To solve this question, we need to analyze the sequence of reactions given in the image. Let's break it down step by step:
The starting compound is a dihalide: 2,3-dibromoheptane.
The first step involves treatment with \(\text{NaNH}_2\), which is a strong base. This leads to the elimination of both bromine atoms, resulting in the formation of an alkyne.
Reacting with a strong base like \(\text{NaNH}_2\) typically results in a double elimination to form an alkyne.
The alkyne formed has the structure: \(\text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_2-\text{CH}_3\), which is 2-heptyne.
The next step involves another treatment with \(\text{NaNH}_2\), followed by reaction with an alkyl halide: \(\text{CH}_3\text{Br}\).
Deprotonation by \(\text{NaNH}_2\) creates an acetylide ion, which can act as a nucleophile to attack the \(\text{Br}\) on the alkyl halide substituting \(\text{Br}\).
This results in the formation of the compound 5-methylhex-2-yne.
Therefore, the correct product \(Y\) is 5-methylhex-2-yne. This aligns with the correct answer given:
