Write the reactions of D-glucose which cannot be explained by its open chain structure. How can cyclic structu

NCERT Exemplar · Q49

Q. Write the reactions of D-glucose which cannot be explained by its open chain structure. How can cyclic structure of glucose explain these reactions?

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Concept understanding — Ring Chain Tautomerism

Ring-Chain Tautomerism: The Shape-Shifting Molecules

Imagine a molecule that can exist in two forms: one with a straight chain of atoms, and another where the chain curls around and joins its own tail to form a ring. That is exactly what ring-chain tautomerism is — a special kind of dynamic equilibrium where a molecule flips between an open-chain structure and a cyclic (ring) structure.

The Intuition

Think of a piece of string. You can lay it out straight, or you can tie its two ends together to make a loop. The string itself hasn't changed — it's the same material — but its shape is completely different. In ring-chain tautomerism, the molecule does something similar: a functional group at one end of the chain reacts with another part of the same molecule, forming a ring. The open form and the ring form are tautomers — they interconvert rapidly under normal conditions.

The most common example involves a molecule that has both a carbonyl group ( $\ce{C=O}$ ) and a hydroxyl group ( $\ce{-OH}$ ) somewhere along the chain. The $\ce{-OH}$ can attack the carbonyl carbon, breaking the $\ce{C=O}$ double bond and forming a new $\ce{C-O}$ single bond, while the oxygen of the carbonyl picks up the hydrogen from the $\ce{-OH}$ . The result? A cyclic hemiacetal or hemiketal.

The Precise Statement

Ring-chain tautomerism is a type of tautomerism in which a molecule exists in equilibrium between an open-chain form (usually containing a carbonyl group) and a cyclic form (usually a hemiacetal, hemiketal, or lactol), formed by an intramolecular nucleophilic addition of a hydroxyl, amino, or thiol group to the carbonyl carbon.

The equilibrium is reversible and depends on:

Ring size: 5- and 6-membered rings are most stable (least ring strain).
Solvent: Polar solvents often favour the cyclic form.
Temperature: Higher temperatures may shift equilibrium toward the open chain.

A Concrete Example: Glucose

The most famous example is glucose. In its open-chain form, glucose has an aldehyde group ( $\ce{-CHO}$ ) at one end and a hydroxyl group on carbon 5. The $\ce{-OH}$ on carbon 5 attacks the aldehyde carbon, forming a 6-membered ring (pyranose form). This is why glucose in water is mostly in the cyclic form — only about 0.02% exists as the open chain at any moment.

\ce{Open-chain glucose <=> Cyclic glucose (pyranose)}

Note

The cyclic form of glucose is actually a hemiacetal — the product of an aldehyde reacting with an alcohol. The new chiral centre formed at the carbonyl carbon (called the anomeric carbon) gives rise to two stereoisomers: $\alpha$ and $\beta$ .

Other Examples

Hydroxy aldehydes (e.g., 5-hydroxypentanal) exist partly as a cyclic hemiacetal.
Sugars like fructose show ring-chain tautomerism between a ketone and a cyclic hemiketal (furanose or pyranose).
Certain drugs (e.g., some barbiturates) can exhibit this behaviour, affecting their biological activity. …

Why this formula?

Ring-Chain Tautomerism: Understanding the "Why" Behind the Equilibrium

What Is Ring-Chain Tautomerism?

Ring-chain tautomerism is a special type of dynamic equilibrium where a molecule exists in two interconvertible forms:

Open-chain form (usually with a carbonyl group and a nucleophilic group like –OH, –NH₂, –SH)
Cyclic (ring) form (formed by intramolecular attack of the nucleophile on the carbonyl carbon)

The key idea: the same molecule can close into a ring or open into a chain, and the position of equilibrium depends on ring stability and steric/electronic factors.

The Core Formula: Ring Size Preference

The 5- and 6-Membered Ring Rule

Statement: For most ring-chain tautomeric systems, 5- and 6-membered rings are strongly favoured over smaller (3,4) or larger (7+) rings.

