This question involves identifying the product of a chemical reaction, specifically the reaction of an alkyne with sodium amide (NaNH₂) in liquid ammonia (NH₃), followed by reaction with a primary alkyl halide. This is a classic method for the alkylation of terminal alkynes.

Step-by-Step Explanation:

Step 1: Formation of the Acetylide Ion
The terminal alkyne (shown in the first image) reacts with the strong base NaNH₂. Sodium amide deprotonates the terminal alkyne, which has an acidic proton. The reaction is:

R - C \equiv C - H + NaNH 2 \to R - C \equiv C - {Na}^{+} + NH 3

This forms the sodium acetylide salt, which is a good nucleophile.

Step 2: Nucleophilic Substitution (S_N2 Reaction)
The nucleophilic acetylide ion then attacks the primary alkyl halide (R'-X) in an S_N2 reaction. The halogen (X) is displaced, and a new carbon-carbon bond is formed, resulting in an internal alkyne.

R - C \equiv C - {Na}^{+} + R' - X \to R - C \equiv C - R' + NaX

Final Answer:
The product is an internal alkyne where the alkyl group from the halide is attached to the terminal carbon of the original alkyne. Looking at the options, the correct product is the one with the structure of an internal alkyne: CH₃-C≡C-CH₂CH₃. This corresponds to the second option.

Related Topics & Formulae:

Acidity of Terminal Alkynes: The proton on a terminal alkyne is more acidic than in alkenes or alkanes due to the higher s-character of the sp-hybridized carbon. This allows it to be deprotonated by strong bases like NaNH₂ or organolithium reagents.

S_N2 Mechanism: This bimolecular nucleophilic substitution is favored for primary alkyl halides. The nucleophile attacks the carbon from the backside, leading to inversion of configuration.

Key Formula: The general reaction for alkyne alkylation is:
$HC \equiv CR + 1$ NaNH₂/NH₃(l) → NaC≡CR
$2$ NaC≡CR + R'X → RC≡CR' + NaX