Chapter 18: Amines

Amines

Overview

• Amines are ammonia derivatives where 1, 2, or 3 H atoms are replaced by alkyl or aryl groups.
• Found naturally in proteins, vitamins, alkaloids, and hormones.

Solved Examples

• Ex 1: Identify the type of amine in CH₃NH₂.
  Answer: Primary amine (one alkyl group attached to N).

Practice Set

• Q1: What is an amine?
  A. A hydrocarbon
  B. Ammonia derivative
  C. Alcohol
  D. Ketone
• Q2: Amines are found in which natural compounds?
  A. Proteins
  B. Vitamins
  C. Alkaloids
  D. All of the above
• Q3: Which group replaces hydrogen in amines?
  A. Halogens
  B. Alkyl/Aryl groups
  C. Oxygen
  D. Sulfur

Answer Key

• Q1: B - Amines are derivatives of ammonia.
• Q2: D - Amines occur in proteins, vitamins, and alkaloids.
• Q3: B - Alkyl or aryl groups replace hydrogen atoms in amines.

Memory Aids

• Amines = Ammonia + Alkyl/Aryl groups
• Natural occurrence: Proteins, Vitamins, Alkaloids

Diazonium Salts

Properties and Naming

• Arenediazonium salts have formula ArN₂⁺X^-, where Ar = aryl group (e.g., C₆H₅), X = Cl^-, Br^-, NO₃^-, HSO₄^-, BF₄^-.
• Named by adding "diazonium" suffix to parent hydrocarbon + anion name (e.g., benzenediazonium chloride).
• Primary aliphatic amines form unstable alkyldiazonium salts.
• Primary aromatic amines form arenediazonium salts stable at 273-278 K.
• Arenediazonium salts are more stable due to positive charge delocalization over benzene ring.

Solved Examples

• Ex 1: Name C₆H₅N₂⁺Cl^-.
  Answer: Benzenediazonium chloride.

Practice Set

• Q1: What is the general formula of arenediazonium salts?
  A. RNH₂
  B. ArN₂⁺X^-
  C. RNO₂
  D. ArOH
• Q2: Which salt is more stable?
  A. Alkyldiazonium salt
  B. Arenediazonium salt
  C. Both equal
  D. None
• Q3: Benzenediazonium hydrogensulphate has formula:
  A. C₆H₅N₂⁺HSO₄^-
  B. C₆H₅NH₂
  C. C₆H₅NO₂
  D. C₆H₅Cl

Answer Key

• Q1: B - Arenediazonium salts have formula ArN₂⁺X^-.
• Q2: B - Arenediazonium salts are more stable due to resonance.
• Q3: A - Benzenediazonium hydrogensulphate is C₆H₅N₂⁺HSO₄^-.

Memory Aids

• Diazonium salts: ArN₂⁺X^-
• Stability: Arenediazonium > Alkyldiazonium
• Named by parent hydrocarbon + diazonium + anion

Structure of Amines

Geometry and Hybridization

• Nitrogen in amines is sp³ hybridized.
• Amines have pyramidal shape due to lone pair on nitrogen.
• Bond angle C–N–E (E = C or H) is less than 109.5°, e.g., 108° in trimethylamine.

Solved Examples

• Ex 1: Describe the shape of trimethylamine.
  Answer: Pyramidal due to lone pair on N.

Practice Set

• Q1: What is the hybridization of nitrogen in amines?
  A. sp
  B. sp²
  C. sp³
  D. sp³d
• Q2: What is the shape of amines?
  A. Linear
  B. Pyramidal
  C. Planar
  D. Tetrahedral
• Q3: Why is the bond angle in amines less than 109.5°?
  A. Lone pair repulsion
  B. Double bonds
  C. No lone pairs
  D. Hydrogen bonding

Answer Key

• Q1: C - Nitrogen is sp³ hybridized in amines.
• Q2: B - Amines have pyramidal shape due to lone pair.
• Q3: A - Lone pair repulsion reduces bond angle below 109.5°.