Why does this hold? — The Derivation of Preference

1. Angle Strain (Baeyer Strain Theory)

When a ring forms, bond angles deviate from the ideal tetrahedral angle ( $109.5^\circ$ for sp³ carbons).

3-membered ring: Bond angles ≈ $60^\circ$ → severe angle strain ( $\Delta H \approx +115 \text{ kJ/mol}$ )
4-membered ring: Bond angles ≈ $90^\circ$ → significant strain ( $\Delta H \approx +110 \text{ kJ/mol}$ )
5-membered ring: Bond angles ≈ $108^\circ$ → very little strain (nearly ideal)
6-membered ring: Bond angles ≈ $109.5^\circ$ (chair conformation) → strain-free

Result: The ring form is only stable when angle strain is minimal — hence 5 and 6 are favoured.

2. Entropy Factor (Ring-Closing Probability)

The probability of the two ends of the chain meeting to form a ring depends on chain length.

For a chain of $n$ atoms, the effective concentration of the reactive ends is proportional to $1/n^{3/2}$ (from random-walk statistics).
Shorter chains (n=3,4) have higher effective concentration → easier to close, BUT the ring is too strained.
Longer chains (n≥7) have lower effective concentration → harder to close, AND the ring has transannular strain (steric repulsion across the ring).

Optimum: n=5 or 6 gives the best balance — low strain + reasonable closing probability.

The Equilibrium Constant Expression

For a general ring-chain tautomerism:

\text{Open-chain} \rightleftharpoons \text{Cyclic}

The equilibrium constant $K$ is:

K = \frac{[\text{Cyclic}]}{[\text{Open-chain}]}

Why does $K$ depend on ring size?

From thermodynamics:

\Delta G^\circ = -RT \ln K

And:

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

$\Delta H^\circ$ is dominated by strain energy (angle strain + torsional strain + transannular strain)
$\Delta S^\circ$ is negative for ring closure (one molecule → one molecule, but loss of conformational freedom)

For 5- and 6-membered rings:

$\Delta H^\circ$ is small (low strain) → favourable
$\Delta S^\circ$ is moderately negative → slightly unfavourable
Net: $\Delta G^\circ$ is negative → $K > 1$ (ring form dominates)

For 3- and 4-membered rings:

$\Delta H^\circ$ is large positive (high strain) → very unfavourable
$\Delta S^\circ$ is still negative
Net: $\Delta G^\circ$ is positive → $K \ll 1$ (open-chain dominates)

The "Anomeric Effect" in Ring-Chain Tautomerism (Special Case)

In sugar chemistry, ring-chain tautomerism shows an additional effect: …

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D-glucose fails to show certain reactions expected from an open-chain aldehyde (like the Schiff test, addition of NaHSO₃, and the formation of a stable bisulfite adduct) because it exists predominantly as a cyclic hemiacetal (pyranose form). The cyclic structure explains these anomalies by showing that the aldehyde group is tied up in an internal hemiacetal ring, which is in equilibrium with a tiny amount of open-chain form — too small to give positive tests for a free aldehyde.

The Concept: Ring-Chain Tautomerism in Glucose

The open-chain structure of D-glucose (an aldohexose) has a free aldehyde group at C1. If that were the whole story, glucose should behave exactly like a typical aldehyde — giving a pink colour with Schiff’s reagent, forming a bisulfite addition compound, and reducing Fehling’s solution readily. But in reality, glucose does not give a positive Schiff test, does not form a stable bisulfite adduct, and yet it does reduce Fehling’s solution. This contradiction was the clue that led chemists to realise glucose exists mostly in a cyclic form.

The key insight is ring-chain tautomerism: the aldehyde group (–CHO) at C1 reacts intramolecularly with the –OH group at C5 (or C4) to form a cyclic hemiacetal. This ring is the dominant species in solution. The open-chain aldehyde is present only in trace amounts (less than 0.1% at equilibrium). So reactions that require a free aldehyde group (like Schiff test or bisulfite addition) fail because the equilibrium concentration of the open-chain form is too low to give a visible result. But reactions that irreversibly consume the open-chain form (like Fehling’s reduction) keep pulling the equilibrium toward the open chain, so they do occur.

