WAEC Chemistry 2019 Paper

Concept:Hydrogenation of an alkyne takes place in two steps: first a triple bond is reduced to a double bond, then to a single bond.Explanation:Ethyne, $\mathrm{C}_2\mathrm{H}_2$, contains a triple bond between the two carbon atoms.When one molecule of hydrogen is added, ethyne gets hydrogenated to ethene, $\mathrm{C}_2\mathrm{H}_4$. So $X$ is ethene.This can be shown as:$\mathrm{C}_2\mathrm{H}_2 + \mathrm{H}_2 \rightarrow \mathrm{C}_2\mathrm{H}_4$Ethene still has a double bond, so it can react with another $\mathrm{H}_2$ molecule to form ethane, $\mathrm{C}_2\mathrm{H}_6$. So $Y$ is ethane.This second step is:$\mathrm{C}_2\mathrm{H}_4 + \mathrm{H}_2 \rightarrow \mathrm{C}_2\mathrm{H}_6$Hence, $X$ and $Y$ respectively are ethene and ethane.Answer:$X$ = ethene and $Y$ = ethane.Correct option: B.