Preparation of Amines

Preparation of amines by alkylation of ammonia

The first reaction we'll look at today is the preparation of amines from ammonia. This involves alkylation: adding an alkyl group to a molecule. You might know that amines are ammonia derivatives, where one or more of the hydrogen atoms have been replaced by an organic hydrocarbon group. In alkylation, we are doing just that – replacing a hydrogen atom with an organic hydrocarbon alkyl chain.

There are two ways of alkylating ammonia:

• Reacting ammonia with a halogenoalkane (also known as a haloalkane).
• Reacting ammonia with an alcohol.

For your exams, you only need to know about the reaction of ammonia with a halogenoalkane. We'll therefore focus most of our attention on that.

Alkylation of ammonia using a halogenoalkane

If you mix an excess of strong, concentrated ammonia with a halogenoalkane in ethanolic solution and heat the mixture, a nucleophilic substitution reaction occurs. Here, the ammonia molecule acts as a nucleophile by attacking the halogenoalkane's partially positively charged carbon atom.

Although it isn't negatively charged, the ammonia molecule can act as a nucleophile because its nitrogen atom has a partial negative charge and a lone pair of electrons. Meanwhile, the halogenoalkane can be attacked because it has an electron-deficient carbon atom with a partial positive charge.

Fig. 1 - Ammonia can act as a nucleophile thanks to its partial charges and lone pair of electrons

The overall reaction uses up two ammonia molecules and produces both a primary amine and an ammonium salt. Here's the general equation.

$$ RX + 2NH_3 \rightarrow RNH_2 + NH_4X $$

The mechanism occurs in two steps.

1. In the first step, one of the ammonia molecules attacks the halogenoalkane using its lone pair of electrons. It replaces the halogen atom, forming a new molecule with an \(NH_3^+\) group.
2. In the second step, the second ammonia molecule removes a hydrogen ion from the new molecule we just formed, resulting in a primary amine and an ammonium ion. The ammonium ion then reacts with the halogen kicked out in the first step to form an ammonium salt.

Fig. 2 - Nucleophilic substitution of a halogenoalkane using ammonia

For example, reacting strong, concentrated ammonia with bromoethane produces ethylamine and ammonium bromide.

$$ CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br $$

Preparation of amines from alcohols

Reacting ammonia with a halogenoalkane is just one way of alkylating ammonia. Another method involves reacting ammonia with an alcohol. This is the method most commonly used in industry. It produces a primary amine and water.

$$ ROH + NH_3 \rightarrow RNH_2 + H_2O $$

For example, reacting propanol with ammonia produces propylamine and water.

$$ CH_3CH_2CH_2OH + NH_3 \rightarrow CH_3CH_2CH_2NH_2 + H_2O $$

Method of preparing amines

Above, we learnt how ammonia reacts with halogenoalkanes in a nucleophilic substitution reaction to produce a primary amine and an ammonium salt. Ammonia is able to act as a nucleophile because it has a nitrogen atom with a partial negative charge and a lone pair of electrons.

But if we look at the products of the reaction, we can see that primary amines also have a nitrogen atom with a partial negative charge and a lone pair of electrons. This means that primary amines can also act as a nucleophile. Provided there is an excess of the halogenoalkane, the reaction can continue. But this time, the primary amine attacks the halogenoalkane, forming a secondary amine. It happens again – the secondary amine attacks the halogenoalkane, forming a tertiary amine, which attacks the halogenoalkane once more to form a quaternary ammonium ion. The quaternary ammonium ion finally reacts with the halogen atom kicked out to form a quaternary ammonium salt.

Therefore, this method of amine production results in a solution that contains a mixture of primary, secondary, and tertiary amines as well as quaternary ammonium salts. However, we can alter the reaction conditions to either favour the production of primary amines, or the production of quaternary ammonium salts.

• Using an excess of ammonia favours the production of primary amines.
• Using an excess of the haloalkane favours the production of quaternary ammonium salts.

If we consider the example we looked at earlier, which was the reaction between bromoethane and ammonia, we initially produce a primary amine – ethylamine. But this can react further to give a secondary amine (diethylamine), a tertiary amine (triethylamine), and a quaternary ammonium salt (tetraethylammonium bromide). The equations are given below.

\begin{gather} CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br \\ CH_3CH_2Br + CH_3CH_2NH_2 + NH_3 \rightarrow { (CH_3CH_2) }_2NH + NH_4Br \\ CH_3CH_2Br + { (CH_3CH_2) }_2NH + NH_3 \rightarrow { (CH_3CH_2) }_3N + NH_4Br \\ CH_3CH_2Br + { (CH_3CH_2) }_3N \rightarrow { (CH_3CH_2) }_4Br \end{gather}

Here's what the different molecules look like.

Fig. 3 - Primary, secondary and tertiary amines, and quaternary ammonium salts

Preparation of amines by reduction of nitriles

Consider the molecule \(CH_3CN\). This is ethanenitrile. It is an example of a nitrile, an organic molecule with the \(C\equiv N\) functional group.

Fig. 4 - Ethanenitrile, an example of a nitrile

Nitriles can be reduced to produce primary amines. This is done using either a strong reducing agent or using hydrogen gas in the presence of a metal catalyst.

