HookAn 18-year-old's failed experiment dyed the world purple
Over the Easter holiday of 1856, an eighteen-year-old chemistry student named William Perkin was trying to synthesise quinine — the only antimalarial the British Empire had — in a makeshift laboratory at the top of his parents' house in east London. The attempt failed completely. Oxidising aniline gave him a flask of black sludge, but when he washed it with ethanol the solution turned a violent, beautiful purple that dyed silk and refused to fade in light or washing. He patented it before his nineteenth birthday, opened a factory, and 'mauveine' — the first synthetic dye — made him rich enough to retire at thirty-six. Queen Victoria wore mauve; Punch diagnosed the country with 'mauve measles'; and the German firms that industrialised his idea grew into the modern chemical industry.
The molecule under all of it, aniline — phenylamine to AQA — is an amine: ammonia with one or more hydrogens swapped for carbon. Everything an amine does traces back to the lone pair on its nitrogen. Point that lone pair at a proton and the amine is a Brønsted-Lowry base — which is why lemon juice kills the smell of fish, protonating volatile trimethylamine into an odourless, non-volatile salt. Point it at a δ+ carbon and the amine is a nucleophile, the role that builds fabric softeners and paracetamol. This section covers how amines are made (two aliphatic routes — one messy, one clean — and a reduction route to phenylamine), how basic they are (with real \(K_b\) numbers), and what that lone pair does next.
MechanismTwo routes to an aliphatic amine — the messy one and the clean one
First, a classification rule that trips students because it is the opposite of the alcohols rule: amines are classified at nitrogen. Count the carbons bonded directly to N — one is a primary amine, two secondary, three tertiary. So 2-aminopropane is a primary amine even though its \(-\text{NH}_2\) sits on a secondary carbon, while propan-2-ol is a secondary alcohol.
Route one: warm a halogenoalkane with an excess of ammonia dissolved in ethanol. It is nucleophilic substitution: \(\text{CH}_3\text{CH}_2\text{Br} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Br}\). Two ammonia molecules do two different jobs — the first attacks the δ+ carbon and displaces the bromide; the second acts as a base and removes a proton from the \(\text{CH}_3\text{CH}_2\text{NH}_3^+\) intermediate. The problem is that the product is a better nucleophile than ammonia (its alkyl group pushes electron density onto the nitrogen), so ethylamine attacks the remaining bromoethane to give diethylamine, which reacts again to give triethylamine, which reacts one final time to give the quaternary salt tetraethylammonium bromide. One flask, four nitrogen products, and a separation problem. A large excess of ammonia biases the outcome towards the primary amine; a large excess of halogenoalkane runs it all the way to the quaternary salt.
Route two avoids the mess entirely: make a nitrile and reduce it. \(\text{CH}_3\text{CH}_2\text{CN} + 2\text{H}_2 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\), using hydrogen with a nickel catalyst (LiAlH₄ in dry ether does the same reduction in the lab). The nitrogen ends up carrying two hydrogens and no spare electrophile to attack, so the product is a clean primary amine. The price: an extra step — and an extra carbon, because if the nitrile was made from a halogenoalkane with ethanolic KCN, the CN carbon has joined the chain.
Design a two-step synthesis of butylamine, \(\text{CH}_3(\text{CH}_2)_3\text{NH}_2\), starting from 1-bromopropane.
Step 1 — count carbons before anything else. The target has four; the starting material has three. You need to add one, and the only chain-extending move in this toolkit is the cyanide route.
Step 2 — substitute: warm 1-bromopropane with potassium cyanide in ethanol. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{KBr}\) — butanenitrile, now four carbons.
Step 3 — reduce: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + 2\text{H}_2 \rightarrow \text{CH}_3(\text{CH}_2)_3\text{NH}_2\) with a nickel catalyst. Pure primary butylamine, no mixture.
The contrast that shows understanding: starting from 1-bromobutane (already four carbons), you would instead use excess ethanolic ammonia in one step and accept the mixture. The carbon count chooses the route — check it first, every time.
MechanismPhenylamine — you cannot bolt an NH₂ straight onto the ring
Aromatic amines cannot be made the aliphatic way. Ammonia is a nucleophile, and benzene's electron-rich delocalised ring repels nucleophiles — there is no direct substitution to be had. The route in is the one Perkin's era worked out at industrial scale: put a nitrogen on the ring as \(-\text{NO}_2\) by nitration (section 3.3.10), then reduce it.
The AQA reducing system is tin and concentrated hydrochloric acid, heated under reflux: \(\text{C}_6\text{H}_5\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O}\), where \([\text{H}]\) is the accepted shorthand for the reducing agent. There is a sting in the conditions: because phenylamine is a base and the flask is full of acid, the product you actually hold at the end of reflux is the phenylammonium salt, not the amine. The final step — adding an excess of sodium hydroxide to deprotonate the salt and liberate free phenylamine — is the mark students forget more than any other in this topic.
Industry runs the same reduction catalytically with hydrogen gas. And while Perkin's dyes are the romantic application, they are no longer the main one: most of the world's phenylamine now feeds the manufacture of polyurethanes — the foam in mattresses, car seats and building insulation. The eighteen-year-old's failed quinine experiment turned into a molecule the modern world sleeps on.
