Why Aniline Is a Much Weaker Base Than Methylamine

Amines are derivatives of ammonia in which alkyl or aryl groups replace hydrogen atoms. They give spoiled fish its smell, form the backbone of many medicines and dyes, and act as bases because of the lone pair on nitrogen. How readily that lone pair is shared — and how many groups crowd the nitrogen — shapes their chemistry.

This lesson covers naming and preparing aliphatic and aromatic amines, the basicity of amines and what affects it, and chemical tests that distinguish primary, secondary and tertiary amines.

Amines are named as alkanamines and classified as primary, secondary or tertiary by how many carbon groups are bonded to nitrogen, and they are prepared by reducing nitro compounds, nitriles or amides, by ammonolysis of alkyl halides, and by the Gabriel and Hoffmann methods.

Naming and classes:

- Primary, $\mathrm{RN}{\mathrm{H}}_2$: $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ is methanamine, or methylamine
- Secondary, ${\mathrm{R}}_2\mathrm{NH}$: $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}$ is N-methylmethanamine
- Tertiary, ${\mathrm{R}}_3\mathrm{N}$: $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}$ is N,N-dimethylmethanamine
- ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$ is benzenamine, commonly called aniline

Methods of preparation:

- Reduction of nitro compounds with iron or tin and HCl, or catalytic hydrogen: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_2+6[\mathrm{H}]\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+2{\mathrm{H}}_2\mathrm{O}$
- Ammonolysis of alkyl halides — heating with ammonia gives a mixture of primary, secondary and tertiary amines and a quaternary salt, which is the method's drawback
- Reduction of nitriles with $\mathrm{LiAl}{\mathrm{H}}_4$: $\mathrm{RCN}\to \mathrm{RC}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$, adding one carbon
- Reduction of amides with $\mathrm{LiAl}{\mathrm{H}}_4$: $\mathrm{RCON}{\mathrm{H}}_2\to \mathrm{RC}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$, keeping the same carbons
- Gabriel phthalimide synthesis — gives pure primary aliphatic amines
- Hoffmann bromamide degradation — gives a primary amine with one carbon fewer: $\mathrm{RCON}{\mathrm{H}}_2+\mathrm{B}{\mathrm{r}}_2+4NaOH\to \mathrm{RN}{\mathrm{H}}_2+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+2NaBr+2{\mathrm{H}}_2\mathrm{O}$

Worked example — from propanamide:

- Hoffmann degradation gives ethanamine, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$, with one carbon fewer
- Reduction with $\mathrm{LiAl}{\mathrm{H}}_4$ gives propan-1-amine, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$, with the same number of carbons

An everyday example. The rubber in bicycle and scooter tyres is toughened using chemicals made from aniline, which industry produces by reducing nitrobenzene.

The substance. Gabriel synthesis cannot make aniline — aryl halides do not undergo the nucleophilic substitution with phthalimide that the method depends on.

Amines are basic because the lone pair on nitrogen can accept a proton; alkyl groups raise basicity by pushing electrons onto nitrogen, while in aromatic amines the lone pair is delocalised into the ring, so aniline is a much weaker base than aliphatic amines.

$$\mathrm{RN}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{RN}{\mathrm{H}}_3^{+}+\mathrm{O}{\mathrm{H}}^{-}{K}_b=\frac{[\mathrm{RN}{\mathrm{H}}_3^{+}][\mathrm{O}{\mathrm{H}}^{-}]}{[\mathrm{RN}{\mathrm{H}}_2]}$$

- A smaller $\mathrm{p}{K}_b$ means a stronger base
- Approximate $\mathrm{p}{K}_b$ values: methanamine 3.38, ammonia 4.75, aniline 9.38

Factors that decide basicity:

- Inductive effect — alkyl groups push electrons onto nitrogen, so aliphatic amines are stronger bases than ammonia
- Solvation — in water, the ammonium ion formed is stabilised by hydrogen bonding, and more N-H hydrogens mean better solvation
- Steric hindrance — bulky groups crowd the nitrogen and hinder both protonation and solvation
- Because these effects compete, the order in water is $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}>\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2>(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}>\mathrm{N}{\mathrm{H}}_3$ for methyl amines, but $({\mathrm{C}}_2{\mathrm{H}}_5{)}_2\mathrm{NH}>({\mathrm{C}}_2{\mathrm{H}}_5{)}_3\mathrm{N}>{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2>\mathrm{N}{\mathrm{H}}_3$ for ethyl amines
- Resonance in aryl amines — aniline's lone pair is spread into the benzene ring; electron-withdrawing groups such as $-\mathrm{N}{\mathrm{O}}_2$ weaken it further, while $-\mathrm{C}{\mathrm{H}}_3$ and $-\mathrm{OC}{\mathrm{H}}_3$ strengthen it

Worked example. Aniline's $\mathrm{p}{K}_b$ is $9.38-3.38=6.0$ units higher than methanamine's, so its $K_b$ is about $10^{6.0}$, or a million, times smaller.

An everyday example. Squeezing lemon over fried fish in coastal Kerala kitchens cuts the fishy smell, because citric acid turns volatile amines into non-volatile ammonium salts.

The substance. In the gas phase the order changes — with no water to solvate the ions, tertiary amines become the strongest bases, since only the inductive effect matters.

The Hinsberg test separates the three classes by how they react with benzenesulphonyl chloride, the carbylamine test picks out primary amines by a foul-smelling isocyanide, and nitrous acid gives a different result with each class.

Hinsberg test — benzenesulphonyl chloride, ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{S}{\mathrm{O}}_2\mathrm{Cl}$, then alkali:

- Primary amines give a sulphonamide that dissolves in alkali, because its remaining N-H hydrogen is acidic
- Secondary amines give a sulphonamide that is insoluble in alkali, because it has no N-H hydrogen
- Tertiary amines do not react

Carbylamine test — heating with chloroform and alcoholic KOH:

$$\mathrm{RN}{\mathrm{H}}_2+\mathrm{CHC}{\mathrm{l}}_3+3KOH\to \mathrm{RNC}+3KCl+3{\mathrm{H}}_2\mathrm{O}$$

Only primary amines, aliphatic or aromatic, give an isocyanide with a very unpleasant smell.

Nitrous acid, made from sodium nitrite and HCl in the cold:

- Primary aliphatic amines release nitrogen gas briskly and form alcohols
- Primary aromatic amines at 273-278 K form diazonium salts
- Secondary amines give yellow, oily N-nitrosamines
- Tertiary aliphatic amines form soluble salts with no visible change

**Worked example — three isomers of ${\mathrm{C}}_3{\mathrm{H}}_9\mathrm{N}$:

- Propan-1-amine: sulphonamide soluble in alkali and a positive carbylamine test — primary
- N-Methylethanamine: sulphonamide insoluble in alkali and a negative carbylamine test — secondary
- N,N-Dimethylmethanamine: no reaction with Hinsberg's reagent — tertiary

An everyday example. Sulphonamide antibacterial medicines prescribed in India are made by the same reaction behind the Hinsberg test: a sulphonyl chloride joining to an amine.

The substance. The carbylamine test works only for amines with two N-H hydrogens to lose** — secondary amines have one and tertiary amines none.