Memory Aids

• Amines: sp³ N, pyramidal shape
• Bond angle < 109.5° due to lone pair

Classification

Explanation

• Amines are classified based on the number of alkyl/aryl groups attached to the nitrogen atom.
• Primary amines (1°): One alkyl/aryl group attached to NH₂ (RNH₂).
• Secondary amines (2°): Two alkyl/aryl groups attached to NH (R₂NH).
• Tertiary amines (3°): Three alkyl/aryl groups attached to N (R₃N), no hydrogen on nitrogen.
• Examples:
  • Primary: Methanamine (CH₃-NH₂)
  • Secondary: N-Methyl methanamine (CH₃-NH-CH₃)
  • Tertiary: N,N-Dimethyl methanamine (CH₃-N(CH₃)-CH₃)

Solved Examples

• Ex 1: Identify the type of amine: CH₃-NH₂.
  Solution: One alkyl group attached to NH₂ → Primary amine.
• Ex 2: Identify the type of amine: CH₃-NH-CH₃.
  Solution: Two alkyl groups attached to NH → Secondary amine.
• Ex 3: Identify the type of amine: CH₃-N(CH₃)-CH₃.
  Solution: Three alkyl groups attached to N, no H → Tertiary amine.

Practice Set

• Q1: Which of the following is a primary amine?
  A. CH₃-NH-CH₃
  B. CH₃-NH₂
  C. N(CH₃)₃
  D. NH₃
• Q2: Identify the secondary amine:
  A. CH₃-NH₂
  B. CH₃-NH-CH₃
  C. N(CH₃)₃
  D. NH₃
• Q3: Which amine has no hydrogen attached to nitrogen?
  A. Primary amine
  B. Secondary amine
  C. Tertiary amine
  D. Ammonia

Answer Key

• Q1: B. CH₃-NH₂ - One alkyl group attached to NH₂ defines primary amine.
• Q2: B. CH₃-NH-CH₃ - Two alkyl groups attached to NH defines secondary amine.
• Q3: C. Tertiary amine - No hydrogen attached to nitrogen.

Memory Aids

• Primary amine = 1 alkyl group + NH₂.
• Secondary amine = 2 alkyl groups + NH.
• Tertiary amine = 3 alkyl groups, no H on N.
• Remember: "1° = 1 R, 2° = 2 R, 3° = 3 R" attached to N.

Nomenclature

Common System Naming

• Aliphatic amines named by prefixing alkyl group to "amine" (e.g., methylamine).

IUPAC System Naming

• Amines named as alkanamines.
• For secondary and tertiary amines, if alkyl groups are identical, use prefixes di-, tri- before alkyl name.

Solved Examples

• Ex 1: Name CH₃NH₂ - Methanamine (simple alkylamine).
• Ex 2: Name (CH₃)₂NCH₃ - N,N-Dimethyl methanamine (two methyl groups on N).

Practice Set

• Q1: What is the IUPAC name of CH₃NH₂?
  A. Methylamine
  B. Methanamine
  C. Ethylamine
  D. Methanamine N,N-dimethyl
• Q2: How is a tertiary amine with three identical alkyl groups named?
  A. Alkylamine
  B. Alkanamine
  C. Tri-alkylamine
  D. N,N-Dialkylamine
• Q3: What prefix is used when two identical alkyl groups are attached to nitrogen?
  A. Mono-
  B. Di-
  C. Tri-
  D. Tetra-

Answer Key

• Q1: B - Methanamine is the correct IUPAC name for CH₃NH₂.
• Q2: C - Tertiary amines with three identical alkyl groups use tri- prefix.
• Q3: B - Di- prefix is used for two identical alkyl groups on nitrogen.

Memory Aids

• Common system: Alkyl + amine (e.g., methylamine).
• IUPAC: Alkanamine; use di-, tri- for multiple identical alkyl groups.
• Look for N-substitution to apply prefixes.

Preparation from Nitro Compounds

Explanation

• Nitro compounds (R–NO₂) are reduced to amines (R–NH₂).
• Reduction agents: H₂ with Ni, Pd, Pt catalysts or metals in acid (e.g., Sn + HCl).
• Commercially, iron scrap and HCl are used; FeCl formed hydrolyses to release HCl, requiring only a small amount of acid.
• Examples: Nitroethane → Ethanamine, Nitrobenzene → Aniline.

Solved Examples

• Ex 1: Reduce nitrobenzene using H₂/Pd catalyst → Aniline (C₆H₅NH₂).

Practice Set

• Q1: Which catalyst is NOT used for reduction of nitro compounds to amines?
  • A. Nickel
  • B. Palladium
  • C. Copper
  • D. Platinum
• Q2: What is the product of reduction of nitroethane?
  • A. Ethanolamine
  • B. Ethanamine
  • C. Ethylamine
  • D. Ethane
• Q3: Why is iron scrap used commercially with HCl in nitro compound reduction?
  • A. To increase temperature
  • B. To generate HCl in situ
  • C. To act as catalyst
  • D. To absorb hydrogen

Answer Key

• Q1: C - Copper is not used; Ni, Pd, Pt are common catalysts.
• Q2: B - Reduction of nitroethane gives ethanamine (primary amine).
• Q3: B - FeCl hydrolyses to release HCl, reducing the need for added acid.

