Amines and Diazonium Salts
Amines are the nitrogen analogues of alcohols, and almost everything about them follows from one feature: the lone pair on nitrogen. It makes an amine basic, makes it a nucleophile, activates a benzene ring it is attached to, and — as you saw with Friedel–Crafts — occasionally ruins a reaction by complexing with the catalyst. The chapter's hardest question is the basicity ordering, which looks like it should be a simple trend and is not, because gas-phase and aqueous-phase orders genuinely differ. The second half is diazonium chemistry, which matters out of proportion to its size because it is the gateway to almost every aromatic substituent you cannot install directly.
Preparing Amines Without Getting a Mixture
| Method | Product | Note |
|---|---|---|
| Ammonolysis of an alkyl halide | A mixture of 1°, 2°, 3° and the quaternary salt | Poor control — the product is itself a nucleophile |
| Gabriel phthalimide synthesis | Pure primary amine | Fails for aromatic amines — aryl halides do not do SN2 |
| Hofmann bromamide degradation | Primary amine with one carbon fewer | From an amide with Br₂ and NaOH |
| Reduction of a nitrile | Primary amine with one carbon more | LiAlH₄ or catalytic hydrogenation |
| Reduction of a nitro compound | Aromatic amine | Sn/HCl or Fe/HCl — the industrial route to aniline |
🧠 Count the carbons to choose the method
Three of these routes change the chain length, and that alone usually decides which one a question wants. Hofmann bromamide loses a carbon, since the carbonyl carbon of the amide is expelled as carbonate. Nitrile reduction gains a carbon, since the cyanide brought one in. Gabriel synthesis keeps the count the same and gives a clean primary amine. So if you are asked to make propylamine from butanamide, that is Hofmann; from ethyl bromide, that is via the nitrile; and if the question stresses "pure primary amine, same number of carbons", that is Gabriel. Reading the carbon count first turns a four-way guess into a single choice.Diazonium Salts — the Aromatic Junction Box
Treating aniline with NaNO₂ and HCl at 273–278 K gives benzenediazonium chloride. The low temperature is not a detail: above about 278 K the salt decomposes to phenol and nitrogen, so an answer that omits the ice bath has described a failed experiment. Aromatic diazonium salts are stabilised by resonance with the ring, which is exactly why aliphatic diazonium salts cannot be isolated at all.
| Reagent | Group installed | Name |
|---|---|---|
| CuCl/HCl or CuBr/HBr | −Cl or −Br | Sandmeyer |
| Cu powder with HX | −Cl or −Br | Gattermann |
| KI | −I | No catalyst needed |
| HBF₄ then heat | −F | Balz–Schiemann |
| Warm water | −OH | Hydrolysis to phenol |
| H₃PO₂ or ethanol | −H | Removes the group entirely |
| Phenol or aniline in alkali | −N=N−Ar | Azo coupling, giving a dye |
🎯 Why diazotisation is the only way to make some aromatic compounds
Consider making iodobenzene. Direct iodination of benzene barely works, because iodine is the least reactive halogen and the reaction is reversible. But aniline diazotised and treated with KI gives iodobenzene cleanly. The same logic covers fluorobenzene, which direct fluorination cannot give at all without destroying the ring. Even more usefully, the hydrogen replacement with hypophosphorous acid lets you use an amino group purely as a temporary director: install it, use it to steer a substituent to the position you want, then remove it altogether. That is why diazonium chemistry is worth far more marks than its page count suggests — it is the standard answer to "how would you synthesise this substitution pattern".Key takeaways
- Everything about amines follows from the nitrogen lone pair — basicity, nucleophilicity and ring activation.
- Gas phase: 3° > 2° > 1°. Aqueous: 2° > 1° > 3° > NH₃, because solvation and sterics oppose induction.
- Aniline is far weaker than ammonia because its lone pair is delocalised into the ring.
- Carbylamine identifies a primary amine; Hinsberg separates all three classes.
