7.5.1 Preparation of Amines

Amines are organic compounds derived from ammonia ($NH₃$) where one or more hydrogen atoms are replaced by alkyl or aryl groups. There is some H-bonding between molecules, but weaker than in alcohols.

• The presence of the lone pair of electrons on the N atom is the centre. of most amine chemistry
• This means they act as nucleophiles and bases to greater or less extents.They are classified based on the number of alkyl or aryl groups attached to the nitrogen: Primary Amines: One hydrogen replaced (e.g., methylamine, $CH₃NH₂$).

Primary aliphatic amines function as Brønsted-Lowry bases because the lone pair of electrons on the nitrogen atom can readily form a dative covalent bond with an H⁺ ion, allowing the amine to accept a proton. They are considered weak bases since they produce only a low concentration of hydroxide ions in solution.

Secondary Amines: Two hydrogens replaced (e.g., dimethylamine, ($CH₃$)$₂NH$).

Tertiary Amines: Three hydrogens replaced (e.g., trimethylamine, ($CH₃$)$₃N$).

Quaternary Ammonium Salts: All four hydrogen atoms are replaced by alkyl groups, resulting in a positively charged nitrogen (e.g., tetraethylammonium bromide, ($CH₃CH₂$)$₄N⁺Br⁻$).

Ammonia can be regarded as the simplest amine. Quaternary ammonium ions are not amines and they do not possess alone pair of electrons on the N.

They are named after the alkyl group(s) attached to the N atom, with the ending -amine. For example:

Preparation of Primary Aliphatic Amines

Primary aliphatic amines can be synthesized through:

• Nucleophilic substitution of halogenoalkanes with ammonia
• Reduction of nitriles

Reaction of Halogenoalkanes with Ammonia

This method involves a nucleophilic substitution reaction where ammonia ($NH₃$) acts as a nucleophile. The lone pair of electrons on the nitrogen atom allows ammonia to attack the carbon attached to the halogen in the halogenoalkane, replacing the halogen atom and forming an amine.

Reaction Conditions:

• The halogenoalkane is heated with excess ammonia in ethanol under pressure.
• This reaction produces a primary amine along with ammonium halide.

Limitations:

• A mixture of products (primary, secondary, tertiary amines, and quaternary ammonium salts) is often formed because the primary amine produced can further react with the halogenoalkane.
• Using excess ammonia helps maximize the formation of the primary amine by minimizing further substitutions.

Formation of a Primary Amine

The formation of a primary amine occurs in two steps.

1. Step 1: A salt is produced—in this example, ethylammonium bromide. This compound is similar to ammonium bromide, but one hydrogen in the ammonium ion is replaced by an ethyl group.

$$CH₃CH₂Br+NH₃→CH₃CH₂NH₃⁺Br⁻$$

2. Step 2: The ethylammonium bromide can undergo a reversible reaction with excess ammonia present in the mixture.

$$CH₃CH₂NH₃⁺Br⁻+NH₃⇌CH₃CH₂NH₂+NH₄⁺Br⁻$$

Ammonia removes a hydrogen ion from the ethylammonium ion, resulting in the formation of a primary amine, ethylamine.

Increasing the concentration of ammonia in the mixture helps to drive the reaction forward, favoring the production of the primary amine.

Reduction of Nitriles

Nitriles ($R-CN$) can be reduced to primary amines by breaking the triple bond between carbon and nitrogen.

Methods for Reducing Nitriles:

3. Using Lithium Aluminium Hydride ($LiAlH₄$):

• LiAlH₄ is a strong reducing agent used in dry ether.
• After reduction, the intermediate is treated with dilute acid to produce the amine.

4. Catalytic Hydrogenation:

• Nitriles can also be reduced by hydrogen gas (H₂) in the presence of a metal catalyst like palladium, platinum, or nickel.
• This reaction is carried out under elevated temperature and pressure.

Formation of a Secondary Amine

The reaction sequence does not stop at the primary amine. Ethylamine can further react with bromoethane in the same two-step process:

Step 1: This forms a salt known as diethylammonium bromide. This salt is similar to ammonium bromide, but with two hydrogens replaced by ethyl groups.

$$CH₃CH₂Br+CH₃CH₂NH₂→(CH₃CH₂) 2 ​ NH₂⁺Br⁻$$

Step 2: There is a reversible reaction between diethylammonium bromide and excess ammonia in the mixture.

$$(CH₃CH₂) 2 ​ NH₂⁺Br⁻+NH₃⇌(CH₃CH₂) 2 ​ NH+NH₄⁺Br⁻$$

The ammonia removes a hydrogen ion from the diethylammonium ion, yielding a secondary amine, diethylamine, which has two alkyl groups attached to the nitrogen.

Formation of a Tertiary Amine

The reaction can proceed further, as diethylamine can react with bromoethane in a similar manner:

Step 1: This produces triethylammonium bromide.

$$CH₃CH₂Br+(CH₃CH₂) 2 ​ NH→(CH₃CH₂) 3 ​ NH⁺Br⁻$$

Step 2: In the presence of excess ammonia, a reversible reaction occurs in which ammonia removes a hydrogen ion from the triethylammonium ion, resulting in a tertiary amine, triethylamine, which has three alkyl groups attached to the nitrogen.

$$(CH₃CH₂) 3 ​ NH⁺Br⁻+NH₃⇌(CH₃CH₂) 3 ​ N+NH₄⁺Br⁻$$

Formation of a Quaternary Ammonium Salt

In the final stage, triethylamine reacts with another molecule of bromoethane, producing tetraethylammonium bromide, a quaternary ammonium salt where all four hydrogens have been replaced by ethyl groups.

At this point, the reaction sequence stops, as the nitrogen has no hydrogen atoms left to be removed.

Note: This entire reaction sequence can be complex to memorize, but understanding the mechanisms behind each step can make it easier to follow the progression from primary amine to quaternary ammonium salt.

Preparation of Aromatic Amines

Aromatic amines, such as phenylamine (aniline), can be synthesized through the reduction of nitro compounds. This process is especially significant in the production of dyes and other aromatic compounds.

Process for Preparing Aromatic Amines:

5. Nitration of Benzene:

• Benzene is nitrated by reacting it with concentrated nitric acid ($HNO₃$) and concentrated sulfuric acid ($H₂SO₄$) at temperatures between 50-55°C, forming nitrobenzene ($C₆H₅NO₂$).

6. Reduction of Nitrobenzene:

• Nitrobenzene is then reduced using tin ($Sn$) and concentrated hydrochloric acid ($HCl$) under reflux conditions, resulting in the formation of phenylammonium ions ($C₆H₅NH₃⁺$).

7. Formation of Phenylamine:

• Finally, sodium hydroxide ($NaOH$) is added to the reaction mixture, which removes a proton from the phenylammonium ion, producing phenylamine ($C₆H₅NH₂$). Overall Reaction:

$${C₆H₅NO₂ + 6[H]} \rightarrow \text{C₆H₅NH₂ + 2H₂O}$$