Consider the following reaction :

The product Y formed is :

Step 1: Formation of X (Dehydrohalogenation)

The starting material is 1,1-dibromobutane. When treated with 2 equivalents of sodamide ($$\text{NaNH}_2$$), it undergoes a double dehydrohalogenation.

• Reaction: Elimination reaction (specifically $$\beta$$-elimination).
• Intermediate X: 1-Butyne ($$\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}$$).
• Mechanism: Strong base $$\text{NaNH}_2$$ removes protons and eliminates $$\text{Br}^-$$ twice to form a triple bond at the terminal position.

Step 2: Formation of Y (Alkylation of Terminal Alkyne)

Terminal alkynes are acidic. This step involves two sub-steps:

1. Deprotonation: $$\text{NaNH}_2$$ removes the acidic terminal proton from 1-butyne to form an acetylide ion ($$\text{CH}_3\text{CH}_2\text{C}\equiv\text{C}^-$$).
2. Nucleophilic Substitution ($$S_N2$$): The acetylide ion acts as a nucleophile and attacks isopropyl bromide (2-bromopropane).

• Reaction: Alkyne Alkylation (Nucleophilic Substitution).
• Product Y: 2-methylhex-3-yne.
• Mechanism: The carbon chain is extended by connecting the terminal alkyne carbon to the isopropyl group.