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
This extends the IUPAC nomenclature you met at AS. 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 seen |
|---|---|---|---|
| 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) |
Primary (1°)
ethanamine — one carbon group on N.
Secondary (2°)
N-methylethanamine — two carbon groups on N.
Tertiary (3°)
N,N-dimethylethanamine — three carbon groups on N.
Quaternary ammonium ion
tetramethylammonium ion — four groups, permanent positive charge, no lone pair.
Name the amine (CH3)2CHNHCH2CH3.
Step 1 — list the carbon groups on the nitrogen: a propan-2-yl group (three carbons, attached through its middle carbon) and an ethyl group (two carbons).
Step 2 — the longest chain is the parent: three carbons beats two, so the parent is propan-2-amine (the N sits on carbon 2).
Step 3 — everything else becomes an N-substituent: the ethyl group is written N-ethyl at the front.
N-ethylpropan-2-amine
Exam questions — naming & recognising amines, as AQA asked it
An incomplete equation for Step 1 in the reaction between bromoethane and an amine is shown.
Complete the equation. In Step 2 of this reaction, the product of Step 1 forms a secondary amine. Name the secondary amine formed.
Show answer
M1 — the missing reactant is CH3NH2 (methylamine), shown as a displayed or abbreviated structural formula. 1 mark
M2 — the secondary amine is N-methylethanamine. 1 mark
The mark scheme also allows: N-methyl ethylamine, N-methyl aminoethane, N-methyl N-ethylamine, methyl ethylamine or methyl ethanamine (alkyl groups may be written in either order).
Prilocaine is used as an anaesthetic in dentistry. Figure 1 shows the structure of prilocaine.
Draw a circle around any chiral centre(s) in Figure 1.
Show answer
One circled C atom only — the C attached to CH3, C=O, H and NH (the carbon between the carbonyl group and the second nitrogen). 1 mark
That carbon carries four different groups, so it is the only chiral centre. Circling any additional atom loses the mark.
Identify the functional group(s) in the prilocaine molecule. Tick (✓) the box(es) corresponding to the functional group(s).
- ☐ Amide
- ☐ Amine
- ☐ Ester
- ☐ Ketone
Show answer
Two ticks only — for amine and amide. 1 mark
The left-hand nitrogen is bonded directly to a C=O, so that whole unit is an amide, not an amine plus a ketone. The right-hand nitrogen sits between two carbon chains with no carbonyl attached — a secondary amine.
Prilocaine is completely hydrolysed in the human body to give a mixture of products. Draw the structures of the two organic products formed in the complete hydrolysis of prilocaine in acidic conditions.
Show answer
M1 — choosing the correct bond to hydrolyse (the amide C–N bond). 1 mark
M2 and M3 — the correct structures of the two products: the carboxylic acid and, because the hydrolysis is in acidic conditions, the phenylammonium ion C6H5NH3+. 2 marks
The mark scheme allows the protonated amino acid for M2, and allows C6H5NH3+ drawn with the + outside a square bracket. The free amine C6H5NH2 is wrong here — in acid, the basic amine is protonated. Full coverage of amide hydrolysis is on the carboxylic acids & derivatives page.
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+) — a Brönsted–Lowry base, the proton acceptor you met in acids and bases. 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.
Strongest base
The alkyl group is electron-releasing — it pushes electron density onto N.
The reference
No alkyl push, no delocalisation — ammonia sits in the middle.
Weakest base
The lone pair is delocalised into the ring — drawn into the π system.
Explain why ethylamine is a stronger base than ammonia, and why phenylamine is a weaker base than ammonia. (4 marks)
Step 1 — ethylamine, the availability point: the lone pair on the N in ethylamine is more available to accept a proton (than in ammonia).
Step 2 — ethylamine, the reason: because the alkyl group is electron-releasing (a positive inductive effect), pushing electron density onto the nitrogen.
Step 3 — phenylamine, the availability point: the lone pair on the N in phenylamine is less available to accept a proton.
Step 4 — phenylamine, the reason: because the lone pair is delocalised into the benzene ring.
Every answer is those two moves per amine: how available is the lone pair, then the named effect that makes it so. “Alkyl groups give it more electrons” on its own scores the reason mark at best — it never scores the availability mark.
Exam questions — base strength, as AQA asked it
Explain why 3-aminopentane is a stronger base than ammonia.
Show answer
M1 — in 3-aminopentane the lone pair on N is more available (or: accepts H+ better / the protonated N is more stable). 1 mark
M2 — because of the alkyl electron pushing / inductive effect. 1 mark
The marks are independent, and the converse argument for ammonia is allowed. Note there are two alkyl groups pushing onto this nitrogen — but the mark scheme wants the effect named, not counted.
Figure 2 shows two amines, P and Q.
Explain why P is a stronger base than Q.
Show answer
M1 — the lone pair on nitrogen in P is more available, or more able to accept protons/H+. 1 mark
M2 — more alkyl groups are electron releasing/donating, OR a greater (positive) inductive effect of the alkyl groups. 1 mark
Here the comparison is between two aliphatic amines, so it is the number of alkyl groups that decides it — P (a secondary amine, two alkyl groups) beats Q (a primary amine, one).
Source: AQA A-Level Chemistry past papers.
Preparation & reactions
Making amines
Amines are prepared by three routes, and which one you use depends on what you are starting from:
- Primary aliphatic amine from a halogenoalkane: heat with an excess of ammonia in ethanol (nucleophilic substitution). The excess ammonia favours the primary amine by making further substitution less likely.
CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br
- Primary amine from a nitrile: reduce with H2 and a nickel catalyst (or LiAlH4). Because the nitrile came from KCN attacking a halogenoalkane (nucleophilic substitution), this route 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. The nitration that makes nitrobenzene is on the aromatic chemistry page.
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O
- For a nitrile, AQA accepts H2 with Ni, Pt or Pd, or LiAlH4 (in dry ether) — but never NaBH4, and never Sn/HCl or Fe/HCl.
- For an aromatic nitro compound it is the other way round: Sn (or Fe) with concentrated HCl, or H2 with a metal catalyst.
- Mark schemes apply the list principle: an extra wrong reagent alongside a right one loses the mark.
Amines as nucleophiles: the mechanism
The nitrogen lone pair attacks the δ+ carbon of a halogenoalkane and the halide leaves — the same nucleophilic substitution you met at AS, where halogenoalkanes react with hydroxide, cyanide and ammonia. The first product is an alkylammonium salt; a second ammonia then removes an H+ from the nitrogen to free the amine.
The same molecule plays both roles, one after the other:
- The first NH3 is the nucleophile — its lone pair attacks the δ+ carbon and displaces the halide.
- The second NH3 is a base — it removes the H+ from the positively charged nitrogen of the alkylammonium salt, freeing the neutral amine.
Suggest how to prepare butylamine, CH3CH2CH2CH2NH2, (a) from 1-bromobutane and (b) from 1-bromopropane.
Step 1 — count the carbons. Butylamine has four. 1-Bromobutane has four; 1-bromopropane has three, so route (b) must add a carbon.
Step 2 — same carbon count → direct substitution: heat 1-bromobutane with an excess of ammonia in ethanol.
CH3CH2CH2CH2Br + 2NH3 → CH3CH2CH2CH2NH2 + NH4Br
Step 3 — one carbon short → go via the nitrile: KCN in aqueous ethanol gives CH3CH2CH2CN, then reduce with H2/Ni.
CH3CH2CH2CN + 4[H] → CH3CH2CH2CH2NH2
Step 4 — sanity-check the products: both routes end at the same primary amine; only the starting carbon count decides between them.
The ladder to quaternary ammonium salts
The primary amine formed still has a lone pair — and it is an even better nucleophile than ammonia. So it attacks more halogenoalkane, and the reaction climbs a ladder of successive substitutions:
Control the product by choosing what is in excess: excess ammonia favours the primary amine (any halogenoalkane molecule is most likely to meet an NH3); excess halogenoalkane drives substitution all the way to the quaternary ammonium salt. Either way the real mixture contains every rung.
Interactive — climb the substitution ladder
A quaternary ammonium salt has a permanent positive charge on nitrogen. Give it a long hydrocarbon tail and the cationic head binds to negatively charged surfaces (wet hair, fabric), so these salts are used as cationic surfactants — fabric softeners and hair conditioners.
Ammonia and primary amines also attack acyl chlorides and acid anhydrides by nucleophilic addition–elimination, giving amides and N-substituted amides. It is the same lone pair doing the attacking — the mechanism is drawn in full on the carboxylic acids & derivatives page.
Exam questions — the halogenoalkane route, as AQA asked it
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
M1 — both organic compounds correct, as structural (not molecular) formulae. 1 mark
M2 — balanced. 1 mark
Four NH3 because each C–Br end needs one ammonia to substitute and a second to remove the HBr. The mark scheme allows one correct structural formula with the other as a molecular formula of type XC6H12X.
Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane.
Suggest the structure of a cyclic secondary amine that can be formed as a by-product in this reaction.
Show answer
M1 — the curly arrow from the lone pair on NH3 to the carbon bonded to Br. 1 mark
M2 — the arrow from the C–Br bond to the Br (printed in the question). 1 mark
M3 — the alkylammonium intermediate with the + on N. 1 mark
The loss of H+ need not show the second NH3 — but using Br− to remove the H is penalised, as are incorrect partial charges in M1. SN1 is allowed. Structural formulae (Br(CH2)6NH2 etc.) are fine.
The cyclic secondary amine 1 mark — the NH2 at the far end of the chain can attack its own C–Br carbon, closing a seven-membered ring (azepane); the mark scheme also allows the fourteen-membered ring formed from two molecules:
Justify the statement that there are no chiral centres in 3-aminopentane.
Show answer
No carbon atom is attached to four different groups — carbon 3 carries the NH2, an H and two identical ethyl groups. 1 mark
The mark scheme also allows “the central carbon has two alkyl groups” or “the molecule is symmetrical”.
This question is about 2-bromopropane. Define the term electronegativity. Explain the polarity of the C–Br bond in 2-bromopropane.
Show answer
M1 — electronegativity is the (relative) tendency of an atom to attract a pair of electrons / the electron density in a covalent bond. 1 mark
M2 — Br is more electronegative than C (or vice versa). 1 mark
M3 — so Br is δ− and C is δ+. 1 mark
Outline the mechanism for the reaction of 2-bromopropane with an excess of ammonia.
Show answer
M1 — lone pair on N and its arrow to the δ+ carbon. 1 mark M2 — arrow from the C–Br bond to Br. 1 mark
M3 — the structure of the alkylammonium intermediate, + on N. 1 mark M4 — loss of H+, arrow from the N–H bond back to N. 1 mark
M4 is penalised if Br− is shown removing the H+ — use a second ammonia. SN1 is allowed.
Draw the skeletal formula of the main organic species formed in the reaction between a large excess of 2-bromopropane and ammonia. Give a use for the organic product.
Show answer
M1 — the quaternary ammonium ion, tetra(propan-2-yl)ammonium, with the + on N (allowed outside square brackets). 1 mark
M2 — use: (hair) conditioner / (cationic) surfactant / disinfectant / fabric softener. 1 mark
CH3CHBrCH2CH3 reacts with NH3. Draw the skeletal formula of the major organic product formed when
- an excess of NH3 is used
- an excess of CH3CHBrCH2CH3 is used.
Show answer
M1 — excess NH3: butan-2-amine. 1 mark M2 — excess halogenoalkane: the quaternary ammonium salt, tetra(butan-2-yl)ammonium bromide. 1 mark
If the answers are drawn non-skeletal, the mark scheme penalises once only.
Source: AQA A-Level Chemistry past papers.
The nitrile route: adding a carbon
When the target amine has one more carbon than your halogenoalkane, substitute with KCN (in aqueous ethanol) first, then reduce the nitrile. Two exam habits pay here: never write HCN (and never add acid to a cyanide — it kills both marks), and reduce with H2/Ni or LiAlH4, never NaBH4.
Exam questions — the nitrile route, as AQA asked it
1,6-Diaminohexane can also be formed in a two-stage synthesis starting from 1,4-dibromobutane. Suggest the reagent and a condition for each stage in this alternative synthesis.
Show answer
M1 — Stage 1 reagent: KCN or NaCN. 1 mark
M2 — Stage 1 condition: aqueous alcohol. 1 mark
M3 — Stage 2 reagent and condition: H2 and Ni (or Pt or Pd). 1 mark
Not HCN — that loses M1 and M2, as does any mention of acid. M2 depends on a correct M1; M3 is only accessible if a cyanide was used in stage 1. LiAlH4 (in dry ether) is allowed for stage 2; NaBH4, Sn/HCl and Fe/HCl are not. Heat, reflux and pressure are ignored.
Figure 1 shows a two-step synthesis to make amine G.
[space for the Step 1 mechanism] Step 1 (CH3)2CHCN + Br−
(CH3)2CHCN + 2H2 Step 2 Amine G
Complete Figure 1 by drawing the mechanism for Step 1 and the displayed formula of amine G.
Show answer
M1 — the structure of 2-bromopropane as the starting material. 1 mark
M2 — the two correct curly arrows: lone pair on the cyanide carbon to the δ+ carbon; C–Br bond to Br. 1 mark
M3 — amine G drawn as a fully displayed structure of 2-methylpropan-1-amine. 1 mark
Acrylonitrile, H2C=CHCN, can be used as a starting material for the synthesis of butane-1,4-diamine, as shown in this reaction scheme.
Identify the reagent that is warmed with isomer W in reaction 2. State the other reaction condition needed.
Show answer
M1 — KCN or NaCN. 1 mark
M2 — aqueous AND ethanol (alcohol). 1 mark
Acid with the cyanide is penalised in M1.
State the reagent and reaction conditions needed for reaction 3. Give an equation for reaction 3.
Show answer
M1 — H2 and Ni (or Pt or Pd). 1 mark
M2 — the equation: 1 mark
NCCH2CH2CN + 4H2 → H2N(CH2)4NH2
LiAlH4 in dry ether is allowed (but not NaBH4); the equation may be written with 8[H]. Note each C≡N needs 2H2, so the dinitrile takes 4H2.
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 chain that needs one more carbon → go via the nitrile (KCN, then reduce); an aromatic nitro compound → reduce with Sn and concentrated HCl to the aromatic amine. Multi-step schemes then chain these with reactions from earlier topics — nitration, free-radical chlorination, acylation.
- Base-strength answers that stop at “alkyl groups give it more electrons” — you must link the effect to the availability of the N lone pair. Examiner reports flag exactly this: students know the slogan but omit the lone-pair step, and it costs the mark.
- Mechanism names are marks: nucleophilic substitution with halogenoalkanes, nucleophilic addition–elimination with acyl chlorides. Examiner reports note fewer students than expected could name the latter, despite frequent past-paper exposure.
- Forgetting that the halogenoalkane route gives a mixture — further substitution means lower yield of the primary amine unless ammonia is in excess. Examiner reports list “explain why one route yields multiple amines” among the weakest answers.
- Reducing agents swapped: NaBH4 does not reduce nitriles (use H2/Ni or LiAlH4); Sn/HCl is for nitro compounds, not nitriles.
- A quaternary product needing more of one group than the starting materials can supply (like two methyls from one methylamine) is impossible — the MCQ trick below.
Exam questions — amines in synthesis, as AQA asked it
This question is about the reaction scheme shown.
State the reagents needed for step 1 and the reagents needed for step 2.
Show answer
M1 — step 1: concentrated HNO3. 1 mark M2 — step 1: concentrated H2SO4. 1 mark
M3 — step 2: Sn and HCl (allow Fe and HCl, or Ni and H2). 1 mark
If “concentrated” is missing from both acids, only one of M1/M2 is awarded.
Give the name of the mechanism for the reaction in step 3.
Show answer
(Nucleophilic) addition–elimination. 1 mark
Step 3 is the amine attacking an acyl chloride (or anhydride) to form the amide — the mechanism on the carboxylic acids & derivatives page. Examiner reports single this naming mark out as one that too few students score.
Name the reagent for step 4. State a necessary condition for step 4.
Show answer
M1 — chlorine (Cl2 allowed). 1 mark M2 — UV light (allow sunlight, or a high temperature above 300 °C). 1 mark
Step 4 is free-radical substitution of the CH3 side-chain — AS chemistry from alkanes, resurfacing in an A2 scheme.
Amine A is formed in step 2 and amine B is formed in step 5. Explain why the yield of B in step 5 is less than the yield of A in step 2.
Show answer
M1 — in step 5, further substitution occurs / other amine products form (the primary amine attacks more halogenoalkane). 1 mark
M2 — in step 2 only one amine can form. 1 mark
Explain why amine B is a stronger base than amine A.
Show answer
M1 — in B the alkyl group is electron donating (positive inductive effect); OR in A the lone pair on N is (partially) delocalised into the ring. 1 mark
M2 — so in B the lone pair on N is more available / in A it is less available. 1 mark
B’s nitrogen hangs off a CH2 — an aliphatic amine — while A’s nitrogen is bonded straight to the ring. One carbon of separation is the whole difference.
Two steps in the synthesis of an aromatic amine are shown.
State the reagent(s) needed for Step 2.
Show answer
Sn/HCl (allow H2 with Pt or Ni, or HCl with Fe). 1 mark
References to NaOH used after Sn/HCl are ignored — but NaOH used at the same time as Sn/HCl is penalised (it would neutralise the acid).
State a possible use for the amine formed in Step 2.
Show answer
Manufacture of dyes / (cationic) surfactants / fabric softener (hair or fabric conditioner also allowed). 1 mark
Paracetamol is a medicine commonly used to relieve mild pain. Traditionally, paracetamol has been made industrially in a three-step synthesis from phenol.
Name the mechanism of the reaction in Step 1.
Show answer
Electrophilic substitution — both words needed. 1 mark
“Nitration” is ignored — it names the reaction, not the mechanism.
Complete the equation for the reaction in Step 2.
4-nitrophenol + __________ → 4-aminophenol + __________
Show answer
+ 3H2 … + 2H2O 1 mark
6[H] is allowed in place of 3H2. Reducing NO2 to NH2 always releases the two oxygens as two waters — the same 6[H] + 2H2O pattern as nitrobenzene → phenylamine.
In theory, either ethanoyl chloride or ethanoic anhydride could be used in Step 3. In practice, ethanoic anhydride is used in the industrial synthesis rather than ethanoyl chloride. Give one reason why ethanoyl chloride is not used in the industrial synthesis.
Show answer
Ethanoyl chloride is corrosive / forms a strong acid, HCl (fumes) / is vulnerable to hydrolysis / is dangerous to use / reacts violently (extremely exothermic) / is more expensive. Any one. 1 mark
“Toxic”, “harmful” and “hazardous” are ignored — too vague. The acylation mechanism itself (Step 3) is drawn on the carboxylic acids & derivatives page.
In a newer two-step process, phenol is oxidised to hydroquinone, which reacts with ammonium ethanoate to form paracetamol. Calculate the mass, in kg, of hydroquinone (Mr = 110.0) needed to produce 250 kg of paracetamol.
Show answer
M1 — Mr of paracetamol (C8H9NO2) = 151.0 1 mark
M2 — amount of paracetamol (= amount of hydroquinone, 1 : 1): 1 mark
n = 250 × 103 g151.0 g mol−1 = 1655.6 mol
M3 — mass of hydroquinone: 1 mark
m = 1655.6 × 110.0 = 182 119 g = 182 kg
The mark scheme also allows the mass-ratio route: 110 g hydroquinone forms 151 g paracetamol, so 250 × 110 / 151.0 = 182 kg. Minimum 2 significant figures; with the Mr values used the wrong way round, M2 can still be scored.
Source: AQA A-Level Chemistry past papers.
Capstone quiz — four past-paper questions
Four real AQA multiple-choice questions on amines. Pick one answer each — the reasoning appears once you commit.
Methylamine reacts with bromoethane by substitution to produce a mixture of products. Which compound is not a possible product of this reaction?
Which compound is the strongest base?
Which statement about HOCH2CH(NH2)COOH is correct?
Which forms a polymer with ClOC(CH2)8COCl?
Source: AQA A-Level Chemistry past papers.
- 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 or LiAlH4 — never NaBH4; the nitrile route adds a carbon); aromatic by reducing a nitro compound (Sn + conc HCl, then NaOH) — phenylamine → dyes.
- Mechanism: nucleophilic substitution — N lone pair attacks the δ+ carbon, halide leaves, a second NH3/amine removes H+. Substitution climbs 1°→2°→3°→quaternary; excess NH3 → primary, excess halogenoalkane → quaternary.
- Quaternary ammonium salts (R4N+, no lone pair) are cationic surfactants — fabric softeners and hair conditioners.
- With acyl chlorides/anhydrides: amines give amides / N-substituted amides by nucleophilic addition–elimination.