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State Board · Class 12 · Chemistry

Lesson 13: Amines

7 concepts · 56 practice questions with answers · Free to read

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Classification and nomenclature

From daily life

The smell of stale fish (trimethylamine) and the aniline behind old blue dyes are both amines.

Simple definition

Amines are NH₃ with H replaced by alkyl/aryl groups; 1°, 2°, 3° by the number of carbons on N.

Amine classesAmines: classes and names1°CH₃CH₂NH₂ethanamine2°(CH₃)₂NHN-methylmethanamine3°(CH₃)₃NN,N-dimethylmethanamineAromatic 1°C₆H₅NH₂benzenamine (aniline)BenzylicC₆H₅CH₂NH₂phenylmethanamine1°: –NH₂ · 2°: >NH · 3°: N bears three carbonsName: replace the final e of the alkane by 'amine'

In the figure : Classes and names.

Rule 1: Classification — 1° RNH₂, 2° R₂NH, 3° R₃N. (Unlike alcohols, count carbons on N itself.)

  • Easy : CH₃NH₂ is 1°.
  • Medium : (CH₃)₂NH is 2°.
  • Exam-type : (CH₃)₃N is 3°.

Rule 2: IUPAC naming — Replace e with 'amine'; extra groups on N take 'N-'. CH₃NHC₂H₅ = N-methylethanamine.

  • Easy : CH₃CH₂CH₂NH₂ = propan-1-amine.
  • Medium : (CH₃)₂CHNH₂ = propan-2-amine.
  • Exam-type : C₆H₅NH₂ = benzenamine (aniline).

Rule 3: Structure — N is sp³ and pyramidal with one lone pair; C–N–C ≈ 108°. The lone pair gives basicity.

  • Easy : NH₃ angle is 107°.
  • Medium : Lone pair = basicity.
  • Exam-type : Aniline's N is nearly planar.

Reasoning: class and name of (C₂H₅)₂NCH₃

  • Step 1: Three carbons on N (two ethyl + one methyl) ⇒ 3°.
  • Step 2: Parent chain is ethane.
  • Step 3: N-ethyl-N-methylethanamine.

CET level (MCQ)

  • CET 1) Name of C₆H₅NH₂? 1) benzenamine 2) phenol 3) benzylamine 4) toluene — Answer: 1).
  • CET 2) Class of (CH₃)₂NH? 1) 1° 2) 2° 3) 3° 4) 4° — Answer: 2).

Real-life use: Dyes, drugs, pesticides, nylon.

Remember

Amine: carbons on N give the class; suffix 'amine'.

Common mistake

Counting carbons on the C–NH₂ carbon, as for alcohols.

CET trap

Alcohol: carbons on the C–OH; amine: carbons on N.

Shortcut

Memory clue: "degree = carbons on N".

Spot the difference

ClassFormulaN–H
1°RNH₂2
2°R₂NH1
3°R₃N0

Practice questions

Easy
  1. 1What are 1°, 2°, 3° amines?
  2. 2Name CH₃NHC₂H₅.
  3. 3IUPAC name of aniline.
Medium
  1. 4Hybridisation of N?
  2. 5Type of (CH₃)₃N?
  3. 6Formula of propan-2-amine.
Challenging
  1. 7What does the lone pair give?
  2. 8Benzylamine vs aniline.
Show answers (try them yourself first)
  1. 11, 2 or 3 carbons on N.
  2. 2N-methylethanamine.
  3. 3Benzenamine.
  4. 4sp³.
  5. 53°.
  6. 6(CH₃)₂CHNH₂.
  7. 7Basicity.
  8. 8NH₂ on ring vs on CH₂.

Preparation of amines

From daily life

Factories make aniline from nitrobenzene for drugs and dyes.

Simple definition

Routes: nitro reduction, nitrile/amide reduction, Hofmann bromamide, Gabriel, ammonolysis.

PreparationPreparation of aminesRX + excess NH₃Hofmann ammonolysis ⇒ mixtureRCN + LiAlH₄ / H₂, NiRCH₂NH₂ (C +1)RCONH₂ + LiAlH₄RCH₂NH₂ (same C)RCONH₂ + Br₂ + 4NaOHHofmann bromamide ⇒ RNH₂ (C –1)phthalimide-K + RX ⇒ NaOHGabriel ⇒ pure 1°ArNO₂ + Sn/HCl or H₂, PdArNH₂ (aniline)Gabriel and Hofmann bromamide give only 1° amines

In the figure : Preparation routes.