Memory Aids

• Reduction of R–NO₂ → R–NH₂ using H₂/Ni, Pd, Pt or Sn/HCl.
• Iron scrap + HCl generates HCl in situ for commercial use.

Preparation from Alkyl Halides

Explanation

• Alkyl halides (R–X) react with ethanolic ammonia to give a mixture of primary, secondary, tertiary amines and quaternary ammonium salts.
• This process is called Hoffmann ammonolysis (cleavage of C–X bond by NH₃).
• Excess ammonia favors primary amine formation.
• Order of halide reactivity: R–I > R–Br > R–Cl.
• Reaction is nucleophilic substitution; ammonium salt formed is treated with strong base to get amine.
• Aryl halides do not undergo this reaction.
• Tertiary butyl bromide prefers elimination over substitution with NH₃.

Solved Examples

• Ex 1: React CH₃Br with excess NH₃ → CH₃NH₂ (primary amine).

Practice Set

• Q1: What is the main product when excess alkyl halide is used in Hoffmann ammonolysis?
  • A. Primary amine
  • B. Secondary amine
  • C. Tertiary amine
  • D. Quaternary ammonium salt
• Q2: Which halide reacts fastest in ammonolysis?
  • A. R–Cl
  • B. R–Br
  • C. R–I
  • D. Aryl halide
• Q3: What happens when tertiary butyl bromide reacts with NH₃?
  • A. Substitution to amine
  • B. Elimination
  • C. No reaction
  • D. Polymerization

Answer Key

• Q1: C - Excess alkyl halide gives mainly tertiary amine.
• Q2: C - R–I is most reactive.
• Q3: B - Tertiary butyl bromide undergoes elimination, not substitution.

Memory Aids

• Hoffmann ammonolysis: R–X + NH₃ → mixture of amines + quaternary salt.
• Excess NH₃ → primary amine; excess R–X → tertiary amine.
• Reactivity order: I > Br > Cl; aryl halides do not react.

Preparation from Nitriles

Explanation

• Nitriles (R–CN) are reduced to primary amines (R–CH₂NH₂).
• Reducing agents: LiAlH₄ or catalytic hydrogenation (H₂/Ni).
• Reduction by Na(Hg) and C₂H₅OH is called Mendius reduction.
• Example: CH₃Cl + KCN → CH₃CN (ethanenitrile), then reduced to ethylamine.

Solved Examples

• Ex 1: Reduce ethanenitrile with LiAlH₄ → Ethylamine.

Practice Set

• Q1: Which reagent is used for reduction of nitriles to primary amines?
  • A. NaBH₄
  • B. LiAlH₄
  • C. H₂SO₄
  • D. KMnO₄
• Q2: Mendius reduction involves reduction by:
  • A. Na(Hg) and C₂H₅OH
  • B. LiAlH₄
  • C. H₂/Ni
  • D. Sn and HCl
• Q3: What is the product when CH₃CN is reduced?
  • A. Methylamine
  • B. Ethylamine
  • C. Propylamine
  • D. Aniline

Answer Key

• Q1: B - LiAlH₄ is a strong reducing agent for nitriles.
• Q2: A - Mendius reduction uses Na(Hg) and ethanol.
• Q3: B - CH₃CN reduction gives ethylamine (2 carbon atoms).

Memory Aids

• Nitrile reduction adds two hydrogens to form primary amine.
• Mendius reduction = Na(Hg) + C₂H₅OH.

Preparation from Amides

Explanation

• Alkanamides (R-CONH₂) are reduced to primary amines using LiAlH₄.
• Hoffmann bromamide degradation: Primary amides treated with Br₂ and alkali give primary amines with one less carbon.
• Mechanism involves formation of bromoamide, isocyanate intermediate, then amine.

Solved Examples

• Ex 1: Reduce ethanamide with LiAlH₄ → Ethanamine.
• Ex 2: Treat benzamide with Br₂ and KOH → Aniline (one carbon less).

Practice Set

• Q1: What is the product of Hoffmann bromamide degradation?
  • A. Primary amine with same carbon number
  • B. Primary amine with one less carbon
  • C. Secondary amine
  • D. Tertiary amine
• Q2: Which reagent reduces amides to amines?
  • A. NaBH₄
  • B. LiAlH₄
  • C. H₂/Ni
  • D. Br₂/KOH
• Q3: What intermediate forms in Hoffmann bromamide degradation?
  • A. Isocyanate
  • B. Nitrile
  • C. Amide
  • D. Ammonium salt

Answer Key

• Q1: B - Primary amine with one less carbon is formed.
• Q2: B - LiAlH₄ reduces amides to amines.
• Q3: A - Isocyanate intermediate forms during reaction.

