An alkene C5H10 gives 2-methylbutan-1-ol on hydroboration-oxidation ($\mathrm{BH}{}_3$/THF, then $\mathrm{H}{}_2\mathrm{O}{}_2$/$\mathrm{NaOH}$), but gives 2-methylbutan-2-ol on acid-catalysed hydration. The alkene is

For 2-methylbut-1-ene, $\mathrm{CH}{}_2=\mathrm{C}(\mathrm{CH}{}_3)\mathrm{CH}{}_2\mathrm{CH}{}_3$, anti-Markovnikov addition puts OH on the terminal $\mathrm{CH}{}_2$ (2-methylbutan-1-ol) and Markovnikov addition puts OH on C-2 (2-methylbutan-2-ol). 3-Methylbut-1-ene would give 3-methylbutan-1-ol, 2-methylbut-2-ene would give 3-methylbutan-2-ol, and pent-1-ene would give pentan-1-ol.