How Aniline Becomes the Colour in Your Clothes

The lone pair on nitrogen lets amines accept protons, react with acids and attack electrophiles. Aniline, the simplest aryl amine, can also be turned into diazonium salts — reactive intermediates behind the bright colours of many dyes.

This part covers the basic strength of amines, their reactions and the Hinsberg test, substitution in aniline, and diazonium salts.

Amines are bases because nitrogen's lone pair accepts a proton; electron-releasing alkyl groups increase basicity, so in the gas phase tertiary > secondary > primary > ammonia, but in water solvation and steric effects change that order, and aryl amines are much weaker bases because their lone pair is delocalised into the ring.

Measuring basicity: a smaller p$K_b$ means a stronger base.

Gas phase — only the inductive effect matters:

$$({\text{CH}}_3{)}_3\text{N}>({\text{CH}}_3{)}_2\text{NH}>{\text{CH}}_3{\text{NH}}_2>{\text{NH}}_3$$

In water, three factors combine:

- Inductive effect — more alkyl groups push more electron density onto nitrogen
- Solvation — ammonium ions with more N–H bonds are stabilised better by hydrogen bonding with water
- Steric hindrance — bulky groups crowd the nitrogen

For methyl amines: (CH${}_3$)${}_2$NH > CH${}_3$NH${}_2$ > (CH${}_3$)${}_3$N > NH${}_3$.

Aryl amines. Aniline's lone pair is shared with the ring, so aniline (p$K_b=9.38$) is a far weaker base than methylamine (p$K_b=3.38$).

Worked example. The ratio of base strengths is

$$\frac{K_b({\text{CH}}_3{\text{NH}}_2)}{K_b({\text{C}}_6{\text{H}}_5{\text{NH}}_2)}=10^{9.38-3.38}=10^6$$

so methylamine is a million times stronger as a base than aniline.

An everyday example. Many medicines are sold as amine hydrochloride salts, which dissolve in water far better than the free bases.

The substance. Tertiary amines are not the strongest bases in water despite having the most alkyl groups — their ions are poorly solvated.

Amines act as nucleophiles in alkylation and acylation, primary amines give foul-smelling isocyanides in the carbylamine test, nitrous acid treats aliphatic and aromatic primary amines differently, and the Hinsberg test separates the three classes by how their products with benzenesulphonyl chloride behave in alkali.

Key reactions:

- Alkylation — with alkyl halides, eventually giving quaternary salts
- Acylation — primary and secondary amines form amides with acid chlorides or anhydrides; aniline gives acetanilide
- Carbylamine reaction — primary amines, aliphatic or aromatic, with CHCl${}_3$ and alcoholic KOH give isocyanides with an offensive smell
- Nitrous acid — aliphatic primary amines give alcohols and N${}_2$ gas; aniline at $273$ to $278$ K gives benzenediazonium chloride

Hinsberg test with benzenesulphonyl chloride:

- Primary — the sulphonamide still has an acidic N–H, so it dissolves in alkali
- Secondary — the sulphonamide has no N–H and is insoluble in alkali
- Tertiary — no reaction

Worked example. An amine C${}_3$H${}_9$N does not react with benzenesulphonyl chloride and gives no carbylamine smell. It has no N–H at all, so it must be tertiary: trimethylamine, (CH${}_3$)${}_3$N.

An everyday example. Paracetamol contains an amide group formed by acylating an aryl amine, the same kind of reaction that makes acetanilide.

The substance. Acylation tames an amine's reactivity, which is why aniline is often acylated before further substitution.

**The –NH${}_2$ group strongly activates the ring towards ortho and para attack, so aniline brominates at once to 2,4,6-tribromoaniline; nitration and sulphonation are complicated because acid protonates –NH${}_2$; and Friedel-Crafts reactions fail because aniline binds the AlCl${}_3$ catalyst.

Bromination. Bromine water gives a white precipitate of 2,4,6-tribromoaniline** immediately. For a single substitution, convert aniline to acetanilide first, brominate to p-bromoacetanilide, then hydrolyse to p-bromoaniline.

Nitration. Direct nitration oxidises some aniline to tarry products, and strong acid turns much of it into the anilinium ion, which directs to the meta position. Protecting –NH${}_2$ as acetanilide gives mainly the para nitro product.

Sulphonation. Aniline with conc. H${}_2$SO${}_4$ forms anilinium hydrogensulphate, which on heating to $453$ to $473$ K gives sulphanilic acid.

Friedel-Crafts. Aniline, a Lewis base, donates its lone pair to AlCl${}_3$, placing a positive charge on nitrogen that deactivates the ring.

Worked example. Direct nitration gives para, meta and ortho nitroanilines in roughly $51:47:2$ parts. Almost half is meta, because much of the aniline reacts as the meta-directing anilinium ion.

An everyday example. Sulphanilic acid is used to make sulpha drugs and azo dyes.

The substance. **Protecting –NH${}_2$ as an amide is the key trick** — it moderates activation and prevents protonation, so single para products form.

**Benzenediazonium chloride is made by treating aniline with NaNO${}_2$ and HCl at $273$ to $278$ K; its –N${}_2^{+}$ group can be replaced by Cl, Br, CN, I, F, H, OH or NO${}_2$, and it couples with phenols and aryl amines to form brightly coloured azo compounds.

Diazotisation:**

$${\text{C}}_6{\text{H}}_5{\text{NH}}_2+{\text{NaNO}}_2+2\text{HCl}\to {\text{C}}_6{\text{H}}_5{\text{N}}_2^{+}{\text{Cl}}^{-}+\text{NaCl}+2{\text{H}}_2\text{O}$$

The salt is used at once because it decomposes on warming.

**Replacement with loss of N${}_2$:

- Sandmeyer — CuCl/HCl, CuBr/HBr or CuCN/KCN give chloro-, bromo- or cyanobenzene
- Gattermann — copper powder with HCl or HBr gives chloro- or bromobenzene
- KI gives iodobenzene; HBF${}_4$ then heat gives fluorobenzene
- H${}_3$PO${}_2$ or ethanol gives benzene; warm water gives phenol

Azo coupling. With phenol in alkali the salt gives orange p-hydroxyazobenzene; with aniline it gives yellow p-aminoazobenzene, the –N=N– group linking the rings at the para position.

Worked example.** Direct bromination of benzene cannot give 1,3,5-tribromobenzene. Instead, brominate aniline to 2,4,6-tribromoaniline, diazotise, and replace –N${}_2^{+}$ by H using H${}_3$PO${}_2$ — the bromines, once ortho and para to –NH${}_2$, end up meta to one another.

An everyday example. Many yellow, orange and red dyes on printed cotton fabrics are azo dyes made by diazonium coupling.

The substance. Diazonium salts place groups where direct substitution cannot, because –NH${}_2$ directs first and then disappears.