Carbohydrates
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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Solution
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.
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About This Question
- Subject
- chemistry
- Chapter
- biomolecules
- Topic
- carbohydrates
- Difficulty
- Medium
- Year
- 2025
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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