Latest
  • Admissions openClass 11 Science, 2027‑28: JEE, NEET and MHT‑CET with junior college and hostel under one roofApply now
  • IMGSAT 2027Free scholarship and admission test for Class 10 students, every Saturday and Sunday at our Nashik campusRegister
  • Free Foundation 2026‑27Evening classes for Class 10 in Physics, Chemistry, Maths and Biology, taught by our IITian and doctor facultyJoin free

+91 70303 00666

Chemistry · Class 12 · Chapter 9

Amines

Amines are organic relatives of ammonia, and the lone pair on nitrogen explains nearly all of their chemistry: their basicity, their reactions with electrophiles and the strong activation of the ring in aniline. Diazonium salts, made from aniline, are the main tool for putting groups on a benzene ring that cannot be introduced directly.

In this chapter: structure and classification of amines, nomenclature, methods of preparation, physical properties, basic character and its order, alkylation, acylation, carbylamine reaction, reaction with nitrous acid and with benzenesulphonyl chloride (Hinsberg test), electrophilic substitution in aniline, and the preparation and reactions of diazonium salts.

Structure, classification and nomenclature

Replacing one, two or three hydrogen atoms of NH3 by alkyl or aryl groups gives primary (1°, RNH2), secondary (2°, R2NH) and tertiary (3°, R3N) amines. Nitrogen is sp3 hybridised; three orbitals form bonds and the fourth holds the lone pair, so amines are pyramidal. Because of the lone pair, the C-N-C angle is less than 109.5° (108° in trimethylamine).

AmineCommon nameIUPAC name
CH3NH2MethylamineMethanamine
(CH3)2NHDimethylamineN-Methylmethanamine
(CH3)3NTrimethylamineN,N-Dimethylmethanamine
C6H5NH2AnilineAniline or benzenamine
C6H5NHCH3N-MethylanilineN-Methylbenzenamine
H2N(CH2)6NH2HexamethylenediamineHexane-1,6-diamine

Preparation

  • Reduction of nitro compounds: H2 over Ni, Pd or Pt, or metal and acid (Sn + HCl, Fe + HCl). Iron scrap and HCl is preferred, because the FeCl2 formed is hydrolysed to release HCl, so only a little acid is needed to start the reaction.
  • Ammonolysis of alkyl halides: heating R-X with ethanolic NH3 in a sealed tube at 373 K. The primary amine formed can react further, so a mixture of 1°, 2° and 3° amines and the quaternary ammonium salt results; a large excess of NH3 favours the primary amine. The free amine is released from its salt by a strong base. Halide reactivity: RI > RBr > RCl.
  • Reduction of nitriles: LiAlH4 or catalytic hydrogenation gives a primary amine with one carbon more than the alkyl halide used to make the nitrile (a way of stepping up the carbon chain).
  • Reduction of amides: LiAlH4 followed by water gives an amine with the same number of carbons.
  • Gabriel phthalimide synthesis: phthalimide with ethanolic KOH gives the potassium salt, which is heated with an alkyl halide; alkaline hydrolysis releases a primary amine. It gives pure 1° aliphatic amines, but not aromatic amines, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
  • Hoffmann bromamide degradation: an amide with Br2 and aqueous or ethanolic NaOH gives a primary amine with one carbon less: RCONH2 + Br2 + 4NaOH → RNH2 + Na2CO3 + 2NaBr + 2H2O. The alkyl or aryl group migrates from the carbonyl carbon to nitrogen.

Physical properties

Lower aliphatic amines are gases with a fishy smell; aniline is a liquid, colourless when pure but it darkens on storage because air oxidises it. Lower aliphatic amines dissolve in water by hydrogen bonding; solubility falls as the hydrocarbon part grows. Among isomers, boiling points follow 1° > 2° > 3°: primary amines have two N-H hydrogens for intermolecular hydrogen bonding, secondary have one and tertiary none.

Basic character

The lone pair on nitrogen accepts a proton: RNH2 + H2O ⇌ RNH3+ + OH−. A larger Kb (smaller pKb) means a stronger base.

AminepKb (aqueous)AminepKb (aqueous)
CH3NH23.38C2H5NH23.29
(CH3)2NH3.27(C2H5)2NH3.00
(CH3)3N4.22(C2H5)3N3.25
NH34.75C6H5NH2 (aniline)9.38
  • Aliphatic amines are stronger bases than ammonia, because the +I effect of alkyl groups raises electron density on nitrogen and stabilises the substituted ammonium ion.
  • In the gas phase, only the inductive effect matters: 3° > 2° > 1° > NH3.
  • In water, the ammonium ion is also stabilised by hydrogen bonding with water (solvation), which is greatest for the 1° ion (three N-H) and least for the 3° ion; bulky groups also hinder solvation and approach. The combined result is (CH3)2NH > CH3NH2 > (CH3)3N > NH3, and (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3.
  • Aryl amines are much weaker bases than ammonia. In aniline the lone pair is delocalised into the ring, so it is less available; the anilinium ion has fewer resonance structures than aniline, so protonation costs stability. Electron-releasing groups (-OCH3, -CH3) on the ring increase basicity; electron-withdrawing groups (-NO2, -SO3H, -COOH, -X) decrease it.

