In this video, we're going to take a look at Friedel–Crafts alkylation. Here in this reaction, benzene reacts with an alkyl halide, and what happens is that one of the hydrogens on the benzene ring is substituted with an alkyl group. To do this, a catalyst of aluminum X₃ must be used, and this has to contain the same halogen as the reagent.

So basically what we mean by this, if we take a look at this Friedel–Crafts reaction down below, this is CH₃X. This X could represent either bromine or chlorine. If it's bromine, then this has to be AlBr₃. If this X is chlorine, then this has to be AlCl₃. So the halogens have to match between these two.

Now in this reaction, all that happens is that one of the hydrogens on benzene gets substituted out, and the alkyl portion of our alkyl halide comes in. The alkyl portion of our alkyl halide here would be methyl, so we wind up making methyl benzene as our final answer.