Reaction of $\mathrm{HBr}$ with propene in the presence of peroxide gives $n$-propyl bromide. This addition reaction is an example of anti-Markownikoffs addition reaction.


Related Theory
Step 1: Peroxide undergoes fission to give free radicals.
$$
\mathrm{R}-\mathrm{O}-\mathrm{O}-\mathrm{R} \longrightarrow 2 \mathrm{R}-\dot{\mathrm{O}}
$$
Step 2: $\mathrm{HBr}$ combines with free radical to form bromine free radical.
$$
R-\dot{O}+\mathrm{HBr} \longrightarrow \mathrm{R}-\mathrm{OH}+\dot{\mathrm{B}}
$$
Step 3: $\dot{B}$ attacks the double bond of the alkene to form a more stable free radical.

Step 4: More stable free radical attacks the $\mathrm{HBr}$.
$$
\begin{aligned}
\mathrm{CH}_3 \dot{\mathrm{C}} \mathrm{CHH}_2 \mathrm{Br}+\mathrm{HBr} \longrightarrow & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}+\dot{\mathrm{B}} \\
& \text { n-propylbromide }
\end{aligned}
$$
Step 5: $\dot{B} r+\dot{B} r \longrightarrow B r_2$