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Question #44471

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from 1-methylcyclohex-1-ene (\textbf{A}):

\textbf{A} →2. \ceMe2S1. \ceO3,\ceCH2Cl2,196 K\xrightarrow[\text{2. } \ce{Me2S}]{\text{1. } \ce{O3}, \ce{CH2Cl2}, 196\text{ K}} \textbf{B}

\textbf{B} →2. \ceH3O+1. \ceMg(Hg),THF\xrightarrow[\text{2. } \ce{H3O^+}]{\text{1. } \ce{Mg(Hg)}, \text{THF}} \textbf{C}

\textbf{C} →\ceHIO4,aq. THF,298 K\xrightarrow{\ce{HIO4}, \text{aq. THF}, 298\text{ K}} \textbf{D}

Identify the major product \textbf{D}.

Options

  • A.

    6-Oxoheptanal

  • B.

    Heptane-1,6-diol

  • C.

    1-Methylcyclohexane-1,2-diol

  • D.

    Heptanedioic acid

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

Ozonolysis Fragmentation Patterns|Glycol Periodic Acid Cleavage|Tautomerism Conditions

Why This Question Is Tricky

Common Mistakes

The trap tests the student's ability to distinguish between standard reductions and reductive couplings.

Students who mistake \ce{Mg(Hg)} for a simple hydride donor (like \ce{NaBH4}) will predict the formation of heptane-1,6-diol.

Since \ce{HIO4} strictly requires a vicinal diol to form its cyclic periodate intermediate, the acyclic 1,6-diol would be inert.

Recognizing the Pinacol coupling is essential, as it generates the cyclic vicinal diol required to make the \ce{HIO4} step chemically active, ultimately creating a 'connectivity loop' that regenerates the original ozonolysis product.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterHydrocarbons - Class 11 Part 2
ConceptOzonolysis Fragmentation Patterns
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree