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Question #44958

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The Question

D-Arabinose undergoes a Kiliani-Fischer synthesis to yield two C2-epimeric aldohexoses, $\mathbf{X}$ and $\mathbf{Y}$ . The aldohexose $\mathbf{X}$ forms a $\beta$ -pyranose conformation in which all non-anomeric hydroxyl groups and the hydroxymethyl group are in equatorial positions.

Both $\mathbf{X}$ and $\mathbf{Y}$ are independently subjected to the following reaction sequence:

Excess $Ac_2O$ , Pyridine
$HBr$ in $CH_3COOH$ (yielding the thermodynamically stable $\alpha$ -glycosyl bromide)
$CH_3OH$ , $Ag_2CO_3$ (Koenigs-Knorr glycosidation)
Catalytic $CH_3ONa$ in $CH_3OH$ (Zemplén deacetylation)

Let the final major isolated methyl glycosides obtained from $\mathbf{X}$ and $\mathbf{Y}$ be $\mathbf{M_X}$ and $\mathbf{M_Y}$ , respectively. Which of the following correctly identifies $\mathbf{M_X}$ and $\mathbf{M_Y}$ ?

Options

A.
$\mathbf{M_X}$ = Methyl $\beta$ -D-glucopyranoside; $\mathbf{M_Y}$ = Methyl $\alpha$ -D-mannopyranoside
B.
$\mathbf{M_X}$ = Methyl $\alpha$ -D-glucopyranoside; $\mathbf{M_Y}$ = Methyl $\beta$ -D-mannopyranoside
C.
$\mathbf{M_X}$ = Methyl $\beta$ -D-glucopyranoside; $\mathbf{M_Y}$ = Methyl $\beta$ -D-mannopyranoside
D.
$\mathbf{M_X}$ = Methyl $\alpha$ -D-glucopyranoside; $\mathbf{M_Y}$ = Methyl $\alpha$ -D-mannopyranoside

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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

Glycosidic Linkage Geometry|Disaccharide Hydrolysis Products|Kiliani-Fischer Chain Extension

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Kiliani-Fischer epimerization and Neighboring Group Participation (NGP) in determining glycosidic linkage geometry.

Students often memorize that the Koenigs-Knorr reaction yields $\beta$ -glycosides via simple $S_N2$ inversion of the $\alpha$ -bromide.

However, the reaction is actually governed by NGP from the C2-acetate, which enforces a 1,2-trans geometry.

Because Kiliani-Fischer creates C2 epimers, the C2-acetate in D-glucose is $\alpha$ (directing $\beta$ -attack), while in D-mannose it is $\beta$ (directing $\alpha$ -attack).

Failing to account for this stereochemical bifurcation at C2 leads to the incorrect assumption that both sugars will form $\beta$ -glycosides.

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