An aromatic compound 'A' ( C 7 H 6 O 2 ) on treatment with aqueous ammonia and heating forms compound 'B', which on heating with Br 2 and KOH yields compound 'C' ( C 6 H 7 N ). Identify A, B, and C with equations, and write how 'C' is converted into benzene.

Identification (NCERT Ex 9.10): A = Benzoic acid ( C 6 H 5 COOH ) B = Benzamide ( C 6 H 5 CONH 2 ) — formed via ammonium benzoate upon heating with NH 3 C = Aniline / Benzenamine ( C 6 H 5 NH 2 ) — formed by Hoffmann bromamide degradation of B with Br 2 /KOH Conversion of Aniline (C) to Benzene: Diazotise with NaNO 2 + HCl at 273–278 K to form C 6 H 5 N 2 + Cl - , then reduce with H 3 PO 2 + H 2 O (or C 2 H 5 OH ) to obtain Benzene ( C 6 H 6 ) .

NCERT Plus Two (Class 12) Chemistry • Chapter 9

Amines

Q: Describe Hinsberg's test to distinguish between primary, secondary, and tertiary amines with chemical equations.

Treat the amine with benzenesulphonyl chloride ( C 6 H 5 SO 2 Cl , Hinsberg's reagent) followed by aqueous KOH : Primary (1°) amine: Forms N -alkylbenzenesulphonamide ( C 6 H 5 SO 2 NHR ), which has an acidic N–H hydrogen and is soluble in aqueous KOH . Secondary (2°) amine: Forms N,N -dialkylbenzenesulphonamide ( C 6 H 5 SO 2 NR 2 ), which has no N–H hydrogen and is insoluble in aqueous KOH . Tertiary (3°) amine: Does not react with benzenesulphonyl chloride.

Q: Explain (i) Carbylamine reaction and (ii) Hoffmann bromamide degradation reaction with balanced equations.

(i) Carbylamine reaction: Aliphatic or aromatic 1° amines heated with chloroform and ethanolic KOH form foul-smelling isocyanides (carbylamines): R–NH 2 + CHCl 3 + 3 KOH (alc.) ──(Δ)──▶ R–NC + 3 KCl + 3 H 2 O (ii) Hoffmann bromamide degradation: Treating a primary amide with Br 2 in aqueous/ethanolic NaOH yields a 1° amine with one carbon atom less: R–CONH 2 + Br 2 + 4 NaOH ──▶ R–NH 2 + Na 2 CO 3 + 2 NaBr + 2 H 2 O

Q: Give reasons: (i) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. (ii) Aniline does not undergo Friedel-Crafts reaction.

(i) Gabriel limitation: Aryl halides do not undergo nucleophilic substitution ( S N 2 ) with the potassium phthalimide anion (due to partial double-bond character of the C–X bond by resonance). (ii) No Friedel-Crafts: Aniline (Lewis base) reacts with anhydrous AlCl 3 (Lewis acid catalyst) to form a salt ( C 6 H 5 NH 2 + AlCl 3 - ), giving nitrogen a positive charge that strongly deactivates the benzene ring.

Q: Although the –NH 2 group is ortho and para directing, direct nitration of aniline gives a substantial amount (47%) of m -nitroaniline. Why? How can pure p -nitroaniline be obtained?

Reason: In the strongly acidic nitrating mixture (conc. HNO 3 + conc. H 2 SO 4 ), aniline is protonated to form the anilinium ion ( C 6 H 5 NH 3 + ) , which is strongly meta-directing , producing 47% m -nitroaniline. Preparation of pure p -nitroaniline: Protect –NH 2 by acetylation with acetic anhydride to form acetanilide , nitrate at 288 K to get p -nitroacetanilide (major), and hydrolyse to yield pure p -nitroaniline .

Q: What is diazotisation? Write equations for the conversion of benzenediazonium chloride into (a) chlorobenzene (Sandmeyer) and (b) p -hydroxyazobenzene (Orange dye).

Diazotisation: Reaction of aniline with nitrous acid ( NaNO 2 + 2 HCl ) at 273–278 K to form benzenediazonium chloride ( C 6 H 5 N 2 + Cl - ). (a) Sandmeyer: C 6 H 5 N 2 + Cl - ──(CuCl / HCl)──▶ C 6 H 5 Cl + N 2 ↑ (b) Coupling: C 6 H 5 N 2 + Cl - + C 6 H 5 OH ──(OH - )──▶ p-HO–C 6 H 4 –N=N–C 6 H 5 (Orange dye) + Cl - + H 2 O

Q: How will you accomplish the following conversions? (i) Aniline into p -bromoaniline (ii) Ethanoic acid into methanamine (iii) Aniline into 1,3,5-tribromobenzene

Aniline → p -bromoaniline: Acetylate with (CH 3 CO) 2 O / pyridine → Acetanilide → Br 2 / CH 3 COOH → p -bromoacetanilide → OH - or H + hydrolysis → p -bromoaniline . Ethanoic acid → methanamine: CH 3 COOH + NH 3 , Δ → CH 3 CONH 2 (Ethanamide) ──(Br 2 / NaOH)──▶ CH 3 NH 2 (Methanamine). Aniline → 1,3,5-tribromobenzene: C 6 H 5 NH 2 + 3 Br 2 (aq) → 2,4,6-tribromoaniline → diazotise ( NaNO 2 /HCl, 273–278 K ) → reduce with H 3 PO 2 + H 2 O → 1,3,5-tribromobenzene .

