Amines and Diazonium Salts
Nitrogen's lone pair at work — the basic, nucleophilic amines, and the diazonium salt that turns aniline into a gateway to almost any substituted benzene
- How amines are classified as 1°, 2° and 3°, and how aniline differs from alkyl amines.
- Preparation by reduction, ammonolysis, the Gabriel synthesis and Hofmann degradation.
- Why aliphatic amines are more basic than ammonia but aniline is less basic.
- The carbylamine and Hinsberg tests, and reactions with nitrous acid.
- Diazotisation and why aryl diazonium salts are stable.
- The Sandmeyer replacement reactions and the azo coupling that gives dyes.
Classifying Amines
Amines are derivatives of ammonia in which one, two or three hydrogens are replaced by alkyl or aryl groups — giving primary (\(\ce{RNH2}\)), secondary (\(\ce{R2NH}\)) and tertiary (\(\ce{R3N}\)) amines. The nitrogen keeps a lone pair, and that lone pair is the source of nearly all amine chemistry: it makes amines both basic and nucleophilic.
| Class | Structure | Example |
|---|---|---|
| Primary (1°) | \(\ce{R-NH2}\) | ethylamine, aniline |
| Secondary (2°) | \(\ce{R2NH}\) | dimethylamine |
| Tertiary (3°) | \(\ce{R3N}\) | trimethylamine |
Preparation of Amines
Amines are made by reducing nitrogen-containing groups, by alkylating ammonia, or by two specialised routes that give clean products: the Gabriel synthesis (pure \(1^\circ\) aliphatic amines) and the Hofmann bromamide degradation (a \(1^\circ\) amine with one fewer carbon).
| Method | Reaction | Note |
|---|---|---|
| Reduction of nitro | \(\ce{C6H5NO2 + 6[H] ->[Sn/HCl] C6H5NH2}\) | aniline from nitrobenzene |
| Reduction of nitrile | \(\ce{RCN + 4[H] ->[Ni] RCH2NH2}\) | gains one carbon |
| Ammonolysis | \(\ce{R-X + NH3 -> RNH2 + HX}\) | gives a mixture (1°/2°/3°) |
| Gabriel synthesis | phthalimide route | pure 1° aliphatic only |
| Hofmann degradation | \(\ce{RCONH2 + Br2 + 4KOH -> RNH2 + ...}\) | loses one carbon |
Basicity of Amines
An amine is basic because its lone pair can accept a proton. Anything that makes that lone pair more available increases basicity; anything that ties it up decreases it. This single principle explains why aliphatic amines beat ammonia while aniline trails it.
Alkyl groups donate electrons (\(+I\)), making aliphatic amines stronger bases than ammonia. In aniline the lone pair is delocalised into the ring, so aniline is a weaker base. In solution the gas-phase order (\(3^\circ>2^\circ>1^\circ\)) is upset by solvation and steric crowding around the bulky \(3^\circ\) nitrogen.
Reactions of Amines
The lone pair drives the chemistry: amines form salts with acids, are acylated to amides, and react with nitrous acid in a way that depends sharply on class.
| Reaction | 1° amine | 2° amine | 3° amine |
|---|---|---|---|
| With acid chloride | amide | amide | no reaction |
| With \(\ce{HNO2}\) (aliphatic) | alcohol + \(\ce{N2}\) | N-nitrosamine | salt (no \(\ce{N2}\)) |
| With \(\ce{HNO2}\) (aromatic) | diazonium salt (0–5 °C) | N-nitrosamine | ring nitrosation |
Distinguishing 1°, 2° and 3° Amines
The Hinsberg test sorts all three classes in one experiment, using benzenesulphonyl chloride and then alkali.
A \(1^\circ\) amine's sulphonamide still has an acidic \(\ce{N-H}\), so it dissolves in alkali; a \(2^\circ\) amine's has none and stays as an insoluble solid; a \(3^\circ\) amine has no \(\ce{N-H}\) to react at all. Together with the carbylamine test (specific to \(1^\circ\)), the three classes are easily told apart.
