
$\mathop {C{H_3} - }\limits_\delta \mathop {C{H_2} - }\limits_\gamma \mathop {CH = }\limits_\beta \mathop {C{H_2}}\limits_\alpha $
$(E)$
$C{H_2} = CH{(C{H_2})_8}COOH + HBr\xrightarrow{{peroxide}}....$
$[Figure]$ $\xrightarrow{{B{r_2}/hv}}$
$C{H_2} = CH - C{H_3} + HBr \to C{H_3}CHBr - C{H_3}$