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Question #44978

Difficulty: 8/10 | 🟢 Free

The Question

A $2.0 \text{ mol}$ sample of an equimolar mixture of D-fructose and Guanine is treated with an excess of phenylhydrazine (\ce{PhNHNH2}) in an acetic acid buffer at $373 \text{ K}$ . Assuming the reaction proceeds to completion, what is the total number of moles of phenylhydrazine consumed and the total number of moles of ammonia (\ce{NH3}) produced, respectively?

Options

A.
$3.0 \text{ mol}, 1.0 \text{ mol}$
B.
$2.0 \text{ mol}, 0.0 \text{ mol}$
C.
$4.0 \text{ mol}, 1.0 \text{ mol}$
D.
$6.0 \text{ mol}, 2.0 \text{ mol}$

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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

Osazone Formation Logic|Nitrogenous Base Keto-Enol Tautomerism

Why This Question Is Tricky

Common Mistakes

This problem tests the distinction between localized reactive carbonyls in carbohydrates and delocalized, resonance-stabilized lactam carbonyls in nucleobases.

The primary trap is treating Guanine's keto-form as an active electrophilic site for nucleophilic addition-elimination (hydrazone formation).

A secondary trap is forgetting the redox stoichiometry of Fischer's osazone mechanism, where the second equivalent of phenylhydrazine acts as an oxidant, releasing exactly one equivalent of ammonia.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Biomolecules - Class 12
Concept	Osazone Formation Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free