Classification, Gabriel Phthalimide & Hoffmann Bromamide, Aqueous vs Gas Basicity Order, Carbylamine & Hinsberg Tests, Diazonium & Azo Dyes

📄 Official NCERT Textbook PDF

Topic 9.1 – 9.3: Structure, Classification (1°, 2°, 3°) & IUPAC Nomenclature of Amines

Every time your heart races during an exam surprise, adrenaline (a 2° amine) surges through your bloodstream; when a dentist numbs your tooth with Novocain or an allergy pill like Benadryl (a 3° amine) calms a sneeze, amines are at work! Picture an ammonia molecule (NH₃) as a three-legged tripod carrying a buzzing lone-pair umbrella on top. Swap one, two, or all three hydrogen legs for carbon chains or benzene rings, and you build primary (1°), secondary (2°), and tertiary (3°) amines—the nitrogen backbone of proteins, vitamins, alkaloids, and colourful dyes!

Amines are organic derivatives of ammonia (NH₃) obtained by replacing one, two, or all three hydrogen atoms with alkyl (R) and/or aryl (Ar) groups.

1. Structure of Amines (9.1)

Like ammonia, the nitrogen atom in amines is trivalent and carries one unshared pair of electrons.
Nitrogen is sp³ hybridised with a pyramidal geometry: three sp³ orbitals overlap with H or C orbitals, while the fourth holds the unshared lone pair.
Because lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion, the C–N–E angle (E = C or H) is compressed below the tetrahedral 109.5°—for example, it is 108° in trimethylamine, (CH₃)₃N.

2. Classification (9.2)

Primary (1°): One H replaced → R–NH₂ or Ar–NH₂ (e.g., CH₃NH₂, C₆H₅NH₂).
Secondary (2°): Two H replaced → R–NH–R' (e.g., CH₃NHC₂H₅).
Tertiary (3°): All three H replaced → R–NR'R'' (e.g., (CH₃)₃N).
Simple vs Mixed: 2° and 3° amines are simple if all attached alkyl/aryl groups are identical, and mixed if different.

3. Common & IUPAC Nomenclature (9.3, Table 9.1)

1° Aliphatic: Drop final -e of alkane and add -amine → alkanamine (CH₃NH₂ = methanamine). When two or more –NH₂ groups exist, retain the letter 'e': H₂N–CH₂–CH₂–NH₂ = ethane-1,2-diamine.
2° & 3° Amines: Use locant N for substituents on nitrogen; longest carbon chain forms the parent alkanamine.
Arylamines: Simplest is C₆H₅NH₂ (Aniline, also accepted IUPAC name; systematically Benzenamine).

Formula	Class	Common Name	IUPAC Name
`CH₃CH₂NH₂`	1°	Ethylamine	Ethanamine
`(CH₃)₂CHNH₂`	1°	Isopropylamine	Propan-2-amine
`CH₃NHCH₂CH₃`	2° (mixed)	Ethylmethylamine	N-Methylethanamine
`(CH₃CH₂)₃N`	3° (simple)	Triethylamine	N,N-Diethylethanamine
`CH₂=CH–CH₂–NH₂`	1°	Allylamine	Prop-2-en-1-amine
`H₂N–(CH₂)₆–NH₂`	1° diamine	Hexamethylenediamine	Hexane-1,6-diamine
`C₆H₅NHCH₃`	2° (aryl)	N-Methylaniline	N-Methylbenzenamine

Isomerism in C₄H₁₁N (NCERT Intext 9.2): Amines exhibit chain (butan-1-amine vs 2-methylpropan-1-amine), position (butan-1-amine vs butan-2-amine), metamerism (C₂H₅NHC₂H₅ vs CH₃NHC₃H₇), and functional isomerism (1° vs 2° vs 3° amines)!

Do NOT classify amines by the degree of the carbon atom! In (CH₃)₃C–NH₂ (tert-butylamine / 2-methylpropan-2-amine), nitrogen is attached to a tertiary carbon, yet it is a PRIMARY (1°) amine because only ONE hydrogen of NH₃ has been replaced! Also remember: ethane-1,2-diamine keeps the letter 'e'.

Topic 9.4.1 – 9.4.4: Preparation of Amines (Part 1): Reduction of Nitro, Nitriles & Amides, and Ammonolysis

Suppose you need to synthesise a primary amine in a factory or research lab. You have four classic toolboxes: strip oxygens off a nitro compound using scrap iron and a tiny drop of acid that keeps recycling itself; fire excess ammonia at an alkyl halide in a sealed pressure tube at 373 K; stretch a carbon chain by one extra carbon using cyanide followed by reduction; or strip the carbonyl oxygen off an amide using LiAlH₄!

