JEE Chemistry 2026 Question with Solution

Given: A vicinal dibromide is treated with excess $\text{NaNH}_2$ to form intermediate $X$, and then with $\text{NaNH}_2$ followed by isopropyl bromide to form $Y$.

Find: The product $Y$.

Vicinal dihalides give alkynes on treatment with excess $\text{NaNH}_2$, and terminal alkynes can be alkylated through acetylide ion formation.

Step 1: Formation of alkyne intermediate $X$.

The given compound is a vicinal dibromide. On treatment with excess sodium amide $\left(\text{NaNH}_2\right)$, 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:

$$\text{5-methylhex-2-yne}$$

Therefore, the correct option is C.