$C{H_3} - CH = C{H_2} + HBr\xrightarrow{{{{({C_6}{H_5}CO)}_2}{O_2}}}$
$\mathop {C{H_3} - }\limits_\delta \mathop {C{H_2} - }\limits_\gamma \mathop {CH = }\limits_\beta \mathop {C{H_2}}\limits_\alpha $
$(E)$