Rule 1: Reduction — ArNO₂ + Sn/HCl ⇒ ArNH₂. RCN + LiAlH₄ ⇒ RCH₂NH₂ (C +1). RCONH₂ + LiAlH₄ ⇒ RCH₂NH₂ (same C).

  • Easy : 12.3 g nitrobenzene ⇒ 9.3 g aniline.
  • Medium : CH₃CN ⇒ C₂H₅NH₂.
  • Exam-type : CH₃CONH₂ ⇒ C₂H₅NH₂.

Rule 2: Hofmann bromamide — RCONH₂ + Br₂ + 4NaOH ⇒ RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O. One carbon fewer.

  • Easy : CH₃CONH₂ ⇒ CH₃NH₂.
  • Medium : 5.9 g acetamide ⇒ 3.1 g CH₃NH₂.
  • Exam-type : Benzamide ⇒ aniline.

Rule 3: Gabriel and ammonolysis — Potassium phthalimide + RX ⇒ N-alkyl ⇒ NaOH ⇒ pure 1° RNH₂. RX + NH₃ gives a 1°/2°/3°/4° mixture.

  • Easy : Gabriel fails with ArX.
  • Medium : Gabriel gives pure 1°.
  • Exam-type : Ammonolysis ⇒ mixture.

Calculation: aniline from 12.3 g nitrobenzene

  • Step 1: n = 12.3 ÷ 123.1 = 0.1 mol.
  • Step 2: 1 : 1 ⇒ 0.1 mol aniline.
  • Step 3: M = 93.1 ⇒ 9.3 g.

CET level (MCQ)

  • CET 1) Carbon count in Hofmann bromamide? 1) one less 2) same 3) one more 4) double — Answer: 1).
  • CET 2) Pure 1° amine from? 1) Gabriel 2) ammonolysis 3) reduction 4) nitration — Answer: 1).

Real-life use: Drugs, dyes, polymers.

Remember

Nitrile +1 C; Hofmann –1 C; Gabriel = pure 1°.

Common mistake

Assuming the carbon count is unchanged in Hofmann.

CET trap

Gabriel and Hofmann give only 1° amines.

Shortcut

Memory clue: "Hofmann loses one C".

Spot the difference

RouteC change
RCN + LiAlH₄+1
Hofmann–1
Amide + LiAlH₄0

Practice questions

Easy
  1. 1Write Hofmann bromamide.
  2. 2Gabriel synthesis.
  3. 3Nitrobenzene ⇒ aniline.
Medium
  1. 4Drawback of ammonolysis.
  2. 5CH₃CN ⇒ ?
  3. 65.9 g acetamide ⇒ ?
Challenging
  1. 7Why Gabriel fails with ArX?
  2. 8Two reducing agents.
Show answers (try them yourself first)
  1. 1RCONH₂ ⇒ RNH₂ (–1 C).
  2. 2Phthalimide + RX ⇒ RNH₂.
  3. 3Sn/HCl.
  4. 4A mixture.
  5. 5C₂H₅NH₂.
  6. 63.1 g.
  7. 7SN2 does not occur on aryl halides.
  8. 8LiAlH₄, Sn/HCl.

Physical properties

From daily life

Methylamine is a gas, but butylamine is a liquid — N–H···N hydrogen bonds.

Simple definition

1° and 2° amines form N–H···N bonds, so b.p. exceeds alkanes but stays below alcohols (N is less polar than O).

Boiling pointsC₄H₁₁N isomers and boiling points (K)273butane3103° amine3292° amine3511° amine391butan-1-ol1° > 2° > 3° (fewer H-bonds) · amines < alcohols (N less polar)3° amines have no N–H

In the figure : Boiling points.

Rule 1: Boiling point — Same M: 1° > 2° > 3° (no N–H in 3°); amines < alcohols.

  • Easy : Butan-1-amine 351 K.
  • Medium : Diethylamine 329 K.
  • Exam-type : The 3° isomer 310 K.

Rule 2: Solubility — Small amines mix with water (N···H–O); solubility falls as carbons rise; aniline is slightly soluble.

  • Easy : CH₃NH₂ is miscible.
  • Medium : Hexylamine is poorly soluble.
  • Exam-type : Aniline ≈ 3.6 g/100 mL.

Rule 3: Smell and others — Lower amines smell fishy; aniline darkens in air.

  • Easy : (CH₃)₃N is the fishy smell.
  • Medium : Aniline: colourless, darkens.
  • Exam-type : Higher amines are solids.

