2-Methylpropene reacts with $\mathrm{HBr}$ in the presence of a peroxide. The major product is

$\mathrm{Br}{}^{\bullet }$ adds to the less substituted carbon C1 to give the tertiary radical $(\mathrm{CH}{}_3){}_2\mathrm{C}{}^{\bullet }-\mathrm{CH}{}_2\mathrm{Br}$. This radical abstracts H from $\mathrm{HBr}$ to give 1-bromo-2-methylpropane. This is the anti-Markovnikov product.