DataBase strength is lone-pair availability — and the numbers settle it
An amine is a Brønsted-Lowry base because its nitrogen lone pair can accept a proton: \(\text{CH}_3\text{CH}_2\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-\), forming the ionic salt ethylammonium chloride — soluble, crystalline and odourless, which is exactly why acid kills amine smells. Add sodium hydroxide and the reaction reverses, releasing the free amine again.
How strong a base depends on one thing AQA words very precisely: the availability of the lone pair. The data draw a three-rung ladder at 25 °C. Ethylamine has \(K_b = 5.6 \times 10^{-4}\ \text{mol dm}^{-3}\); ammonia has \(K_b = 1.8 \times 10^{-5}\); phenylamine limps in at about \(4.2 \times 10^{-10}\). An aliphatic amine is roughly thirty times more basic than ammonia; phenylamine is tens of thousands of times less basic than ammonia.
The explanations are mirror images. In ethylamine, the alkyl group has a positive inductive effect — it pushes electron density towards nitrogen, so the lone pair is denser, more available, and better at holding the proton once captured. In phenylamine, the lone pair overlaps with the ring's π system and is partially delocalised into the ring: less available, so a much weaker base. One structural change, six orders of magnitude in \(K_b\).
Arrange ammonia, butylamine and phenylamine in order of increasing base strength, using the data, and justify the order.
Step 1 — the order from \(K_b\): phenylamine (\(\approx 4.2 \times 10^{-10}\)) < ammonia (\(1.8 \times 10^{-5}\)) < butylamine (\(\approx 4 \times 10^{-4}\), typical of primary aliphatic amines).
Step 2 — the scale of it: \(\dfrac{4 \times 10^{-4}}{4.2 \times 10^{-10}} \approx 10^{6}\) — a million-fold spread in basicity across one functional group.
Step 3 — the two-sentence justification that earns full marks: the butyl group's positive inductive effect pushes electron density onto nitrogen, making its lone pair the most available for accepting a proton; in phenylamine the lone pair is partially delocalised into the aromatic ring, making it the least available. Write 'available' — not 'more reactive', not 'more electrons' — because availability of the lone pair is the phrase the mark scheme is built on.
MechanismThe same lone pair, aimed at carbon — quats and paracetamol
Swap the target from a proton to a δ+ carbon and the base becomes a nucleophile. With halogenoalkanes, amines run the same substitution chemistry as ammonia — which is precisely why route one in this section gives mixtures. Each substitution puts another alkyl group on nitrogen: primary to secondary to tertiary, and finally to the quaternary ammonium salt, where nitrogen holds four carbon groups and a permanent positive charge. No lone pair remains, so the chemistry stops there.
Quaternary ammonium salts with one or two long hydrocarbon chains are cationic surfactants, and they are the reason this reaction appears on shopping receipts. Wet fabric and wet hair carry slight negative surface charges; the positive nitrogen head binds to them, the hydrocarbon tails face outward, and the surface turns soft, smooth and anti-static. That is fabric conditioner and hair conditioner in one sentence — and benzalkonium chloride, the antibacterial in wipes and eye drops, is the same idea again.
Amines also attack acyl chlorides and acid anhydrides — the nucleophilic addition-elimination chemistry of section 3.3.9 — to give N-substituted amides. Two moles of amine are needed per mole of acyl chloride because the second mops up the HCl: \(\text{CH}_3\text{COCl} + 2\text{CH}_3\text{NH}_2 \rightarrow \text{CH}_3\text{CONHCH}_3 + \text{CH}_3\text{NH}_3^+\text{Cl}^-\), the organic product being N-methylethanamide. The flagship example is paracetamol: acylate 4-aminophenol with ethanoic anhydride and the nitrogen is converted to an amide, giving N-(4-hydroxyphenyl)ethanamide — paracetamol — with the anhydride chosen over the acyl chloride for the same reasons as in aspirin manufacture: cheaper, safer to handle, and no corrosive HCl by-product.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Conditions are where this topic pays or punishes. Write 'excess ammonia, dissolved in ethanol' for the direct route; 'KCN in ethanol, then hydrogen with a nickel catalyst' for the nitrile route; and 'tin and concentrated hydrochloric acid, heat under reflux, then add excess NaOH' for phenylamine. That final NaOH step — liberating the amine from its salt — is the single most commonly dropped mark in 3.3.11. The \(6[\text{H}]\) shorthand is accepted for the reduction; balance it with \(2\text{H}_2\text{O}\).
Base-strength answers live or die on one word: availability. State the order, quote \(K_b\) values if given, then explain with the two mechanisms — alkyl groups push electron density onto nitrogen (more available lone pair), the phenyl ring delocalises the lone pair into itself (less available). 'Phenylamine has fewer electrons' scores nothing.
In mechanisms, show two molecules of ammonia or amine: one attacking the δ+ carbon, the second removing the proton — examiners specifically look for the deprotonation step. With acyl chlorides, name the mechanism nucleophilic addition-elimination, use the 2:1 amine ratio, and name products properly: N-methylethanamide, with the N- locant, not 'methyl ethanamide'.
Finally, applications are cheap marks AQA recycles: quaternary ammonium salts as cationic surfactants in fabric and hair conditioners, and the low yield of the ammonia route explained by further substitution. If a question asks why the yield of primary amine is poor, the answer is the product's own nucleophilicity — write it as a mechanism story, not a shrug.