Reasoning: why 1° amine b.p. > 3° isomer?

  • Step 1: Same M (73).
  • Step 2: The 1° amine has two N–H, so more H-bonds.
  • Step 3: The 3° has none ⇒ 310 K < 351 K.

CET level (MCQ)

  • CET 1) Highest b.p. among C₄H₁₁N isomers? 1) 1° 2) 2° 3) 3° 4) all same — Answer: 1).
  • CET 2) No intermolecular N–H···N? 1) (CH₃)₃N 2) CH₃NH₂ 3) (CH₃)₂NH 4) none — Answer: 1).

Real-life use: Drug solubility, solvents.

Remember

b.p.: 1° > 2° > 3°; amines < alcohols.

Common mistake

Assuming intermolecular H-bonds in 3° amines.

CET trap

3° amines accept H-bonds from water but not from each other.

Shortcut

Memory clue: "more N–H, higher b.p.".

Spot the difference

TypeN–Hb.p. (K)
1°2351
2°1329
3°0310

Practice questions

Easy
  1. 1Why amine b.p. < alcohol?
  2. 2Why low b.p. for 3°?
  3. 3Which amines mix with water?
Medium
  1. 4What happens to aniline in air?
  2. 5Speciality of (CH₃)₃N.
  3. 6Solubility as C increases?
Challenging
  1. 7What is N–H···N?
  2. 8Compare 1° and 2°.
Show answers (try them yourself first)
  1. 1N is less polar.
  2. 2No N–H.
  3. 3Small amines.
  4. 4It darkens.
  5. 5Fishy smell.
  6. 6It decreases.
  7. 7Intermolecular H-bond.
  8. 81° has the higher b.p.

Basicity of amines

From daily life

Lemon juice removes fish smell — the acid turns the amine into an odourless salt.

Simple definition

Amines donate a lone pair to H⁺ (Lewis bases). Lower pKb ⇒ stronger base. In water: 2° > 1° ≈ 3° > NH₃ > aniline.

BasicitypKb (aqueous): lower ⇒ stronger base(C₂H₅)₂NH3.00(CH₃)₂NH3.27C₂H₅NH₂3.29CH₃NH₂3.38(C₂H₅)₃N3.25(CH₃)₃N4.22NH₃4.75C₆H₅NH₂9.38Aliphatic: 2° > 1° ≈ 3° > NH₃ > anilineAniline's lone pair is delocalised into the ring ⇒ weak base

In the figure : pKb comparison.

Rule 1: Aliphatic amines — +I groups raise electron density on N ⇒ stronger than NH₃. In water 3° amines are poorly solvated, so (CH₃)₃N < (CH₃)₂NH.

  • Easy : (C₂H₅)₂NH pKb 3.00.
  • Medium : CH₃NH₂ 3.38.
  • Exam-type : NH₃ 4.75.

Rule 2: Aromatic amines — In aniline the lone pair is delocalised into the ring, so pKb = 9.38; –NO₂ lowers basicity further, –CH₃/–OCH₃ raise it.

  • Easy : Aniline < NH₃.
  • Medium : p-toluidine > aniline.
  • Exam-type : p-nitroaniline < aniline.

Rule 3: Salts and easy separation — RNH₂ + HCl ⇒ RNH₃⁺Cl⁻ (water-soluble); NaOH regenerates the amine.

  • Easy : Aniline + HCl ⇒ anilinium chloride.
  • Medium : Salt + NaOH ⇒ amine.
  • Exam-type : Acid–base extraction.

Reasoning: why is aniline weaker than NH₃?

  • Step 1: N's lone pair conjugates with the ring.
  • Step 2: Less electron density on N.
  • Step 3: Harder to accept H⁺ ⇒ pKb 9.38 vs 4.75.

CET level (MCQ)

  • CET 1) Weakest base? 1) aniline 2) CH₃NH₂ 3) NH₃ 4) (CH₃)₂NH — Answer: 1).
  • CET 2) Strongest methylamine in water? 1) (CH₃)₂NH 2) (CH₃)₃N 3) CH₃NH₂ 4) NH₃ — Answer: 1).

Real-life use: Drug salts, separations, pH.

Remember

Lower pKb = stronger; aniline 9.38; 2° best.

Common mistake

Assuming 3° amines are the strongest in water (less solvation).

CET trap

In the gas phase the order is 3° > 2° > 1°; in water it changes.

Shortcut

Memory clue: "+I strengthens; the ring weakens".

