$ group.

Statement I: Vanillin will react with NaOH and also with Tollen’s reagent.

The phenolic $-OH$ group reacts with NaOH:

$$\text{Vanillin-OH} + \text{NaOH} \rightarrow \text{Vanillin-O}^-\text{Na}^+ + \text{H}_2\text{O}$$

The aldehyde group gives Tollen’s test:

$$\text{R-CHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\text{OH}^- \rightarrow \text{R-COO}^- + 2\text{Ag} + 4\text{NH}_3 + 2\text{H}_2\text{O}$$

So, Statement I is correct.

Statement II: Vanillin will undergo self-aldol condensation very easily.

Aldol condensation requires at least one $\alpha$-hydrogen on the carbon adjacent to the carbonyl group. In vanillin, the aldehyde group is attached directly to the aromatic ring, so there is no $\alpha$-hydrogen available. Therefore, vanillin cannot form the enolate needed for self-aldol condensation.

Hence, Statement II is incorrect.

Aromatic aldehydes without $\alpha$-hydrogen generally undergo Cannizzaro reaction in strong base instead of aldol condensation.

Therefore, the correct option is A.