\(Ph - CH_3 \xrightarrow[hv]{C{{l}_{2}}} Ph - CH_2 -Cl\)
$H_2C = CH - CH_2 - I \xrightarrow[CC{{l}_{4}}]{HI(excess)}$
$(1)$ $CH_3 - C \equiv C - CH_3$ $(2)$ $CH_3CH_2CH_2CH_3$
$(3)$ $CH_3CH_2C \equiv CH$ $(4)$ $CH_3CH = CH_2$