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Question #44961

Difficulty: 8/10 | 🟢 Free

The Question

An unknown carbohydrate $\mathbf{X}$ ( $C_{12}H_{22}O_{11}$ ) exhibits the following chemical behavior: I. It does not reduce Tollens' reagent and fails to exhibit mutarotation in aqueous solution at $298\text{ K}$ . II. Upon heating with excess phenylhydrazine in an acetate buffer ( $\text{pH} \approx 4.5$ ) at $100^\circ\text{C}$ for 45 minutes, it yields a yellow crystalline osazone $\mathbf{Y}$ . III. Exhaustive methylation of $\mathbf{X}$ using $CH_3I/Ag_2O$ followed by mild acid hydrolysis yields two distinct tetra-O-methylated monosaccharides, $\mathbf{M_1}$ and $\mathbf{M_2}$ . IV. Structural analysis reveals that $\mathbf{M_1}$ possesses exactly one primary methoxy group, whereas $\mathbf{M_2}$ possesses exactly two primary methoxy groups.

Which of the following correctly identifies $\mathbf{X}$ and the mechanistic rationale for the formation of $\mathbf{Y}$ ?

Options

A.
$\mathbf{X}$ is Sucrose. $\mathbf{Y}$ forms because the highly labile fructofuranosyl glycosidic bond undergoes hydrolysis under the mildly acidic and thermal conditions, releasing monosaccharides that subsequently form the osazone.
B.
$\mathbf{X}$ is Trehalose. $\mathbf{Y}$ forms because the $\alpha,\alpha-1,1$ -glycosidic linkage undergoes direct C1-C2 keto-enol tautomerization with phenylhydrazine, bypassing the need for hydrolysis.
C.
$\mathbf{X}$ is Sucrose. $\mathbf{Y}$ forms because the fructofuranose ring opens to expose a reactive C2 ketone, reacting with phenylhydrazine while the glycosidic bond remains intact.
D.
$\mathbf{X}$ is Maltose. $\mathbf{Y}$ forms because the free hemiacetal group at C1 of the reducing end reacts directly with phenylhydrazine, while the non-reducing end remains intact.

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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

Reducing Sugar Criteria|Osazone Formation Logic|Tollens vs. Fehling Specificity

Why This Question Is Tricky

Common Mistakes

The question tests the structural criteria for reducing sugars and a specific chemical exception regarding sucrose in the osazone test.

Students often memorize the heuristic 'non-reducing sugars do not form osazones' and might incorrectly eliminate sucrose upon seeing osazone formation.

The hidden variable is the kinetic lability of the fructofuranosyl glycosidic bond under the mildly acidic and thermal conditions of the phenylhydrazine reagent, leading to in situ hydrolysis (inversion) and subsequent osazone formation.

Furthermore, the methylation data rigorously proves the identity of sucrose by differentiating the primary hydroxyl count in the resulting monosaccharide derivatives, preventing guesswork.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Reducing Sugar Criteria
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free