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Question #45008

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The Question

Disaccharide $\textbf{X}$ ( $\ce{C12H22O11}$ ) is completely hydrolyzed by $\alpha$ -glucosidase to yield exclusively D-glucose. In a buffered aqueous solution at 298 K (pH 7.4), $\textbf{X}$ exhibits mutarotation and gives a positive test with a mild, non-denaturing $\ce{Cu^{2+}}$ oxidant. When $\textbf{X}$ is incubated with a high-affinity binding protein $\textbf{E}$ , a stable non-covalent complex $[\textbf{E}-\textbf{X}]$ is formed. Addition of the same $\ce{Cu^{2+}}$ oxidant to the $[\textbf{E}-\textbf{X}]$ complex yields no reaction, despite the oxidant freely diffusing into the binding pocket. Identify the correct statement regarding the identity of $\textbf{X}$ and the fundamental mechanistic reason for the negative oxidation test.

Options

A.
$\textbf{X}$ is Maltose. The active site of $\textbf{E}$ locks the reducing end in a cyclic hemiacetal conformation, preventing the acid/base-catalyzed ring opening to the obligate acyclic aldehyde intermediate.
B.
$\textbf{X}$ is Cellobiose. The active site of $\textbf{E}$ selectively stabilizes the $\beta$ -anomer, which is kinetically inert to $\ce{Cu^{2+}}$ oxidation due to equatorial steric hindrance.
C.
$\textbf{X}$ is Sucrose. The protein $\textbf{E}$ shields the $\alpha, \beta-(1 \rightarrow 2)$ glycosidic linkage, preventing its spontaneous hydrolysis into reactive monosaccharides.
D.
$\textbf{X}$ is Maltose. The binding pocket forces $\textbf{X}$ into a rigid acyclic aldehyde conformation, which cannot form the cyclic enediol transition state required for electron transfer.

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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

Enzyme Specificity 'Lock and Key'|Reducing Sugar Criteria|Tollens vs. Fehling Specificity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Tollens/Fehling Accessibility Constraint' by bridging organic carbohydrate chemistry with biochemical kinetics.

Students often memorize that 'hemiacetals are reducing sugars' without understanding the underlying mechanism: oxidation requires equilibrium with the open-chain aldehyde form.

The trap lies in assuming that a chemically reducing sugar will always react with an oxidant.

By placing the sugar in an enzyme's active site that restricts conformational changes (mutarotation), the hidden variable-steric and electronic locking of the cyclic form-prevents the formation of the obligate acyclic intermediate, rendering the test negative.

Distractors exploit confusion over anomer reactivity, incorrect enzymatic cleavage specificity, and fundamental misunderstandings of the oxidation transition state.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Enzyme Specificity 'Lock and Key'
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free