Question

When a monosaccharide forms a cyclic hemiacetal, the carbon atom that contained the carbonyl group is identified as the … carbon atom, because

• A. The carbonyl group is drawn to the right
• B. The carbonyl group is drawn to the left
• C. Acetal forms bond to an -OR and an —OH
• D. Anomeric, its substituents can assume an \( \alpha \) or \( \beta \) position

Hint:

In monosaccharides, the anomeric carbon is the one that was part of the carbonyl group and forms the cyclic structure.

Answer 1

The Correct Option is D

Step 1: Understanding hemiacetal formation.
When a monosaccharide forms a cyclic hemiacetal, the carbon atom that was part of the carbonyl group becomes the anomeric carbon. This carbon can assume either an \( \alpha \)- or \( \beta \)-anomeric form.

Step 2: Conclusion.
The correct identification is that the anomeric carbon can assume either an \( \alpha \)- or \( \beta \)-position, corresponding to option (4).