How about starting with a 3 carbon alkyne? NaNH2 can remove the terminal hydrogen so the alkyne can react with something like ethyl bromide through substitution. Now, we have 2-pentyne. You can hydrogenate this with excess H2 to get it to be an alkane. With the alkane, perhaps a radical reaction? If you use Cl2 (heat and light) you won't get the primary halide as your major product, but you'll still get it in some form (maybe 45% vs its isomer... which would be 55%). After this, it's just a substitution reaction with benzene and the primary halide which im sure you were taught about in class.