JEE Main 2024 — Biomolecules Question with Solution

$\mathrm{Glucose}\underset{∆}{\overset{{\mathrm{NaHCO}}_3}{\to }}\mathrm{no} \mathrm{reaction}$

$\mathrm{Glucose}\underset{∆}{\overset{{\mathrm{HNO}}_3}{\to }}\mathrm{saccharic} \mathrm{acid}$

Glucose, on oxidation with nitric acid, gives saccharic acid.

$\mathrm{Glucose}\underset{∆}{\overset{\mathrm{HI}}{\to }}\mathrm{n}-\mathrm{hexane}$

Glucose, on prolonged heating with hot HI, gives n-Hexane.

$\mathrm{Glucose}\underset{∆}{\overset{{\mathrm{Br}}_2}{\to }}\mathrm{Gluconic} \mathrm{acid}$

Glucose reacts with bromine water to form gluconic acid.

 Bromine water is a mild oxidising agent which selectively oxidizes aldehyde to carboxylic acid only. It doesn’t oxidise alcohol or ketone.

Hence option B is the answer.