Question

Given below are two statements:
Statement I: Vanillin will react with NaOH and also with Tollen’s reagent. 
 
Statement II: Vanillin will undergo self-aldol condensation very easily. 

In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Statement I is correct but Statement II is incorrect
• B. Statement I is incorrect but Statement II is correct
• C. Both Statement I and Statement II are incorrect
• D. Both Statement I and Statement II are correct

Hint:

Aromatic aldehydes generally do not undergo self-aldol condensation due to the lack of an active hydrogen atom, which is required for the reaction.

Answer 1

The Correct Option is A

This problem presents two statements about the chemical reactivity of vanillin. We need to analyze the functional groups present in the vanillin molecule to determine the validity of each statement.

Concept Used:

1. Acidity of Phenols: The hydroxyl (-OH) group attached directly to a benzene ring is phenolic. Phenols are acidic (pKa ≈ 10) and react with strong bases like sodium hydroxide (NaOH) in an acid-base neutralization reaction to form a sodium phenoxide salt and water.

2. Tollen's Test for Aldehydes: Tollen's reagent, an ammoniacal silver nitrate solution ( \( [\text{Ag(NH}_3)_2]^+ \) ), is a mild oxidizing agent used to distinguish aldehydes from ketones. Aldehydes are oxidized to carboxylate anions, while the Ag⁺ ions are reduced to metallic silver, forming a characteristic silver mirror.

3. Aldol Condensation: This reaction requires an aldehyde or ketone to possess at least one α-hydrogen (a hydrogen atom on the carbon adjacent to the carbonyl group). In the presence of a base, the α-hydrogen is abstracted to form a resonance-stabilized enolate ion, which then acts as a nucleophile. Aldehydes lacking an α-hydrogen cannot undergo self-aldol condensation.

Step-by-Step Solution:

Step 1: Analyze the structure of vanillin. Vanillin is 4-hydroxy-3-methoxybenzaldehyde. Its structure consists of a benzene ring with three functional groups attached:

• An aldehyde group (-CHO)
• A hydroxyl group (-OH)
• A methoxy group (-OCH₃)

The hydroxyl group is phenolic because it is directly bonded to the aromatic ring.

 

Step 2: Evaluate Statement I. "Statement I: Vanillin will react with NaOH and also with Tollen's reagent."

The phenolic -OH group is acidic and will react with the strong base NaOH through an acid-base reaction. \[ \text{Vanillin-OH} + \text{NaOH} \rightarrow \text{Vanillin-O}^-\text{Na}^+ + \text{H}_2\text{O} \] The aldehyde -CHO group is susceptible to oxidation. It will be oxidized by Tollen's reagent, a test characteristic of aldehydes. \[ \text{R-CHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\text{OH}^- \rightarrow \text{R-COO}^- + 2\text{Ag(s)} + 4\text{NH}_3 + 2\text{H}_2\text{O} \] Since vanillin contains both of these functional groups, it reacts with both reagents. Thus, Statement I is correct.

Step 3: Evaluate Statement II. "Statement II: Vanillin will undergo self-aldol condensation very easily."

Self-aldol condensation requires the presence of an α-hydrogen. The α-carbon is the carbon atom adjacent to the carbonyl carbon of the aldehyde group. In vanillin, the aldehyde group is attached directly to a carbon atom of the benzene ring. This ring carbon is the α-carbon.

This α-carbon is bonded to the carbonyl carbon, two other carbons within the aromatic ring, and has its fourth valence electron participating in the aromatic π-system. There are no hydrogen atoms attached to this α-carbon.

Because vanillin lacks an α-hydrogen, it cannot form an enolate ion, which is the necessary intermediate for aldol condensation. Therefore, it cannot undergo this reaction. Instead, aldehydes without α-hydrogens undergo the Cannizzaro reaction in the presence of a strong base. Thus, Statement II is incorrect.

Final Conclusion:

Statement I is correct as vanillin has a phenolic group to react with NaOH and an aldehyde group to react with Tollen's reagent. Statement II is incorrect as vanillin lacks an α-hydrogen, making self-aldol condensation impossible.

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