A hydrocarbon $\mathrm{C}{}_5\mathrm{H}{}_{10}$ gives 2-methylbutan-2-ol on oxymercuration-demercuration and a secondary alcohol on hydroboration-oxidation. The hydrocarbon is

Oxymercuration gives Markovnikov OH at the more substituted carbon, so a tertiary alcohol needs a branch carbon in the double bond. 2-Methylbut-1-ene would give a primary alcohol on hydroboration, whereas 2-methylbut-2-ene gives 3-methylbutan-2-ol (secondary). 3-Methylbut-1-ene gives a secondary (not tertiary) alcohol on oxymercuration, and pent-2-ene gives no tertiary alcohol.