The alkene shown is treated with (i) $\mathrm{HBr}$; (ii) $\mathrm{Mg}$ in dry ether; (iii) $\mathrm{D}{}_2\mathrm{O}$ (D = deuterium). Which structure is the final product?

The alkene is vinylcyclohexane. $\mathrm{HBr}$ adds with a hydride shift to give 1-bromo-1-ethylcyclohexane. The Grignard reagent formed at the tertiary carbon is protonated (here deuterated) by $\mathrm{D}{}_2\mathrm{O}$, so deuterium ends up on the same ring carbon, giving 1-deuterio-1-ethylcyclohexane.