Spot the difference

AminepKb
(C₂H₅)₂NH3.00
NH₃4.75
C₆H₅NH₂9.38

Practice questions

Easy
  1. 1Why are amines basic?
  2. 2Why is aniline weak?
  3. 3Meaning of pKb?
Medium
  1. 4Role of the +I effect?
  2. 5Amine + HCl?
  3. 6p-nitroaniline compared.
Challenging
  1. 7Order among methylamines.
  2. 8Salt with NaOH?
Show answers (try them yourself first)
  1. 1A lone electron pair.
  2. 2Resonance.
  3. 3Lower ⇒ stronger base.
  4. 4Raises electron density on N.
  5. 5RNH₃⁺Cl⁻.
  6. 6Weaker base.
  7. 72° > 1° > 3° > NH₃.
  8. 8Amine is regenerated.

Tests and reactions of amines

From daily life

Labs use chemical tests to tell 1°, 2° and 3° amines apart.

Simple definition

Hinsberg (C₆H₅SO₂Cl), carbylamine, HNO₂ and acylation separate 1°, 2° and 3° amines.

TestsTests for aminesHinsberg C₆H₅SO₂Cl1° ⇒ dissolves in KOH; 2° ⇒ insoluble; 3° ⇒ no reactionCarbylamine CHCl₃ + KOHonly 1° ⇒ foul smell (RNC)HNO₂ (NaNO₂ + HCl)1° aliphatic ⇒ ROH + N₂ ↑HNO₂, 273–278 K1° aromatic ⇒ diazonium saltdiazonium + β-naphtholorange-red azo dye(CH₃CO)₂O / CH₃COCl1°, 2° ⇒ amide; 3° ⇒ none

In the figure : Tests.

Rule 1: Hinsberg test — 1° ⇒ sulphonamide dissolves in KOH; 2° ⇒ insoluble; 3° ⇒ no reaction.

  • Easy : C₂H₅NH₂ ⇒ soluble.
  • Medium : (C₂H₅)₂NH ⇒ insoluble.
  • Exam-type : (C₂H₅)₃N ⇒ none.

Rule 2: Carbylamine — RNH₂ + CHCl₃ + 3KOH ⇒ RNC (foul smell) + 3KCl + 3H₂O; only 1° (aliphatic and aromatic).

  • Easy : Aniline ⇒ phenyl isocyanide.
  • Medium : 2° ⇒ none.
  • Exam-type : 3° ⇒ none.

Rule 3: Nitrous acid and acylation — 1° aliphatic + HNO₂ ⇒ ROH + N₂↑. 1° aromatic + HNO₂ (273–278 K) ⇒ diazonium. 1°/2° + (CH₃CO)₂O ⇒ amide.

  • Easy : C₂H₅NH₂ ⇒ C₂H₅OH + N₂.
  • Medium : Aniline ⇒ C₆H₅N₂⁺Cl⁻.
  • Exam-type : Aniline ⇒ acetanilide.

Reasoning: unknown amine is carbylamine +, dissolves in KOH

  • Step 1: Carbylamine positive ⇒ 1°.
  • Step 2: Hinsberg sulphonamide dissolves in KOH ⇒ 1° confirmed.
  • Step 3: N₂ with HNO₂ ⇒ aliphatic.

CET level (MCQ)

  • CET 1) Carbylamine test is given by? 1) 1° 2) 2° 3) 3° 4) all — Answer: 1).
  • CET 2) 3° amine + Hinsberg reagent ⇒ ? 1) no reaction 2) soluble 3) ppt 4) gas — Answer: 1).

Real-life use: Lab identification, drug testing.

Remember

Hinsberg: 1° soluble, 2° insoluble, 3° none; carbylamine = 1°.

Common mistake

Swapping the Hinsberg results of 1° and 2°.

CET trap

The leftover N–H in the 1° sulphonamide is acidic, so it dissolves in KOH.

Shortcut

Memory clue: "one dissolves, two stays as sludge".

Spot the difference

Test1°2°3°
Carbylamine+––
Hinsbergsolubleinsolublenone
HNO₂N₂ / ArN₂⁺nitrosaminesalt

Practice questions

Easy
  1. 1Describe Hinsberg.
  2. 2Describe carbylamine.
  3. 31° aliphatic + HNO₂.
Medium
  1. 4Aniline + HNO₂ (273 K).
  2. 5What is acylation?
  3. 6Why does the 1° sulphonamide dissolve?
Challenging
  1. 7Why no reaction for 3°?
  2. 8Smell of isocyanide.
Show answers (try them yourself first)
  1. 1C₆H₅SO₂Cl, KOH.
  2. 2CHCl₃ + KOH ⇒ foul smell.
  3. 3ROH + N₂.
  4. 4Diazonium salt.
  5. 5Acyl group on N.
  6. 6Acidic leftover N–H.
  7. 7No N–H.
  8. 8Foul.

