Answer
With $H−C≡C-H$ (acetylene) 1) $NaNH_2$ 2) $CH_3CH_2CH_2Br$ can be used to synthesize $H−C≡C-CH_2CH_2CH_3$
Then 1) $NaNH_2$ is added to the $H−C≡C-CH_2CH_2CH_3$ to form $CH_3−C≡C-CH_2CH_2CH_3$ using $CH_3I$
Work Step by Step
1) $NaNH_2$ deprotonates $H−C≡C-H$ and forms $H−C≡C:^-$. This attacks the 2) $CH_3CH_2CH_2Br$ to give $H−C≡C-CH_2CH_2CH_3$.
Then again 1) $NaNH_2$ deprotonates $H−C≡C-CH_2CH_2CH_3$ to form $^-:C≡C-CH_2CH_2CH_3$. This attacks $CH_3I$ to form the product.