How to synthesise 2-aminoethyl benzene Starting with benzene, describe the synthesis of 2-phenylethan-1-amine providing all the reagents and conditions.


I am not really great at organic chemistry and my knowledge is pretty limited. 

I am currently thinking of forming toluene by Grignard reagent and then reacting with $\ce{Cl2}$ to form benzyl chloride. Then react with sodium cyanide to form benzyl cyanide and then react it with $\ce{HCl}$ and $\ce{H2O}$ to form an amide. Then reduce it with $\ce{LiAlH4}$ to form the compound. 

I am not sure if that is correct as it seems a bit complicated for the synthesis of a seemingly simple compound. Also I am not sure on the exact conditions for the chlorination step. 

I was also thinking of acylation with ehtanoyl chloride followed by Wolff-Kishner reduction to form ethyl benzene. However I am not sure on how to from the $\ce{NH2}$ group.
Several thoughts:


I can't see how you form toluene with a Grignard - it's certainly possible though
Radical chlorination of toluene with $\ce{Cl2}$ isn't the cleanest reaction.
To reduce the nitrile to the amine, you don't need to go through the amide.


Also, from ethylbenzene, it's very hard to introduce that amine group cleanly. You effectively just have a compound with C-H groups and you have to somehow introduce some kind of functional group into the non-benzylic position. Try to think backwards instead of forward!

The amine you want can be obtained by direct reduction of a nitrile either by simple $\ce{LiAlH4}$ reduction or hydrogenation:



That arrow essentially means "can be made from".

Of course, that's easy to make using a nucleophilic substitution reaction with $\ce{CN-}$ and benzyl chloride (for example):



To get from benzene to benzyl chloride, you could use a Blanc chloromethylation. So your forward synthesis would be:



and you might want to use catalytic $\ce{KI}$ in the nucleophilic substitution step. Of course, for such a simple compound, there are tons of ways to synthesise it. The amine could be made from reduction of an azide, or by a reductive amination, and so on. I strongly doubt my proposed route is the most efficient. Still, I think it should work.

Theoretically, you could use a Friedel-Crafts alkylation and a radical bromination to get to benzyl bromide instead:



I don't like the alkylation step though.