n-Hexane is obtained when glucose is

The correct answer is Option (1) → Prolonged heated with HI

The open-chain structure of glucose was worked out from its reactions, each of which reveals one feature of the molecule. On prolonged heating with HI, glucose is reduced completely: every –OH group and the –CHO group are replaced by hydrogen, and n-hexane is formed. Since n-hexane is an unbranched chain, this shows that the six carbon atoms of glucose are linked in a straight chain.

    \( \displaystyle \ce{OHC-(CHOH)4-CH2OH} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-CH2-CH2-CH2-CH2-CH3} \)

Why the other options are incorrect

• Option (2) Reacted with hydroxylamine. — Incorrect. Glucose forms an oxime, which shows that it has a carbonyl group:

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH + NH2OH -> HON=CH-(CHOH)4-CH2OH + H2O} \)

• Option (3) Treated with mild oxidising agent like bromine water. — Incorrect. Glucose is oxidised to gluconic acid, which shows that the carbonyl group is an aldehyde group:

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{HOOC-(CHOH)4-CH2OH} \)

• Option (4) Treated with acetic anhydride. — Incorrect. Glucose forms glucose pentaacetate, which shows that it has five –OH groups:

      \( \displaystyle \ce{OHC-(CHOH)4-CH2OH + 5(CH3CO)2O -> OHC-(CHOCOCH3)4-CH2OCOCH3 + 5CH3COOH} \)

NCERT (2026-27 edition):
Glucose – …

Glucose – “… 3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C = O) in glucose.
4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group. …
5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups.”