Memory Aids

• Hoffmann bromamide degradation shortens carbon chain by one.
• LiAlH₄ reduces amides to primary amines.

Gabriel Phthalimide Synthesis

Explanation

• Potassium phthalimide reacts with alkyl halides to form N-alkyl phthalimide.
• Alkaline hydrolysis of N-alkyl phthalimide produces primary amines.
• Used to prepare primary amines without secondary/tertiary amines.
• 1° aromatic amines (aniline, toluidines) cannot be prepared as aryl halides do not react under mild conditions.

Solved Examples

• Ex 1: React potassium phthalimide with ethyl bromide → N-ethyl phthalimide, then hydrolyze → ethylamine.

Practice Set

• Q1: What is the main product of Gabriel synthesis?
  • A. Secondary amine
  • B. Tertiary amine
  • C. Primary amine
  • D. Quaternary ammonium salt
• Q2: Why can't aromatic amines be prepared by Gabriel synthesis?
  • A. Aryl halides do not undergo nucleophilic substitution
  • B. Reaction is too slow
  • C. Phthalimide is unstable
  • D. Aromatic amines are unstable
• Q3: Which reagent is used to hydrolyze N-alkyl phthalimide?
  • A. HCl
  • B. NaOH
  • C. KOH
  • D. NH₃

Answer Key

• Q1: C - Gabriel synthesis produces primary amines.
• Q2: A - Aryl halides do not undergo nucleophilic substitution under mild conditions.
• Q3: B - NaOH is used for hydrolysis.

Memory Aids

• Gabriel synthesis: potassium phthalimide + R–X → N-alkyl phthalimide → hydrolysis → primary amine.
• Does not work for aryl halides (no nucleophilic substitution).

Physical Properties

Physical State

• Lower aliphatic amines: gases with ammoniacal smell.
• Primary amines (≥3 C atoms): liquids with fishy odour.
• Lower aromatic amines: liquids with unpleasant odour.
• Higher aromatic amines: solids, odourless.

Colour

• Pure amines are colourless.
• Get coloured due to atmospheric oxidation.

Solubility

• Amines are soluble in water due to hydrogen bonding.
• Solubility decreases with increasing alkyl size.
• Higher and aromatic amines are insoluble in water but soluble in organic solvents.

Boiling Points

• Amines are polar and form intermolecular hydrogen bonds (except tertiary amines).
• Exist as associated molecules, showing higher boiling points than hydrocarbons of similar molecular mass.
• Boiling points are lower than alcohols and carboxylic acids of comparable molecular mass.
• Order of boiling points: 1° > 2° > 3° amines due to number of hydrogen atoms available for bonding.

Solved Examples

• Ex 1: Identify physical state of amine with 2 carbon atoms.
  Answer: Gas with ammoniacal smell (lower aliphatic amine).
• Ex 2: Compare boiling points of 1° and 3° amines.
  Answer: 1° amine has higher boiling point due to more hydrogen bonding.

Practice Set

• Q1: What is the physical state of primary amines with three or more carbon atoms?
  A. Gas
  B. Liquid
  C. Solid
  D. Colourless gas
• Q2: Why do pure amines get coloured on exposure to air?
  A. Reaction with water
  B. Atmospheric oxidation
  C. Hydrogen bonding
  D. Polymerization
• Q3: Which order correctly represents boiling points of isomeric amines?
  A. 3° > 2° > 1°
  B. 1° > 2° > 3°
  C. 2° > 1° > 3°
  D. 1° > 3° > 2°

Answer Key

• Q1: B - Primary amines with ≥3 carbons are liquids with fishy odour.
• Q2: B - Amines get coloured due to atmospheric oxidation.
• Q3: B - Boiling points decrease from 1° to 3° due to hydrogen bonding availability.

Memory Aids

• Lower aliphatic amines: gas + ammoniacal smell.
• Boiling point order: 1° (two H) > 2° (one H) > 3° (no H).
• Pure amines colourless; colour on air due to oxidation.