- Hofmann loses a carbon, nitrile reduction gains one, Gabriel keeps the count — read the count to pick the route.
- Ammonolysis gives a mixture, because the amine produced is itself a nucleophile that reacts again.
- Diazotisation needs 273–278 K; above that the salt decomposes to phenol and nitrogen.
- Diazonium salts install −I and −F, and can be removed with H₃PO₂ so an amino group works as a temporary director.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The correct order of basicity of methylamines in aqueous solution is:
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Answer: B — (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
In water three factors compete: inductive donation favours more alkyl groups, but solvation of the protonated ion needs N–H bonds and steric bulk hinders approach of both proton and water. The trimethyl compound has only one N–H left and three crowding groups, so it falls to third place. The first option is the gas-phase order, where solvation plays no part and induction alone decides.Aniline is a much weaker base than ammonia because:
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Answer: B — Its lone pair is delocalised into the benzene ring and so is less available
The nitrogen lone pair in aniline is conjugated with the aromatic pi system and spends much of its time spread over the ring, so it is far less available to accept a proton. Protonation also destroys that stabilising delocalisation, which raises the energy cost further. Aniline does have a lone pair — it uses it to activate the ring towards electrophiles, which is the flip side of the same delocalisation.Which method gives a pure primary amine with the same number of carbon atoms as the starting halide?
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Answer: B — Gabriel phthalimide synthesis
Gabriel synthesis protects the nitrogen inside a phthalimide ring so it can react only once, then releases a clean primary amine with the carbon count unchanged. Ammonolysis gives a mixture because the amine formed is itself nucleophilic and reacts again. Hofmann degradation loses a carbon and nitrile reduction gains one, so neither preserves the count — which is exactly how these four are usually told apart.Diazotisation of aniline must be carried out at 273–278 K because above this temperature:
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Answer: B — The diazonium salt decomposes to phenol and nitrogen
Benzenediazonium chloride is only marginally stable and above about 278 K it hydrolyses, releasing nitrogen gas and leaving phenol. Since the whole point of making the salt is to use it as a stepping stone to some other substituent, letting it decompose destroys the synthesis. This is why every diazotisation procedure specifies an ice bath, and why the temperature is part of the answer rather than an incidental practical detail.The best route to iodobenzene from benzene is:
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Answer: B — Nitration, reduction, diazotisation, then KI
Direct iodination is impractical because iodine is the least reactive halogen and the reaction is reversible, so the four-step route through the diazonium salt is standard: nitrate, reduce to aniline, diazotise in the cold, then treat with potassium iodide, which needs no copper catalyst. Halogen exchange fails because aryl halides do not undergo nucleophilic substitution, and Friedel–Crafts would attach a methyl group rather than iodine.The carbylamine test is positive for:
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Answer: A — Primary amines only
Chloroform with alcoholic KOH generates dichlorocarbene, which reacts with a primary amine to give a foul-smelling isocyanide; secondary and tertiary amines cannot form that product because they lack the two hydrogens on nitrogen that the mechanism needs. This makes the test a single-step identification of the primary class, complementing the Hinsberg test, which separates all three.Hofmann bromamide degradation of propanamide gives:
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Answer: B — Ethylamine
The reaction expels the carbonyl carbon as carbonate, so the amine produced has one carbon fewer than the amide — propanamide with three carbons gives ethylamine with two. Propylamine would keep the count and butylamine would gain one, which is what nitrile reduction does; recognising which way the carbon count moves is the quickest way to answer any question of this type.Treating a diazonium salt with hypophosphorous acid (H₃PO₂) replaces the diazonium group with:
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Answer: B — −H
Hypophosphorous acid reduces the diazonium group away and leaves a hydrogen in its place, removing the substituent entirely. That is unexpectedly useful in synthesis: an amino group can be installed purely to direct another substituent to a chosen position and then be deleted, which is how substitution patterns that no direct route can reach are built up. Warm water would give the phenol and CuCl the chloride instead.