An optically active hydrocarbon X has molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{12}.$ X on catalytic hydrogenation gives optically inactive ${\mathrm{C}}_6{\mathrm{H}}_{14}$. X could be 

3-Methyl-1-pentene.

Reason: $X$ must be an alkene $C_6H_{12}$ with a stereogenic center.
CH₂=CH–CH(CH₃)–CH₂–CH₃ is chiral (C-3 has four different groups).
Hydrogenation removes the C=C, giving 3-methylpentane, where C-3 then has two identical ethyl groups, so the product $C_6H_{14}$ is optically inactive.