JEE Main Chemistry · Chapter 17

Organic Compounds Containing Nitrogen

Amines — preparation, basicity, Hinsberg and carbylamine tests; diazonium salts and azo coupling.

NCERT Class 12JEE Main: 1 Q per shift 8 min read

1. Chapter at a Glance

Organic compounds containing nitrogen — mainly amines and diazonium salts — are derivatives of ammonia in which hydrogen atoms are replaced by alkyl or aryl groups. Amines are basic because of the lone pair on nitrogen. Aromatic diazonium salts are extremely useful intermediates for making substituted benzenes and azo dyes. JEE Main asks about basicity order, distinguishing tests (Hinsberg, carbylamine), named preparations (Gabriel, Hoffmann bromamide) and diazonium reactions.

What JEE asks most
  • Basic strength of aliphatic and aromatic amines in water and gas phase.
  • Hinsberg's test and the carbylamine test to identify 1°, 2°, 3° amines.
  • Gabriel phthalimide synthesis and Hoffmann bromamide degradation.
  • Sandmeyer, Gattermann and coupling reactions of benzenediazonium chloride.

2. Classification and Structure

  • Primary (1°) RNH₂, secondary (2°) R₂NH, tertiary (3°) R₃N; quaternary ammonium salts R₄N⁺X⁻.
  • Nitrogen is sp³ hybridised with a pyramidal shape; the C–N–C angle in trimethylamine is about 108°.
  • IUPAC: alkanamine (CH₃CH₂NH₂ = ethanamine); N-substituted names for 2° and 3° amines (N-methylethanamine). Aniline is benzenamine.
  • 1° and 2° amines form H-bonds, so their boiling points are higher than 3° amines of the same mass: 1° > 2° > 3°.

3. Preparation of Amines

MethodReactionNote
Reduction of nitro compoundsArNO₂ + Fe/HCl (or Sn/HCl, H₂/Pd) → ArNH₂Fe/HCl preferred (FeCl₂ hydrolyses and regenerates HCl)
Ammonolysis of alkyl halidesRX + NH₃ → RNH₂ → R₂NH → R₃N → R₄N⁺X⁻Gives a mixture; excess NH₃ favours 1°
Reduction of nitrilesRCN + LiAlH₄ (or H₂/Ni) → RCH₂NH₂Adds one carbon (ascending series)
Reduction of amidesRCONH₂ + LiAlH₄ → RCH₂NH₂Same number of carbons
Gabriel phthalimide synthesisPhthalimide + KOH, then RX, then hydrolysis → RNH₂Pure 1° aliphatic amines only; not aniline (aryl halides do not react)
Hoffmann bromamide degradationRCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂OAmine has one carbon less

4. Basic Character

Amines are Lewis and Brønsted bases. Basic strength is measured by Kb (larger Kb, smaller pKb = stronger base).

  • Aliphatic amines are stronger than NH₃ (+I effect of alkyl groups).
  • In the gas phase (only +I matters): 3° > 2° > 1° > NH₃.
  • In water, solvation of the cation (H-bonding) and steric effects also matter: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ and (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.
  • Aniline is a much weaker base than NH₃ (pKb ≈ 9.4) because its lone pair is delocalised into the benzene ring. Electron-donating groups (–CH₃, –OCH₃) at para increase its basicity; electron-withdrawing groups (–NO₂) decrease it.
  • Benzylamine (C₆H₅CH₂NH₂) behaves like an aliphatic amine — its lone pair is not conjugated with the ring.

5. Reactions and Tests of Amines

  • Acylation: 1° and 2° amines react with acid chlorides or anhydrides to give amides. Acetylation of aniline gives acetanilide, used to "protect" and moderate the –NH₂ group.
  • Carbylamine test: only 1° amines (aliphatic and aromatic) with CHCl₃ + alc. KOH give isocyanides with a very foul smell.
  • Hinsberg's test (benzenesulphonyl chloride, C₆H₅SO₂Cl): 1° amine gives a sulphonamide soluble in alkali (acidic N–H); 2° amine gives an insoluble sulphonamide; 3° amine does not react.
  • Nitrous acid (NaNO₂ + HCl): 1° aliphatic amines give N₂ gas and alcohols; 1° aromatic amines at 273–278 K give diazonium salts; 2° amines give yellow N-nitrosoamines.
  • Electrophilic substitution of aniline: –NH₂ is strongly activating, o/p directing. Bromine water gives 2,4,6-tribromoaniline at once. Direct nitration gives a lot of meta product (about 47%) because in acid aniline becomes the anilinium ion (–NH₃⁺, meta directing); acetylation first gives p-nitro product mainly. Aniline does not undergo Friedel–Crafts reactions (it forms a salt with AlCl₃).
  • Sulphonation of aniline gives sulphanilic acid, which exists as a zwitterion.

6. Diazonium Salts

Benzenediazonium chloride, C₆H₅N₂⁺Cl⁻, is made by diazotisation: aniline + NaNO₂ + HCl at 273–278 K. Aryl diazonium salts are stabilised by resonance with the ring, but they decompose above about 5 °C, so they are used immediately.

C₆H₅N₂⁺Cl⁻ CuCl/HCl → C₆H₅ClKI → C₆H₅ICuCN → C₆H₅CN H₂O, warm→ C₆H₅OHHBF₄, Δ→ C₆H₅F H₃PO₂ → C₆H₆CuBr/HBr → C₆H₅Brphenol → azo dye
Uses of benzenediazonium chloride: replacement of N₂ by Cl, Br, CN (Sandmeyer), I, F (Balz–Schiemann), OH or H, and coupling to form coloured azo compounds.
  • Sandmeyer: CuCl/HCl, CuBr/HBr, CuCN/KCN. Gattermann: Cu powder with HCl or HBr.
  • Coupling: with phenol (weakly alkaline) gives p-hydroxyazobenzene (orange); with aniline (weakly acidic) gives p-aminoazobenzene (yellow). The –N=N– group is the chromophore of azo dyes; the diazonium ion is a weak electrophile and attacks the para position.
  • Replacement by H with H₃PO₂ (or ethanol) is used to remove an –NH₂ group after it has directed a substitution.

7. Master Table

Test / reaction1° amine2° amine3° amine
Hinsberg (C₆H₅SO₂Cl)Product soluble in KOHProduct insoluble in KOHNo reaction
Carbylamine (CHCl₃/KOH)Foul-smelling isocyanideNoNo
HNO₂Aliphatic: N₂ + alcohol; aromatic: diazonium saltN-Nitrosoamine (yellow oil)Salt (no N₂)
AcylationYesYesNo (no N–H)

8. Mind Map

Nitrogen Compounds
Preparation
  • Reduction (nitro, nitrile, amide)
  • Gabriel
  • Hoffmann bromamide
Basicity
  • +I effect
  • Solvation, sterics
  • Aniline weak
Tests
  • Hinsberg
  • Carbylamine
  • Nitrous acid
Diazonium
  • Diazotisation 0–5 °C
  • Sandmeyer, Balz–Schiemann
  • Azo coupling

9. Tips and Tricks

Shortcuts
  • Carbon count check: nitrile reduction → +1 C; Hoffmann bromamide → −1 C; amide reduction → same.
  • For "which is the weakest base", look for a lone pair in resonance with a ring or a C=O (amides are almost neutral).
  • To get p-nitroaniline: acetylate aniline → nitrate → hydrolyse.
  • Diazonium questions: first identify the reagent, then simply replace –N₂⁺ with the new group.
Mistakes students make
  • Trying to make aniline by Gabriel synthesis — aryl halides do not undergo SN2.
  • Forgetting that 2° amines do not give the carbylamine test.
  • Using gas-phase basicity order for aqueous questions.

10. Important One-liners

  • Aniline is less basic than ammonia.
  • Diazotisation is carried out at 0–5 °C.
  • The carbylamine test identifies primary amines.
  • Hoffmann bromamide degradation removes one carbon atom.
  • Acetanilide nitration mainly gives the para product.
  • Azo dyes contain the –N=N– group linking two aromatic rings.

