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Question #44957

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The Question

A naturally occurring disaccharide $\mathbf{X}$ (\ce{C12H22O11}) exhibits the following properties: I. It does not decolorize bromine water and fails to exhibit mutarotation in an aqueous solution at $298\text{ K}$ . II. Exhaustive methylation of $\mathbf{X}$ with excess \ce{CH3I}/\ce{Ag2O} followed by dilute aqueous \ce{HCl} hydrolysis yields a single chiral carbohydrate derivative $\mathbf{Y}$ . III. Enzymatic hydrolysis of $\mathbf{X}$ with yeast $\alpha$ -glucosidase yields only D-glucose, whereas emulsin ( $\beta$ -glucosidase) produces no reaction.

Identify the exact glycosidic linkage in $\mathbf{X}$ and the IUPAC name of $\mathbf{Y}$ .

Options

A.
$\mathbf{X}$ : $\alpha$ -D-glucopyranosyl- $\alpha$ -D-glucopyranoside ; $\mathbf{Y}$ : 2,3,4,6-tetra- $O$ -methyl-D-glucopyranose
B.
$\mathbf{X}$ : $\alpha$ -D-glucopyranosyl-(1 $\rightarrow$ 4)- $\alpha$ -D-glucopyranose ; $\mathbf{Y}$ : 2,3,6-tri- $O$ -methyl-D-glucopyranose
C.
$\mathbf{X}$ : $\alpha$ -D-glucopyranosyl- $\beta$ -D-fructofuranoside ; $\mathbf{Y}$ : 1,3,4,6-tetra- $O$ -methyl-D-fructofuranose
D.
$\mathbf{X}$ : $\beta$ -D-glucopyranosyl- $\beta$ -D-glucopyranoside ; $\mathbf{Y}$ : 2,3,4,6-tetra- $O$ -methyl-D-glucopyranose

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Glycosidic Linkage Geometry|D-Glucose Anomeric Equilibrium|Reducing Sugar Criteria

Why This Question Is Tricky

Common Mistakes

The problem tests the 'Anomeric Equilibrium Constraint' in carbohydrate chemistry.

Students often memorize that disaccharides like maltose and lactose are reducing, but fail to structurally deduce why a 1,1'-linkage (as in trehalose) locks both anomeric carbons, preventing the ring-opening required for mutarotation and oxidation.

A secondary trap involves the regiochemistry of exhaustive methylation: students must distinguish between acid-stable ether linkages formed at C2, C3, C4, C6 and the acid-labile acetal linkage at C1, recognizing that hydrolysis regenerates the C1 hemiacetal to yield a single tetra-methylated product rather than a fully methylated open chain.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Glycosidic Linkage Geometry
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free