Chemistry · Biomolecules · NEET
Glucose does have an aldehyde group in its open-chain form. But in solution, glucose mostly exists as a closed ring (the cyclic hemiacetal or pyranose form). In this ring, the -CHO carbon (C1) is joined to an -OH from C5, so there is no truly free -CHO group most of the time. Schiff's test and the NaHSO3 addition are very sensitive and need a free aldehyde right away, so both fail. This is one of the reasons NCERT says the simple open-chain structure could NOT explain all of glucose's properties.
The ring and the open-chain form are in equilibrium. A tiny amount of the open-chain aldehyde is always present. HCN (gives cyanohydrin), NH2OH (gives oxime) and Tollen's reagent (silver mirror) are slower reactions. As the small open-chain amount reacts, more ring opens up to replace it, so the reaction slowly goes to completion. Schiff's and NaHSO3 are too fast/sensitive to wait for this, so they do not work.
Schiff's reagent and NaHSO3 are quick spot-tests that demand a free -CHO instantly - glucose cannot supply it, so they are negative. HCN and NH2OH are addition reactions that can slowly pull glucose through the small open-chain equilibrium, so they are positive and confirm a carbonyl group is present. So glucose is a bit of a puzzle: it proves it has a carbonyl group (via oxime and cyanohydrin) yet fails the fast free-aldehyde tests.
When glucose is acetylated, its five -OH groups become -OCOCH3. This includes the -OH on the ring oxygen path, which locks glucose permanently in the ring form. Now the ring can no longer open, so there is NO free -CHO at all. Since NH2OH (which makes an oxime) needs a carbonyl group, and none can form, the pentaacetate does not react with hydroxylamine. NCERT uses this fact to say there is no free -CHO group in the cyclic structure.
Yes, all these facts point to the same conclusion: glucose is mainly cyclic. When the ring closes, C1 (the old aldehyde carbon, now called the anomeric carbon) can have its new -OH pointing two ways. This gives two crystal forms, alpha and beta, called anomers. The failure of Schiff's/NaHSO3 and the existence of anomers together forced chemists to accept the cyclic (pyranose) structure over the plain open-chain one.
The reagents with which glucose does NOT react to give the corresponding test/product are: A. Tollen's reagent B. Schiff's reagent C. HCN D. NH2OH E. NaHSO3. Choose the correct option from those given below:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Glucose is a reducing sugar, so it gives a positive Tollen's test (silver mirror) and a positive Fehling's test. These slower reactions can use the small amount of open-chain aldehyde present at equilibrium, so they work even though Schiff's and NaHSO3 fail.
Glucose fails only two tests: Schiff's reagent test and the NaHSO3 (sodium hydrogensulphite) addition test. Both need a free -CHO group, which the mainly-cyclic glucose cannot readily supply. This is the exact pair asked in NEET 2024.
It proves that glucose does not exist as a plain open-chain aldehyde. Instead it is mostly in a cyclic (pyranose) hemiacetal form where the -CHO is locked up. This, along with the alpha and beta anomers, is the key evidence for the cyclic structure of glucose in NCERT.
Fructose is a ketohexose and also exists mainly in a cyclic (furanose) form, so it too does not give a normal free-carbonyl spot test. But fructose is still a reducing sugar because in basic medium it can rearrange to an aldose and give Tollen's/Fehling's positive.
