The chemistry of amines flows from one feature: the lone pair on nitrogen. It makes amines basic (a proton acceptor), nucleophilic (attacks electrophilic carbon and acyl groups), and, in aniline, it feeds electron density into the ring.
Basic Character
An amine accepts a proton to form a substituted ammonium ion. Basic strength is measured by $K_b$ (or $pK_b$); a larger $K_b$ / smaller $pK_b$ means a stronger base. Three factors decide it:
- +I (inductive) effect of alkyl groups pushes electron density onto N, increasing availability of the lone pair → aliphatic amines are stronger bases than ammonia.
- Solvation / stabilisation of the conjugate acid by water (H-bonding to the ammonium ion). More N–H bonds give better solvation.
- Steric hindrance: bulky groups around N hinder both protonation and solvation.
Gas phase vs aqueous order
In the gas phase (no solvent) only the inductive effect operates, so basicity rises steadily: 3° > 2° > 1° > NH3. In aqueous solution, solvation and steric effects compete with +I, giving the irregular experimental order:
(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (ethyl), and (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (methyl).
Aromatic amines are weak bases
In aniline, the nitrogen lone pair is delocalised into the benzene ring (resonance). The lone pair is less available, so aniline ($pK_b \approx 9.4$) is a far weaker base than ammonia ($pK_b = 4.75$) or aliphatic amines. Electron-withdrawing groups (–NO2) further weaken it; electron-releasing groups (–CH3, –OCH3) strengthen it. Order: aliphatic amine > NH3 > aniline > nitroaniline.
Key Reactions
Alkylation: amines react with R–X to climb the series 1° → 2° → 3° → quaternary salt.
Acylation: 1° and 2° amines react with acyl chlorides/anhydrides to give amides (e.g. aniline + (CH3CO)2O → acetanilide). 3° amines have no N–H, so they do not acylate.
Carbylamine (isocyanide) test: a 1° amine + CHCl3 + alc. KOH gives a foul-smelling isocyanide (R–NC). Only 1° amines respond, so it distinguishes 1° from 2°/3°.
Reaction with nitrous acid (HNO2, from NaNO2 + HCl): 1° aliphatic amines give an unstable diazonium salt that decomposes evolving N2 (giving alcohols); 1° aromatic amines give a stable diazonium salt at 0–5°C; 2° amines give yellow N-nitrosamines; 3° amines simply form soluble salts. This is a powerful diagnostic.
Hinsberg test (benzenesulphonyl chloride, C6H5SO2Cl): 1° amines give an N-sulphonamide soluble in KOH (the N–H is acidic); 2° amines give a sulphonamide insoluble in KOH (no N–H left); 3° amines do not react. This distinguishes all three classes.
Electrophilic substitution in aniline
The –NH2 group is strongly activating and ortho/para-directing. Bromination of aniline in water gives 2,4,6-tribromoaniline (white precipitate) directly — it is hard to stop at mono-substitution. To get p-bromoaniline, the amine is first protected by acetylation (the bulky, less-activating acetamido group), then brominated, then hydrolysed. Nitration is done after acetylation to avoid oxidation by HNO3. Aniline does not undergo Friedel–Crafts reactions because the Lewis acid AlCl3 binds the basic N.
Arrange the following in increasing order of basic strength in aqueous solution: C2H5NH2, (C2H5)2NH, (C2H5)3N, NH3.
Solution- In water, basicity is set by a balance of +I effect (raises it), solvation of the cation (raises it), and steric hindrance (lowers it).
- NH3 has no alkyl groups, so it is the weakest.
- For ethyl amines the experimental order is (C2H5)2NH > (C2H5)3N > C2H5NH2: the 2° amine has the best compromise of +I and solvation; the 3° amine suffers steric/solvation loss.
- So increasing order: NH3 < C2H5NH2 < (C2H5)3N < (C2H5)2NH.
Answer: NH3 < C2H5NH2 < (C2H5)3N < (C2H5)2NH.
Why is aniline a weaker base than methylamine? Support with the role of resonance.
Solution- Basicity depends on availability of the nitrogen lone pair to accept a proton.
- In methylamine the +I effect of CH3 pushes electron density onto N, making the lone pair more available.
- In aniline the lone pair is delocalised into the benzene ring through resonance, so it is less available for protonation.
- Also, the anilinium ion (after protonation) loses this resonance stabilisation, making protonation unfavourable.
Answer: Resonance delocalisation of the N lone pair into the ring makes it less available, so aniline ($pK_b \approx 9.4$) is a much weaker base than methylamine ($pK_b \approx 3.4$).
An organic compound A on warming with CHCl3 and alcoholic KOH gives a foul-smelling compound. Identify the class of A and write the reaction with ethanamine.
Solution- The foul smell on heating with chloroform and alcoholic KOH is the carbylamine (isocyanide) test, positive only for 1° amines.
- Therefore A is a primary amine (aliphatic 1° or aromatic 1°).
- The reaction: R–NH2 + CHCl3 + 3KOH → R–NC + 3KCl + 3H2O.
- For ethanamine: C2H5NH2 + CHCl3 + 3KOH → C2H5NC (ethyl isocyanide) + 3KCl + 3H2O.
Answer: A is a primary amine; ethanamine gives ethyl isocyanide (C2H5NC).
Describe how the Hinsberg test distinguishes 1°, 2° and 3° amines.
Solution- All are treated with benzenesulphonyl chloride (C6H5SO2Cl), then aqueous KOH.
- A 1° amine gives N-alkylbenzenesulphonamide with one N–H left; this N–H is acidic, so it dissolves in KOH (clear solution).
- A 2° amine gives N,N-dialkylbenzenesulphonamide with no N–H; it is insoluble in KOH (precipitate remains).
- A 3° amine has no N–H to react and gives no sulphonamide (no reaction).
Answer: 1° → product soluble in KOH; 2° → product insoluble in KOH; 3° → no reaction.
Why does bromination of aniline give 2,4,6-tribromoaniline, and how is p-bromoaniline obtained selectively?
Solution- –NH2 is a strong activating, o/p-directing group; it raises ring electron density so much that bromine water substitutes at all three available o/p positions at once.
- This gives 2,4,6-tribromoaniline (white ppt), and the reaction is hard to stop at mono-bromo.
- To get p-bromoaniline, first acetylate: aniline + (CH3CO)2O → acetanilide. The acetamido group is bulkier and less activating, so bromination now gives mainly p-bromoacetanilide.
- Acid (or base) hydrolysis then removes the acetyl group to give p-bromoaniline.
Answer: Strong activation gives the 2,4,6-tribromo product; protecting –NH2 by acetylation, then brominating and hydrolysing, gives p-bromoaniline.
Give the products when (i) a 1° aromatic amine, (ii) a 2° amine, and (iii) a 3° aliphatic amine each react with nitrous acid (NaNO2 + HCl) at low temperature.
Solution- Nitrous acid (HNO2) is generated in situ from NaNO2 and dilute HCl at 0–5°C.
- (i) 1° aromatic amine (e.g. aniline) → a stable arenediazonium salt, C6H5N2+Cl– (diazotisation).
- (ii) 2° amine (e.g. (CH3)2NH) → a yellow oily N-nitrosamine, (CH3)2N–NO.
- (iii) 3° aliphatic amine → only a water-soluble ammonium-type salt; no stable nitroso product.
Answer: (i) stable diazonium salt; (ii) N-nitrosamine; (iii) soluble salt only.