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.


1 Answer 1


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:

Forward 1

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:

enter image description here

I don't like the alkylation step though.


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