Watch out

Common Pitfall

Many students think glucose never has a free aldehyde group. That’s wrong. The open-chain form exists in equilibrium — it’s just that the equilibrium lies heavily toward the cyclic form. Reactions that trap the open-chain form (like oxidation) proceed because Le Chatelier’s principle shifts the equilibrium.

Reactions Not Explained by the Open-Chain Structure

Let’s list the specific reactions that contradict the open-chain model, then see how the cyclic structure resolves each one.

1. Schiff’s Test (Negative)

What the open-chain structure predicts: A free –CHO group should react with Schiff’s reagent (fuchsin-sulfurous acid) to give a pink-magenta colour.

What actually happens: D-glucose gives no colour with Schiff’s reagent.

Why the cyclic structure explains it: The –CHO group is not free — it is part of the hemiacetal ring. The equilibrium concentration of the open-chain aldehyde is far too low to produce a visible colour change. The Schiff test requires a significant concentration of free aldehyde to work; trace amounts are insufficient.

Note

Other reducing sugars like fructose also fail the Schiff test for the same reason — they exist predominantly in cyclic forms.

2. No Stable Bisulfite Addition Compound

What the open-chain structure predicts: Aldehydes add sodium bisulfite (NaHSO₃) to form crystalline bisulfite addition products. Glucose should form such a solid.

What actually happens: Glucose does not form a stable, isolable bisulfite addition compound.

Why the cyclic structure explains it: The equilibrium between cyclic and open-chain forms lies so far toward the cyclic form that the open-chain aldehyde concentration is negligible. Even if a tiny amount of bisulfite adduct forms, it cannot be isolated because the equilibrium constantly shifts. Moreover, the cyclic hemiacetal itself does not add bisulfite — only the free aldehyde does.

Tip

Insight

This is a classic example of how kinetic vs thermodynamic control works: the bisulfite addition is reversible, and the cyclic form is thermodynamically more stable. So the system prefers to stay cyclic.

3. Does Not Give 2,4-DNP Test (or Gives It Very Slowly)

What the open-chain structure predicts: Aldehydes react with 2,4-dinitrophenylhydrazine to form a yellow-orange precipitate.

What actually happens: Glucose gives the test only after prolonged heating, and the yield is poor.

Why the cyclic structure explains it: The reaction requires the open-chain form. Since it is present in trace amounts, the reaction is very slow. Heating shifts the equilibrium slightly toward the open chain, but the test is still not as clean as for a typical aldehyde.

4. Does Not Show Aldehyde Proton in NMR (in D₂O)

What the open-chain structure predicts: The aldehyde proton (–CHO) should appear as a singlet near δ 9–10 ppm in proton NMR.

What actually happens: In D₂O solution, the NMR spectrum of D-glucose shows no signal near δ 9–10 ppm. Instead, signals appear in the δ 4–5 ppm region corresponding to the anomeric proton (the hemiacetal –OH and –H at C1).

Why the cyclic structure explains it: The aldehyde group is gone — it has become a hemiacetal carbon (C1) with an –OH and an –H. The anomeric proton appears at δ ~4.5–5.5 ppm, consistent with a cyclic hemiacetal.

\text{Open-chain aldehyde} \rightleftharpoons \text{Cyclic hemiacetal (pyranose)}

How the Cyclic Structure Explains These Reactions — Step by Step

Formation of the cyclic hemiacetal: The –OH group on C5 attacks the carbonyl carbon (C1) of the open-chain form. This creates a six-membered ring (pyranose) with an oxygen bridge. The new –OH group at C1 is called the anomeric hydroxyl. …

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Concept: Cyclic Structure of Glucose (Haworth Projection)

The open-chain structure of glucose (an aldohexose) fails to explain several key reactions. The cyclic hemiacetal structure (pyranose form) resolves these anomalies.