Chemical reactions of amines

  • Salt formation: amines react with acids to form salts, e.g. anilinium chloride; the free amine is regenerated with NaOH.
  • Alkylation: with alkyl halides, 1° amines give 2° and 3° amines and finally quaternary ammonium salts.
  • Acylation: 1° and 2° amines react with acid chlorides, anhydrides or esters to give amides. Pyridine, a stronger base, removes the HCl formed. With benzoyl chloride the reaction is called benzoylation: CH3NH2 + C6H5COCl → C6H5CONHCH3 + HCl. Tertiary amines have no N-H and are not acylated.
  • Carbylamine reaction: aliphatic and aromatic primary amines only, heated with chloroform and ethanolic KOH, give isocyanides with a very unpleasant smell: R-NH2 + CHCl3 + 3KOH → R-NC + 3KCl + 3H2O. It is a test for 1° amines.
  • Nitrous acid (made in situ from NaNO2 and HCl): 1° aliphatic amines give unstable diazonium salts, which break down to alcohols with quantitative evolution of N2 (used to estimate amino acids and proteins). Aromatic 1° amines at 273 to 278 K give stable diazonium salts. Secondary and tertiary amines react in other ways.
  • Benzenesulphonyl chloride (Hinsberg's reagent): distinguishes 1°, 2° and 3° amines, as shown below. It is now often replaced by p-toluenesulphonyl chloride.
Hinsberg test for primary, secondary and tertiary amineswww.iitmedicoguide.comAmine + C₆H₅SO₂Cl(Hinsberg's reagent)1° amine, RNH₂2° amine, R₂NH3° amine, R₃NC₆H₅SO₂NHRN-alkylbenzene-sulphonamideN-H is acidicsoluble in alkaliC₆H₅SO₂NR₂N,N-dialkylbenzene-sulphonamideno H on nitrogeninsoluble in alkaliNo reactionno N-H to replacewww.iitmedicoguide.com
The strongly electron-withdrawing sulphonyl group makes the remaining N-H of the primary amine product acidic, so it dissolves in alkali. The secondary amine product has no N-H and stays insoluble.

Electrophilic substitution in aniline

The -NH2 group is strongly activating and ortho, para-directing.

  • Bromination: with bromine water at room temperature, aniline gives a white precipitate of 2,4,6-tribromoaniline at once. To get a monosubstituted product, the activating effect is reduced by acetylation (with acetic anhydride) to acetanilide; after bromination, hydrolysis of 4-bromoacetanilide gives 4-bromoaniline. The acetyl group pulls the nitrogen lone pair partly towards the carbonyl oxygen by resonance.
  • Nitration: direct nitration gives tarry oxidation products as well as nitroanilines. In strongly acidic medium much of the aniline is protonated to the anilinium ion, which is meta-directing, so a large amount of meta product forms (about 51% para, 47% meta, 2% ortho). Acetylation before nitration protects the group, and the para-nitro derivative is the major product.
  • Sulphonation: aniline with concentrated H2SO4 forms anilinium hydrogensulphate, which on heating at 453 to 473 K gives sulphanilic acid (4-aminobenzenesulphonic acid), existing mainly as a zwitterion.
  • No Friedel-Crafts reaction: aniline forms a salt with the Lewis acid AlCl3; nitrogen then carries a positive charge and strongly deactivates the ring.
Worked example: How would you convert aniline into 4-nitroaniline in good yield?
Solution: direct nitration fails because the amino group is oxidised and the anilinium ion directs meta. (1) Acetylate aniline with acetic anhydride (pyridine) to get acetanilide. (2) Nitrate with conc. HNO3 and conc. H2SO4 at about 288 K; the NHCOCH3 group is still ortho, para-directing but only moderately activating, and the para product 4-nitroacetanilide is major. (3) Hydrolyse the amide with acid or alkali to get 4-nitroaniline.

Diazonium salts

Arenediazonium salts, ArN2+X−, are made by diazotisation: aniline is treated with NaNO2 and HCl at 273 to 278 K.

C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl− + NaCl + 2H2O (273 to 278 K)

Arenediazonium ions are stabilised by resonance with the ring, so they survive in cold solution; alkyldiazonium salts decompose at once. Even benzenediazonium chloride is not stored; it is used immediately. Benzenediazonium fluoroborate is water insoluble and stable at room temperature.

Reactions of benzenediazonium chloridewww.iitmedicoguide.comC₆H₅NH₂ + NaNO₂ + 2HCl273-278 KC₆H₅N₂⁺Cl⁻benzenediazonium chlorideCuCl / HCl (Sandmeyer)C₆H₅ClCuBr / HBr (Sandmeyer)C₆H₅BrCuCN / KCNC₆H₅CNKIC₆H₅IHBF₄, then heatC₆H₅FH₃PO₂ (or CH₃CH₂OH)C₆H₆H₂O, warmC₆H₅OHHBF₄; NaNO₂, Cu, heatC₆H₅NO₂phenol, OH⁻ (orange dye)p-Hydroxyazobenzeneaniline, H⁺ (yellow dye)p-AminoazobenzeneN₂ is lost (replaced)N₂ kept (coupling)www.iitmedicoguide.com
Most reactions of the diazonium group replace it and release N₂. In coupling reactions the -N=N- link is kept and joins two rings at the para position, giving coloured azo dyes.

With copper powder and the corresponding halogen acid instead of cuprous salts, the replacement by Cl or Br is called the Gattermann reaction. Diazonium salts matter because they give aryl fluorides and iodides, which cannot be made by direct halogenation, and cyanobenzene, which cannot be made from chlorobenzene by nucleophilic substitution. Coupling with phenol is carried out in alkaline medium and with aniline in mildly acidic medium; the diazonium ion is the electrophile and attacks the para position.

Common mistakes: (1) Writing the gas-phase order (3° > 2° > 1°) for basicity in water. (2) Using Gabriel synthesis to make aniline. (3) Forgetting that Hoffmann bromamide degradation removes one carbon. (4) Saying secondary amines give the carbylamine test; only primary amines do. (5) Predicting only ortho and para products for direct nitration of aniline, ignoring the meta product from the anilinium ion.

JEE and NEET focus

  • Preparation methods and which amine each gives: nitro reduction, nitrile and amide reduction, Gabriel synthesis, Hoffmann bromamide degradation.
  • Basicity orders in gas phase and in water, the reasons behind them, and the effect of ring substituents on aniline.
  • Distinguishing 1°, 2° and 3° amines: carbylamine test, Hinsberg test, nitrous acid.
  • Aniline reactions: bromination, protection by acetylation, nitration products, sulphanilic acid, no Friedel-Crafts.
  • Diazotisation conditions, Sandmeyer and Gattermann reactions, replacement by I, F, H, OH, and coupling to azo dyes.

Practice questions

The strongest base in aqueous solution is:

  1. CH3NH2
  2. (CH3)2NH
  3. (CH3)3N
  4. C6H5NH2
Show answer
B. pKb 3.27, the lowest; the balance of inductive, solvation and steric effects favours the 2° amine.

Which amine cannot be prepared by Gabriel phthalimide synthesis?

  1. Methanamine
  2. Ethanamine
  3. Aniline
  4. Propan-1-amine
Show answer
C. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion.

Propanamide on Hoffmann bromamide degradation gives:

  1. Propan-1-amine
  2. Ethanamine
  3. Butan-1-amine
  4. Propanenitrile
Show answer
B. CH3CH2CONH2 → CH3CH2NH2; one carbon is lost as carbonate.

The carbylamine test is given by:

  1. Primary amines only
  2. Secondary amines only
  3. Tertiary amines only
  4. All amines
Show answer
A. Both H atoms on nitrogen are needed to form R-NC.

In the Hinsberg test, the product that dissolves in aqueous KOH comes from:

  1. A primary amine
  2. A secondary amine
  3. A tertiary amine
  4. Both primary and secondary amines
Show answer
A. The N-alkylbenzenesulphonamide has an acidic N-H.

Direct nitration of aniline gives a substantial amount of m-nitroaniline because:

  1. -NH2 is meta-directing
  2. The anilinium ion formed in acid is meta-directing
  3. The meta position is least hindered
  4. Nitration is a nucleophilic substitution
Show answer
B. -NH3+ withdraws electrons and directs meta.

Benzenediazonium chloride reacts with hypophosphorous acid (H3PO2) to give:

  1. Phenol
  2. Chlorobenzene
  3. Benzene
  4. Aniline
Show answer
C. The diazonium group is replaced by hydrogen, with loss of N2.

Benzenediazonium chloride couples with phenol in alkaline medium to give:

  1. Diphenyl ether
  2. p-Hydroxyazobenzene
  3. p-Aminoazobenzene
  4. Salicylic acid
Show answer
B. An orange azo dye forms by electrophilic attack at the para position of phenol.
Call WhatsApp Apply
Chat with us on WhatsApp