Q.Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

The open-chain aldehyde structure predicts normal aldehyde behaviour — yet D-glucose fails several such tests, and NCERT lists exactly three facts that the open-chain structure cannot explain, all resolved by the cyclic hemiacetal structure.

1. Despite having the aldehyde group, glucose does not give Schiff's test and does not form the hydrogensulphite addition product with $NaHSO_3$.
2. The pentaacetate of glucose does not react with hydroxylamine, indicating the absence of a free –CHO group (acetylation of the anomeric –OH locks the ring). …

The Core Idea: Why the Open Chain Fails

D-glucose was long represented as a straight-chain aldehyde (structure I, the Fischer projection). An aldehyde should give the full set of characteristic aldehyde reactions — but glucose does not. The resolution: in reality one of the –OH groups (the –OH at C-5) adds to the –CHO group, forming a six-membered cyclic hemiacetal. The aldehyde is tied up in the ring, and a new stereocentre — the anomeric carbon (C1) — is created. The two cyclic forms exist in equilibrium with a small amount of the open-chain structure.

Let's go through each observation the open chain cannot explain.

1. No Schiff's Test, No Bisulphite Addition Product

An open-chain aldehyde should give:

• a pink-magenta colour with Schiff's reagent, and
• a crystalline hydrogensulphite addition product with $NaHSO_3$.

D-glucose gives neither — impossible if a free aldehyde group were present. In the cyclic hemiacetal, the –CHO is no longer free.

2. The Pentaacetate Does Not React with Hydroxylamine

Free glucose reacts with hydroxylamine (via its open-chain equilibrium form) to give an oxime. But glucose pentaacetate does not react with hydroxylamine at all — indicating the absence of a free –CHO group.

Why? Acetylation converts all five –OH groups of cyclic glucose to acetate esters — including the anomeric –OH at C1. With the anomeric position esterified, the ring can no longer open to regenerate the aldehyde: ring–chain tautomerism is abolished. This is direct evidence that glucose's normal state is the ring, with the "aldehyde" existing only as the ring-opening product of the free hemiacetal.

3. Two Crystalline Forms — $\alpha$ and $\beta$ (Anomers)

Glucose is found to exist in two different crystalline forms:

• $\alpha$-D-glucose (m.p. 419 K, i.e. 146°C) — obtained by crystallisation from a concentrated solution at 303 K.
• $\beta$-D-glucose (m.p. 423 K, i.e. 150°C) — obtained by crystallisation from a hot, saturated aqueous solution at 371 K.

The open chain has no stereocentre at C1 — it is just an aldehyde carbon — so it cannot account for two isomers differing only at C1. In the cyclic form, C1 becomes a new chiral centre (the anomeric carbon), giving the two anomers.

4. Related Evidence: Mutarotation and the Two Methyl Glycosides

These follow from the same cyclic structure and are standard extensions:

• Mutarotation: when either pure anomer is dissolved in water, its specific rotation changes gradually ($+112°$ for $\alpha$, $+18.7°$ for $\beta$) until both reach the same equilibrium value ($+52.5°$), as the anomers interconvert through the open-chain form. A single open-chain structure cannot show this. …

Method: Testing the Open-Chain Structure Against Observed Behaviour

This method lists what an open-chain aldehyde structure predicts, checks each prediction against what D-glucose actually does, and keeps only the failures — those are the reactions the open chain cannot explain.

Steps

1. Recall the open-chain structure and its predictions

• D-glucose drawn as $CHO–(CHOH)_4–CH_2OH$ has a free aldehyde at C1.
• Predictions: positive Schiff's test, bisulphite addition product, free –CHO behaviour in every derivative, a single form of the compound.

2. List the observations that contradict these predictions

• No Schiff's test — glucose does not restore the pink colour of Schiff's reagent.
• No hydrogensulphite addition product — glucose does not form the crystalline adduct with $NaHSO_3$.
• The pentaacetate does not react with hydroxylamine — indicating the absence of a free –CHO group in the acetylated compound (free glucose does form an oxime via its small open-chain fraction; the locked pentaacetate cannot).
• Two crystalline forms exist — $\alpha$-D-glucose (m.p. 419 K) and $\beta$-D-glucose (m.p. 423 K), which also show mutarotation in solution; a single open-chain molecule cannot give two such isomers.

3. Explain each anomaly with ring–chain tautomerism

• The –OH at C-5 adds to the –CHO group, giving a six-membered cyclic hemiacetal (pyranose ring).
• C1 becomes a new chiral centre — the anomeric carbon — producing the $\alpha$ and $\beta$ forms, in equilibrium with a trace of the open chain. …

Here are the common mistakes students make when enumerating the reactions of D-glucose that its open-chain structure cannot explain — with precise corrections.

1. Mistake: Listing all reactions of glucose instead of only the anomalous ones

The Error: Students list general reactions (reduction to sorbitol, oxidation to gluconic acid, oxime formation) that the open-chain aldehyde explains perfectly well. This wastes time and misses the point of the question.

The Fix: Enumerate only the observations the open chain cannot explain. NCERT's own list has three:

• No Schiff's test and no hydrogensulphite ($NaHSO_3$) addition product, despite the supposed aldehyde group.
• The pentaacetate of glucose does not react with hydroxylamine — showing the absence of a free –CHO group.
• Two crystalline forms exist — $\alpha$ (m.p. 419 K) and $\beta$ (m.p. 423 K).

Mutarotation and the formation of two isomeric methyl glucosides are valid supporting additions.

2. Mistake: Claiming glucose fails to add HCN or form an oxime

The Error: Students extend the "no free aldehyde" logic too far and write that glucose does not add HCN or does not react with hydroxylamine.

The Fix: The book states the opposite for free glucose: it reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give a cyanohydrin — these are printed as evidence for the carbonyl group in glucose. The small open-chain fraction present in equilibrium reacts and is continuously replenished. The hydroxylamine anomaly belongs to the pentaacetate, whose locked ring cannot open — never to free glucose.

Keep the contrast: free glucose + $NH_2OH$ → oxime ✓; glucose pentaacetate + $NH_2OH$ → no reaction ✗ (this is the cyclic-structure evidence).

3. Mistake: Confusing mutarotation with ordinary optical activity

The Error: "Glucose is optically active, which proves the ring." Wrong — the open-chain form is also optically active (four chiral carbons).

The Fix: The anomaly is mutarotation — the change of specific rotation with time (from $+112°$ for $\alpha$ or $+18.7°$ for $\beta$ to the equilibrium $+52.5°$) as the two anomers interconvert through the open chain. A single open-chain structure has nothing to interconvert.

4. Mistake: Forgetting that the open chain does explain some reactions

The Error: Claiming all reactions of glucose need the cyclic structure.

The Fix: Reactions that trap the trace open-chain form proceed normally and are not anomalies: oxidation by bromine water to gluconic acid, reduction of Fehling's/Tollens' reagents, oxime formation, HCN addition, osazone formation. Do not list these.

5. Mistake: Calling the $\alpha$/$\beta$ forms "epimers"

The Error: "Glucose forms two epimers at C1." …