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Carbohydrates

Mediumchemistry

Open-chain D-glucose reacts with hydroxylamine to give an oxime and adds one molecule of hydrogen cyanide, yet a freshly prepared solution shows mutarotation. Which combination of features accounts for all three observations?

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About This Question

Subject
chemistry
Chapter
biomolecules
Topic
carbohydrates
Difficulty
Medium
Year
2025
Tags
glucose structuremutarotationhemiacetaloxime formationanomers

Solution

Correct Answer:

A free aldehyde group together with a cyclic hemiacetal in equilibrium

Glucose behaves as a typical aldehyde toward several reagents while also showing ring-specific behaviour, and reconciling both requires recognising the open-chain and cyclic forms coexist. Oxime formation with hydroxylamine and the addition of one HCN molecule are classic carbonyl reactions, proving the presence of one free aldehyde (-CHO) group in the open-chain aldohexose. Mutarotation, the gradual change in optical rotation of a fresh solution, arises because the open-chain aldehyde recyclises through its C5 hydroxyl to form a hemiacetal, generating alpha and beta anomers at the new C1 stereocentre that interconvert via the open form. Thus the equilibrium between the open-chain aldehyde and the cyclic hemiacetal explains all three facts at once. A ketone-only description fails because glucose forms an oxime characteristic of aldehydes; a carboxylic acid end is absent in glucose itself; and two aldehyde groups are not present. This is precisely the NCERT account of glucose structure. As a check, the failure of glucose to give the Schiff aldehyde test and its existence mainly as the pyranose ring are both consistent with a small equilibrium concentration of the reactive open-chain aldehyde.

This medium difficulty chemistry question is from the chapter biomolecules, covering the topic of carbohydrates. It appeared in the 2025 exam.

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