On prolonged heating with HI , glucose forms

The correct answer is Option (3) → n-Hexane

The open chain structure of glucose was worked out from a set of reactions, each of which reveals one feature of the molecule:

• Prolonged heating with HI gives n-hexane: the six carbon atoms are in a straight chain.

• Reaction with hydroxylamine (oxime) and HCN (cyanohydrin): a carbonyl group is present.

• Oxidation by bromine water gives gluconic acid: the carbonyl group is an aldehyde group.

• Acetylation with acetic anhydride gives a pentaacetate: five –OH groups are present.

• Oxidation by nitric acid gives saccharic acid: one of the –OH groups is a primary alcohol group.

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

HI removes all the oxygen atoms and reduces the chain completely. So glucose on prolonged heating with HI forms n-hexane.

Why the other options are incorrect

• Option (1) Starch — Incorrect. Starch is a polysaccharide made of a large number of glucose units. HI does not join glucose units together; it removes the oxygen atoms and reduces the chain.

• Option (2) Gluconic acid — Incorrect. It is formed by oxidation of glucose with bromine water.

• Option (4) Fructose — Incorrect. Fructose is an isomer of glucose, \( \ce{C6H12O6} \), with a ketone group. HI does not turn glucose into another sugar; the product has no oxygen at all.

NCERT (2026-27 edition):
Glucose – …

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