The given reaction suggests that sodium acetylide reacts with an alkyl halide to yield the final product. Here, compound A is formed as sodium acetylide:
CH3 – C ≡ CH + Na → CH3 – C ≡ CNa
Then, it reacts with compound B (1-bromopropane):
CH3 – C ≡ CNa + CH3CH2CH2Br → CH3 – C ≡ C – CH2CH2CH3 + NaBr