This question involves the reaction between a terminal alkyne and a strong base (sodium amide, NaNH $2$ ), followed by reaction with benzyl bromide (PhCH $2$ Br).

Step 1: Deprotonation of Terminal Alkyne
Terminal alkynes (R-C≡C-H) are acidic due to the sp-hybridized carbon. Sodium amide (NaNH $2$ ) is a strong base that deprotonates the terminal alkyne to form an acetylide anion:

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

Step 2: Nucleophilic Substitution with Benzyl Bromide
The sodium acetylide anion acts as a strong nucleophile. It undergoes an S $N_{2}$ reaction with benzyl bromide (an alkyl halide with a good leaving group, Br $-$ ). The benzyl carbon is primary and unhindered, making it highly reactive towards S $N_{2}$ substitution:

$Ph - C \equiv C - {Na}^{+} + Ph - CH 2 - Br \to Ph - C \equiv C - CH 2 - Ph + NaBr$

The final product is 1,3-diphenylpropyne (Ph-C≡C-CH $2$ -Ph).

Why not the other options?
- No reaction: Incorrect, as terminal alkynes readily react with strong bases and alkyl halides.
- Ph-C≡C-Ph: This is the symmetric alkyne, which would form if phenyl bromide (PhBr) was used instead of benzyl bromide. But PhBr is unreactive towards S $N_{2}$ due to the sp $2$ hybridized carbon and poor leaving group ability in aryl halides.
- The structure with a triple bond between the two phenyl rings is not formed.

Final Answer: The product is $Ph - C \equiv C - CH 2 - Ph$ .

Related Topics & Formulae

Acidity of Terminal Alkynes: The pK $a$ of terminal alkynes is ~25, making them more acidic than alkenes and alkanes. This is due to the high s-character of the sp-hybridized carbon, which stabilizes the conjugate base (acetylide ion).

Nucleophilic Substitution (S $N_{2}$ ): Acetylide ions are strong nucleophiles and react with primary alkyl halides via S $N_{2}$ mechanism to form extended alkynes. Benzyl halides are especially reactive due to the benzylic position, which stabilizes the transition state.

Important Note: Acetylide ions cannot be used with secondary or tertiary alkyl halides as elimination (E2) predominates over substitution.