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nmdcat.online June 27, 2026

The reducing power of a monosaccharide in Benedict’s test is due to the presence of a

A. Phosphate group attached to the 5' carbon
B. Free anomeric carbon with an aldehyde or ketone group
✓
C. Nitrogenous base linked to the 1' carbon
D. Branching structure created by α-1,6-glycosidic bonds

📝 Explanation

The test detects the free carbonyl group (C=O) at the anomeric carbon of a reducing sugar. This group can be oxidized, thereby reducing the Cu²⁺ in Benedict's reagent to Cu⁺, forming a colored precipitate. Non-reducing sugars lack this free group.

📖 Additional Information

  • Phosphate group attached to the 5' carbon
  • Free anomeric carbon with an aldehyde or ketone group
  • Nitrogenous base linked to the 1' carbon
  • Branching structure created by α-1,6-glycosidic bonds

The test detects the free carbonyl group (C=O) at the anomeric carbon of a reducing sugar. This group can be oxidized, thereby reducing the Cu²⁺ in Benedict's reagent to Cu⁺, forming a colored precipitate. Non-reducing sugars lack this free group.

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