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Question #44995

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of a naturally occurring, non-reducing disaccharide X\mathbf{X} (\ceC12H22O11\ce{C12H22O11}) is hydrolyzed with dilute \ceH2SO4\ce{H2SO4} to yield a single D-aldohexose Y\mathbf{Y}. Y\mathbf{Y} forms an osazone identical to that formed by D-fructose. Y\mathbf{Y} is subjected to the following reaction sequence: Y→2. \ceH3O+, Δ3. \ceNa(Hg), \ceH2O, pH 3−51. \ceHCNP+Q\mathbf{Y} \xrightarrow[2. \ \ce{H3O^+}, \ \Delta \quad 3. \ \ce{Na(Hg)}, \ \ce{H2O}, \ pH \ 3-5]{1. \ \ce{HCN}} \mathbf{P} + \mathbf{Q} where P\mathbf{P} and Q\mathbf{Q} are isomeric aldoheptoses. Subsequent oxidation of P\mathbf{P} and Q\mathbf{Q} with warm concentrated \ceHNO3\ce{HNO3} yields heptaric acids P′\mathbf{P'} and Q′\mathbf{Q'}, respectively. Which of the following correctly describes the stereochemical relationship between P\mathbf{P} and Q\mathbf{Q}, and the optical activity of P′\mathbf{P'} and Q′\mathbf{Q'}?

Options

  • A.

    C2-epimers; exactly one meso compound

  • B.

    C2-epimers; both optically active

  • C.

    C3-epimers; exactly one meso compound

  • D.

    Enantiomers; both optically inactive

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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

Disaccharide Hydrolysis Products|Kiliani-Fischer Chain Extension|D-Glucose Anomeric Equilibrium

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to integrate carbohydrate identification with stereochemical transformations.

The 'Hidden Variable' is the symmetry of the resulting heptaric acids.

Students often assume that chain extension of a chiral molecule like D-glucose will exclusively yield chiral derivatives (Option B).

The trap is failing to draw the full 7-carbon Fischer projection to recognize that one of the C2-epimers perfectly aligns its new stereocenter with the existing asymmetric carbons to create a meso compound (plane of symmetry at C4).

Furthermore, if a student incorrectly identifies Y\mathbf{Y} as D-mannose, stereochemical analysis would incorrectly lead them to Option B, rewarding only those who correctly identify Trehalose.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptDisaccharide Hydrolysis Products
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree