Answer : (C) Assertion is true, but Reason is false
Step 1 — Test the Assertion
Hydroxylamine $(H_2N-OH)$ reacts only with a free carbonyl group, forming an oxime.
Glucose exists mainly in the cyclic hemiacetal form, in which C-1 carries an $-OH$ group. A small amount of the open-chain form is nevertheless in equilibrium with it, which is why ordinary glucose does give an oxime.
When glucose is acetylated, all five $-OH$ groups, including the one at C-1, become $-OCOCH_3$ groups. The ring is now locked shut and can never reopen, so no free $-CHO$ group can be produced.
With no carbonyl group available, the pentaacetate does not react. The assertion is true.
Step 2 — Test the Reason
The reason states that this observation indicates the presence of a free $-CHO$ group in glucose. That is the wrong conclusion.
The failure to react proves exactly the opposite — it shows that in the pentaacetate there is no free $-CHO$ group, and hence that glucose does not exist wholly in the open-chain aldehyde form. This was one of the key pieces of evidence for the cyclic structure of glucose.
The reason is false.
Step 3 — State the corrected reason
It indicates the absence of a free $-CHO$ group in glucose pentaacetate, and hence that glucose exists in a cyclic hemiacetal form.
Step 4 — Conclude
Assertion true, Reason false $\Rightarrow$ option (C).
Two other observations point the same way : glucose does not give Schiff's test, and it does not form a bisulphite addition product with $NaHSO_3$.
