To solve this problem, we need to understand the reaction sequence given:$\\$1. Start with acetylene $(H-C \equiv C-H). \\$2. The first step involves the use of $\mathrm{NaNH_2}$ in liquid ammonia, which is a strong base.$\\$ This base will deprotonate the terminal hydrogen of acetylene, forming an acetylide ion:$\\$ $H-C \equiv C^-$ (acetylide ion).$\\$3. The acetylide ion then reacts with $\mathrm{CH_3CH_2Br}$ (ethyl bromide) in an $S_N2$ reaction.$\\$ This results in the formation of a new carbon-carbon bond:$\\$ $H-C \equiv C-CH_2CH_3$ (1-butyne).$\\$4. The compound formed, 1-butyne, is denoted as $X. \\$5. The second part of the reaction sequence repeats the same steps:$\\$ a. Deprotonation of the terminal hydrogen of 1-butyne using $\mathrm{NaNH_2}$ in liquid ammonia,$\\$ forming a new acetylide ion:$\\$ $H-C \equiv C-CH_2CH_2^-$.$\\$ b. Reaction of this acetylide ion with another equivalent of $\mathrm{CH_3CH_2Br}$ (ethyl bromide).$\\$ This results in the formation of a new carbon-carbon bond:$\\$ $H-C \equiv C-CH_2CH_2CH_2CH_3$ (3-hexyne).$\\$6. The compound formed, 3-hexyne, is denoted as $Y. \\$Therefore, the correct answer is:$\\$Option 1: $X =$1-Butyne; $Y =$3-Hexyne.