Which of the following reaction conditions confirm the presence of six carbons and five hydroxyl groups in glucose, respectively?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Prolonged heating with HI and acetylation with acetic anhydride.
The open chain structure of glucose was worked out from a set of reactions, each of which proves one feature:
Prolonged heating with HI gives n-hexane, so the six carbon atoms form a straight chain:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow[\text{heat}]{\ce{HI}} \ce{CH3-(CH2)4-CH3} \)
Acetylation with acetic anhydride gives glucose pentaacetate, so there are five –OH groups:
\( \displaystyle \ce{C6H12O6 + 5(CH3CO)2O -> C6H7O(OCOCH3)5 + 5CH3COOH} \)
Oxidation with bromine water gives gluconic acid, which proves an aldehyde group:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)
Oxidation with nitric acid gives saccharic acid, which proves a primary alcohol group:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \)
Why the other options are incorrect
Option (1) Heating with HI and oxidation with HNO3 — Incorrect. HI proves the six-carbon chain, but \( \ce{HNO3} \) oxidation proves the primary alcohol group, not five –OH groups.
Option (2) Heating with Br2, water and oxidation. — Incorrect. Bromine water oxidation proves the aldehyde group; it says nothing about the number of carbons or –OH groups.
Option (3) Oxidation with HNO3 and heating with HI. — Incorrect. The order is wrong, and \( \ce{HNO3} \) oxidation does not count –OH groups.
NCERT (2026-27 edition):
Glucose – “… 2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain. … 5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.”