Chemical Reactions

Overview of Primary Amine Reactions

• Primary amines (RNH₂) undergo alkylation with alkyl halides (R'X) to form secondary, tertiary amines, and quaternary ammonium salts.
• Acylation with acid chlorides (R'COCl) or anhydrides forms amides (RNH-COR').
• Benzoylation with benzoyl chloride (C₆H₅COCl) produces benzamides (RNHCOC₆H₅).
• Carbylamine test: Primary amines react with chloroform and alcoholic KOH to form isocyanides (RNC), confirming primary amines.
• Hinsberg’s test: Primary amines react with benzenesulfonyl chloride to form soluble sulfonamides in alkali.
• Nitrous acid reaction: Primary amines react with HNO₂ to give alcohols, nitrogen gas, and water.

Solved Examples

• Ex 1: Alkylation of methylamine (CH₃NH₂) with ethyl bromide (C₂H₅Br) forms ethylmethylamine.
• Ex 2: Carbylamine test with an unknown amine producing foul-smelling isocyanide confirms primary amine.

Practice Set

• Q1: Which reagent is used in the carbylamine test?
  • A. Chloroform and alcoholic KOH
  • B. Benzoyl chloride
  • C. Nitrous acid
  • D. Alkyl halide
• Q2: Primary amines react with benzenesulfonyl chloride to form?
  • A. Insoluble sulfonamides
  • B. Soluble sulfonamides
  • C. Quaternary ammonium salts
  • D. Isocyanides
• Q3: Reaction of primary amines with nitrous acid produces?
  • A. Alcohols and nitrogen gas
  • B. N-nitrosamines
  • C. Nitrite salts
  • D. Diazonium salts

Answer Key

• Q1: A - Chloroform and alcoholic KOH produce isocyanides in carbylamine test.
• Q2: B - Primary amines form soluble sulfonamides with benzenesulfonyl chloride.
• Q3: A - Primary amines react with nitrous acid to give alcohols and nitrogen gas.

Memory Aids

• Carbylamine test positive only for primary amines.
• Hinsberg’s test solubility distinguishes 1°, 2°, and 3° amines.
• Nitrous acid reaction: primary amines → alcohols; aromatic primary amines → diazonium salts.

Basic Character of Amines

Key Points on Basicity

• Amines are basic due to lone pair on nitrogen.
• Aliphatic amines are stronger bases than ammonia due to +I effect of alkyl groups.
• Basicity affected by inductive, steric, solvation, and resonance effects.
• Gaseous phase basicity order: 3° amine > 2° amine > 1° amine > NH₃.
• Aqueous phase basicity varies due to solvation; e.g., (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₃ > NH₃.
• Aromatic amines are weaker bases due to resonance delocalization of lone pair.

Solved Examples

• Ex 1: Compare basicity of methylamine and aniline; methylamine is stronger base due to no resonance.
• Ex 2: Explain why 3° amines are more basic than 1° in aqueous phase due to steric hindrance reducing solvation.

Practice Set

• Q1: Why are aliphatic amines stronger bases than ammonia?
  • A. Resonance effect
  • B. +I effect of alkyl groups
  • C. Steric hindrance
  • D. Solvation effect
• Q2: Order of basicity in gaseous phase is?
  • A. 1° > 2° > 3° > NH₃
  • B. 3° > 2° > 1° > NH₃
  • C. NH₃ > 1° > 2° > 3°
  • D. 2° > 3° > 1° > NH₃
• Q3: Aromatic amines are weaker bases because?
  • A. Lone pair is localized on nitrogen
  • B. Lone pair is delocalized over benzene ring
  • C. They have no lone pair
  • D. They are sterically hindered

Answer Key

• Q1: B - +I effect of alkyl groups increases electron density on nitrogen.
• Q2: B - Gaseous phase basicity order is 3° > 2° > 1° > NH₃.
• Q3: B - Resonance delocalizes lone pair, reducing availability for protonation.

Memory Aids

• Basicity in aqueous phase affected by solvation and steric hindrance.
• Aromatic amines weaker bases due to resonance.
• 3° amines more basic than 1° in aqueous phase due to steric effects.

Distinction Between 1°, 2° and 3° Amines

Hinsberg’s Test

• Primary amines react with benzenesulfonyl chloride to form N-alkylbenzenesulfonamides soluble in alkali.
• Secondary amines form N,N-dialkylbenzenesulfonamides insoluble in alkali.
• Tertiary amines do not react.

Solved Examples

• Ex 1: Identify amine type by solubility of Hinsberg’s test product in alkali.

