JEE Chemistry 2023 Question with Solution

Given: X is an L-tetrose with two chiral carbons, gives positive Schiff's test, and on acetylation gives triacetate.

Find: the correct structure of X.

A positive Schiff's test shows that X contains an aldehyde group, so X is an aldotetrose. Formation of triacetate means there are three alcoholic -OH groups available for acetylation in the open-chain form consistent with the given carbohydrate structure.

From the solution, oxidation with $\mathrm{HNO_3}$ gives a dicarboxylic acid A, so both terminal groups are oxidizable as expected for an aldotetrose. Reduction with $\mathrm{NaBH_4}$ gives polyol B, and B is stated to be optically active. Therefore, the reduced product cannot be a meso tetritol.

The solution working identifies the structure that satisfies all these conditions as the L-tetrose shown in option (4).

Therefore, the correct option is D.

Given: X is an L-isomer of tetrose and reacts as described.

Find: which option matches X.

1. Schiff's test positive implies presence of $-CHO$, so X must be an aldose.
2. Two chiral carbons matches an aldotetrose Fischer projection.
3. Acetylation gives triacetate confirms three hydroxyl groups are acetylated.
4. On oxidation with $\mathrm{HNO_3}$, the aldehyde end and terminal alcohol end give the corresponding dicarboxylic acid.
5. On reduction with $\mathrm{NaBH_4}$, the aldehyde group becomes $-CH_2OH$ to form a tetritol that is optically active.

Among the given structures, the one consistent with the L-configuration and the reaction products shown in the solution is option (4).

Hence, X = option D.