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Question #44954

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The Question

D-Arabinose undergoes Kiliani-Fischer chain extension (1. $HCN$ , 2. $H_2/Pd-BaSO_4$ , $H_2O$ ) to yield two diastereomeric aldohexoses, $X$ and $Y$ . Upon dissolution in water at 298 K, both $X$ and $Y$ undergo mutarotation to reach their respective anomeric equilibria. For epimer $X$ , the $\beta$ -pyranose form is thermodynamically dominant ( $K_{eq} (\alpha \rightleftharpoons \beta) > 1$ ). For epimer $Y$ , the $\alpha$ -pyranose form is thermodynamically dominant ( $K_{eq} (\alpha \rightleftharpoons \beta) < 1$ ). Identify $X$ and $Y$ , and the primary stereochemical reason for the anomalous anomeric equilibrium of $Y$ .

Options

A.
$X$ = D-Glucose, $Y$ = D-Mannose; The $\alpha$ -anomer of $Y$ is favored due to the anomeric effect and the destabilization of its $\beta$ -anomer by a severe gauche interaction between the equatorial C1-OH and axial C2-OH ( $\Delta 2$ effect).
B.
$X$ = D-Mannose, $Y$ = D-Glucose; The $\alpha$ -anomer of $Y$ is favored because it allows all hydroxyl groups to occupy equatorial positions, minimizing 1,3-diaxial repulsions.
C.
$X$ = D-Glucose, $Y$ = D-Mannose; The $\alpha$ -anomer of $Y$ is favored primarily due to strong intramolecular hydrogen bonding between the axial C1-OH and the axial C3-OH in the pyranose ring.
D.
$X$ = D-Galactose, $Y$ = D-Talose; The $\alpha$ -anomer of $Y$ is favored because the $\beta$ -anomer suffers from severe 1,3-diaxial interactions between the C1-OH and C5-CH_2OH groups.

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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

Epimerization vs Anomerization|Kiliani-Fischer Chain Extension|D-Glucose Anomeric Equilibrium

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of carbohydrate chain extension (Kiliani-Fischer) and advanced conformational analysis of cyclic hemiacetals.

Students often assume that for all D-aldohexoses, the $\beta$ -anomer is thermodynamically more stable due to equatorial preference (as seen in D-Glucose).

The trap lies in failing to recognize that epimerization at C2 (yielding D-Mannose) introduces an axial hydroxyl group.

This alters the steric environment, causing the $\beta$ -anomer to suffer from a $\Delta 2$ gauche interaction.

Consequently, the anomeric effect dominates, and the $\alpha$ -anomer becomes the major species at equilibrium.

This highlights that epimers do not share identical anomeric equilibrium constants ( $K_{eq}$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Epimerization vs Anomerization
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free