JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Biomolecules

The key facts are:

1. X cannot be oxidized by bromine water → X is a non-reducing sugar (no free anomeric carbon).
2. Acid hydrolysis gives a laevorotatory solution.

Sucrose is a non-reducing disaccharide (glucose + fructose linked via both anomeric carbons). Upon hydrolysis:
$\text{Sucrose}\stackrel{{\mathrm{H}}^{+}}{\to }\text{Glucose}(+52.5°)+\text{Fructose}(-92°)$

The mixture is laevorotatory because fructose has a larger negative rotation. Sucrose itself cannot be oxidized by ${\mathrm{Br}}_2/{\mathrm{H}}_2\mathrm{O}$.

Options (B), (C), and (D) are reducing sugars (they have a free anomeric OH) and would be oxidized by bromine water.

Therefore, X is sucrose, option (A).