The presence of carbonyl group in glucose is confirmed by treating it…

The correct answer is Option (3) → NH₂OH

The open chain structure of glucose was worked out from its reactions. A carbonyl group (>C=O) is shown by reactions in which the C=O group adds a reagent: glucose reacts with hydroxylamine to form an oxime, and adds HCN to form a cyanohydrin:

    \( \displaystyle \ce{CHO-(CHOH)4-CH2OH + NH2OH -> CH(=N-OH)-(CHOH)4-CH2OH + H2O} \)

Why the other options are incorrect

• Option (1) HI — Incorrect. On prolonged heating with HI, glucose gives n-hexane; this shows that the six carbon atoms are in a straight chain.

      \( \ce{C6H12O6} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-(CH2)4-CH3} \)

• Option (2) Mild Oxidising agent Br₂ water — Incorrect. Bromine water oxidises glucose to gluconic acid; this shows that the carbonyl group is an aldehyde group, once its presence is known.

      \( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\text{Br}_2\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)

• Option (4) HNO₃ — Incorrect. Nitric acid oxidises glucose to saccharic acid, showing a primary alcohol (–\( \ce{CH2OH} \)) group.

      \( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \)

NCERT (2026-27 edition):
Glucose – …