Electrophilic Substitution of Aniline
In aniline the \(\ce{-NH2}\) group is strongly activating and ortho/para-directing — so much so that reactions can run out of control or be spoiled by the basic nitrogen.
| Reaction | Outcome |
|---|---|
| Bromine water | 2,4,6-tribromoaniline (white ppt) — instant tri-substitution |
| Nitration (direct) | problematic — \(\ce{-NH2}\) is protonated/oxidised |
| Nitration (after acetylation) | mainly \(p\)-nitroaniline, then hydrolyse back |
Diazonium Salts: Preparation
Treating a primary aromatic amine with nitrous acid at low temperature gives a diazonium salt (\(\ce{ArN2+ X-}\)) — the diazotisation reaction. The aryl diazonium ion is stabilised by resonance with the ring, so it survives at \(0\!-\!5\,^\circ\text{C}\); the aliphatic version is too unstable to isolate.
The low temperature is essential — above about \(5\,^\circ\text{C}\) the salt decomposes to phenol and nitrogen. Its stability over the alkyl analogue comes from delocalisation of the positive charge into the benzene ring.
Reactions of Diazonium Salts
The diazonium salt is one of organic chemistry's most useful intermediates because the \(\ce{-N2+}\) group is an excellent leaving group. Its reactions fall into two camps: replacement (nitrogen leaves) and coupling (nitrogen stays, forming a dye).
| Reaction | Reagent | Product |
|---|---|---|
| Sandmeyer | \(\ce{CuCl}/\ce{CuBr}/\ce{CuCN}\) | \(\ce{ArCl},\ \ce{ArBr},\ \ce{ArCN}\) |
| Balz–Schiemann | \(\ce{HBF4}\), then heat | \(\ce{ArF}\) |
| With \(\ce{KI}\) | \(\ce{KI}\) | \(\ce{ArI}\) |
| Hydrolysis | warm \(\ce{H2O}\) | \(\ce{ArOH}\) (phenol) |
| With \(\ce{H3PO2}\) | hypophosphorous acid | \(\ce{ArH}\) (removes the group) |
| Azo coupling | phenol / aniline | azo dye (\(\ce{-N=N-}\)) |
Putting It to Work
Problem. How would you prepare a pure primary aliphatic amine free of 2° and 3° products?
Solution. Ammonolysis gives a mixture; use the clean route:
Problem. Arrange \(\ce{NH3},\ \ce{CH3NH2},\ \ce{C6H5NH2}\) by increasing basicity, and explain.
Solution. Alkyl \(+I\) raises basicity; ring resonance lowers it:
Problem. An amine heated with \(\ce{CHCl3}\) and alc. \(\ce{KOH}\) gives a foul-smelling product. What class is it?
Solution. The carbylamine test is positive only for primary amines:
Problem. At what temperature is aniline diazotised, and why must it be kept there?
Solution. The salt decomposes to phenol above ~5 °C:
Problem. What forms when benzenediazonium chloride is treated with \(\ce{CuCN}\)?
Solution. Sandmeyer replaces \(\ce{-N2+}\) by \(\ce{-CN}\):
Problem. What is formed when benzenediazonium chloride couples with phenol, and what type of compound is it?
Solution. Coupling keeps the nitrogen, forming an azo linkage:
Chapter Summary
1°/2°/3° amines; nitrogen's lone pair makes them basic and nucleophilic.
Reduction, ammonolysis (mixture), Gabriel (pure 1°), Hofmann (one C less).
Aliphatic > \(\ce{NH3}\) > aniline; aniline's lone pair is tied up by the ring.
Carbylamine (1° only); Hinsberg sorts 1°/2°/3°; \(\ce{HNO2}\) varies by class.
\(\ce{ArNH2 + HNO2}\) at 0–5 °C → stable aryl diazonium salt.
Replacement (Sandmeyer etc.) installs many groups; coupling gives azo dyes.
Problems
For each item, first decide whether it concerns amine structure/basicity, an identification test, or diazonium chemistry — then apply the relevant idea. Difficulty rises down the list.
- Classify amines and explain why the lone pair makes them basic.
- How is aniline prepared from nitrobenzene? Write the reaction.
- Why does the Gabriel synthesis fail for aromatic amines?
- Explain why methylamine is a stronger base than ammonia, but aniline is weaker.
- Describe the carbylamine test and state which amines respond.
- Outline the Hinsberg test and how it distinguishes 1°, 2° and 3° amines.
- Give the products of 1° aliphatic and 1° aromatic amines with nitrous acid.
- Why must aniline be acetylated before nitration?
- Write the diazotisation of aniline and explain the temperature requirement.
- Write the Sandmeyer reactions that convert benzenediazonium chloride to chlorobenzene and benzonitrile.
- How would you convert aniline to fluorobenzene?
- Write the coupling of benzenediazonium chloride with phenol and name the product class.