Assertion (A): The pentaacetate of glucose does not react with H₂N – OH.
Reason (R): It indicates the presence of free – CHO group in glucose.
Assertion (A): The pentaacetate of glucose does not react with H₂N – OH.
Reason (R): It indicates the presence of free – CHO group in glucose.
Options
Correct option: (C) — Assertion is true, but Reason is false.
Why A is true:
- Glucose pentaacetate does not react with hydroxylamine (H₂N–OH).
- Hydroxylamine reacts with a free carbonyl (–CHO) to form an oxime; the pentaacetate gives no such reaction.
Why R is false:
- The failure to react actually indicates the absence of a free –CHO group in glucose pentaacetate (the aldehyde exists in the cyclic, acetylated form).
- The Reason wrongly states it indicates the presence of a free –CHO group — the opposite of the truth.
- Hence: true assertion, false reason → code (C).
Marking Scheme
- 11 mark: correct code (C) — Assertion true (pentaacetate does not react with H₂N–OH), Reason false (it indicates absence, not presence, of free –CHO).
- 2Full mark for (C); no partial credit.
Hint
If a free –CHO were present it WOULD react with hydroxylamine; since it does not react, the –CHO must be absent — the reason states the opposite.
Quick Oral Answer
The assertion is true because glucose pentaacetate does not react with hydroxylamine, but the reason is false: its non-reaction shows the absence, not the presence, of a free –CHO group, so the answer is (C).
Analysis & Explanation
Concept:
Hydroxylamine (H₂N–OH) is a classic test reagent for a free carbonyl group; an aldehyde or ketone reacts to form an oxime. Open-chain glucose does have a –CHO group and reacts with hydroxylamine to give an oxime.
The pentaacetate evidence:
- When all five –OH groups of glucose are acetylated, the resulting pentaacetate does not react with hydroxylamine.
- This shows that in the pentaacetate the aldehyde is locked in the cyclic (hemiacetal-derived) form and no free –CHO is available.
- Therefore the correct interpretation is absence, not presence, of a free –CHO group.
Exam trap:
Students recall that glucose 'has an aldehyde group' and mark the reason true. The subtlety is that the pentaacetate's non-reaction proves the aldehyde is masked in the cyclic form — key evidence for glucose's ring structure.
Real-world:
This pentaacetate experiment historically helped establish that glucose exists predominantly in a cyclic (pyranose) form rather than the open-chain aldehyde form in the solid state.
Common Mistakes
- 1Marking the reason true because glucose is known to contain an aldehyde group — the pentaacetate's non-reaction proves the –CHO is masked (absent as free group).
- 2Confusing 'presence' with 'absence' of the free –CHO group when interpreting a negative test.
- 3Assuming hydroxylamine reacts with hydroxyl groups; it reacts with carbonyl (–CHO/–C=O) groups to give oximes.
Interesting Facts
The failure of glucose pentaacetate to react with hydroxylamine was among the pieces of evidence used to show glucose exists in a cyclic form rather than as an open-chain aldehyde.
The same masked-carbonyl idea explains why glucose does not give the Schiff's base test and shows mutarotation — the –CHO is tied up in a hemiacetal ring.
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Frequently Asked Questions
Why does glucose pentaacetate not react with hydroxylamine?
Because in the pentaacetate the aldehyde group is locked in the cyclic (acetylated hemiacetal) form, leaving no free –CHO group. Hydroxylamine only reacts with a free carbonyl to form an oxime, so no reaction occurs.
What does this non-reaction tell us about glucose structure?
It shows that glucose does not contain a free aldehyde group in its acetylated form and therefore exists largely in a cyclic (ring) structure rather than as an open-chain aldehyde — important historical evidence for the ring form of glucose.