Practice Set

• Q1: Hinsberg’s test product soluble in alkali indicates?
  • A. Primary amine
  • B. Secondary amine
  • C. Tertiary amine
  • D. Quaternary ammonium salt
• Q2: Which amine does not react with Hinsberg’s reagent?
  • A. Primary
  • B. Secondary
  • C. Tertiary
  • D. Quaternary
• Q3: Hinsberg’s test distinguishes amines based on?
  • A. Solubility of sulfonamide
  • B. Color change
  • C. Gas evolution
  • D. Precipitate formation

Answer Key

• Q1: A - Soluble sulfonamide indicates primary amine.
• Q2: C - Tertiary amines do not react with Hinsberg’s reagent.
• Q3: A - Test is based on solubility differences of sulfonamides.

Memory Aids

• Hinsberg’s test: Primary soluble, secondary insoluble, tertiary no reaction.
• Carbylamine test positive only for primary amines.
• Azo dye test positive only for primary aromatic amines.

Electrophilic Substitution Reactions of Arylamines

Key Reactions of Aniline

• Aniline undergoes electrophilic substitution; —NH₂ is ortho/para directing and activating.
• Bromination forms tribromoaniline or p-bromoaniline.
• Nitration forms p-nitroaniline.
• Sulfonation with H₂SO₄ forms sulphanilic acid (zwitter ion).
• Diazotization with NaNO₂/HCl forms diazonium salt.
• Reaction with chloroform and KOH forms phenyl isocyanide.
• Oxidation forms p-benzoquinone.
• Aniline forms salt with AlCl₃ during alkylation, deactivating electrophilic substitution; no Friedel-Crafts reaction.

Solved Examples

• Ex 1: Bromination of aniline yields tribromoaniline due to activating —NH₂ group.
• Ex 2: Explain why aniline does not undergo Friedel-Crafts alkylation.

Practice Set

• Q1: The —NH₂ group in aniline directs substitution to?
  • A. Meta position
  • B. Ortho and para positions
  • C. Only para position
  • D. No substitution
• Q2: Why does aniline not undergo Friedel-Crafts reaction?
  • A. It is too reactive
  • B. Forms salt with AlCl₃ deactivating ring
  • C. No lone pair on nitrogen
  • D. It is a weak base
• Q3: Diazonium salt formation requires?
  • A. NaNO₂ and HCl at 0-5 °C
  • B. Chloroform and KOH
  • C. Bromine water
  • D. Concentrated H₂SO₄

Answer Key

• Q1: B - —NH₂ is ortho/para directing.
• Q2: B - Aniline forms salt with AlCl₃, deactivating electrophilic substitution.
• Q3: A - Diazotization requires NaNO₂ and HCl at low temperature.

Memory Aids

• —NH₂ group activates and directs ortho/para substitution.
• Aniline salt formation with AlCl₃ prevents Friedel-Crafts reaction.
• Diazonium salts form at 0-5 °C with NaNO₂/HCl.

Basicity of Substituted Anilines

Effect of Substituents

• Electron releasing groups (-OCH₃, -CH₃) increase electron density on nitrogen, enhancing basicity.
• Electron withdrawing groups (-NO₂, -SO₃H, -COOH, -X) decrease basicity.
• pK_b values indicate strength: lower pK_b means stronger base.

Solved Examples

• Ex 1: Compare pK_b of p-methylaniline (9.21) and p-nitroaniline (13.0); methyl group increases basicity.

Practice Set

• Q1: Which group increases basicity of aniline?
  • A. -NO₂
  • B. -SO₃H
  • C. -CH₃
  • D. -COOH
• Q2: Lower pK_b value indicates?
  • A. Weaker base
  • B. Stronger base
  • C. No change
  • D. Acidic nature
• Q3: Electron withdrawing groups cause?
  • A. Increase in basicity
  • B. Decrease in basicity
  • C. No effect
  • D. Increase in acidity

Answer Key

• Q1: C - Electron donating -CH₃ increases basicity.
• Q2: B - Lower pK_b means stronger base.
• Q3: B - Electron withdrawing groups decrease basicity.

Memory Aids

• Electron donating groups increase basicity; withdrawing groups decrease it.
• pK_b values help compare basic strengths.

Anilinium Ion Stability

Key Points

• Anilinium ion (protonated aniline) has fewer resonance structures (2) than aniline (5), making it less stable.
• Protonation reduces electron delocalization on nitrogen.

Solved Examples

• Ex 1: Explain why anilinium ion is less stable than aniline based on resonance.