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

Which amine gives a foul-smelling compound on heating with chloroform and alcoholic KOH?

  1. (CH₃)₂NH
  2. (CH₃)₃N
  3. C₆H₅NH₂
  4. C₆H₅NHCH₃
Show solution

The carbylamine test is given only by primary amines. Aniline is a primary amine and forms phenyl isocyanide (C₆H₅NC).

Answer: (C)

Q2Medium

Acetamide is treated with Br₂ and NaOH. The amine formed is:

  1. Ethanamine
  2. Methanamine
  3. Propanamine
  4. Aniline
Show solution

Hoffmann bromamide degradation removes the carbonyl carbon: CH₃CONH₂ → CH₃NH₂ (methanamine).

Answer: (B)

Q3Easy

Which compound is the weakest base?

  1. CH₃NH₂
  2. NH₃
  3. C₆H₅NH₂
  4. (CH₃)₂NH
Show solution

In aniline the nitrogen lone pair is delocalised into the benzene ring and is less available for protonation. So aniline is the weakest base here.

Answer: (C)

Q4Medium

Benzenediazonium chloride is warmed with CuCN/KCN. The product is:

  1. Benzonitrile
  2. Phenyl isocyanide
  3. Benzamide
  4. Chlorobenzene
Show solution

Sandmeyer reaction replaces –N₂⁺ with –CN, giving C₆H₅CN (benzonitrile).

Answer: (A)

Q5Hard

Direct nitration of aniline with a nitrating mixture gives a large amount of m-nitroaniline because:

  1. –NH₂ is a meta director
  2. Aniline is protonated to anilinium ion, which is meta directing
  3. Nitration is a nucleophilic reaction
  4. The para position is blocked
Show solution

In strong acid, aniline becomes C₆H₅NH₃⁺. The –NH₃⁺ group has no lone pair and withdraws electrons (−I), so it directs meta. –NH₂ itself is o/p directing.

Answer: (B)

Q6Hard

An amine reacts with benzenesulphonyl chloride to give a product insoluble in aqueous KOH. The amine is:

  1. Ethanamine
  2. N-Methylethanamine
  3. N,N-Dimethylethanamine
  4. Aniline
Show solution

A secondary amine gives an N,N-disubstituted sulphonamide with no acidic N–H, so it does not dissolve in alkali. N-Methylethanamine is secondary. Primary amines (ethanamine, aniline) give alkali-soluble products; the tertiary amine does not react.

Answer: (B)

Q7Numerical

How many isomeric amines (1°, 2° and 3° together) have the formula C₄H₁₁N?

Show solution

Primary (4): butan-1-amine, butan-2-amine, 2-methylpropan-1-amine, 2-methylpropan-2-amine. Secondary (3): N-methylpropan-1-amine, N-methylpropan-2-amine, N-ethylethanamine. Tertiary (1): N,N-dimethylethanamine. Total 8.

Answer: 8

Q8Medium

Gabriel phthalimide synthesis cannot be used to prepare:

  1. Ethanamine
  2. Benzylamine
  3. Aniline
  4. Propan-1-amine
Show solution

The key step is SN2 attack of the phthalimide anion on an alkyl halide. Aryl halides such as chlorobenzene do not undergo this substitution, so aniline cannot be made. Benzyl halides react easily, so benzylamine can be made.

Answer: (C)

12. Quick Revision Summary

Remember before the exam
  • Basicity in water: 2° > 1° > 3° > NH₃ (methyl); aniline is weak.
  • Gabriel → pure 1° aliphatic amines; Hoffmann bromamide → one carbon less.
  • Carbylamine: 1° only; Hinsberg distinguishes 1°, 2°, 3°.
  • Diazotisation at 0–5 °C; Sandmeyer (CuX, CuCN), KI, HBF₄, H₂O, H₃PO₂.
  • Coupling with phenol → orange azo dye.
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