Glucose reacts with hydroxylamine, and it also adds a molecule of hydrogen cyanide. Which of the following correctly gives the products of these two reactions and the conclusion drawn from them about the structure of glucose?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Oxime and cyanohydrin; glucose contains a carbonyl group (>C=O)
Hydroxylamine and hydrogen cyanide both add to a carbonyl group. So these two reactions of glucose show that its open chain has a >C=O group.
With hydroxylamine, the carbonyl group of glucose forms an oxime and a molecule of water is lost:
\( \displaystyle \ce{HOCH2-(CHOH)4-CHO + NH2OH -> HOCH2-(CHOH)4-CH=N-OH + H2O} \)
Hydrogen cyanide adds across the C=O bond and gives a cyanohydrin:
\( \displaystyle \ce{HOCH2-(CHOH)4-CHO + HCN -> HOCH2-(CHOH)4-CH(OH)CN} \)
Why the other options are incorrect
Option (1) Oxime and cyanohydrin; glucose contains an aldehydic group — Incorrect.
The products are right, but ketones also form oximes and cyanohydrins, so these reactions cannot tell an aldehyde from a ketone.
The aldehydic nature of the carbonyl group is shown by bromine water, which oxidises glucose to gluconic acid.
Option (2) Semicarbazone and cyanohydrin; glucose contains a carbonyl group — Incorrect. A semicarbazone is formed with semicarbazide, not with hydroxylamine; hydroxylamine gives an oxime.
Option (3) Oxime and gluconic acid; glucose contains a primary alcoholic group — Incorrect. Gluconic acid comes from oxidation by bromine water, not from HCN. The primary alcoholic group is shown by oxidation with nitric acid, which gives saccharic acid.
NCERT (2026-27 edition):
10.1.2.1 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.”