NEET UG Chemistry · Chapter 19

Amines

Classification, preparation, basicity, tests for amines and diazonium salts.

NCERT Class 12NEET: 1 Q per paper 9 min read

1. Chapter at a Glance

Amines are organic derivatives of ammonia in which one, two or three hydrogens are replaced by alkyl or aryl groups. They are found in proteins, vitamins, alkaloids, hormones and dyes. This chapter covers their classification, naming, preparation, basic character, chemical reactions and tests, and the important chemistry of diazonium salts. NEET usually asks one question, often on basicity order, the Hofmann bromamide reaction, the carbylamine test, the Hinsberg test or diazonium reactions.

What NEET asks most
  • Preparation: reduction of nitro compounds, Gabriel phthalimide, Hofmann bromamide.
  • Basicity order in water and in the gas phase; aniline vs alkylamines.
  • Carbylamine test (1° amines) and Hinsberg test (1°, 2°, 3°).
  • Bromination and nitration of aniline; protecting –NH₂ by acetylation.
  • Diazonium salts: Sandmeyer, Gattermann, coupling reactions.

2. Classification, Structure and Naming

  • Primary (RNH₂), secondary (R₂NH) and tertiary (R₃N) amines; quaternary ammonium salts (R₄N⁺X⁻) have no lone pair.
  • Nitrogen is sp³ with a lone pair, giving a pyramidal shape (C–N–C ≈ 108° in trimethylamine).
  • IUPAC names use the suffix -amine: CH₃CH₂NH₂ is ethanamine; (CH₃)₂NH is N-methylmethanamine; C₆H₅NH₂ is aniline (benzenamine).
  • 1° and 2° amines form hydrogen bonds, so they boil higher than 3° amines of similar mass, but lower than alcohols (N is less electronegative than O).

3. Preparation of Amines

MethodReactionNEET Note
Reduction of nitro compoundsArNO₂ → ArNH₂ (Sn/HCl or Fe/HCl or H₂/Pd)Fe/HCl preferred industrially
Ammonolysis of alkyl halidesRX + NH₃ → RNH₂ + R₂NH + R₃N + R₄N⁺X⁻Gives a mixture
Reduction of nitrilesRCN → RCH₂NH₂ (LiAlH₄ or H₂/Ni)Adds one carbon (ascent)
Reduction of amidesRCONH₂ → RCH₂NH₂ (LiAlH₄)Same carbon number
Gabriel phthalimide synthesisPhthalimide + KOH, then RX, then hydrolysis → RNH₂Pure 1° aliphatic amines only; not aromatic amines
Hofmann bromamide degradationRCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂OAmine has one carbon less

4. Basic Character

  • Amines are Lewis and Brønsted bases because of the lone pair on nitrogen: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. A stronger base has a larger Kb (smaller pKb).
  • Aliphatic amines are stronger bases than ammonia (+I effect of alkyl groups).
  • In water, the order is not simply 3° > 2° > 1°, because solvation (hydrogen bonding of the ammonium ion) and steric hindrance also matter. For methyl amines: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. For ethyl amines: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.
  • In the gas phase (no solvation), the order is 3° > 2° > 1° > NH₃.
  • Aniline is a much weaker base than ammonia: its lone pair is delocalised into the benzene ring (resonance), and the sp² carbon is electron-withdrawing. Electron-donating groups (–CH₃, –OCH₃) on the ring increase basicity; electron-withdrawing groups (–NO₂) decrease it.
(CH₃)₂NHCH₃NH₂(CH₃)₃NNH₃C₆H₅NH₂ basicity in water (higher bar = stronger base)
Basicity of methyl amines in water. Aniline is far weaker than ammonia because its lone pair is shared with the ring.

5. Chemical Reactions of Amines

  • Salt formation with acids: RNH₂ + HCl → RNH₃⁺Cl⁻ (water-soluble).
  • Acylation: 1° and 2° amines react with acid chlorides or anhydrides to form amides (replacing N–H). Tertiary amines do not react.
  • Carbylamine test: only 1° amines (aliphatic and aromatic) + CHCl₃ + alcoholic KOH give isocyanides with a very foul smell.
  • With nitrous acid (NaNO₂ + HCl): 1° aliphatic amines give alcohols and N₂ gas (unstable diazonium). 1° aromatic amines at 273–278 K give stable diazonium salts.
  • Hinsberg test (benzenesulphonyl chloride): 1° amine gives a sulphonamide soluble in alkali (acidic N–H remains); 2° amine gives a sulphonamide insoluble in alkali; 3° amine does not react.

