Identify the incorrect statement for glucose

The correct answer is Option (3) → Respond to Schiff's test

The open-chain structure of glucose, \( \ce{CH2OH(CHOH)4CHO} \), explains most of its reactions, but not all. Glucose actually exists mainly as a cyclic hemiacetal, in which the –CHO group has reacted with the –OH group at C-5. There is very little free aldehyde, so glucose does not give Schiff’s test and does not form the hydrogensulphite addition product, although mild oxidising agents can still oxidise it.

• Option (1) Responds to Tollen's Test — Shown by glucose. Glucose reduces Tollens’ reagent to a silver mirror and is oxidised to gluconate:

      \( \displaystyle \ce{CH2OH(CHOH)4CHO + 2[Ag(NH3)2]+ + 3OH- -> CH2OH(CHOH)4COO- + 2Ag v + 4NH3 + 2H2O} \)

• Option (2) On reduction with HI it forms n-hexane — Shown by glucose. Prolonged heating with HI removes all the oxygen and gives n-hexane, which shows that the six carbon atoms form a straight chain:

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

• Option (3) Respond to Schiff's test — Not shown by glucose. Glucose exists mainly as the cyclic hemiacetal, with very little free –CHO group, so it does not restore the pink colour of Schiff’s reagent.

• Option (4) Respond to Fehling's test — Shown by glucose. Glucose reduces the \( \ce{Cu^2+} \) of Fehling’s solution to a red precipitate of \( \ce{Cu2O} \):

      \( \displaystyle \ce{CH2OH(CHOH)4CHO + 2Cu^2+ + 5OH- -> CH2OH(CHOH)4COO- + Cu2O v + 3H2O} \)

NCERT (2026-27 edition):
Cyclic Structure of Glucose – …

Structure of Glucose – “… On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.”