Unfortunately, the primary amine product is also a powerful nucleophile, and so some of it will attack a second molecule of the alkyl halide. If the primary amine is desired, one way to avoid this is to use a large excess of NH3., so that the reaction rate with NH3 greatly exceeds the rate with the amine. In a synthesis question, we would not show the mechanism, and we would simply write the reaction as:
In some cases this “use excess” approach may be unsuitable, in which case it is common to use a surrogate for the NH3 which cannot react further. One traditional approach uses phthalimide, and is called the Gabriel synthesis; a more common approach in modern synthesis is to use sodium azide followed by reduction. Sodium azide (NaN3) is cheap (it is used in air bags!), and the azide ion (N3–) is an excellent nucleophile for SN2. The product is an alkyl azide (RN3), which can be reduced in a separate step to RNH2 using excess H2/Pd or LiAlH4. The mechanism for the second step is complex and not necessary for you to learn at this point. An example synthesis is shown below to illustrate this approach. (You could do this on an exam, but unfortunately alkyl azides with small alkyl groups can be explosive, so I wouldn’t recommend running this actual reaction in the lab or on a chemical plant!)
In a synthesis question we use (i) and (ii) to show that the reagents are added in two separate steps:
The reaction of amines with alkyl halides
As with the NH3 reaction described above, when amines are reacted it is common to obtain mixtures because of reactions with more than one molecule of alkyl halide Under some circumstances it may be possible to drive the reaction as was seen in the preparation of primary amines. It is even possible to end up with four alkyl groups on one positively charged nitrogen – a quaternary ammonium salt. For example, consider a reaction between ethylamine (a primary amine) and bromoethane (a primary alkyl halide):
In the first stage of the reaction, you get the salt of a secondary amine formed. For example if you started with ethylamine and bromoethane, you would get diethylammonium bromide
In the presence of excess ethylamine in the mixture, there is the possibility of a reversible reaction. The ethylamine removes a hydrogen from the diethylammonium ion to give free diethylamine – a secondary amine.
But it doesn’t stop here! The diethylamine also reacts with bromoethane – in the same two stages as before. This is where the reaction would start if you reacted a secondary amine with a halogenoalkane.
In the first stage, you get triethylammonium bromide.
There is again the possibility of a reversible reaction between this salt and excess ethylamine in the mixture.
The ethylamine removes a hydrogen ion from the triethylammonium ion to leave a tertiary amine – triethylamine.
Making a quaternary ammonium salt
The final stage! The triethylamine reacts with bromoethane to give tetraethylammonium bromide – a quaternary ammonium salt (one in which all four hydrogens have been replaced by alkyl groups).
This time there isn’t any hydrogen left on the nitrogen to be removed. The reaction stops here.