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Carbohydrates

Easychemistry

Sucrose does not reduce Tollens' reagent or Fehling's solution even though it is a disaccharide built from glucose and fructose units. What structural feature explains this behaviour?

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About This Question

Subject
chemistry
Chapter
biomolecules
Topic
carbohydrates
Difficulty
Easy
Year
2025
Tags
sucrosenon-reducing sugarglycosidic linkageanomeric carbonhemiacetal

Solution

Correct Answer:

Both anomeric carbons are locked in the glycosidic linkage

A reducing sugar must possess a free anomeric carbon, that is, a hemiacetal or hemiketal group that can open into an aldehyde or alpha-hydroxy ketone capable of reducing mild oxidising reagents. In sucrose, the glycosidic bond forms between the C1 anomeric carbon of alpha-D-glucose and the C2 anomeric carbon of beta-D-fructose. Because both anomeric centres are simultaneously committed to the linkage, neither unit can revert to an open-chain carbonyl form in solution. Consequently sucrose shows no mutarotation and gives a negative test with Tollens' and Fehling's reagents, defining it as a non-reducing sugar. The option claiming fructose has no open-chain keto form is false, since free fructose is reducing because of its alpha-hydroxy ketone. The statement that ring sizes differ is irrelevant to reducing ability, and the idea that only glucose tautomerism is blocked ignores that both carbons are tied up. This matches the NCERT discussion of glycosidic linkages in carbohydrates. As a consistency check, maltose and lactose, which retain one free anomeric carbon, are indeed reducing, confirming that the locked-carbon explanation is the correct distinguishing feature.

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

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