Glucose does not give reaction with which of the following?

The correct answer is Option (4) → NaHSO₃

The open-chain structure of glucose has an aldehyde group and five –OH groups, \( \ce{HOCH2(CHOH)4CHO} \). In solution, however, glucose exists almost entirely as the cyclic hemiacetal (α- and β-pyranose forms), with only a trace of the open-chain aldehyde. Reagents that pull the small open-chain fraction along still react; the hydrogensulphite addition does not take place.

• Option (1) NH₂–OH — Reacts with glucose. Hydroxylamine gives an oxime, which shows the carbonyl group.

      \( \displaystyle \ce{HOCH2(CHOH)4CHO + NH2OH -> HOCH2(CHOH)4CH=NOH + H2O} \)

• Option (2) HCN — Reacts with glucose. Glucose adds HCN to give a cyanohydrin.

      \( \displaystyle \ce{HOCH2(CHOH)4CHO + HCN -> HOCH2(CHOH)4CH(OH)CN} \)

• Option (3) (CH₃CO)₂O — Reacts with glucose. Acetylation gives glucose pentaacetate, which shows the five –OH groups.

      \( \displaystyle \ce{HOCH2(CHOH)4CHO + 5(CH3CO)2O} \xrightarrow{\ce{ZnCl2}} \ce{CH3COOCH2(CHOCOCH3)4CHO + 5CH3COOH} \)

• Option (4) NaHSO₃ — Does not react with glucose. Despite having the aldehyde group, glucose does not form the hydrogensulphite addition product with \( \ce{NaHSO3} \), and it does not give Schiff’s test. These facts led to the cyclic structure, in which the –CHO group is tied up as a hemiacetal.

NCERT (2026-27 edition):
Structure of 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. … 5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups.”

Cyclic Structure of Glucose – …