An amine is ammonia with one or more hydrogens swapped for carbon groups. That nitrogen still carries a lone pair, and everything on this page flows from it: the lone pair accepts a proton (so amines are bases) and attacks electron-poor carbon (so amines are nucleophiles). We start with the part students most often get wrong — the naming.
Classes & nomenclature
Amines are classified by how many carbon groups are on the nitrogen:
- Primary (1°) — one carbon group: R–NH2
- Secondary (2°) — two: R2NH
- Tertiary (3°) — three: R3N
- Quaternary ammonium ion — four, giving a permanent positive charge: R4N+
The IUPAC rule
Name amines the same way throughout: take the longest carbon chain attached to the nitrogen as the parent, and add the ending –amine with a locant. Any other group on the nitrogen is written as an N-substituent at the front. So CH3CH2NH2 is ethanamine, and CH3CH2NHCH3 — ethyl chain as parent, methyl on the N — is N-methylethanamine.
Learn one consistent system — the IUPAC –amine names — but recognise the older “–ylamine” style, because AQA mark schemes accept both.
| Structure | Class | IUPAC name | Also accepted |
|---|---|---|---|
| CH3NH2 | 1° | methanamine | methylamine |
| CH3CH2NH2 | 1° | ethanamine | ethylamine |
| (CH3)2CHNH2 | 1° | propan-2-amine | 1-methylethylamine / isopropylamine |
| CH3CH2NHCH3 | 2° | N-methylethanamine | N-methylethylamine / methylethylamine |
| (CH3CH2)2NH | 2° | N-ethylethanamine | diethylamine |
| (CH3CH2)3N | 3° | N,N-diethylethanamine | triethylamine |
| C6H5NH2 | 1° aromatic | phenylamine | aniline (older trivial name) |
🧪 Exam-style questions
Bromoethane reacts with methylamine (CH3NH2) to form a secondary amine, CH3CH2NHCH3. Give the IUPAC name of this secondary amine, and state its class.
Show answer
N-methylethanamine (AQA also accepts N-methylethylamine / methylethylamine). 1 mark
It is a secondary (2°) amine — two carbon groups on the nitrogen. 1 mark
Source: AQA A-Level Chemistry past papers.
Why some amines are stronger bases
Amines are weak bases: the nitrogen lone pair accepts a proton (H+). The more available that lone pair, the stronger the base. This single idea explains the whole order:
primary aliphatic amine > ammonia > primary aromatic amine
- Aliphatic amine > ammonia: the alkyl groups are electron-releasing (a positive inductive effect). They push electron density onto the nitrogen, so its lone pair is more available to accept a proton — a stronger base than ammonia.
- Aromatic amine < ammonia: in phenylamine the nitrogen lone pair is delocalised into the benzene ring. It is drawn into the π system and so is less available to accept a proton — a weaker base than ammonia.
🧪 Exam-style questions
Which compound is the strongest base?
Explain why 3-aminopentane (CH3CH2CH(NH2)CH2CH3) is a stronger base than ammonia.
Show answer
In 3-aminopentane the lone pair on nitrogen is more available (it accepts a proton more readily). 1 mark
This is because the alkyl groups are electron-releasing / have a positive inductive effect, pushing electron density onto the nitrogen. 1 mark
Source: AQA A-Level Chemistry past papers.
Preparation & reactions
Making amines
There are three routes you need, chosen by what you are starting from:
- Primary aliphatic amine from a halogenoalkane: heat with an excess of ammonia in ethanol (nucleophilic substitution). The excess ammonia helps stop the amine reacting further.
CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br
- Primary amine from a nitrile: reduce with H2 and a nickel catalyst (or LiAlH4). This adds a carbon to the chain.
CH3CN + 4[H] → CH3CH2NH2
- Aromatic amine from a nitro compound: reduce nitrobenzene with tin and concentrated hydrochloric acid (then add NaOH to free the amine) → phenylamine, used to make dyes.
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
Amines as nucleophiles
The nitrogen lone pair attacks the δ+ carbon of a halogenoalkane, displacing the halide (nucleophilic substitution). A second ammonia (or amine) then removes H+ to give the free amine. But that product still has a lone pair, so it can react again — the reaction climbs a ladder:
NH3 → 1° amine → 2° amine → 3° amine → quaternary ammonium salt
Control the product by choosing what is in excess: excess ammonia favours the primary amine; excess halogenoalkane drives it all the way to the quaternary ammonium salt.
A quaternary ammonium salt has a permanent positive charge on nitrogen and a long hydrocarbon tail. The cationic head binds to negatively charged surfaces (hair, fabric), so these salts are used as cationic surfactants — fabric softeners and hair conditioners.
Ammonia and primary amines also react with acyl chlorides and acid anhydrides by nucleophilic addition–elimination, giving amides and N-substituted amides. That mechanism is drawn in full on the carboxylic acids & derivatives page.
🧪 Exam-style questions
Methylamine reacts with bromoethane by nucleophilic substitution to produce a mixture of products. Which is not a possible product of this reaction?
Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia.
Show answer
Br(CH2)6Br + 4NH3 → H2N(CH2)6NH2 + 2NH4Br 2 marks
(One mark for both organic structures correct; one for balancing. Each C–Br needs one NH3 to substitute and one to remove the HBr.)
1,6-Diaminohexane can also be made in two stages from 1,4-dibromobutane. Suggest the reagent and a condition for each stage.
Show answer
Stage 1 (make the dinitrile): KCN or NaCN, in aqueous ethanol. 1 mark reagent 1 mark condition
Stage 2 (reduce the nitrile groups): H2 with a Ni (or Pt/Pd) catalyst. 1 mark
(The nitrile route adds two carbons, converting a 4-carbon dibromide into the 6-carbon diamine.)
Source: AQA A-Level Chemistry past papers.
Choosing the route
Synthesis questions usually hinge on picking the right preparation. Match the starting material to the method: a halogenoalkane → react with excess ammonia (same carbon count); a shorter chain that needs one more carbon → go via the nitrile (KCN, then reduce); an aromatic nitro compound → reduce with Sn/HCl to the aromatic amine.
- Base-strength answers that stop at “more electrons” — you must say the lone pair is more (or less) available, and why (alkyl inductive effect; or delocalisation into the ring).
- Forgetting that the halogenoalkane route gives a mixture — excess ammonia is needed to favour the primary amine.
- Using NaBH4 to reduce a nitrile — it does not work; use H2/Ni or LiAlH4.
- A quaternary product needing more of one group than the starting materials can supply (like two methyls from one methylamine).
Examiner reports note that base-strength marks are lost to vague answers: name the effect (positive inductive effect, or delocalisation into the ring) and link it explicitly to the availability of the nitrogen lone pair.
- Classes: 1° (RNH2), 2° (R2NH), 3° (R3N), quaternary ammonium (R4N+).
- Naming: parent = longest chain on N + –amine; other groups are N-substituents (e.g. N-ethylethanamine). AQA also accepts the –ylamine forms (diethylamine).
- Base strength: primary aliphatic amine > ammonia > primary aromatic amine. Alkyl groups push electrons onto N (lone pair more available); in phenylamine the lone pair delocalises into the ring (less available).
- Preparation: 1° aliphatic from NH3 + halogenoalkane (excess NH3) or by reducing a nitrile (H2/Ni); aromatic by reducing a nitro compound (Sn + conc HCl, then NaOH).
- As nucleophiles: substitution with halogenoalkanes climbs 1°→2°→3°→quaternary; quaternary ammonium salts are cationic surfactants. With acyl chlorides/anhydrides they give amides by addition–elimination.