Method: Anomaly-Resolution Analysis

Step 1: Identify reactions unexplained by open-chain structure

Does not react with NaHSO₃ (sodium bisulfite) — aldehydes typically form addition products.
Does not give Schiff’s test — a test for free aldehyde groups.
Shows mutarotation — optical rotation changes on dissolving in water.
Forms two crystalline forms (α and β) with different melting points and rotations.

Step 2: Explain using cyclic hemiacetal formation

The aldehyde group ( $-CHO$ ) at C1 reacts with the $-OH$ group at C5 to form a six-membered pyranose ring (hemiacetal).
This creates a new chiral centre at C1 (called the anomeric carbon), giving two stereoisomers:
- α-D-glucose: OH at C1 is trans to CH₂OH at C5 (axial in chair form).
- β-D-glucose: OH at C1 is cis to CH₂OH at C5 (equatorial in chair form).

Step 3: Map each anomaly to cyclic structure

Anomaly	Open-chain fails because	Cyclic structure explains
No reaction with NaHSO₃	Aldehyde is absent in cyclic form	$-CHO$ is tied up in hemiacetal ring
No Schiff’s test	Free aldehyde not available	Hemiacetal lacks free carbonyl
Mutarotation	Only one form predicted	Ring opens to aldehyde, then recloses to α or β — equilibrium mixture

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Here are the common mistakes students make when answering this question, along with how to avoid each.

Mistake 1: Listing the wrong reactions

Students often list general reactions of glucose (like reduction to sorbitol or oxidation to gluconic acid) that are explained by the open-chain structure.

Why it’s wrong: The question specifically asks for reactions that cannot be explained by the open-chain structure. The open-chain aldehyde form explains oxidation, reduction, and osazone formation perfectly.
How to avoid: Memorize the two key anomalies:
1. Failure to give the Schiff’s test (or react with NaHSO₃).
2. Existence of two isomeric forms (α and β) with different optical rotations (mutarotation).

Mistake 2: Confusing mutarotation with optical activity

Students say “glucose shows optical activity” as an anomaly.

Why it’s wrong: The open-chain structure also has a chiral carbon and is optically active. The anomaly is not optical activity itself, but the change in specific rotation over time (mutarotation).
How to avoid: Be precise. Say: “Freshly prepared solution of α-D-glucose has a specific rotation of +112°, which gradually changes to +52.5° (equilibrium). The open-chain structure cannot explain this change.”

Mistake 3: Vague explanation of the cyclic structure

Students say “glucose forms a ring” without showing which carbon bonds to the aldehyde group.

Why it’s wrong: The exam expects you to know that the –OH of C5 attacks the carbonyl carbon (C1), forming a hemiacetal (pyranose ring).
How to avoid: Draw or describe:
- C1 (carbonyl) + C5–OH → cyclic hemiacetal.
- This creates a new chiral centre at C1 → two anomers (α and β).
- In the cyclic form, the aldehyde group is masked (no free –CHO), so no Schiff’s test.

Mistake 4: Forgetting to mention the anomeric carbon

Students explain mutarotation but don’t mention the anomeric carbon (C1).

Why it’s wrong: Mutarotation is the interconversion of α and β anomers via the open-chain form. Without naming the anomeric carbon, the explanation is incomplete. …

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Q.Write the reactions of D-glucose which cannot be explained by its open chain structure. How can cyclic structure of glucose explain these reactions?

The Concept: Ring-Chain Tautomerism in Glucose

Reactions Not Explained by the Open-Chain Structure

1. Schiff’s Test (Negative)

2. No Stable Bisulfite Addition Compound

3. Does Not Give 2,4-DNP Test (or Gives It Very Slowly)

4. Does Not Show Aldehyde Proton in NMR (in D₂O)

How the Cyclic Structure Explains These Reactions — Step by Step

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