All the alkenes of molecular formula C5H10 (counting each geometrical isomer separately) are treated with D2 over a platinum catalyst, which adds one…

Enumerate the acyclic alkenes of $\mathrm{C_5H_{10}}$: pent-1-ene, $(E)$-pent-2-ene, $(Z)$-pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene and 2-methylbut-2-ene (six in all). Adding $\mathrm{D_2}$ (syn addition) gives: $\mathrm{CH_2D\text{-}CHD\text{-}C_2H_5}$ (one stereocentre $\mathrm{CHD}$, with four different groups $\mathrm{H, D, CH_2D, C_3H_7}$), $\mathrm{CH_3CHD\text{-}CHDCH_2CH_3}$ for both 2-pentenes (two stereocentres, ends different), $\mathrm{CH_2D\text{-}CD(CH_3)C_2H_5}$ (stereocentre has $\mathrm{D, CH_2D, CH_3, C_2H_5}$), $\mathrm{CH_2D\text{-}CHD\text{-}CH(CH_3)_2}$ and $\mathrm{(CH_3)_2CD\text{-}CHD\text{-}CH_3}$. Every product carries at least one stereocentre because the two isotopes make the carbons inequivalent, and none has an internal mirror plane, so all six products are chiral (with $\mathrm{H_2}$ none would be, since pentane and 2-methylbutane are achiral). Answer: 6.