The reaction of glucose with which of the following reagent confirms the presence of five hydroxyl groups?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Acetic anhydride
The open chain structure of glucose, \( \ce{CHO-(CHOH)4-CH2OH} \), was worked out from a set of reactions, each of which reveals one feature of the molecule:
Prolonged heating with HI: gives n-hexane, so the six carbon atoms are in a straight chain.
Hydroxylamine: gives an oxime, so glucose has a carbonyl group (C=O).
Bromine water: a mild oxidising agent, gives gluconic acid (a six-carbon acid), so the carbonyl group is an aldehyde group.
Acetic anhydride: acetylation gives glucose pentaacetate, so there are five –OH groups:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH + 5(CH3CO)2O -> CHO-(CHOCOCH3)4-CH2OCOCH3 + 5CH3COOH} \)
Nitric acid: oxidation gives saccharic acid (a dicarboxylic acid), so there is a primary alcoholic (–CH2OH) group.
So the presence of five hydroxyl groups is confirmed by the reaction with acetic anhydride.
Why the other options are incorrect
Option (1) Hydroxylamine — Incorrect. The oxime shows the carbonyl group.
Option (2) Bromine water — Incorrect. Gluconic acid shows that the carbonyl group is an aldehyde.
Option (4) Hydroiodic acid — Incorrect. n-hexane shows the straight chain of six carbons.
NCERT (2026-27 edition):
Glucose – …
