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Question #45005

Difficulty: 8/10 | 🟢 Free

The Question

A non-reducing trisaccharide X\mathbf{X} (C18H32O16C_{18}H_{32}O_{16}) is subjected to the following enzymatic and chemical analyses at 298 K and pH 7.4:

  1. Complete exhaustive methylation of X\mathbf{X} followed by acid hydrolysis yields equimolar amounts of:

    • 2,3,4,6-tetra-OO-methyl-D-galactose
    • 2,3,4,6-tetra-OO-methyl-D-glucose
    • 2,3,6-tri-OO-methyl-D-glucose
  2. Hydrolysis of X\mathbf{X} with lactase (β\beta-galactosidase) yields D-galactose and a disaccharide Y\mathbf{Y}. Y\mathbf{Y} does not reduce Tollens' reagent and is completely resistant to maltase (α\alpha-glucosidase).

  3. Hydrolysis of X\mathbf{X} with emulsin (β\beta-glucosidase) yields D-glucose and a disaccharide Z\mathbf{Z}. Z\mathbf{Z} reduces Tollens' reagent and undergoes mutarotation.

Based on these observations, identify the correct IUPAC structure of the trisaccharide X\mathbf{X}.

Options

  • A.

    OO-\beta-D-galactopyranosyl-(1\rightarrow4)-OO-\beta-D-glucopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside

  • B.

    OO-\beta-D-galactopyranosyl-(1\rightarrow4)-OO-\alpha-D-glucopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside

  • C.

    OO-\alpha-D-galactopyranosyl-(1\rightarrow4)-OO-\beta-D-glucopyranosyl-(1\leftrightarrow1)-\alpha-D-glucopyranoside

  • D.

    OO-\beta-D-glucopyranosyl-(1\rightarrow4)-OO-\beta-D-galactopyranosyl-(1\leftrightarrow1)-\beta-D-glucopyranoside

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Enzyme Specificity 'Lock and Key'|Glycosidic Linkage Geometry|Disaccharide Hydrolysis Products

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to integrate methylation analysis with the strict stereospecificity of glycosidase enzymes.

The 'Hidden Variable' is the anomeric configuration of the (1↔\leftrightarrow1) linkage.

Students often overlook that a non-reducing disaccharide like Y\mathbf{Y} can have α,α\alpha,\alpha, α,β\alpha,\beta, or β,β\beta,\beta configurations.

The resistance to maltase (an α\alpha-glucosidase) acts as the critical negative evidence, proving the absence of an α\alpha-linkage and confirming the β,β\beta,\beta geometry.

Furthermore, it tests the understanding that enzymatic cleavage of a non-reducing sugar can expose a latent anomeric carbon, generating a reducing sugar (Z\mathbf{Z}) that undergoes mutarotation.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptEnzyme Specificity 'Lock and Key'
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree