The products $A$ and $B$ obtained in the following reactions, respectively are $3\mathrm{ROH}+\mathrm{PCl}{}_3\to 3\mathrm{RCl}+A$ $\mathrm{ROH}+\mathrm{PCl}{}_5\to \mathrm{RCl}+\mathrm{HCl}+B$

Solution & Explanation

With $\mathrm{PCl}{}_3$, three moles of alcohol convert to alkyl chloride and the phosphorus by-product is phosphorous acid: $3\mathrm{ROH}+\mathrm{PCl}{}_3\to 3\mathrm{RCl}+\mathrm{H}{}_3\mathrm{PO}{}_3$ ($A=\mathrm{H}{}_3\mathrm{PO}{}_3$). With $\mathrm{PCl}{}_5$, each alcohol gives the chloride plus $\mathrm{HCl}$ and phosphorus oxychloride: $\mathrm{ROH}+\mathrm{PCl}{}_5\to \mathrm{RCl}+\mathrm{HCl}+\mathrm{POCl}{}_3$ ($B=\mathrm{POCl}{}_3$).