Question

Which of the following explains the reducing end of polysaccharides?
(A). End with free epimeric carbon
(B). End with free anomeric carbon
(C). End with free carboxylic carbon
(D). End with free carbonyl carbon
Choose the correct answer from the options given below:

• (A) and (D) only.
• (B) and (D) only.
• (B) and (C) only.
• (C) and (D) only.

Hint:

The reducing capability of a carbohydrate depends directly on having a free anomeric carbon.
This carbon can open up to form a free carbonyl group, which donates electrons to reduce other compounds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Saccharides are classified as reducing or non-reducing based on their ability to act as reducing agents in chemical tests (such as Benedict's or Fehling's tests).

Step 2: Detailed Explanation:
A sugar possesses reducing properties only if it has a reactive, openable ring structure containing a hemiacetal or hemiketal group:
- Anomeric carbon (B): The carbon derived from the carbonyl carbon of the open-chain form. If this carbon is not locked in a glycosidic bond, it is "free" and can undergo mutarotation to open the ring.
- Carbonyl carbon (D): When the ring opens at the free anomeric carbon, it regenerates a free aldehyde or ketone (carbonyl) group. This active carbonyl group reduces metal ions like \(Cu^{2+}\) in chemical tests.
Therefore, the reducing end is characterized by a free anomeric carbon (B) that can open to expose a free carbonyl carbon (D).

Step 3: Final Answer:
The correct answer is (B) and (D) only, corresponding to option (B).