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Question #44949

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The Question

Pure $\alpha$ -D-glucopyranose (specific rotation $+112^\circ$ ) is dissolved in a dilute $\ce{NaOD / D2O}$ solution at $298\text{ K}$ and allowed to reach thermodynamic equilibrium. The final specific rotation of the mixture is observed to be $\theta$ . In a separate experiment, the same sugar in neutral $\ce{H2O}$ reaches an equilibrium specific rotation of $+52.7^\circ$ .

Which of the following statements correctly describes the sites of deuterium incorporation in the D-glucopyranose molecules isolated from the alkaline equilibrium mixture, and the relationship between $\theta$ and $+52.7^\circ$ ?

Options

A.
Deuterium is incorporated at the hydroxyl groups and C-2 only; $\theta < +52.7^\circ$ due to the reversible formation of D-mannose and D-fructose.
B.
Deuterium is incorporated at the hydroxyl groups, C-1, and C-2; $\theta < +52.7^\circ$ due to the reversible formation of D-mannose and D-fructose.
C.
Deuterium is incorporated at the hydroxyl groups and C-1 only; $\theta > +52.7^\circ$ due to the irreversible formation of D-mannose.
D.
Deuterium is incorporated at the hydroxyl groups, C-1, and C-2; $\theta = +52.7^\circ$ because the base only accelerates mutarotation without altering the final equilibrium.

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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

D-Glucose Anomeric Equilibrium|Epimerization vs Anomerization

Why This Question Is Tricky

Common Mistakes

The question tests the student's understanding of the Lobry de Bruyn-van Ekenstein transformation beyond mere memorization.

The 'Hidden Variable' is the acidity of the protons.

Students typically know that the aldehyde proton (C-1 of glucose) is non-acidic and should not exchange.

However, they fail to realize that the reversible isomerization to D-fructose converts this C-1 carbon into an $\alpha$ -carbon relative to the C-2 ketone.

This makes the C-1 protons acidic, allowing H/D exchange.

When the molecule isomerizes back to D-glucose, the deuterium is trapped at the C-1 aldehyde position.

Furthermore, the question tests the thermodynamic impact of this epimerization/isomerization on the specific rotation, distinguishing it from simple mutarotation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	D-Glucose Anomeric Equilibrium
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free