Alkylation of benzene can be accomplished by treating benzene with haloalkane (RX) in the presence of $\mathrm{AlCl}_{3} .$ The reaction is known as a Friedel-Crafts alkylation reaction. (The reaction is named after Charles Friedel, a French chemist, and James M. Crafts, an American chemist, who discovered this method of making alkylbenzenes in $1877 .$ ) An example of a Friedel-Crafts alkylation reaction is shown below: The mechanism for this reaction involves the following steps. First, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}$ and $\mathrm{AlCl}_{3}$ react in a Lewis acid-base reaction to form an adduct, $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{Cl}-\mathrm{AlCl}_{3},$ in which a chlorine atom is
bonded to both carbon and aluminum. The adduct then dissociates to $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}^{+},$ a carbocation, and $\mathrm{AlCl}_{4}^{-}$. The carbocation acts as an electrophile in a reaction with benzene, forming an arenium ion. Finally, a proton is removed from the arenium ion by $\mathrm{AlCl}_{4}^{-},$ yielding an alkylbenzene, $\mathrm{HCl},$ and $\mathrm{AlCl}_{3}$ Write chemical equations for the elementary processes involved in forming cumene and HCl from $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}$ and benzene. Use curved arrows to show the movement of electrons.