Question
Glucose exists predominantly in cyclic (hemiacetal) form rather than the open-chain aldehyde form. (a) Draw the open-chain Fischer projection of D-glucose. (b) Explain the formation of - and -D-glucopyranose. (c) Why does glucose not give a Schiff’s test despite being an aldehyde?
Solution — Step by Step
D-Glucose: 6-carbon aldohexose. Fischer projection has at C1, at C6, and the configuration at C5 is D (the OH is on the right).
C1: CHO; C2: H, OH (right); C3: HO (left), H; C4: H, OH (right); C5: H, OH (right); C6: .
The C5-OH attacks the C1 aldehyde carbon intramolecularly, forming a six-membered ring with one oxygen (the pyranose form). This generates a new chiral centre at C1, called the anomeric carbon.
If the new OH at C1 is trans to the at C5, it is the -anomer (axial OH). If cis, it is the -anomer (equatorial OH).
In aqueous solution, both forms interconvert via the open chain — this is mutarotation. Equilibrium: , , open chain.
Schiff’s reagent reacts with free aldehyde groups. In aqueous glucose, the exists almost entirely in the cyclic hemiacetal form, where C1 is sp, not sp. There simply is not enough free aldehyde at any moment to give a positive Schiff’s test.
Tollens’ and Fehling’s tests, in contrast, succeed because they continuously consume the small amount of open-chain form, shifting the equilibrium and pulling more glucose through the open chain (Le Chatelier).
Glucose cyclises to - and -D-glucopyranose; the cyclic form lacks free , which is why Schiff’s test fails despite glucose being an aldose.
Why This Works
The intramolecular cyclisation is favoured because it forms a stable six-membered ring with minimised ring strain. The hemiacetal is in equilibrium with the open chain, but the open-chain fraction is tiny.
Schiff’s reagent requires a steady-state concentration of free aldehyde to give the colour. Slow reactions like Tollens’ can drain glucose through the trace open-chain population over time; fast colour tests like Schiff’s cannot.
Alternative Method
Think of it kinetically. Schiff’s gives an immediate colour with free . Tollens’ is an oxidation reaction at finite rate — it has time to “pull” glucose out of the cyclic form. The same logic applies to fructose’s reactivity.
Common Mistake
Stating that glucose has no aldehyde group at all. It does — chemically, glucose is an aldose. The correct statement is that the aldehyde is “masked” in the cyclic hemiacetal form. Wording matters in NEET answer keys.