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.

COMEDK 2026Set 2026-M1 markMCQ
$Q. Which among the following is an INCORRECT statement? (A) In D (+) Glucose, D represents the configuration similar to D-Glyceraldehyde (B) In monosaccharides, aldoses are reducing sugars while ketoses are non-reducing sugars (C) Polysaccharides are known as non-sugars as they are not sweet in taste (D) D (-) fructose forma a five membered ring known as furanose which is in analogy with the compound furan$
›Reveal solutionSolution
The key idea is to test your understanding of carbohydrate nomenclature and reducing sugar properties. The incorrect statement is (B), because both aldoses and ketoses are reducing sugars.

Concept & Intuition

This question checks common misconceptions about sugars. The D/L notation refers to the configuration at the chiral carbon farthest from the carbonyl group, not to optical rotation. Reducing sugars are those with a free aldehyde or ketone group that can reduce oxidizing agents — and all monosaccharides, whether aldoses or ketoses, are reducing. The sweetness of polysaccharides is irrelevant to their chemical classification. Let’s examine each option.

Step-by-step reasoning

Option (A): “In D (+) Glucose, D represents the configuration similar to D-Glyceraldehyde.”

The D/L system is based on the configuration of the highest-numbered chiral carbon (the one farthest from the carbonyl). For glucose, that carbon has the same configuration as D-glyceraldehyde (OH on the right in Fischer projection).

The (+) sign indicates dextrorotatory optical rotation, which is independent of D/L configuration.

This statement is correct.

Option (B): “In monosaccharides, aldoses are reducing sugars while ketoses are non-reducing sugars.”

A reducing sugar has a free anomeric carbon that can open to a carbonyl group.

Aldoses have an aldehyde group, which is reducing. Ketoses have a ketone group, but under basic conditions they tautomerize to an aldehyde (via enediol rearrangement), so they also reduce reagents like Benedict’s or Fehling’s.

Therefore, all monosaccharides are reducing sugars. This statement is incorrect.

Option (C): “Polysaccharides are known as non-sugars as they are not sweet in taste.”

Polysaccharides like starch and cellulose are not sweet, but the term “non-sugar” is a historical classification based on taste, not chemical structure. …
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KCET 2025Set D-41 markMCQ

\xrightarrow{\text{Reagent}}

List-I (Reagents)	List-II (Structural prediction)
a. Acetic anhydride	i. Glucose has an aldehyde group
b. Bromine water	ii. Glucose has a straight chain of six carbon atoms
c. Hydroiodic acid	iii. Glucose has five hydroxyl groups
d. Hydrogen cyanide	iv. Glucose has a carbonyl group

›Reveal solution

Each reagent is a classical structural probe: acetic anhydride counts the –OH groups (five), bromine water proves the group is an aldehyde (not a ketone), HI reduction proves an unbranched C₆ chain, and HCN addition proves a carbonyl.

Step 1 — The idea behind the question

The open-chain structure of glucose was deduced, not observed. Each of these four reagents was a specific chemical test whose product revealed one structural feature. To match them, ask for each reagent: what does its product prove?

Step 2 — (a) Acetic anhydride, $\ce{(CH3CO)2O}$

Acetic anhydride acetylates hydroxyl groups: every $\ce{-OH}$ is converted into an ester $\ce{-OCOCH3}$ . Glucose reacts to give glucose penta-acetate:

\ce{C6H12O6 + 5(CH3CO)2O -> C6H7O(OCOCH3)5 + 5CH3COOH}

Exactly five acetyl groups are taken up. Since one acetyl group is consumed per hydroxyl, glucose must contain five –OH groups.

\Rightarrow\ \boxed{\text{a} \to \text{iii (five hydroxyl groups)}}

Step 3 — (b) Bromine water, $\ce{Br2/H2O}$

Bromine water is a mild oxidising agent. Crucially, it oxidises an aldehyde ( $\ce{-CHO}$ ) to a carboxylic acid, but it does not oxidise a ketone. Glucose is oxidised to the six-carbon gluconic acid:

\ce{CHO-(CHOH)4-CH2OH ->[Br2 \text{ water}] COOH-(CHOH)4-CH2OH}\quad \text{(gluconic acid)}

Because it is oxidised by this mild reagent, the carbonyl of glucose must specifically be an aldehyde, not a ketone.

\Rightarrow\ \boxed{\text{b} \to \text{i (aldehyde group)}}

⚠️ Note the contrast with (d). HCN only shows there is some carbonyl (aldehyde or ketone). Bromine water is what pins it down as an aldehyde. That distinction is the whole point of this pair.

Step 4 — (c) Hydroiodic acid, $\ce{HI}$ (with red phosphorus, prolonged heating)

Hot HI/red P is a powerful reducing agent that strips out every oxygen function, replacing each $\ce{C-OH}$ and the $\ce{C=O}$ with $\ce{C-H}$ . Glucose is reduced all the way down to the hydrocarbon $n$ -hexane:

\ce{C6H12O6 ->[HI,\ \text{red P},\ \Delta] CH3-CH2-CH2-CH2-CH2-CH3}\quad (n\text{-hexane})

The product is normal (straight-chain) hexane, not a branched isomer. Since reduction cannot rearrange the carbon skeleton, the six carbons of glucose must already have been in an unbranched straight chain. …

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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{Br}_2$
›Reveal solutionSolution
The key is to recall the structural and chemical properties of carbohydrates: fructose forms a five-membered furanose ring, glucose oxidation with Br₂ water gives gluconic acid (not saccharic), amylose is water-soluble and unbranched, and cellulose is linear with β-D-glucose. Only statement (A) is correct.

Concept & Intuition

This question tests your knowledge of carbohydrate chemistry — specifically ring forms, oxidation reactions, and polymer structure. Each option makes a claim about a common sugar or polysaccharide. To pick the right one, you need to recall the exact ring size of fructose in solution, the specific oxidation product of glucose with mild vs. strong oxidizers, and the branching and monomer type of amylose and cellulose. The classic trap is confusing “furanose” with “pyranose” or mixing up gluconic and saccharic acids.

Step-by-step reasoning

Option (A): Fructose exists as a five membered ring named furanose.

Fructose is a ketohexose. In solution, it predominantly forms a five-membered cyclic hemiacetal (a furanose ring) because the ketone group at C2 reacts with the hydroxyl at C5. This is indeed called a furanose ring (analogous to furan). So this statement is true.

Option (B): Glucose gives Saccharic acid on reaction with Br₂ water.

Bromine water is a mild oxidizing agent. It oxidizes the aldehyde group of glucose to a carboxylic acid, producing gluconic acid, not saccharic acid. Saccharic acid (a dicarboxylic acid) requires oxidation of both ends — that happens with strong oxidizers like nitric acid. So this statement is false.

Option (C): Amylose is a water insoluble branched chain polymer of α-D-(–)-glucose. …
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←Previous questionQ33 — How do you explain the presence of five —OH groups in glucose molecule…→Next questionQ35 — Why must vitamin C be supplied regularly in diet?

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

Step 1 — The idea behind the question

Step 2 — (a) Acetic anhydride, $\ce{(CH3CO)2O}$

Step 3 — (b) Bromine water, $\ce{Br2/H2O}$

Step 4 — (c) Hydroiodic acid, $\ce{HI}$ (with red phosphorus, prolonged heating)

Concept

Why compound (A) is inert to NH2OH …

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