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Question #44399

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following reaction sequence starting from 1-chloro-1-methylcyclopentane:

1-chloro-1-methylcyclopentane→Δ\ceNaOEt/EtOHMajor Product (B)\text{1-chloro-1-methylcyclopentane} \xrightarrow[\Delta]{\ce{NaOEt / EtOH}} \text{Major Product (B)}

(B)→2.\ceMe2S1.\ceO3,195 K(C)\text{(B)} \xrightarrow[2. \ce{Me2S}]{1. \ce{O3}, 195\text{ K}} \text{(C)}

(C)→2.\ceH3O+1.\ceI2 (excess),\ceNaOH,ΔYellow precipitate+(D)\text{(C)} \xrightarrow[2. \ce{H3O+}]{1. \ce{I2} \text{ (excess)}, \ce{NaOH}, \Delta} \text{Yellow precipitate} + \text{(D)}

Identify the IUPAC name of the final organic product (D).

Options

  • A.

    Pentanedioic acid

  • B.

    4-Formylbutanoic acid

  • C.

    5-Oxohexanoic acid

  • D.

    Hexanedioic 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

Williamson Ether Synthesis Constraints|E1/E2 Solvent and Base Effects|Steric Hindrance in SN2

Why This Question Is Tricky

Common Mistakes

This problem tests the classic 'Williamson Ether Synthesis' trap where students erroneously predict an S_N2 substitution between a tertiary halide and an alkoxide.

The hidden variable is the kinetic dominance of the E2 pathway due to steric hindrance, yielding an alkene.

Furthermore, it tests a dual-reactivity trap in the final step: students often recognize the haloform reaction for the methyl ketone but fail to account for the simultaneous oxidation of the aldehyde group by the in situ generated hypohalite (\ceNaOI\ce{NaOI}).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterAlcohols, Phenols, and Ethers - Class 12
ConceptWilliamson Ether Synthesis Constraints
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree