$X$ and $Y$ are optically active isomers having formula $\mathrm{C}_5 \mathrm{H}_9 \mathrm{Cl}$. When…
$X$ and $Y$ are optically active isomers having formula $\mathrm{C}_5 \mathrm{H}_9 \mathrm{Cl}$. When treated with one mole of $\mathrm{H}_2, X$ gets converted to an optically inactive compound $Z$, but $Y$ gives an optically active compound $P$. The structure of $X$ and $Y$ respectively are
Solution
Let us consider the formula of $\mathrm{C}_5 \mathrm{H}_9 \mathrm{Cl}$ which is optically active.
Optically active means molecule does not contain plane of symmetry.
So, possible optically inactive structure of $\mathrm{C}_5 \mathrm{H}_9 \mathrm{Cl}$ is
When, ' $X^{\prime}$ react with 1 mole of $\mathrm{H}_2$, it converted into optically inactive compound ' $Z$ '
Same as $(X),(Y)$ is optically active and give compound $(P)$ which is inactive.
When ' $Y$ ' react with 1 mole of $\mathrm{H}_2$, it will give ' $P$ ' which is optically active. So, ' $P$ ' is
