2-Bromo-3-methylbutane is heated with alcoholic KOH and the major alkene formed is then treated with HBr in the presence of an organic peroxide. The…

Alcoholic KOH brings about E2 elimination; by Saytzeff's rule the more substituted alkene $(\mathrm{CH_3})_2\mathrm{C{=}CHCH_3}$ (2-methylbut-2-ene) dominates over 3-methylbut-1-ene. With HBr and peroxide the addition is free-radical (anti-Markovnikov): $\mathrm{Br\cdot}$ adds to the less substituted carbon ($\mathrm{CHCH_3}$) so that the more stable tertiary radical forms, then abstracts H from HBr. This gives $(\mathrm{CH_3})_2\mathrm{CH{-}CHBrCH_3}$, i.e. 2-bromo-3-methylbutane, the starting material. Without peroxide, Markovnikov addition would give 2-bromo-2-methylbutane.