Propene reacts with $\mathrm{HBr}$ in the presence of an organic peroxide. The major product is

In the presence of peroxide $\mathrm{HBr}$ adds by a free radical chain (Kharasch effect). The $\mathrm{Br}$ radical adds to the terminal carbon to give the more stable secondary radical, and $\mathrm{H}$ ends on C-2, so the product is anti-Markovnikov: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{CH}{}_2\mathrm{Br}$.