$\text{ 1-Butene }+\mathrm{HBr}\underset{hv}{\overset{H_2{O}_2}{\to }}\text{ 1-Bromobutane }$

 The above reaction follows

Markovnikov’s Rule predicts that the most abundant product will be the one where the bromide atom adds to the carbon atom with the lesser number of hydrogen atoms.

 if  perform the reaction in a way that favors the forming of free radicals, likein the presence of benzoyl peroxide or hydrogen peroxide, then we can enforce an anti-Markovnikov path which leads to the bromide atom being mostly added to the carbon atom with the greater number of hydrogen atoms.