Reduction of nitriles using a reducing agent

One way of reducing nitriles is by using a strong reducing agent, such as lithium aluminium hydride. The correct IUPAC name is lithium tetrahydridoaluminate (III). That's a bit of a mouthful – you'll probably know it instead as \(LiAlH_4\) We represent it in equations using \([H]\). The reaction takes place in a solution of diethyl ether, and a dilute acid is added at the end. Here's the general equation:

$$ RCN+4[H]\rightarrow RCH_2NH_2 $$

For example, reacting ethanenitrile with LiAlH₄ produces ethylamine.

$$ CH_3CN+4[H]\rightarrow CH_3CH_2NH_2 $$

Reduction of nitriles using hydrogen gas

LiAlH₄ is relatively pricey. This means that reducing nitriles using this method is too expensive for use in industry – a different reaction must be used instead. As an alternative, we can reduce nitriles using hydrogen gas in the presence of a metal catalyst such as palladium, platinum, or nickel. This reaction takes place at a high temperature and pressure.

$$ RCN+2H_2\xrightarrow{\text{metal catalyst}} RCH_2NH_2 $$

For example, we can achieve the exact same product as in the reaction above by reducing ethanenitrile with hydrogen gas and a nickel catalyst.

$$CH_3CN+2H_2\xrightarrow{\text{nickel}} CH_3CH_2NH_2 $$

Preparation of amines by reduction of amides

We've just learnt how nitriles can be turned into amines by reducing them with \(LiAlH_4\). We can do the same thing with amides. Amides are organic molecules with the \(-CONH_2\) or \(-CONHR-\) functional group. They are similar to amines, except one of the carbon atoms directly attached to the nitrogen atom contains a \(C=O\) bond. This means that all amides have an amine group bonded to a carbonyl group. The simplest amide is methanamide, commonly known as formamide.

Fig. 5 - Methanamide

The reaction between an amide and \(LiAlH_4\) also takes place in diethyl ether at room temperature. Once again, a dilute acid is added at the end. Overall, the reaction swaps the carbonyl group's oxygen atom with two hydrogen atoms, resulting in water and an amine. However, the amine varies slightly depending on the type of amide reduced.

• Reducing primary amides with the formula \( RCONH_2\) produces a primary amine.
• Reducing secondary amides with the formula \(RCONHR\) produces a secondary amine.
• Reducing tertiary amide with the formula \(RCONR_2\) produces a tertiary amine.

Here's the general equation.

$$ RCONH_2+4[H]\rightarrow RCH_2NH_2+H_2O $$

For example, reacting methanamide with LiAlH₄ produces the primary amine methylamine.

$$ CHONH_2+4[H]\rightarrow CH_3NH_2+H_2O $$

Similarly, reacting N-methylmethanamide with LiAlH₄ produces the secondary amine dimethylamine.

$$ CHONHCH_3+4[H]\rightarrow CH_3NHCH_3+H_2O $$

Preparation of aromatic amines by reduction of nitro-compounds

All the examples we've looked at before have involved making aliphatic amines. Remember that amines are ammonia derivatives, where one or more hydrogen atoms have been swapped for an organic hydrocarbon group. In aliphatic amines, these organic groups are all open hydrocarbon chains. But in aromatic amines, one or more of these organic groups features an aromatic benzene ring. One example is phenylamine.

Fig. 6 - Phenylamine

When it comes to making aromatic amines, you might be able to guess what we start with: benzene itself. The process involves two stages. We first nitrate benzene into nitrobenzene. We then reduce nitrobenzene into phenylamine. We'll walk you through these stages now.

Nitration of benzene

In the production of an aromatic amine such as phenylamine, the first stage is the nitration of benzene into nitrobenzene. This is an electrophilic substitution reaction. It involves heating benzene with a mixture of concentrated sulphuric and nitric acids at 50°C. We also use reflux to prevent any volatile components escaping.

Firstly, sulphuric and nitric acid react together to produce this reaction's electrophile, \(NO_2^+\)

$$ 2H_2SO_4+HNO_3\rightarrow 2HSO_4^- + NO_2^+ + H_3O^+ $$

The electrophile then reacts with benzene.

1. The electrophile is attracted to benzene's ring of delocalisation and adds on to the molecule. This uses electrons from the ring of delocalisation, making benzene unstable.
2. To restore its ring of delocalisation, benzene kicks out a hydrogen ion. This results in nitrobenzene.

Fig. 7 - Electrophilic substitution of benzene, resulting in nitrobenzene

The hydrogen ion reacts with the \(HSO_4^-\) ion, regenerating sulphuric acid. This means that overall, sulphuric acid acts as a catalyst.

We're now ready to move on to the next stage: reducing nitrobenzene.

Reduction of nitro-compounds

To produce an aromatic amine, all you have to do is reduce nitrobenzene. This synthesis also involves two steps.

1. In the first step, we reduce nitrobenzene using a combination of tin and concentrated hydrochloric acid, heating under reflux. This produces a phenylammonium ion.
2. In the second step, we add sodium hydroxide solution. This turns the phenylammonium ion into our desired product, phenylamine.

Fig. 8 - The preparation of aromatic amines

And that's it! You've made phenylamine. It is an important precursor to many dyes and other industrial compounds. It is also added to rubber.