This question involves the reaction of an alkyne with sodium amide (NaNH₂) followed by reaction with an alkyl halide, which is a standard method for alkylation of terminal alkynes. Let's break it down step by step.

Step 1: Reaction with NaNH₂

The given compound is a terminal alkyne (≡C-H). Sodium amide (NaNH₂) is a strong base. It deprotonates the terminal alkyne to form the corresponding acetylide ion.

Reaction: $HC \equiv C - CH = CH - CH {= CH}_{2} + NaNH \to_{2} Na {[C \equiv C - CH = CH - CH = CH]_{2}}^{-} + NH 3$

So, product A is the sodium acetylide: Na[C≡C-CH=CH-CH=CH₂]

Step 2: Reaction with CH₃CH₂Br

The acetylide ion acts as a nucleophile and undergoes an S_N2 reaction with ethyl bromide (CH₃CH₂Br), resulting in alkylation at the terminal carbon.

Reaction: $Na {[C \equiv C - CH = CH - CH = CH]_{2}}^{-} + CH 3 CH 2 Br \to CH 3 CH 2 - C \equiv C - CH = CH - CH = CH 2 + NaBr$

So, the final product B is CH₃CH₂-C≡C-CH=CH-CH=CH₂

Final Answer

Comparing with the options, the correct products are:

A: NaC≡C-CH=CH-CH=CH₂

B: CH₃CH₂C≡C-CH=CH-CH=CH₂

This corresponds to the first option.

Related Topics & Formulae

Alkylation of Terminal Alkynes: Terminal alkynes (RC≡CH) can be converted to nucleophiles (acetylide ions RC≡C^-) using strong bases like NaNH₂. These ions then react with primary alkyl halides via S_N2 mechanism to form extended alkynes.

Key Reaction: $RC \equiv CH + NaNH 2 \to RC \equiv C -^{+} Na + NH 3$
$RC \equiv C -^{+} Na + R'X \to RC \equiv CR' + NaX$

Note: The alkyl halide must be primary to avoid elimination reactions. The base NaNH₂ is used in liquid NH₃.