Identify the structure of vanillin

Vanillin is 4-hydroxy-3-methoxybenzaldehyde. It contains:

• a phenolic OH group,
• a methoxy group,
• an aldehyde group attached to a benzene ring.

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

(a) Reaction with NaOH

Vanillin has a phenolic OH group. Phenols are weak acids and react with NaOH to form phenoxide salts: $\text{ArOH} + \text{NaOH} \rightarrow \text{ArONa} + \text{H}_2\text{O}$ So, vanillin does react with NaOH.

(b) Reaction with Tollen's reagent

Vanillin also contains an aldehyde group. Aldehydes reduce Tollen's reagent to metallic silver: $\text{RCHO} + 2[\text{Ag}(\text{NH}_3)_2]^+ + 3\text{OH}^- \rightarrow \text{RCOO}^- + 2\text{Ag} + 4\text{NH}_3 + 2\text{H}_2\text{O}$ Therefore, vanillin does react with Tollen's reagent.

Hence, Statement I is correct.

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

For self-aldol condensation, the carbonyl compound must have at least one alpha-hydrogen on the carbon adjacent to the carbonyl carbon.

In vanillin, the aldehyde group is directly attached to the benzene ring: $\text{Ar} - \text{CHO}$ The carbon adjacent to the carbonyl carbon is part of the aromatic ring, so there is no alpha-hydrogen available.

Therefore, vanillin cannot undergo self-aldol condensation easily (in fact, it does not undergo normal self-aldol condensation).

Hence, Statement II is incorrect.

Conclusion

• Statement I: Correct
• Statement II: Incorrect

So, the correct option is: $\boxed{\text{A}}$

Comparison with stored answer

Stored correct answer = A

My derived answer = A

Therefore, they agree.