Question

Consider the following statements:
Statement I: All monosaccharides are reducing sugars.
Statement II: Sucrose can reduce ammoniacal silver nitrate solution.
Choose the correct answer from the options given below.

• A. Both Statement I and Statement II are correct
• B. Both Statement I and Statement II are incorrect
• C. Statement I is correct but Statement II is incorrect
• D. Statement I is incorrect but Statement II is correct

Hint:

To identify a reducing sugar, check for a hemiacetal group in its cyclic structure. A hemiacetal group (a carbon bonded to both an -OH and an -OR group) indicates that the ring can open to form a free aldehyde. In sucrose, the anomeric carbons are linked in an acetal/ketal linkage, so there are no hemiacetal groups.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Carbohydrates are classified as reducing or non-reducing based on their ability to act as mild reducing agents in basic solutions.
Step 2: Key Formula or Approach:
A sugar is classified as "reducing" if it possesses a free hemiacetal or hemiketal group (a free anomeric carbon) that can equilibrate to form a free aldehyde or ketone group, capable of reducing mild oxidizing agents like Tollen's reagent (ammoniacal silver nitrate).
Step 3: Detailed Explanation:
1. Statement I: All monosaccharides (e.g., glucose, fructose, ribose) contain a free anomeric carbon in equilibrium with an open-chain form that contains a carbonyl group. Thus, they are all reducing sugars. (Statement I is True)

2. Statement II: Sucrose is a disaccharide formed by the condensation of $\alpha$-D-glucose and $\beta$-D-fructose. The glycosidic linkage involves the anomeric carbons of both monomer units ($C1$ of glucose and $C2$ of fructose). Since both potential aldehyde and ketone groups are structurally locked in the bond, sucrose has no free hemiacetal/hemiketal groups. Therefore, it is a non-reducing sugar and cannot reduce Tollen's reagent. (Statement II is False)
Step 4: Final Answer:
Statement I is correct, and Statement II is incorrect.