Diazonium salts

From daily life

Many bright azo dyes on clothes start from diazonium salts.

Simple definition

ArNH₂ + NaNO₂ + HCl (273–278 K) ⇒ ArN₂⁺Cl⁻, a versatile intermediate that loses N₂ and takes up a Nu.

DiazoniumReactions of benzenediazonium chlorideC₆H₅N₂⁺Cl⁻CuCl/HCl · CuBr/HBr⇒ C₆H₅Cl · C₆H₅Br (Sandmeyer)Cu, HCl · Cu, HBr⇒ C₆H₅Cl · C₆H₅Br (Gattermann)KI⇒ C₆H₅IHBF₄, heat⇒ C₆H₅F (Balz–Schiemann)CuCN / KCN⇒ C₆H₅CNH₃PO₂ + H₂O⇒ C₆H₆ (removes NH₂)H₂O, warm⇒ C₆H₅OH + N₂phenol, pH 9–10 / aniline, pH 4–5⇒ azo dye (coupling)

In the figure : Diazonium reactions.

Rule 1: Reactions losing N₂ — Sandmeyer: CuCl/CuBr ⇒ ArCl/ArBr. Gattermann: Cu powder + HX. KI ⇒ ArI. HBF₄ ⇒ ArF. H₃PO₂ ⇒ ArH. H₂O, warm ⇒ ArOH.

  • Easy : C₆H₅N₂Cl ⇒ C₆H₅Cl.
  • Medium : ⇒ C₆H₅I (no Cu needed).
  • Exam-type : ⇒ C₆H₆.

Rule 2: Coupling — Phenol (pH 9–10) ⇒ p-hydroxyazobenzene (orange); aniline (pH 4–5) ⇒ p-aminoazobenzene (yellow).

  • Easy : –N=N– is the azo group.
  • Medium : β-naphthol ⇒ orange-red.
  • Exam-type : pH matters.

Rule 3: Stability — Aromatic diazonium salts are stable at 273–278 K (resonance); aliphatic ones decompose at once.

  • Easy : Above ~278 K ⇒ phenol.
  • Medium : Aliphatic ⇒ ROH + N₂.
  • Exam-type : Prepare in ice.

Calculation: diazonium chloride from 9.3 g aniline

  • Step 1: n = 9.3 ÷ 93.1 = 0.1 mol.
  • Step 2: 1 : 1 ⇒ 0.1 mol.
  • Step 3: M = 140.6 ⇒ 14.1 g.

CET level (MCQ)

  • CET 1) Reagent in the Sandmeyer reaction? 1) CuCl/HCl 2) KI 3) H₃PO₂ 4) HBF₄ — Answer: 1).
  • CET 2) Diazonium + H₃PO₂ ⇒ ? 1) benzene 2) phenol 3) aniline 4) toluene — Answer: 1).

Real-life use: Azo dyes, drugs, F/I compounds.

Remember

273–278 K; CuX = Sandmeyer; Cu powder = Gattermann; H₃PO₂ removes N₂⁺.

Common mistake

Swapping Sandmeyer and Gattermann.

CET trap

KI needs no Cu; iodine enters directly.

Shortcut

Memory clue: "Sandmeyer = CuX; Gattermann = Cu powder".

Spot the difference

ReagentProduct
CuClArCl
KIArI
H₃PO₂ArH
H₂O, ∆ArOH

Practice questions

Easy
  1. 1How is a diazonium salt made?
  2. 2Write Sandmeyer.
  3. 3Write Gattermann.
Medium
  1. 4Coupling reaction.
  2. 5What is the azo group?
  3. 6Why prepare in ice?
Challenging
  1. 79.3 g aniline ⇒ ?
  2. 8How to get fluorobenzene?
Show answers (try them yourself first)
  1. 1ArNH₂ + NaNO₂ + HCl, 273 K.
  2. 2CuCl/HCl.
  3. 3Cu powder + HCl.
  4. 4Coupling with phenol/aniline.
  5. 5–N=N–.
  6. 6Decomposes when warm.
  7. 714.1 g.
  8. 8HBF₄, heat.

