1. Amine Classification & Nomenclature
Amines are organic derivatives of ammonia (NH₃) in which one or more hydrogen atoms have been replaced by alkyl or aryl hydrocarbon groups:
- Primary (1 degree) Amine: Nitrogen bonded to ONE carbon group: R-NH₂ (e.g. ethylamine, CH₃CH₂NH₂).
- Secondary (2 degree) Amine: Nitrogen bonded to TWO carbon groups: R₂NH (e.g. dimethylamine, (CH₃)₂NH).
- Tertiary (3 degree) Amine: Nitrogen bonded to THREE carbon groups: R₃N (e.g. trimethylamine, (CH₃)₃N).
- Quaternary (4 degree) Ammonium Salt: Nitrogen bonded to FOUR carbon groups, bearing a permanent positive formal charge: R₄N⁺ X⁻ (e.g. tetramethylammonium chloride).
2. Comparative Base Strength of Amines
Amines act as Bronsted-Lowry bases because the nitrogen atom possesses a lone pair of electrons that can accept a proton (H⁺) via dative covalent bonding:
R-NH₂ + H⁺ ⇌ R-NH₃⁺
The relative strength of an amine as a base depends entirely on the availability of the nitrogen lone pair to accept a proton.
- 1. Primary Aliphatic Amines (e.g. Ethylamine) - Strongest Bases: The alkyl group has an electron-releasing positive inductive effect (+I). This pushes electron density towards the nitrogen atom, increasing the charge density of the nitrogen lone pair and making it more available to accept a proton than in ammonia.
- 2. Ammonia (NH₃) - Intermediate Base: Lacks alkyl groups (no inductive electron release) and lacks an aromatic ring (no delocalisation).
- 3. Aromatic Amines (e.g. Phenylamine) - Weakest Bases: The lone pair of electrons on nitrogen overlaps sideways and delocalises into the benzene ring pi electron system. This significantly reduces the electron density on the nitrogen atom, making the lone pair much less available to accept a proton.
3. Base Strength & Lone Pair Availability Diagram
4. Synthetic Preparation of Amines
Method 1: Nucleophilic Substitution of Haloalkanes
Heating a haloalkane with excess ethanolic ammonia in a sealed tube under pressure. This is nucleophilic substitution, which you met in Unit 2 (Halogenoalkanes: nucleophilic substitution):
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
The primary amine product (ethylamine) also has a lone pair on nitrogen and is an even stronger nucleophile than ammonia. If equimolar quantities are used, ethylamine reacts with remaining bromoethane to form diethylamine, triethylamine, and finally a quaternary ammonium salt. Using a large excess of ammonia ensures that haloalkane molecules collide almost exclusively with ammonia, maximising the yield of the primary amine.
Mechanism and Successive Substitution
The nitrogen lone pair attacks the δ+ carbon. Then a second ammonia molecule removes H⁺ from the nitrogen.
The amine product still has a lone pair on nitrogen, so it is also a nucleophile. It can react with more halogenoalkane in the same way. Each step adds one more alkyl group to the nitrogen:
RNH₂ → R₂NH → R₃N → R₄N⁺X⁻
- RNH₂: primary amine
- R₂NH: secondary amine
- R₃N: tertiary amine
- R₄N⁺X⁻: quaternary ammonium salt (X⁻ is the halide ion)
Method 2: Reduction of Nitriles (Preparation of Pure Primary Amines)
Nitriles can be reduced cleanly to primary amines using lithium tetrahydridoaluminate (LiAlH₄) in dry ether followed by dilute acid, or by catalytic hydrogenation using hydrogen gas over a nickel catalyst (H₂ / Ni):
R-C≡N + 4[H] → R-CH₂-NH₂
For example, propanenitrile is reduced to propylamine (H₂/Ni or LiAlH₄):
CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂
Key Synthetic Advantage: Unlike substitution of a halogenoalkane with ammonia, nitrile reduction gives only the primary amine, with no secondary or tertiary amines or quaternary salt formed by further substitution.
