Biomolecules: Edge Cases and Subtle Traps (3)

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Question

Glucose reacts with HCN\text{HCN} to form a cyanohydrin and with hydroxylamine to form an oxime, suggesting an aldehyde group. But it does not give a Schiff’s test (with sodium bisulphite addition product) the way simple aldehydes do. Why? Many students answer “size”. What’s the precise reason?

Solution — Step by Step

Glucose shows several aldehyde reactions:

  • Reduces Tollens’ reagent (silver mirror).
  • Reduces Fehling’s solution.
  • Forms an oxime with NH2OH\text{NH}_2\text{OH}.
  • Forms a cyanohydrin with HCN.

But glucose:

  • Does not form a bisulphite addition product easily.
  • Does not give 2,4-DNP test directly without warming.

In aqueous solution, glucose exists predominantly as a cyclic hemiacetal — pyranose form. The free aldehyde is present only in trace amounts (<1%< 1\%).

Reagents that drive the open-chain \rightleftharpoons cyclic equilibrium forward (consume the aldehyde irreversibly, like Tollens, Fehling, NH2OH\text{NH}_2\text{OH}) succeed by Le Chatelier shift.

Reagents that need a high concentration of free aldehyde at the moment of attack (like NaHSO3\text{NaHSO}_3, which forms a reversible addition product) fail because the steady-state aldehyde concentration is too low.

The precise reason: glucose exists mainly in cyclic hemiacetal form; only a tiny fraction is the free aldehyde, and bisulphite addition is too reversible to drive the equilibrium.

Why This Works

The cyclic form is thermodynamically favoured because the hemiacetal is more stable than an open-chain aldehyde with a long carbon backbone. Reactions that are essentially irreversible (oxidation by Tollens, oxime formation) consume what little open-chain aldehyde is present, and equilibrium replenishes it.

This is why glucose still tests positive with Tollens but fails with bisulphite — kinetics meets thermodynamics in a beautiful way.

Alternative Method

Frame it as anomeric equilibrium: α\alpha-glucose \rightleftharpoons open-chain \rightleftharpoons β\beta-glucose. The open-chain form is the reactive intermediate, present at 0.02%\sim 0.02\% in solution.

Saying “glucose has no free aldehyde”. This is wrong — there is free aldehyde, just at very low concentration. The right phrasing is “glucose exists predominantly in the cyclic hemiacetal form”.

Common Mistake

Stating that glucose is unreactive toward all aldehyde tests. NEET 2023 had a four-option trap with “fails Tollens” as a wrong distractor. The fix: memorise which tests are reversible (NaHSO3, Schiff’s reagent) and which are irreversible (Tollens, Fehling, oxidation by HNO3).

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