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Question #43997

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The Question

An equilibrium mixture of D-glucose in water at $298\text{ K}$ is lyophilized to yield a solid mixture of anomers. This solid is subjected to two distinct acetylation conditions:\n\nCondition 1: Treatment with excess acetic anhydride ( $\ce{Ac_2O}$ ) and anhydrous sodium acetate ( $\ce{NaOAc}$ ) at $373\text{ K}$ for a short duration, followed by rapid quenching, yielding major product P.\n\nCondition 2: Treatment with excess acetic anhydride ( $\ce{Ac_2O}$ ) and a catalytic amount of anhydrous zinc chloride ( $\ce{ZnCl_2}$ ) at $373\text{ K}$ for an extended period, yielding major product Q.\n\nConsidering the mutarotation kinetics and the geometry of the glycosidic linkage, identify the major products P and Q.

Options

A.
P = $1,2,3,4,6$ -penta- $O$ -acetyl- $\beta$ -D-glucopyranose; Q = $1,2,3,4,6$ -penta- $O$ -acetyl- $\alpha$ -D-glucopyranose
B.
P = $1,2,3,4,6$ -penta- $O$ -acetyl- $\alpha$ -D-glucopyranose; Q = $1,2,3,4,6$ -penta- $O$ -acetyl- $\beta$ -D-glucopyranose
C.
P = $1,2,3,4,6$ -penta- $O$ -acetyl- $\alpha$ -D-glucopyranose; Q = $1,2,3,4,6$ -penta- $O$ -acetyl- $\alpha$ -D-glucopyranose
D.
P = $1,2,3,4,6$ -penta- $O$ -acetyl- $\beta$ -D-glucopyranose; Q = $1,2,3,4,6$ -penta- $O$ -acetyl- $\beta$ -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

Esterification Reversibility|Glycosidic Linkage Geometry|D-Glucose Anomeric Equilibrium

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of kinetic vs. thermodynamic control in carbohydrate chemistry.

Students often assume the thermodynamic product ( $\alpha$ -anomer due to the anomeric effect) is always formed, or that the product ratio strictly mirrors the initial $36:64$ equilibrium.

The 'hidden variable' is the differential reaction kinetics of the anomeric $\ce{-OH}$ groups: the equatorial $\beta$ -hydroxyl reacts faster.

Because base catalysts ( $\ce{NaOAc}$ ) cannot anomerize the resulting ester, the system is kinetically trapped as the $\beta$ -pentaacetate.

Conversely, Lewis acids ( $\ce{ZnCl_2}$ ) allow equilibration via an oxocarbenium ion, yielding the thermodynamically stable $\alpha$ -pentaacetate.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Aldehydes, Ketones, and Carboxylic Acids - Class 12
Concept	Esterification Reversibility
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free