AQA A-Level Organic Chemistry 3.3.11 Amines Amine Properties
3.3.11

Amine Properties

Classification, basicity comparisons, and synthetic routes for aliphatic and aromatic amines.

Amines are organic derivatives of ammonia in which one or more hydrogen atoms have been replaced by alkyl or aryl groups. They play a vital role in organic synthesis, biochemistry, and the manufacturing of dyes and pharmaceuticals.

🔑 Key Principle

The chemical properties of amines: particularly their basicity and nucleophilic character: are entirely governed by the lone pair of electrons on the nitrogen atom. Any factor that increases the electron density on the nitrogen atom increases its basicity, while any factor that decreases this density decreases its basicity.

Classification of Amines

Amines are classified as primary, secondary, or tertiary depending on the number of carbon atoms directly attached to the nitrogen atom. This is different from the classification of alcohols or halogenoalkanes, which depends on the carbon atom holding the functional group.

Primary (1°) Amine

An amine in which one alkyl or aryl group is directly bonded to the nitrogen atom (general formula: \\( \text{RNH}_2 \\)). Example: methylamine, \\( \text{CH}_3\text{NH}_2 \\).

Secondary (2°) Amine

An amine in which two alkyl or aryl groups are directly bonded to the nitrogen atom (general formula: \\( \text{R}_2\text{NH} \\)). Example: dimethylamine, \\( (\text{CH}_3)_2\text{NH} \\).

Tertiary (3°) Amine

An amine in which three alkyl or aryl groups are directly bonded to the nitrogen atom (general formula: \\( \text{R}_3\text{N} \\)). Example: trimethylamine, \\( (\text{CH}_3)_3\text{N} \\).

If a fourth group bonds to the nitrogen atom via a dative covalent bond using the nitrogen lone pair, a quaternary ammonium salt is formed (general formula: \\( \text{R}_4\text{N}^+ \text{X}^- \\)). In this case, the nitrogen atom carries a permanent positive charge, and the compound behaves as an ionic salt.

✏️ Worked Example: Classifying Amines
Classify each of the following compounds as a primary, secondary, or tertiary amine, or a quaternary ammonium salt:
  1. \\( \text{CH}_3\text{CH}_2\text{NHCH}_3 \\)
  2. \\( (\text{CH}_3\text{CH}_2)_4\text{N}^+ \text{Br}^- \\)
  3. \\( \text{C}_6\text{H}_5\text{NH}_2 \\) (phenylamine)

1. Secondary (2°) Amine: The nitrogen atom is directly bonded to two carbon atoms (an ethyl group and a methyl group).

2. Quaternary Ammonium Salt: The nitrogen atom is bonded to four ethyl groups and carries a positive charge, with a bromide counter-ion.

3. Primary (1°) Amine: The nitrogen atom is bonded to only one carbon atom (the benzene ring carbon).

Basicity of Amines

According to the Brønsted-Lowry theory, a base is a proton acceptor. Amines act as weak bases because the nitrogen atom possesses a lone pair of electrons that can accept a proton (\( \text{H}^+ \)) by forming a dative covalent bond.

In aqueous solution, amines establish a dynamic equilibrium by accepting a proton from water: \[ \text{RNH}_2\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{RNH}_3^+\text{(aq)} + \text{OH}^-\text{(aq)} \]

The Basicity Trend

You must be able to compare and explain the relative basic strengths of ammonia, primary aliphatic amines (such as methylamine), and aromatic amines (such as phenylamine):

Basicity Strength Order

Primary Aliphatic Amines > Ammonia > Aromatic Amines

Let us look at why this trend exists by comparing the electron density on the nitrogen atom in each class of compound:

Basicity Comparison in Amines 1. Methylamine (Aliphatic) CH₃ N H H Strongest Base Alkyl group releases electrons (+I effect) Increases electron density on Nitrogen Lone pair is highly available to accept H⁺ 2. Ammonia N H H H Intermediate Base No alkyl groups present Standard electron density on Nitrogen Reference basicity point 3. Phenylamine (Aromatic) N H H Weakest Base Lone pair delocalises into benzene pi system Significantly lowers electron density on N Lone pair is not available to accept H⁺
✏️ Worked Example: Explaining Basicity Differences
Compare the basicity of ethylamine and phenylamine, and explain the difference in their basic strengths.

