Q.Glucose pentaacetate (in which all five $-\text{OH}$ groups are acetylated) does not react with hydroxylamine, $\text{NH}_2\text{OH}$, even though hydroxylamine normally reacts readily with a free aldehyde group. What does this tell you about the actual structure of glucose in solution?

If glucose genuinely existed as the simple open-chain aldehyde structure, its pentaacetate -- in which all five hydroxyl groups are acetylated -- would still retain a completely free, unreacted aldehyde group at C1, since acetylation only affects $-\text{OH}$ groups, not the carbonyl. A free aldehyde group reacts readily with hydroxylamine, $\text{NH}_2\text{OH}$, to form an oxime. The fact that glucose pentaacetate does not react with hydroxylamine at all shows that no free aldehyde group is actually available in the molecule to react.

This is explained by glucose existing predominantly as a cyclic hemiacetal: the $-\text{OH}$ group on C5 reacts intramolecularly with the aldehyde carbon at C1, converting what would have been a free aldehyde into a ring carbon bearing an $-\text{OH}$ group (the anomeric $-\text{OH}$) instead. Once this cyclic form is acetylated at all five hydroxyl positions (including the anomeric one), there is no free carbonyl left anywhere in the molecule for hydroxylamine to attack.

Compare what the open-chain structure predicts (a reactive free aldehyde should survive acetylation) against what is actually observed (no reaction at all), and conclude that the real structure must not have a free aldehyde -- i.e. it is cyclic.

Concluding instead that glucose simply 'has no aldehyde group' -- it does, in the open-chain form, which is why glucose still gives a positive Fehling's/Tollens' test in solution (where the ring can reopen); the correct conclusion is that the acetylated form specifically has no free aldehyde left to react.