6. Electrophilic Substitution in Aniline

  • –NH₂ is strongly activating and ortho/para-directing.
  • Bromination: aniline + bromine water gives a white precipitate of 2,4,6-tribromoaniline at once. To get a mono-bromo product, first acetylate (–NHCOCH₃ is only moderately activating), brominate, then hydrolyse.
  • Nitration: direct nitration gives oxidation products and, in strong acid, about 47% meta product (because –NH₃⁺ forms, which is meta-directing). Protecting by acetylation gives mainly p-nitroaniline.
  • Sulphonation: aniline + conc. H₂SO₄ forms anilinium hydrogensulphate, which on heating gives sulphanilic acid (a zwitterion).
  • Aniline does not undergo Friedel–Crafts reactions: it forms a salt with AlCl₃, which deactivates the ring.

7. Diazonium Salts

  • Prepared by diazotisation: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O at 273–278 K. Arenediazonium salts are stable only in cold solution (resonance with the ring).
ReagentProductName / NEET Note
CuCl/HCl or CuBr/HBrArCl / ArBrSandmeyer reaction
Cu powder + HCl or HBrArCl / ArBrGattermann reaction
CuCN/KCNArCNSandmeyer; route to benzoic acid
KIArINo copper needed
HBF₄, heatArFBalz–Schiemann
H₃PO₂ or CH₃CH₂OHArHRemoves –NH₂ group
Warm waterArOHPhenol
HBF₄ then NaNO₂/CuArNO₂—
Phenol (alkaline) or aniline (acidic)Azo dye (Ar–N=N–Ar′)Coupling: orange (p-hydroxyazobenzene) and yellow (p-aminoazobenzene) dyes

8. Formula and Reaction Master Table

Name / ruleDetails
Hofmann bromamideRCONH₂ + Br₂ + 4NaOH → RNH₂ (one C less)
Gabriel synthesisPhthalimide → N-alkylphthalimide → RNH₂
Carbylamine testRNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O
DiazotisationArNH₂ + NaNO₂ + 2HCl (273–278 K) → ArN₂⁺Cl⁻
Basicity (water, methyl)2° > 1° > 3° > NH₃ > aniline
Basicity (gas phase)3° > 2° > 1° > NH₃
Hinsberg reagentC₆H₅SO₂Cl
Basicity measurestronger base ⇒ larger Kb, smaller pKb

9. Master Summary Table

Test / reaction1° amine2° amine3° amine
CarbylamineFoul-smelling isocyanideNoNo
HinsbergProduct soluble in alkaliProduct insoluble in alkaliNo reaction
AcylationYesYesNo (no N–H)
HNO₂ (aliphatic)Alcohol + N₂N-nitrosamine (yellow oil)Soluble nitrite salt
NEET NoteGabriel gives only 1° aliphatic amines; aniline is made by reducing nitrobenzene

10. Mind Map

Amines
Preparation
  • Nitro reduction
  • Gabriel
  • Hofmann bromamide
Basicity
  • +I effect
  • Solvation
  • Aniline weak
Tests
  • Carbylamine
  • Hinsberg
  • Nitrous acid
Diazonium
  • Sandmeyer, Gattermann
  • Replacement
  • Coupling dyes

11. Tips and Tricks

Shortcuts
  • Hofmann bromamide: carbon count decreases by one. Nitrile reduction: increases by one (relative to the halide used to make the nitrile).
  • To make m-bromoaniline or 1,3,5-tribromobenzene, use diazonium chemistry to add or remove groups.
  • Only aromatic 1° amines give stable diazonium salts.
  • Remember: "Sandmeyer = Cu(I) salts; Gattermann = Cu powder."
Common NEET traps
  • Assuming 3° amine is always the strongest base in water.
  • Using the Gabriel synthesis to make aniline.
  • Expecting a carbylamine reaction from 2° amines.
  • Expecting only o/p products in direct nitration of aniline.

12. Important One-liners

  • Aniline is a weaker base than ammonia and than methylamine.
  • The carbylamine test is given only by primary amines.
  • Benzenesulphonyl chloride is Hinsberg's reagent.
  • Diazonium salts are prepared at 0–5 °C.
  • Azo dyes are made by coupling diazonium salts with phenols or amines.
  • Aniline does not undergo Friedel–Crafts reactions.
  • Acetylation of aniline reduces the activating effect of –NH₂.

13. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

Which amine gives the carbylamine test?

  1. (CH₃)₂NH
  2. (CH₃)₃N
  3. C₆H₅NH₂
  4. C₆H₅N(CH₃)₂
Show solution

Only primary amines give the carbylamine reaction, and aniline is a 1° aromatic amine. The other three are 2° or 3° amines.

Answer: (C)

Q2Easy

Acetamide on treatment with Br₂ and NaOH gives:

  1. Ethanamine
  2. Methanamine
  3. Acetonitrile
  4. Ethanol
Show solution

In the Hofmann bromamide degradation, CH₃CONH₂ loses one carbon (as carbonate) and gives CH₃NH₂. Ethanamine would come from reducing the amide with LiAlH₄.

Answer: (B)

Q3Medium

The correct order of basic strength in aqueous solution is:

  1. (CH₃)₃N > (CH₃)₂NH > CH₃NH₂
  2. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
  3. CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
  4. CH₃NH₂ > (CH₃)₃N > (CH₃)₂NH
Show solution

In water, the +I effect, solvation of the ammonium ion and steric hindrance all work together. The result is 2° > 1° > 3° for methyl amines. Order (A) is the gas-phase order.

Answer: (B)

Q4Medium

Gabriel phthalimide synthesis is used to prepare:

  1. Primary aromatic amines
  2. Primary aliphatic amines
  3. Secondary amines
  4. Tertiary amines
Show solution

The phthalimide anion attacks an alkyl halide by SN2 substitution. Aryl halides do not undergo SN2, so this method gives only pure 1° aliphatic amines, with no 2° or 3° products.

Answer: (B)

Q5Medium

Benzenediazonium chloride reacts with CuCN/KCN to give:

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

In the Sandmeyer reaction with CuCN, –N₂⁺ is replaced by –CN to give C₆H₅CN. Hydrolysis of this nitrile then gives benzoic acid.

Answer: (A)

Q6Medium

A compound reacts with Hinsberg's reagent to give a product that is insoluble in alkali. The compound is a:

  1. Primary amine
  2. Secondary amine
  3. Tertiary amine
  4. Quaternary salt
Show solution

A 2° amine forms an N,N-disubstituted sulphonamide. It has no acidic N–H, so it does not dissolve in alkali. A 1° amine's product dissolves in alkali, and a 3° amine does not react.

Answer: (B)

Q7Hard

Direct nitration of aniline with a mixture of conc. HNO₃ and H₂SO₄ gives a significant amount of the meta product because:

  1. –NH₂ is meta-directing
  2. Aniline is protonated to the anilinium ion, which is meta-directing
  3. Nitration always gives meta products
  4. The ring is deactivated by –NH₂
Show solution

In strong acid, aniline becomes C₆H₅NH₃⁺. The –NH₃⁺ group withdraws electrons and directs substitution to the meta position. Free –NH₂ itself is an ortho/para director that activates the ring.

Answer: (B)

Q8Hard

Which is the weakest base?

  1. p-Toluidine
  2. Aniline
  3. p-Nitroaniline
  4. Benzylamine
Show solution

In p-nitroaniline, –NO₂ pulls the nitrogen lone pair further into the ring through resonance, so it is the weakest base. Benzylamine is strongest, because its –NH₂ is not attached to the ring. The –CH₃ in p-toluidine makes it slightly stronger than aniline.

Answer: (C)

14. Quick Revision Summary

Remember before the exam
  • Prepare: nitro reduction, Gabriel (1° aliphatic), Hofmann bromamide (one C less), nitrile/amide reduction.
  • Basicity: aliphatic > NH₃ > aniline; water order 2° > 1° > 3° (methyl).
  • Carbylamine (1° only); Hinsberg distinguishes 1°, 2°, 3°.
  • Aniline: tribromo with Br₂ water; acetylate to control; no Friedel–Crafts.
  • Diazonium (0–5 °C): Sandmeyer, Gattermann, KI, H₃PO₂, water, coupling to azo dyes.
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