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AIIMS2019Chemistry-Organic Chemistry

AIIMS 2019 Chemistry Alkenes and Alkynes MCQ Question

Type: MCQ-conceptual-Medium-Class 11

Ph−CH2−CH=CH−CH3→(ii) Alk. KOH(i) Br2\text{Ph}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3 \xrightarrow[\text{(ii) Alk. KOH}]{\text{(i) Br}_2}

A

Ph−CH=CH−CH=CH2\text{Ph}-\text{CH}=\text{CH}-\text{CH}=\text{CH}_2

B

Ph−CH2−CH−CH−CH3 ∣∣OHOH\begin{array}{c} \text{Ph}-\text{CH}_2-\text{CH}-\text{CH}-\text{CH}_3 \\ \ |\qquad | \\ \text{OH}\quad \text{OH} \end{array}

C

Ph−CH2−C≡C−CH3\text{Ph} - \text{CH}_2 - \text{C} \equiv \text{C} - \text{CH}_3

D

Ph−C≡C−CH2−CH3\text{Ph} - \text{C} \equiv \text{C} - \text{CH}_2 - \text{CH}_3

Correct Answer

Option C

Detailed Explanation

The reaction sequence involves the bromination of the alkyne, where Br₂ adds across the triple bond of Ph−CH₂−C≡C−CH₃ to form a dibromide intermediate. The subsequent treatment with alcoholic KOH leads to the elimination of HBr, resulting in the formation of the final product, Ph−CH₂−C≡C−CH₃. Other options are not applicable as they do not represent valid intermediates or products in this specific reaction pathway, which focuses on the transformation of the alkyne through halogenation and elimination processes.

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