Question

Which of the following compounds will behave as reducing sugar in an aqueous $\mathrm{KOH}$ solution?

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is C

The explanation for the correct answer:

(C)

1. Reducing sugar is a sugar that can reduce Tollens and Fehling's solution.
2. The compound having aldehyde, ketone, and hemiacetal can act as reducing sugar.
3. In this compound, we have a free hemiacetal group that converts $\mathrm{\alpha }\mathrm{Hydroxy}\mathrm{ketone}.$
4. Hence it acts as reducing sugar.
5. Hence, Option C is correct.

The explanation for the incorrect answer

(A)

1. This compound does not contain any Aldehyde, ketone, or Hemiacetal group.
2. Thus, it cannot behave as reducing sugar.
3. Thus, it is an incorrect answer.

(B)

1. This compound does not contain any Aldehyde, ketone, or Hemiacetal group.
2. Thus, it cannot behave as reducing sugar.
3. Thus, it is an incorrect answer.

(D)

1. This compound does not contain any Aldehyde, ketone, or Hemiacetal group.
2. Thus, it cannot behave as reducing sugar.
3. Thus, it is an incorrect answer.

Thus, C is the compound that will behave as reducing sugar in an aqueous $\mathrm{KOH}$ solution.