Question

Consider the following reaction:

The product \(Y\) formed is:

• A. 2-methylhex-3-yne
• B. 2-methylhex-2-yne
• C. 5-methylhex-2-yne
• D. Isopropylbut-1-yne

Hint:

Vicinal dihalides give alkynes on treatment with excess \(\text{NaNH}_2\), and terminal alkynes can be alkylated via acetylide ions.

Answer 1

The Correct Option is C

Step 1: Formation of alkyne intermediate \(X\).
The given compound is a vicinal dibromide. On treatment with excess sodium amide (\(\text{NaNH}_2\)), double dehydrohalogenation occurs, leading to the formation of a terminal alkyne \(X\).

Step 2: Formation of acetylide ion.
The terminal alkyne reacts with \(\text{NaNH}_2\) to form a sodium acetylide ion due to the acidic nature of the terminal hydrogen.

Step 3: Alkylation of acetylide ion.
The acetylide ion undergoes nucleophilic substitution with isopropyl bromide, resulting in carbon–carbon bond formation and chain extension.

Step 4: Identify the final product.
The alkylation introduces an isopropyl group at the terminal carbon of the alkyne, giving the final product: \[ \text{5-methylhex-2-yne} \]
Final Answer: \[ \boxed{\text{5-methylhex-2-yne}} \]