Electrophilic substitution in aniline

From daily life

Bromine water on aniline instantly gives a white precipitate — a very reactive ring.

Simple definition

–NH₂ is a strong activating o/p director, so mono-substitution is hard; acylation protects –NH₂.

Aniline EASAniline: electrophilic substitutionBr₂ / H₂O⇒ 2,4,6-tribromoaniline (white)(CH₃CO)₂O⇒ acetanilide (protects –NH₂)acetanilide + Br₂⇒ p-bromo (mono) ⇒ H₃O⁺ ⇒ p-bromoanilineHNO₃ + H₂SO₄ (direct)⇒ o, p + ~47 % m (anilinium)H₂SO₄, 453 K⇒ sulphanilic acid (zwitterion)AlCl₃ / Friedel–Crafts⇒ fails (N–AlCl₃ complex)–NH₂ activates o/p; in acid NH₃⁺ directs m

In the figure : Aniline reactions.

Rule 1: Bromination — Br₂/H₂O ⇒ 2,4,6-tribromoaniline (white). For the mono product, acetylate first.

  • Easy : 9.3 g aniline ⇒ 33.0 g tribromo.
  • Medium : Acetanilide ⇒ p-bromo.
  • Exam-type : Then hydrolysis.

Rule 2: Nitration — Direct HNO₃/H₂SO₄ forms anilinium (–NH₃⁺, m-directing) ⇒ o, p and ≈ 47 % m, with oxidation. After protection, p-nitro forms.

  • Easy : Acetanilide ⇒ p-nitro.
  • Medium : Then H₃O⁺ ⇒ p-nitroaniline.
  • Exam-type : Hence –NH₂ is protected.

Rule 3: Sulphonation and FC — Anilinium hydrogen sulphate at 453 K ⇒ sulphanilic acid (zwitterion). Friedel–Crafts fails because N binds AlCl₃.

  • Easy : Sulphanilic acid = p-NH₂C₆H₄SO₃H.
  • Medium : A zwitterion.
  • Exam-type : No FC.

Reasoning: how to make p-bromoaniline

  • Step 1: Aniline + (CH₃CO)₂O ⇒ acetanilide.
  • Step 2: Br₂ ⇒ p-bromoacetanilide (major).
  • Step 3: H₃O⁺ hydrolysis ⇒ p-bromoaniline.

CET level (MCQ)

  • CET 1) Aniline + bromine water ⇒ ? 1) white ppt 2) yellow 3) red 4) none — Answer: 1).
  • CET 2) Why meta product in direct nitration? 1) anilinium ion 2) +I 3) oxidation 4) resonance — Answer: 1).

Real-life use: Drug synthesis (sulpha drugs), dyes.

Remember

–NH₂ = o/p activator; protect by acylation; no FC.

Common mistake

Thinking direct nitration gives only para.

CET trap

In acid ⇒ anilinium ⇒ meta director.

Shortcut

Memory clue: "protect first, substitute next".

Spot the difference

ReagentProduct
Br₂/H₂Otribromo
Ac₂Oacetanilide
H₂SO₄, 453 Ksulphanilic

Practice questions

Easy
  1. 1Bromination of aniline.
  2. 2Why protect –NH₂?
  3. 3Direct nitration.
Medium
  1. 4How is sulphanilic acid made?
  2. 5Why does FC fail?
  3. 6Tribromo from 9.3 g?
Challenging
  1. 7What is anilinium?
  2. 8Route to p-nitroaniline.
Show answers (try them yourself first)
  1. 12,4,6-tribromo.
  2. 2To avoid oxidation and meta products.
  3. 3A mixture of o, p and m.
  4. 4Anilinium sulphate at 453 K.
  5. 5AlCl₃ complexes with N.
  6. 633.0 g.
  7. 7C₆H₅NH₃⁺.
  8. 8Acylate ⇒ nitrate ⇒ H₃O⁺.

Quick revision of the whole lesson

ConceptKey formula / fact
1°/2°/3°carbons on N
Hofmann–1 C
b.p.1° > 2° > 3°
Basicity2° > 1° ≈ 3° > NH₃ > aniline
Hinsberg1° dissolves
Carbylamineonly 1°
SandmeyerCuX
Aniline EASprotect first
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Classes, names.
  • Preparation.
  • H-bonding, b.p.
  • Basicity.
  • Tests.
  • Diazonium.
  • Aniline EAS.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.