Glucose has an aldehyde group but doesn't give Schiff's test or NaHSO3 addition. Plus, what do alpha and beta actually differ in??
Pooja Iyer12th Sample
3 Oct 2026 · 1 min read
Okay drama alert 🎭 glucose is betraying me.
NCERT shows glucose reacting with hydroxylamine to give an oxime and with HCN to give a cyanohydrin, and with bromine water to give gluconic acid, so it clearly has an aldehyde group -CHO. Cool. But then the very next page says glucose does NOT give Schiff's test, does NOT add sodium hydrogen sulphite, and the pentaacetate doesn't react with hydroxylamine. So does glucose have a free –CHO or not??? Pick a side!
Then the cyclic structure. I understand the ring forms between C-1 and C-5, a hemiacetal. But what is the real difference between -D-glucose and -D-glucose? My notes say "OH on C-1 down vs up" but I don't know what makes that a different compound. And D vs L, and d vs l, I mix those too.
Also is it true that a glucose molecule has 16 stereoisomers? Why 16?
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[+][–]Tanvi Deshmukh12thSample· 16d ago
Glucose in solution exists mainly as the cyclic hemiacetal. Only a very small fraction exists as open-chain at any moment. Reactions like Schiff's test and NaHSO3 addition need a decent amount of free –CHO, so they don't go. But oxime formation, HCN addition, and bromine-water oxidation can pull the open-chain form out of the equilibrium: as the open-chain form gets used up, more ring opens (Le Chatelier). So both pages are true.
The pentaacetate has all five –OH groups acetylated, and the ring is "locked" in that form, so it can't open to give –CHO, hence no reaction with hydroxylamine. That's actually a nice proof that the ring form exists.
[+][–]Zoya Khan12thSample· 16d ago
Adding to this, the open chain and ring forms interconvert, which is what explains mutarotation. If you dissolve pure -D-glucose its specific rotation starts at . Pure -form starts at . Either one slowly changes in water to , the equilibrium mixture. This happens because the ring opens briefly and closes again, either way up. That's also why NCERT says the two forms are in equilibrium via the open-chain structure.
[+][–]Aarav Kulkarni11thSample· 16d ago
I read somewhere that alpha and beta differ at C-2, and D/L is about whether it rotates light to the right or left? I might be wrong. Sorry if so.
[+][–]Rahul DubeyDropperSample· 16d ago
Aarav, a few mix-ups, let me sort them.
- D/L has nothing to do with the direction of rotation. It's about the configuration at the highest-numbered chiral carbon (C-5 in glucose) compared to D-glyceraldehyde: –OH on the right in the Fischer projection = D. Direction of rotation is written with small d/l or (+)/(−). D-glucose is (+), but a D sugar can be (−), for example D-fructose is (−).
- α/β differ at C-1, the anomeric carbon. When the ring closes, C-1 becomes a new chiral centre, so you can get two forms. These are anomers.
- Two sugars differing at any other single chiral carbon (like glucose and galactose at C-4, or glucose and mannose at C-2) are epimers.
So C-2 is where glucose and mannose differ, not α and β.
[+][–]Pooja Iyer12thSample· 16d agoAuthor
RAHUL BHAIYA you are a legend. D vs d finally makes sense, writing "D is a family, (+) is a behaviour" in my notebook.
[+][–]Aarav Kulkarni11thSample· 16d ago
ok that was embarrassing but thanks, C-1 for alpha/beta, noted.
[+][–]Kavya NaikDropperSample· 16d ago
Counting isomers: glucose in open-chain has four chiral carbons (C-2, C-3, C-4, C-5). So stereoisomers, which means 8 D-sugars and 8 L-sugars. D-glucose is just one of the 8 D-aldohexoses.
Ring size also shows up in exams. Glucose ring is six-membered (pyran-like, "pyranose"). Fructose ring is five-membered (furan-like, "furanose"), between C-2 and C-5. Glucose pyranose has 5 carbons and 1 oxygen in the ring.
