When benzene is treated with an equimolar mixture of and in the presence of anhydrous in the dark, w — Haloalkanes and Haloarenes Chemistry Question
Question
When benzene is treated with an equimolar mixture of $Cl_2$ and $Br_2$ in the presence of anhydrous $FeCl_3$ in the dark, which of the following is the predominant monohalogenated product, and what is the specific mechanistic reason?
💡 Solution & Explanation
When $Cl_2$ and $Br_2$ are mixed, they rapidly equilibrate to form the interhalogen compound bromine monochloride ($BrCl$). In the presence of a Lewis acid like $FeCl_3$, the catalyst coordinates preferentially to the more electronegative atom in the interhalogen, which is chlorine. This polarizes the bond as $Br^{\delta+} - Cl^{\delta-} \cdots FeCl_3$, effectively generating the bromonium ion ($Br^+$) as the active electrophile. Consequently, electrophilic attack on the benzene ring predominantly yields bromobenzene.