Practice Set

• Q1: Anilinium ion is less stable because?
  • A. More resonance structures
  • B. Fewer resonance structures
  • C. No lone pair on nitrogen
  • D. It is neutral
• Q2: Protonation affects aniline by?
  • A. Increasing electron delocalization
  • B. Decreasing resonance
  • C. No effect
  • D. Increasing basicity
• Q3: Resonance in aniline involves?
  • A. Benzene ring and nitrogen lone pair
  • B. Only benzene ring
  • C. Only nitrogen lone pair
  • D. No resonance

Answer Key

• Q1: B - Anilinium ion has fewer resonance forms, less stable.
• Q2: B - Protonation reduces resonance delocalization.
• Q3: A - Resonance involves benzene ring and nitrogen lone pair.

Memory Aids

• Protonation reduces resonance, lowering stability.
• Aniline has 5 resonance forms; anilinium ion has 2.

Method of Preparation of Diazonium Salts

Explanation

• Benzenediazonium chloride is prepared by reacting aniline (C₆H₅NH₂) with nitrous acid (HNO₂), generated in situ from sodium nitrite (NaNO₂) and hydrochloric acid (HCl).
• The reaction is carried out at low temperature (0-5 °C) to maintain the stability of the diazonium salt.
• This process is called diazotisation — conversion of primary aromatic amines into diazonium salts.
• Diazonium salts are unstable and generally used immediately after preparation.
• Chemical reaction:
  • C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

Solved Examples

• Ex 1: Write the reaction for the preparation of benzenediazonium chloride from aniline.
  • React aniline with NaNO₂ and HCl at 0-5 °C.
  • Product formed: C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride).

Practice Set

• Q1: What is the temperature range for the diazotisation reaction?
  • A. 25-30 °C
  • B. 0-5 °C
  • C. 50-60 °C
  • D. 10-15 °C
• Q2: Which reagent is NOT used in the preparation of benzenediazonium chloride?
  • A. Aniline
  • B. Sodium nitrite
  • C. Hydrochloric acid
  • D. Sulfuric acid
• Q3: What are the by-products formed during the diazotisation of aniline?
  • A. NaCl and H₂O
  • B. NaOH and H₂O
  • C. HCl and NaNO₂
  • D. NH₃ and HCl

Answer Key

• Q1: B. 0-5 °C — Low temperature is required to stabilize the diazonium salt.
• Q2: D. Sulfuric acid — HCl is used, not H₂SO₄.
• Q3: A. NaCl and H₂O — These are the by-products formed in the reaction.

Memory Aids

• Diazotisation = Primary aromatic amine + NaNO₂ + HCl at 0-5 °C → Diazonium salt.
• Keep temperature low to prevent decomposition.
• Diazonium salts are unstable; use immediately.

Physical Properties

Properties of Benzenediazonium Chloride

• Colourless crystalline solid.
• Readily soluble in water.
• Stable when kept cold.
• Reacts with water on warming.
• Decomposes easily in dry state.

Properties of Benzenediazonium Fluoroborate

• Insoluble in water.
• Stable at room temperature.

Solved Examples

• Ex 1: Identify the solubility of benzenediazonium chloride.
  • Answer: Soluble in water.
• Ex 2: State the stability condition of benzenediazonium fluoroborate.
  • Answer: Stable at room temperature.

Practice Set

• Q1: What is the physical state of benzenediazonium chloride?
  • A. Colourless crystalline solid
  • B. Coloured liquid
  • C. Powdered solid
  • D. Gaseous
• Q2: Which of the following is true about benzenediazonium chloride?
  • A. Insoluble in water
  • B. Stable when warm
  • C. Decomposes easily when dry
  • D. Stable at room temperature
• Q3: Benzenediazonium fluoroborate is:
  • A. Water soluble and unstable
  • B. Water insoluble and stable at room temperature
  • C. Water soluble and stable at room temperature
  • D. Water insoluble and unstable

Answer Key

• Q1: A - It is a colourless crystalline solid as per its physical description.
• Q2: C - It decomposes easily in dry state and is stable only when cold.
• Q3: B - Benzenediazonium fluoroborate is water insoluble and stable at room temperature.

Memory Aids

• Benzenediazonium chloride: "Cold stable, water soluble, dry state decomposes".
• Benzenediazonium fluoroborate: "Water insoluble, room temperature stable".

