The number of chlorine atoms present in the organic products $X$ and $Y$ of the following reactions, respectively, are : $\ce{C6H6 + 6 Cl2 ->[\text{anhyd. AlCl3}][\text{dark, cold}]}$ $X$ $\ce{C6H6 + 3 Cl2 ->[\text{UV}][\text{500 K}]}$ $Y$

Solution & Explanation

With anhydrous $\ce{AlCl3}$, benzene undergoes electrophilic aromatic substitution; exhaustive chlorination replaces all six ring hydrogens to give hexachlorobenzene $\ce{C6Cl6}$ ($X$), which contains 6 chlorine atoms. Under UV light, $\ce{Cl2}$ adds across the ring by a free-radical addition: $\ce{C6H6 + 3 Cl2 -> C6H6Cl6}$ (benzene hexachloride / BHC, $Y$), also containing 6 chlorine atoms. Hence 6 and 6.