CBSE2026Class Class 12 · Chemistry1 Marks · Long✅ Verified
Assertion (A) : The pentaacetate of glucose does not react with $H_2N-OH$.
Reason (R) : It indicates the presence of free $-CHO$ group in glucose.
Codes: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
✅ Answer & Solution
Step 1 - Recall the test.
Hydroxylamine ($H_2N-OH$) reacts with a FREE aldehydic $(-CHO)$ group to give an oxime.
So the reaction is a test for a free carbonyl group.
Step 2 - Examine the Assertion (A).
Glucose reacts with acetic anhydride to give glucose PENTAACETATE (all five $-OH$ groups
acetylated). This pentaacetate does NOT form an oxime with $H_2N-OH$.
Assertion (A) is TRUE.
Step 3 - Examine the Reason (R).
Failure to react means there is NO free $-CHO$ group available. Hence the reason as
stated - that it indicates the PRESENCE of a free $-CHO$ group - is exactly the
opposite of the truth.
Reason (R) is FALSE.
Step 4 - Explain the chemistry behind it.
In the solid and in solution, glucose exists mainly in the CYCLIC hemiacetal
(pyranose) form. In this ring form the aldehydic carbon is tied up as a hemiacetal.
On acetylation, the anomeric $-OH$ is also acetylated, locking the ring permanently.
The molecule can then no longer open to release a free $-CHO$ group, so no oxime forms.
This observation is one of the key pieces of evidence for the CYCLIC structure of glucose.
Answer: (C) - Assertion (A) is true, but Reason (R) is false.