Basicity Comparison: Ethylamine is a much stronger base than phenylamine.

Explanation:

  • In ethylamine, the ethyl group is electron releasing due to the positive inductive effect. This pushes electron density towards the nitrogen atom, making the lone pair on the nitrogen more available to accept a proton.
  • In phenylamine, the lone pair of electrons on the nitrogen atom overlaps with the delocalised pi electron system of the benzene ring, becoming partially delocalised into the ring. This reduces the electron density on the nitrogen atom, making the lone pair significantly less available to accept a proton.

Preparation of Primary Aliphatic Amines

There are two primary methods for preparing aliphatic amines. You must understand the reagents, conditions, and chemical trade offs of each route.

Route 1: Halogenoalkanes with Excess Ammonia

Halogenoalkanes undergo nucleophilic substitution when reacted with ammonia. The ammonia acts as a nucleophile, attacking the electron deficient carbon atom bonded to the halogen: \[ \text{R-X} + 2\text{NH}_3 \rightarrow \text{R-NH}_2 + \text{NH}_4\text{X} \]

📝 AQA Examiner Tip

You must explain why excess ammonia is required in this reaction. The product of the reaction is a primary amine (\( \text{RNH}_2 \role="math" \)), which itself has a lone pair on the nitrogen atom and can act as a nucleophile. If excess ammonia is not used, the primary amine will react with remaining halogenoalkane molecules to produce secondary amines, tertiary amines, and eventually quaternary ammonium salts. Using excess ammonia ensures that a halogenoalkane molecule is far more likely to collide with an ammonia molecule than with a newly formed amine molecule, maximising the yield of the primary amine.

Route 2: Reduction of Nitriles

Unlike the halogenoalkane route, reduction of nitriles provides a cleaner synthetic pathway that yields only the primary amine as a product.

Method A: Laboratory Reduction with LiAlH₄
Lithium tetrahydridoaluminate (LiAlH₄) is a powerful reducing agent, represented in equations as 4[H]: \[ \text{R-C}\equiv\text{N} + 4[\text{H}] \rightarrow \text{R-CH}_2\text{NH}_2 \] This reaction is carried out in dry ether solvent, followed by the addition of dilute acid to hydrolyse the intermediate complex.

Method B: Industrial Catalytic Hydrogenation
Industrially, using LiAlH₄ is too expensive. Instead, nitriles are reduced using hydrogen gas in the presence of a transition metal catalyst: \[ \text{R-C}\equiv\text{N} + 2\text{H}_2 \rightarrow \text{R-CH}_2\text{NH}_2 \] Typical catalysts include nickel (Ni), palladium (Pd), or platinum (Pt) at elevated temperatures and pressures.

📝 AQA Examiner Tip

A key advantage of reducing nitriles is that it yields a pure primary amine because no further substitution can occur. However, remember that forming a nitrile from a halogenoalkane (by reacting with KCN) increases the carbon chain length by one carbon atom. Always count your carbon atoms carefully in synthesis questions to check if you need to use this route.

Preparation of Aromatic Amines

Aromatic amines like phenylamine are prepared by the reduction of nitro compounds, such as nitrobenzene. This is a multi step process:

\( \text{Nitrobenzene } (\text{C}_6\text{H}_5\text{NO}_2) \xrightarrow{\text{Sn / conc. HCl}} \text{Phenylammonium chloride} \xrightarrow{\text{NaOH}} \text{Phenylamine } (\text{C}_6\text{H}_5\text{NH}_2) \)

Step 1: Reduction of Nitrobenzene

Nitrobenzene is heated under reflux with tin (Sn) and concentrated hydrochloric acid (HCl). The tin and acid act as the reducing agent (represented as 6[H]): \[ \text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} \]

Because the reaction mixture is highly acidic, the basic phenylamine product immediately reacts with the excess HCl to form a soluble salt: phenylammonium chloride (\( \text{C}_6\text{H}_5\text{NH}_3^+ \text{Cl}^- \)).

Step 2: Liberating the Phenylamine

To obtain the free organic amine, sodium hydroxide (NaOH) solution is added to neutralise the acid and deprotonate the phenylammonium ion: \[ \text{C}_6\text{H}_5\text{NH}_3^+ + \text{OH}^- \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{O} \] The phenylamine can then be separated from the mixture using steam distillation and solvent extraction.

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