My textbook said that when an excess of ethanolic ammonia is heated with a haloalkane ($\ce{R-X}$) one should give the products as being $\ce{R-NH2}$ and $\ce{NH4Cl}$ as opposed to $\ce{RNH3Cl}$. I understand that in reality an equilibrium will exist between the alkylammonium and ammonium cations, but I thought amines were stronger bases than ammonia, so I find it odd that I would be asked to give the ammonium salt and an amine rather than the alkylammonium salt as the major product of this reaction. Have I misunderstood something?


1 Answer 1


You have actually answered the question right in the first sentence of the description.

You are right in saying that $\ce{RNH_3Cl}$ will be formed.

The alkyl halide, $\ce{R-X}$, is the alkylating agent. The reaction medium is basic in nature, so I'd expect the $\ce{S_N2}$ mechanism. So,

$\ce{H3N +RX\to H3NR+X-}$

We know that an excess of ammonia has been used. This simply means that the forward reaction is favoured:

$\ce{H3NR+X- +NH3\rightleftharpoons RNH2 + NH4+X-}$

It's all Le Chatelier's once again - makes extraction simpler, doesn't it?


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