Question

Assertion (A) : The pentaacetate of glucose does not react with \( \mathrm{H_2N-OH} \).

Reason (R) : It indicates the presence of free \( \mathrm{-CHO} \) group in glucose.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

In the pentaacetate every OH is acetylated, so the ring cannot open and no free CHO group is present. The Assertion is true but the Reason is false.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Glucose has five \( \mathrm{-OH} \) groups. Acetic anhydride converts all five into acetate groups, giving glucose pentaacetate.
The test with hydroxylamine checks for a free carbonyl (aldehyde) group.

Step 2: Check the Assertion (A).
Glucose reacts with hydroxylamine to form an oxime, because it can open to the aldehyde form.
In the pentaacetate, the five \( \mathrm{-OH} \) groups include the anomeric \( \mathrm{-OH} \) on C-1 of the ring form.
This \( \mathrm{-OH} \) is now acetylated, so the ring cannot open and no \( \mathrm{-CHO} \) group is available.
So the pentaacetate does not react with \( \mathrm{H_2N-OH} \), and (A) is TRUE.

Step 3: Check the Reason (R).
The reason says the result shows a free \( \mathrm{-CHO} \) group. The opposite is true.
The pentaacetate does not give the oxime, which shows that a free \( \mathrm{-CHO} \) group is absent. This supports the cyclic structure of glucose. So (R) is FALSE.

Step 4: Choose the code.
(A) is true and (R) is false. This matches code (C).

Final Answer:
Assertion is true but Reason is false. \[ \boxed{\text{(C)}} \]