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Question #44476

Difficulty: 8/10 | 🟢 Free

The Question

An organic compound P\textbf{P} (C4H6C_4H_6) exhibits terminal acidity. Two separate experiments are performed on 1 mole of P\textbf{P}:

\textbf{Experiment 1:} P\textbf{P} is subjected to vigorous oxidation with hot acidic KMnO4KMnO_4, yielding a carboxylic acid Q\textbf{Q} and a gas. The reaction consumes xx moles of KMnO4KMnO_4.

\textbf{Experiment 2:} P\textbf{P} is treated with HgSO4/H2SO4HgSO_4 / H_2SO_4 to undergo complete hydration, forming a stable tautomer R\textbf{R}. Compound R\textbf{R} is then subjected to vigorous oxidation with hot acidic KMnO4KMnO_4, yielding exclusively the major product S\textbf{S} as dictated by Popoff's rule. This oxidation step consumes yy moles of KMnO4KMnO_4.

Assuming complete conversion in all steps, what are the values of xx and yy, and the identity of S\textbf{S}?

Options

  • A.

    x=1.6x = 1.6, y=1.2y = 1.2, S\textbf{S} = Acetic acid

  • B.

    x=1.6x = 1.6, y=1.6y = 1.6, S\textbf{S} = Propanoic acid

  • C.

    x=1.2x = 1.2, y=1.2y = 1.2, S\textbf{S} = Acetic acid

  • D.

    x=1.6x = 1.6, y=0.8y = 0.8, S\textbf{S} = Propanoic acid

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Alkynes Terminal Acidity|Tautomerism Conditions|Equivalent Concept in Redox

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of kinetic/thermodynamic organic pathways with rigorous redox stoichiometry.

A common student trap is assuming that since hydration merely adds water (which has no net oxidation state change), the subsequent oxidation of the ketone will consume the same equivalents of oxidant as the direct oxidation of the alkyne (leading to Option B).

However, the regioselectivity of ketone cleavage (Popoff's rule) dictates a different carbon-carbon bond breaking (C2-C3) compared to the alkyne cleavage (C1-C2).

This fundamentally alters the final oxidation states of the products, dropping the n-factor from 8 to 6.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHydrocarbons - Class 11 Part 2
ConceptAlkynes Terminal Acidity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree