Consider compound (A): a cyclic (pyranose) form of glucose, drawn as a Fischer-type projection closed into a r

NCERT Exemplar · Q34

Q. Consider compound (A): a cyclic (pyranose) form of glucose, drawn as a Fischer-type projection closed into a ring by an oxygen that joins the top carbon (the former carbonyl / anomeric carbon) to the ring carbon bearing the terminal -CH2OAc group. The key feature is that the anomeric (top) carbon carries an acetylated oxygen, -OAc, rather than a free hemiacetal -OH; the terminal group is -CH2OAc and the remaining ring carbons carry -OAc or -OH substituents. Why does compound (A) not react with hydroxylamine (NH2OH) to form an oxime?

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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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An oxime forms only from a free carbonyl group reacting with NH2OH. In compound (A) the anomeric hydroxyl is acetylated (-OAc), which locks the cyclic acetal so it cannot open into the open-chain aldehyde. With no free -CHO available, no oxime is produced.

Concept

Glucose exists mainly in the cyclic hemiacetal (pyranose) form but is in equilibrium with a small amount of the open-chain aldehyde. It is that open-chain -CHO that condenses with hydroxylamine to give an oxime:

-CHO + NH2OH -> -CH=N-OH + H2O

The ring can open to the aldehyde only because the anomeric carbon carries a free hemiacetal -OH.

Why compound (A) is inert to NH2OH …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
Starch is a polymer of α-D-glucose, but its amylose component is linear and water‑insoluble, while cellulose is a linear polymer of β-D-glucose. Only statements I and IV are correct.

Concept & Intuition

Polysaccharides are large polymers made of monosaccharide units linked by glycosidic bonds. The key difference between starch and cellulose lies in the stereochemistry of the glucose units (α vs. β) and the resulting chain structure (branched vs. straight). Starch has two components: amylose (linear, α-1,4 links) and amylopectin (branched, α-1,6 links). Cellulose is entirely linear with β-1,4 links. Understanding these structural details lets us judge each statement.

Step‑by‑step reasoning

Statement I: “Starch is a polymer of α‑D(+)-glucose”

Starch is composed entirely of D‑glucose units joined by α‑glycosidic bonds. Both amylose (α‑1,4) and amylopectin (α‑1,4 and α‑1,6) use α‑glucose. This statement is true.

Statement II: “Amylose component of starch is not soluble in water.”

Amylose is a linear, helical polymer. Although it can form a colloidal dispersion in hot water, it is not truly soluble in cold water because the long chains pack tightly via hydrogen bonds. In common biochemical contexts, amylose is considered water‑insoluble. This statement is true.

Statement III: “Amylose is a branched chain polymer of α‑D(+)-glucose.” …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$Q. Sucrose on hydrolysis gives (A) D-(+)-Glucose and D-(+)-Fructose (B) L-(+)-Glucose and D-(+)-Fructose (C) D-(+)-Glucose and D-(-)-Fructose (D) D-(-)-Glucose and D-(-)-Fructose$
›Reveal solutionSolution
Sucrose is a disaccharide made of α-D-glucose and β-D-fructose linked via their anomeric carbons; on hydrolysis it yields D-(+)-glucose and D-(-)-fructose, so the correct choice is (C).

Sucrose is a non‑reducing sugar because the anomeric carbons of both monosaccharide units are involved in the glycosidic bond. When you break that bond (hydrolysis), you get back the two individual monosaccharides. The key is to recall the specific optical rotation of each product — that’s what the “(+)” and “(−)” in the options refer to.

Identify the monosaccharide units in sucrose

Sucrose consists of one molecule of α‑D‑glucopyranose and one molecule of β‑D‑fructofuranose, joined by an α‑1,β‑2 glycosidic linkage (between C‑1 of glucose and C‑2 of fructose). Both anomeric carbons are tied up, so the disaccharide is non‑reducing.

Recall the optical rotation of the free sugars

D‑Glucose in aqueous solution at equilibrium has a specific rotation of $+52.7^\circ$ , so it is D‑(+)‑glucose.

D‑Fructose in aqueous solution at equilibrium has a specific rotation of $-92.4^\circ$ , so it is D‑(−)‑fructose. The “D” refers to the configuration at the chiral centre farthest from the carbonyl group (C‑5 for glucose, C‑5 for fructose), not to the sign of rotation. Both are D‑sugars, but their optical rotations are opposite.

Apply to the hydrolysis products

Hydrolysis of sucrose simply cleaves the glycosidic bond, releasing free glucose and free fructose. No change in configuration occurs at any chiral centre. Therefore the products are exactly the free sugars: D‑(+)‑glucose and D‑(−)‑fructose.

Match with the options

(A) says D‑(+)‑glucose and D‑(+)‑fructose — wrong sign for fructose. …
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$Q. Sucrose on hydrolysis gives (A) D-(+)-Glucose and D-(+)-Fructose (B) L-(+)-Glucose and D-(+)-Fructose (C) D-(+)-Glucose and D-(-)-Fructose (D) D-(-)-Glucose and D-(-)-Fructose$
›Reveal solutionSolution
Sucrose is a disaccharide of α-D-glucose and β-D-fructose; on hydrolysis it yields D-(+)-glucose and D-(-)-fructose, so the correct choice is (C).

Sucrose is a non‑reducing sugar because the anomeric carbons of both monosaccharide units are involved in the glycosidic bond. To know what it breaks into, we need to recall the structure and the optical activity of the products.

Structure of sucrose

Sucrose consists of one α-D-glucopyranose unit and one β-D-fructofuranose unit linked via their anomeric carbons (C1 of glucose and C2 of fructose). This bond is an α-1,2‑β glycosidic linkage.

Hydrolysis breaks the glycosidic bond

In aqueous acid or by the enzyme sucrase, the bond is cleaved, releasing the two monosaccharides intact.

Identity of the glucose unit

The glucose in sucrose is D‑glucose. Its anomeric carbon is α in the disaccharide, but once free it mutarotates to an equilibrium mixture of α and β forms. The equilibrium mixture is dextrorotatory, so it is designated D-(+)-glucose.

Identity of the fructose unit

The fructose in sucrose is D‑fructose, but in the disaccharide it is in the β‑furanose form. Upon hydrolysis, it mutarotates to an equilibrium mixture. The equilibrium mixture of D‑fructose is levorotatory (rotates plane‑polarised light to the left), so it is designated D-(-)-fructose.

Optical activity of the mixture …
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Concept understanding — Ring Chain Tautomerism

Ring-Chain Tautomerism: The Shape-Shifting Molecules

The Intuition

The Precise Statement

A Concrete Example: Glucose

Other Examples

Why this formula?

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

What Is Ring-Chain Tautomerism?

The Core Formula: Ring Size Preference

The 5- and 6-Membered Ring Rule

Why does this hold? — The Derivation of Preference

1. Angle Strain (Baeyer Strain Theory)

2. Entropy Factor (Ring-Closing Probability)

The Equilibrium Constant Expression

Why does $K$ depend on ring size?

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

Concept

Why compound (A) is inert to NH2OH …

Method: Reasoning from Functional-Group Reactivity: Why an Acetylated Sugar Won't Form an Oxime

Core Concept

Steps

Applying it to this question

Concept

Why compound (A) is inert to NH2OH …

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