Presence of carbonyl group in the aldehydic form in glucose is confirmed by the following reagent.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Br2 water
Glucose is oxidised to the six-carbon gluconic acid by a mild oxidising agent, bromine water. Only an aldehyde group is oxidised this easily, so this shows that the carbonyl group of glucose is present as an aldehydic group:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH + Br2 + H2O -> COOH-(CHOH)4-CH2OH + 2HBr} \)
Why the other options are incorrect
Option (1) HI — Incorrect. On prolonged heating it reduces glucose to n-hexane, which proves the straight chain of six carbons.
Option (2) NH2OH — Incorrect. It forms an oxime with any carbonyl group, so it shows that >C=O is present but cannot tell an aldehyde from a ketone.
Option (4) (CH3CO)2 O — Incorrect. Acetylation shows the presence of five –OH groups, not the carbonyl group.
NCERT (2026-27 edition):
Glucose – …
