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Question #44977

Difficulty: 8/10 | 🟢 Free

The Question

An aldopentose P\textbf{P} upon oxidation with dilute HNO3HNO_3 yields an optically active aldaric acid. P\textbf{P} is subjected to Kiliani-Fischer synthesis (using HCNHCN, followed by hydrolysis and reduction) to yield two aldohexoses, Q\textbf{Q} and R\textbf{R}. Oxidation of Q\textbf{Q} with HNO3HNO_3 yields an optically inactive (meso) aldaric acid, whereas R\textbf{R} yields an optically active aldaric acid. Q\textbf{Q} and R\textbf{R} are then individually reacted with 3 equivalents of phenylhydrazine (\cePhNHNH2\ce{PhNHNH2}) at 100∘C100^\circ\text{C} to form osazones OQ\textbf{O}_Q and OR\textbf{O}_R respectively. What is the relationship between OQ\textbf{O}_Q and OR\textbf{O}_R, and how many chiral centers are present in the osazone product?

Options

  • A.

    Identical, 3 chiral centers

  • B.

    Diastereomers, 4 chiral centers

  • C.

    Identical, 4 chiral centers

  • D.

    Enantiomers, 3 chiral centers

Check Your Answer

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

Osazone Formation Logic|Kiliani-Fischer Chain Extension

Why This Question Is Tricky

Common Mistakes

The question tests the deduction of sugar stereochemistry via symmetry in aldaric acids and the mechanistic outcome of the Kiliani-Fischer synthesis.

The core trap exploits osazone formation: students often forget that C-2 epimers (created during chain extension) converge to an identical osazone because the C-2 stereocenter is converted to an sp2sp^2 hybridized carbon.

Additionally, miscounting the chiral centers in the osazone (assuming 4 instead of 3) is a frequent stereochemical trap.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterBiomolecules - Class 12
ConceptOsazone Formation Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree