You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Amines are organic derivatives of ammonia in which one, two, or three of the N–H bonds have been replaced by N–C bonds, giving the general formula RN(H₂), R₂NH, or R₃N. They are important as bases (the lone pair on nitrogen accepts a proton), as nucleophiles (the same lone pair attacks electrophilic carbon), and as synthetic precursors to dyes, drugs, and biological molecules. This lesson develops AQA §3.3.11 in four parts: classification (primary, secondary, tertiary); basicity trends (aliphatic vs ammonia vs aromatic), rationalised through inductive donation, π-delocalisation into the ring, and steric/solvation effects; preparation routes (reduction of nitriles, reduction of nitro-aromatics, amination of halogenoalkanes); and reactions with acids, acyl chlorides, aldehydes/ketones (signposted), and halogenoalkanes. Amino acids and proteins are developed in lesson 6; condensation polymers including nylon-6,6 and Kevlar in lesson 7.
Spec mapping (AQA 7405): This lesson maps to §3.3.11 (amines). Cross-references: §3.3.7 (carbonyls, lesson 0 — amine + aldehyde gives imine, signposted); §3.3.9 (acyl chlorides, lesson 2 — amine + acyl chloride gives amide); §3.3.10 (aromatic chemistry, lesson 4 — nitration of benzene then reduction is the standard route to aniline); §3.3.12 (amino acids, lesson 6); §3.3.13 (condensation polymers, lesson 7). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Classifying amines and recalling preparation routes are AO1. Writing balanced equations for protonation, alkylation, amide formation, and reduction is AO2. Rationalising the basicity ranking aliphatic > NH₃ > aromatic via inductive donation, π-delocalisation of the lone pair, and steric/solvation effects is AO3 — consistently distinguishes A* from B.
Amines are classified by the number of R groups (alkyl or aryl) bonded to N; the remaining N bonds are to H, and N bears one lone pair:
| Class | General formula | Example | Notes |
|---|---|---|---|
| Primary (1°) amine | RN(H₂) | CH₃NH₂ (methylamine) | One R group; two N–H bonds |
| Secondary (2°) amine | R₂NH | (CH₃)₂NH (dimethylamine) | Two R groups; one N–H bond |
| Tertiary (3°) amine | R₃N | (CH₃)₃N (trimethylamine) | Three R groups; no N–H bond |
| Quaternary (4°) ammonium salt | R₄N⁺ X⁻ | (CH₃)₄N⁺Cl⁻ | Four R groups; positively charged; exists only as a salt |
Important contrast with alcohols. In amines, classification counts the R groups on the nitrogen. In alcohols (lesson covered earlier in A-Level), classification counts the R groups on the carbon bearing the –OH. CH₃CH(OH)CH₃ is a secondary alcohol (two R groups on the C–OH carbon) but CH₃CH(OH)CH₂NH₂ is still a primary amine (only one R group on the N). Examiners specifically test this contrast.
A quaternary ammonium ion (R₄N⁺) has no N–H so cannot lose a proton — not a Brønsted base; it exists as a salt with a counter-anion. Long-chain quaternary salts are cationic surfactants (e.g. CTAB in fabric conditioners) — a useful real-world hook beyond A-Level.
Two systems coexist: suffix (IUPAC: methanamine, ethanamine, propan-2-amine) and prefix (traditional: methylamine, ethylamine, isopropylamine). For N-substituents, use the "N-" locant: (CH₃)(CH₂CH₃)NH = N-methylethanamine. AQA accepts either. The aromatic parent C₆H₅NH₂ is called phenylamine or aniline — both accepted.
The defining chemical property of an amine is that the lone pair on nitrogen can accept a proton, making the amine a Brønsted-Lowry base:
R–N(H₂) + H⁺ → R–N⁺(H₃) R–N(H₂) + H₂O ⇌ R–N⁺(H₃) + OH⁻
The position of this equilibrium — and therefore how strong the amine is as a base — depends on two factors: how readily the lone pair attacks H⁺ (which depends on the electron density on N), and how stable the resulting conjugate acid R–N⁺(H₃) (the "ammonium" ion) is in solution.
Basicity is quantified through the pKa of the conjugate acid BH⁺:
BH⁺ ⇌ B + H⁺ Kₐ = [B][H⁺] / [BH⁺]
A higher pKa(BH⁺) means a weaker conjugate acid and therefore a stronger amine base. Always quote pKa(BH⁺) explicitly to avoid ambiguity.
| Base B | pKa(BH⁺) | Relative basicity |
|---|---|---|
| Ammonia, NH₃ | 9.25 | Baseline |
| Methylamine, CH₃NH₂ | 10.6 | Stronger than NH₃ |
| Dimethylamine, (CH₃)₂NH | 10.8 | Strongest of the aliphatic series |
| Trimethylamine, (CH₃)₃N | 9.8 | Weaker than di- and mono-methylamine |
| Aniline (phenylamine), C₆H₅NH₂ | 4.6 | Much weaker than ammonia |
| N,N-Dimethylaniline | 5.1 | Slightly stronger than aniline (alkyl donates) |
The headline ranking at A-Level is therefore: secondary aliphatic > primary aliphatic > NH₃ > tertiary aliphatic (close to NH₃) ≫ aromatic amine.
(a) Aliphatic amines are more basic than ammonia. Alkyl groups are weakly electron-donating through the σ-framework (the "inductive effect"). Donating electron density onto N makes the lone pair more available to attack H⁺ and stabilises the positive charge on N in the conjugate acid R–N⁺(H₃). Both effects raise pKa(BH⁺).
(b) Aromatic amines are much less basic than ammonia. In aniline, the nitrogen lone pair overlaps with the ring π-system — it is partially delocalised into the ring. This delocalisation lowers the energy of the free amine and is lost on protonation (the protonated N has no lone pair left to donate). Protonation is therefore energetically costly, and the equilibrium lies toward the free amine. The pKa(BH⁺) for aniline (4.6) is six orders of magnitude lower than methylamine (10.6).
Common misconception. Students sometimes claim aniline is a weak base "because it has fewer lone pairs" or "because the ring takes the lone pair". The lone pair is still on nitrogen — it is partially shared with the ring (resonance forms place negative charge on ortho/para carbons), and protonation breaks that stabilisation. Quantitatively, the resonance stabilisation lost on protonation accounts for ~30 kJ mol⁻¹.
(c) Tertiary aliphatic amines are less basic than secondary in water — solvation matters. In the gas phase, (CH₃)₃N is the strongest base of the methylamine series, as inductive donation predicts. In water, however, the conjugate acid R–N⁺(H₃) is stabilised by hydrogen bonding from water to its N–H bonds. (CH₃)₃N⁺H has only one N–H (versus three for CH₃N⁺H₃), and bulky methyls hinder water approach. Reduced solvation destabilises the protonated form, lowering pKa(BH⁺); dimethylamine wins in water. A classic A* point — examiners reward candidates who name both the inductive effect and the steric/solvation reversal.
Four routes appear on the AQA specification. The first three give primary amines (R–NH₂); the fourth gives over-substitution mixtures.
R–C≡N + 4[H] → R–CH₂–NH₂
Reagents: LiAlH₄ in dry ether followed by acid work-up; or H₂ over Ni catalyst at high T and P. Strategically powerful because nitriles are made from haloalkane + KCN, extending the chain by one carbon: CH₃CH₂Br → CH₃CH₂CN → CH₃CH₂CH₂NH₂.
Standard preparation of an aromatic primary amine. Nitrobenzene (from nitration of benzene, lesson 4) is reduced to aniline:
C₆H₅–NO₂ + 6[H] → C₆H₅–NH₂ + 2H₂O
Reagents: Sn + concentrated HCl, reflux (Bechamp reduction), then NaOH to liberate the free amine from the anilinium salt; or H₂ over Pt/Ni. Industrially produced on a million-tonne scale; parent of azo dyes, paracetamol, sulfa antibiotics, and MDI for polyurethanes. Combined with lesson 4 nitration, this is the two-step preparation of aniline from benzene.
R–X + 2NH₃ → R–NH₂ + NH₄⁺X⁻
Conditions: excess concentrated NH₃ in ethanol, sealed tube, heat. Mechanism: S_N2 — the N lone pair attacks the δ⁺ carbon, displacing halide; a second NH₃ deprotonates the intermediate.
The newly-formed primary amine is itself a (stronger) nucleophile than NH₃, so it competes for further R–X:
R–NH₂ + R–X → R₂NH + HX R₂NH + R–X → R₃N + HX R₃N + R–X → R₄N⁺X⁻
The product is a mixture. Large excess of NH₃ favours the primary amine; excess R–X favours the quaternary salt; mixtures are separated by distillation.
Practical-skills box. Preparation of a primary aliphatic amine via nitrile reduction: dissolve ~1.0 g of propanenitrile (CH₃CH₂CN) in ~30 cm³ dry diethyl ether under nitrogen, cool in ice, and add solid LiAlH₄ portionwise (~0.5 g total) with stirring (LiAlH₄ reacts violently with water — fume cupboard, dry apparatus). Stir at RT for two hours, quench drop-wise with cold dilute H₂SO₄ (effervescence of H₂), basify with NaOH to pH > 12, extract the free amine (propan-1-amine, CH₃CH₂CH₂NH₂) into ether, dry over Na₂SO₄, and distil. Confirm by IR: a primary amine shows two N–H stretches at ~3300–3500 cm⁻¹ (secondary: one; tertiary: none).
Amines act as (i) Brønsted bases (accept a proton) and (ii) nucleophiles (attack electrophilic carbon — acyl chlorides, carbonyls, halogenoalkanes).
With strong mineral acids, amines form ionic ammonium salts:
CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻ (methylammonium chloride) C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻ (anilinium chloride)
The salts are crystalline solids and highly water-soluble (free amines are often volatile and sparingly soluble). Many pharmaceuticals are formulated as the amine hydrochloride for solubility and shelf-life. NaOH liberates the free amine — the standard "amine work-up".
The most synthetically useful amine reaction. A primary or secondary amine reacts with an acyl chloride to give an amide + HCl:
R–NH₂ + R'–COCl → R'–CO–NH–R + HCl
Exam-board example: CH₃NH₂ + CH₃COCl → CH₃CONHCH₃ + HCl (methylamine + ethanoyl chloride → N-methylethanamide).
Mechanism (addition–elimination): the N lone pair attacks the δ⁺ carbonyl C, breaking the π-bond and forming a tetrahedral intermediate with O⁻ and N⁺H₂R; Cl⁻ is expelled, regenerating C=O; a second amine deprotonates to give the neutral amide. Two equivalents of amine are needed (or a separate base such as pyridine). Vigorous and exothermic at RT — same mechanism developed in lesson 2.
Secondary amines work equally well (N,N-disubstituted amide); tertiary amines have no N–H and cannot form an amide.
R–NH₂ + R'–CHO ⇌ R'–CH=N–R + H₂O
The N lone pair attacks the C=O carbon and water is eliminated to give a C=N imine ("Schiff base"). Mechanistically the same addition-then-dehydration as the 2,4-DNP test (lesson 0) — 2,4-DNP is a substituted hydrazine acting as a primary amine. Imines are intermediates in reductive amination (imine + NaBH₃CN → secondary amine), the modern workhorse for secondary-amine synthesis. Not examined at AQA — signposted for continuity with lesson 0.
Same S_N2 mechanism as route 3 above: the N lone pair attacks the δ⁺ carbon of C–X, displacing halide.
R–NH₂ + R'–X → R–NH–R' + HX
This is the synthetic route to N,N-disubstituted amines — provided the reaction is stopped before further alkylation.
Aniline differs from aliphatic amines in two ways: (i) it is a much weaker base (pKa(BH⁺) = 4.6); (ii) the –NH₂ group is a strongly activating ortho/para director.
Aniline reacts with bromine water at room temperature, with no FeBr₃ catalyst, to give 2,4,6-tribromoaniline as a white precipitate:
C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr
The N lone pair donates into the ring through π-overlap, activating it ~10⁶× relative to benzene; the reaction goes straight to the trisubstituted product. Benzene (lesson 4) needs FeBr₃ and stops at monobromination. Aniline is so reactive that direct Friedel-Crafts fails — AlCl₃ complexes with the lone pair — so the amine is first protected as the amide.
Aniline + cold (< 10 °C) nitrous acid (from NaNO₂ + dilute HCl) gives a diazonium salt:
C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O
Aromatic diazonium salts decompose above ~10 °C, evolving N₂. They couple with electron-rich aromatics (phenol, N,N-dimethylaniline) under mild alkali to give azo dyes (Ar–N=N–Ar'). The extended π-system absorbs visible light: methyl orange, Congo red, and Sudan dyes derive from this chemistry — the foundation of the synthetic-dye industry since Perkin (1856).
Question 1. [13 marks total]
(a) Classify each of the following amines as primary, secondary, or tertiary. [2 marks] (i) (CH₃)₂NCH₂CH₃ (ii) C₆H₅CH₂NH₂
(b) The values of pKa for the conjugate acids of three nitrogen bases are: NH₃ (NH₄⁺, pKa = 9.25), methylamine (CH₃NH₃⁺, pKa = 10.6), and aniline (C₆H₅NH₃⁺, pKa = 4.6).
Explain, in terms of the availability of the nitrogen lone pair and the stability of the conjugate acid, why methylamine is more basic than ammonia, and why aniline is much less basic than ammonia. [5 marks]
(c) Write a balanced equation for the reaction of methylamine with ethanoyl chloride (CH₃COCl), name the organic product, and identify the role of methylamine in the mechanism. [3 marks]
(d) State the reagents and conditions for the two-step preparation of aniline (phenylamine, C₆H₅NH₂) from benzene. [3 marks]
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.