Glucose does not react with:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Sodium bisulphite
The open-chain structure of glucose has an aldehyde group, yet glucose does not form the hydrogensulphite (bisulphite) addition product with NaHSO3, and it does not give Schiff’s test or the 2,4-DNP test. This is because glucose exists almost entirely in the cyclic hemiacetal form, in which the –CHO group is not free; these facts were the evidence for the ring structure.
Option (1) Conc. \(HNO_3\) — Reacts with glucose. Glucose is oxidised by nitric acid; both the –CHO and the terminal –\( \ce{CH2OH} \) become –COOH, giving saccharic acid.
\( \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \)
Option (2) Acetic Anhydride — Reacts with glucose. Acetylation gives glucose pentaacetate, which shows five –OH groups.
\( \ce{CHO-(CHOH)4-CH2OH + 5(CH3CO)2O -> CHO-(CHOCOCH3)4-CH2OCOCH3 + 5CH3COOH} \)
Option (3) Hydroxylamine — Reacts with glucose. Glucose forms an oxime, which shows the carbonyl group.
\( \ce{CHO-(CHOH)4-CH2OH + NH2OH -> CH(=N-OH)-(CHOH)4-CH2OH + H2O} \)
Option (4) Sodium bisulphite — Does not react with glucose. No hydrogensulphite addition product is formed, because the –CHO group is held in the cyclic hemiacetal ring.
NCERT (2026-27 edition):
Cyclic Structure of Glucose – “The structure (I) of glucose explained most of its properties but the following reactions and facts could not be explained by this structure. 1. Despite having the aldehyde group, glucose does not give Schiff’s test and it does not form the hydrogensulphite addition product with NaHSO3.”