Reactions involving displacement of nitrogen (diazo group)

Explanation

• Arenediazonium salts (ArN₂⁺X^-) are key intermediates in aromatic substitution.
• Displacement of N₂ group leads to various substituted aromatic compounds.
• Common reagents and products:
  • KI → ArI (aryl iodide)
  • HBF₄, heat (Δ) → ArF (aryl fluoride) via Balz-Schiemann reaction
  • H₂O, high temperature → ArOH (phenol)
  • NaNO₂, Cu → ArNO₂ (nitrobenzene)
  • NaOH, benzene (C₆H₆) → ArC₆H₅ (diphenyl) via Gomberg reaction
  • Sandmeyer and Gattermann reactions use Cu salts to form ArCl, ArBr, ArCN
  • Conversion back to ArH (aromatic hydrocarbon) is also possible.

Solved Examples

• Ex 1: Convert ArN₂⁺X^- to ArI using KI.
  • Step 1: Add KI to arenediazonium salt.
  • Step 2: ArN₂⁺X^- + KI → ArI + N₂ + KX
• Ex 2: Prepare ArF from arenediazonium salt.
  • Step 1: Treat ArN₂⁺X^- with HBF₄ and heat.
  • Step 2: ArN₂⁺X^- → ArF + N₂ + BF₃

Practice Set

• Q1: Which reagent converts ArN₂⁺X^- to ArI?
  • A. NaOH
  • B. KI
  • C. HBF₄
  • D. CuCl
• Q2: The Balz-Schiemann reaction produces:
  • A. ArCl
  • B. ArF
  • C. ArBr
  • D. ArCN
• Q3: Which reaction forms diphenyl (ArC₆H₅)?
  • A. Sandmeyer reaction
  • B. Gomberg reaction
  • C. Gattermann reaction
  • D. Balz-Schiemann reaction

Answer Key

• Q1: B - KI replaces diazo group with iodine forming ArI.
• Q2: B - Balz-Schiemann reaction converts diazonium salt to aryl fluoride (ArF).
• Q3: B - Gomberg reaction couples arenediazonium salt with benzene to form diphenyl.

Memory Aids

• "KI for Iodine" - KI replaces N₂⁺ with I.
• Balz-Schiemann = ArF formation using HBF₄ and heat.
• Sandmeyer uses Cu salts for ArCl, ArBr, ArCN.
• Gomberg reaction forms diphenyl from ArN₂⁺ and benzene.

Reactions involving retention of diazo group (coupling reactions)

Explanation

• Diazonium salts (ArN₂⁺X^-) can couple with aromatic compounds retaining the diazo group.
• Coupling depends on pH and reagents, producing azo dyes.
• Example: p-aminoazobenzene forms diazonium salt intermediate at low temperature and acidic pH.
• Adding phenol under basic conditions (pH 9-10) forms p-hydroxyazobenzene (orange dye).
• In p-aminophenol, coupling position depends on pH:
  • Acidic (pH 4-5): coupling ortho to amino group.
  • Basic (pH 9-10): coupling ortho to phenolic group.

Solved Examples

• Ex 1: Formation of p-hydroxyazobenzene from p-aminoazobenzene.
  • Step 1: Diazotize p-aminoazobenzene with aniline and acid at 0-5°C, pH 4-5.
  • Step 2: Add phenol and base (OH^-) at pH 9-10, low temperature.
  • Step 3: Diazonium salt couples with phenol forming p-hydroxyazobenzene.
• Ex 2: p-Aminophenol coupling position changes with pH.
  • Acidic pH (4-5): Coupling ortho to -NH₂ group forming 4-amino-3-phenylazophenol.
  • Basic pH (9-10): Coupling ortho to -OH group forming 4-amino-2-phenylazophenol.

Practice Set

• Q1: What is the product when p-aminoazobenzene is diazotized and reacted with phenol at pH 9-10?
  • A. p-Aminoazobenzene
  • B. p-Hydroxyazobenzene
  • C. Phenol
  • D. Aniline
• Q2: In acidic medium, coupling of p-aminophenol occurs at which position?
  • A. Ortho to -OH
  • B. Ortho to -NH₂
  • C. Para to -OH
  • D. Para to -NH₂
• Q3: Which pH favors coupling ortho to phenolic group in p-aminophenol?
  • A. pH 4-5
  • B. pH 7
  • C. pH 9-10
  • D. pH 1-2

Answer Key

• Q1: B - Phenol reacts with diazonium salt at basic pH to form p-hydroxyazobenzene.
• Q2: B - Acidic medium causes coupling ortho to amino (-NH₂) group.
• Q3: C - Basic pH (9-10) favors coupling ortho to phenolic (-OH) group.

Memory Aids

• Coupling reactions retain diazo group forming azo dyes.
• Acidic pH → coupling near amino group.
• Basic pH → coupling near phenol group.
• p-Aminoazobenzene → diazotization → azo dye formation.