The products $A$ and $B$ obtained in the following reactions,respectively,are
$3 ROH + PCl_3 \rightarrow 3 RCl + A$
$ROH + PCl_5 \rightarrow RCl + HCl + B$
$3 ROH + PCl_3 \rightarrow 3 RCl + A$
$ROH + PCl_5 \rightarrow RCl + HCl + B$
- A$POCl_3$ and $H_3 PO_4$
- B$H_3 PO_4$ and $POCl_3$
- C$H_3 PO_3$ and $POCl_3$
- D$POCl_3$ and $H_3 PO_3$
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