Method 3: Reduction of Nitrobenzene to Phenylamine
Nitrobenzene is converted to phenylamine in the laboratory in two stages:
- Reflux nitrobenzene with tin (Sn) and concentrated hydrochloric acid (HCl). In the acid, the amine is present as its salt, phenylammonium chloride (C₆H₅NH₃⁺Cl⁻).
- Add aqueous sodium hydroxide (NaOH) to liberate the free amine base:
C₆H₅NH₃⁺Cl⁻ + OH⁻ → C₆H₅NH₂ + H₂O + Cl⁻
Overall: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O (tin and concentrated hydrochloric acid, then NaOH(aq) to release the free amine)
Use: aromatic amines made by reducing nitro compounds are used to make dyes.
5. Quaternary Ammonium Salts as Cationic Surfactants
When a tertiary amine reacts with a haloalkane, a quaternary ammonium salt is produced:
(CH₃)₃N + CH₃Cl → (CH₃)₄N⁺ Cl⁻ (tetramethylammonium chloride)
Cationic Surfactant Structure & Applications
Quaternary ammonium salts with one or two very long non-polar alkyl hydrocarbon chains (e.g. C₁₆ to C₁₈) act as cationic surfactants:
- Hydrophobic tail: The long non-polar hydrocarbon chain dissolves readily in non-polar grease and oils.
- Hydrophilic head: The positively charged nitrogen cation (N⁺) interacts electrostatically with polar water molecules and negatively charged surfaces.
- Uses: Fabric softeners and hair conditioners. Fabric fibres and wet hair naturally carry negative surface charges. The positive cationic heads bind electrostatically to the fabric or hair surface, leaving the smooth, non-polar hydrocarbon chains facing outward to reduce friction, eliminate static charge, and impart softness.
6. Worked Synthesis & Mechanism
Step 1: Nucleophilic substitution to form a nitrile:
Reflux bromoethane (CH₃CH₂Br) with aqueous-ethanolic potassium cyanide (KCN):
CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻
Product: Propanenitrile (extends chain to 3 carbons).
Step 2: Reduction of the nitrile:
Pass propanenitrile and hydrogen gas over a finely divided nickel catalyst at elevated temperature:
CH₃CH₂CN + 2H₂ → CH₃CH₂CH₂NH₂
Product: Propylamine.
Advantage: nitrile reduction gives only the primary amine, with no secondary or tertiary amines or quaternary salt formed by further substitution.
Stage 1: Nitration of benzene:
C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
Reagents: Concentrated HNO₃ and concentrated H₂SO₄, below 55 °C.
Stage 2: Reduction to phenylamine:
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O
Reagents: Tin (Sn) and concentrated HCl under reflux, followed by addition of NaOH(aq).
7. Reactions with Acyl Chlorides and Acid Anhydrides
Amines are nucleophiles, so they react with acyl chlorides and acid anhydrides by nucleophilic addition-elimination. A primary amine gives an N-substituted amide.
CH₃COCl + CH₃CH₂NH₂ → CH₃CONHCH₂CH₃ + HCl
The product is N-ethylethanamide. The mechanism is on the Carboxylic Acids and Derivatives page.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
8. Practice Questions
Show Answer & Mark Scheme
Answer:
- The ethyl group pushes electron density onto the nitrogen (1).
- So the lone pair on N in ethylamine is more available to accept H⁺ (1).
- In phenylamine, the lone pair on N is delocalised into the benzene ring (1).
- So the lone pair is less available to accept H⁺ (1).
Show Answer & Mark Scheme
Answer:
- Ethylamine is also a nucleophile, so it can react with more bromoethane to give secondary and tertiary amines and a quaternary ammonium salt (1).
- With excess ammonia, bromoethane is much more likely to react with ammonia (1).
Show Answer & Mark Scheme
Answer:
- H₂ with a nickel catalyst, or LiAlH₄ (1).
- CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂ (1).
Show Answer & Mark Scheme
Answer:
- They are cationic surfactants with a positive head (1).
- The positive head is attracted to the negatively charged surface of the fabric (1).