In the following reaction sequence, $X$ and $Z$ respectively are : $\ce{CH3CH2CH2OH + PCl3 -> CH3CH2CH2Cl + }X \ce{ + HCl}$ $\ce{CH3CH2CH2Cl ->[\text{alc. KOH}][\Delta]}$ $Y$ $Y \ce{->[HBr][\text{peroxide}]}$ $Z$

Solution & Explanation

Propan-1-ol with $\ce{PCl3}$ gives 1-chloropropane plus phosphorus oxychloride and HCl, so $X = \ce{POCl3}$ (this single-mole stoichiometry shown gives $\ce{POCl3}$, not $\ce{H3PO3}$). Alcoholic KOH then dehydrohalogenates 1-chloropropane to propene ($Y$). Propene with HBr in the presence of peroxide adds anti-Markovnikov (peroxide effect), placing Br on the terminal carbon to give 1-bromopropane $\ce{CH3CH2CH2Br}$ ($Z$).