1. Reduction of Nitro Compounds

Nitroarenes or nitroalkanes are reduced to primary amines by catalytic hydrogenation (H₂ / Pd, Ni or Pt in ethanol) or active metals in acidic medium (Sn + HCl or Fe + HCl).

Ar–NO₂ + 3 H₂ ──(Pd / Ethanol)──▶ Ar–NH₂ + 2 H₂O
R–NO₂ + 6 [H] ──(Fe + HCl or Sn + HCl)──▶ R–NH₂ + 2 H₂O

Why Fe scrap + HCl is preferred: FeCl₂ formed during the reaction undergoes hydrolysis to release HCl back into the mixture. Hence, only a very small catalytic amount of HCl is needed to initiate the reaction!

2. Ammonolysis of Alkyl Halides

Cleavage of the C–X bond by an ethanolic solution of ammonia in a sealed tube at 373 K yields a primary amine via nucleophilic substitution. However, the 1° amine itself acts as a nucleophile and attacks more R–X, yielding a mixture of 1°, 2°, 3° amines and quaternary ammonium salt (R₄N⁺X^-).

Free amine is liberated from the ammonium salt using a strong base (R–NH₃⁺X^- + NaOH → R–NH₂ + H₂O + NaX).
How to obtain 1° amine as major product: Use a large excess of ammonia.
Order of reactivity of halides: R–I > R–Br > R–Cl.

NCERT Example 9.1:
(i) Ethanolic NH₃ + C₂H₅Cl → mixture of C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, and (C₂H₅)₄N⁺Cl^-.
(ii) C₆H₅CH₂Cl + NH₃ → C₆H₅CH₂NH₂ (Benzylamine) + 2 CH₃Cl → C₆H₅CH₂N(CH₃)₂ (N,N-Dimethylphenylmethanamine) + 2 HCl.

3. Reduction of Nitriles (Ascent of Amine Series)

Nitriles reduced with LiAlH₄, H₂ / Ni, or Na(Hg) / C₂H₅OH produce 1° amines containing one carbon atom more than the starting halide!

NCERT Example 9.2 (Step-Up by +1 Carbon):
(i) CH₃CH₂Cl ──(alc. KCN)──▶ CH₃CH₂CN ──(H₂/Ni or LiAlH₄)──▶ CH₃CH₂CH₂NH₂ (Propan-1-amine)
(ii) C₆H₅CH₂Cl ──(KCN)──▶ C₆H₅CH₂CN ──(H₂/Ni)──▶ C₆H₅CH₂CH₂NH₂ (2-Phenylethanamine)

4. Reduction of Amides (Same Carbon Count)

Amides treated with LiAlH₄ followed by H₂O yield primary amines with the exact same number of carbons: R–CONH₂ ──(i) LiAlH₄, (ii) H₂O──▶ R–CH₂NH₂.

Do NOT confuse the two amide reactions! Reducing R–CONH₂ with LiAlH₄ converts C=O into CH₂ (same number of carbons: R–CH₂NH₂), whereas Hoffmann bromamide (Br₂/NaOH) removes C=O completely (one carbon less: R–NH₂)! Also remember: aryl halides do NOT easily undergo ammonolysis.

Topic 9.4.5 – 9.4.6: Preparation of Amines (Part 2): Gabriel Phthalimide Synthesis & Hoffmann Bromamide Degradation

Because ammonolysis makes a messy soup of 1°, 2°, 3°, and 4° products, chemists invented two laser-accurate routes to pure primary amines! First, Gabriel Phthalimide Synthesis uses phthalimide as a one-seat chaperone: only ONE alkyl group can attach to nitrogen, guaranteeing 100% pure 1° aliphatic amine! Second, Hoffmann Bromamide Degradation acts like chemical scissors—add bromine and alkali to an amide, snip away the C=O carbon, and step smoothly down by one carbon!

5. Gabriel Phthalimide Synthesis (Pure 1° Aliphatic Amines)

Phthalimide is treated with ethanolic KOH to form potassium phthalimide. Heating this salt with an alkyl halide (R–X) forms N-alkylphthalimide, which upon alkaline hydrolysis (aq. NaOH) yields pure primary aliphatic amine + sodium phthalate.

Phthalimide ──(alc. KOH)──▶ K⁺ Phthalimide^- ──(R–X, Δ)──▶ N-Alkylphthalimide ──(aq. NaOH)──▶ R–NH₂ (Pure 1° Amine)

Why Aromatic 1° Amines (like Aniline) CANNOT be prepared by Gabriel synthesis: Aryl halides (C₆H₅–X) have partial double-bond character in the C–X bond due to resonance and do NOT undergo nucleophilic substitution (S_N2) with the phthalimide anion!

6. Hoffmann Bromamide Degradation Reaction (Step-Down by 1 Carbon)

When a primary amide (R–CONH₂ or Ar–CONH₂) is treated with bromine (Br₂) in an aqueous or ethanolic solution of sodium hydroxide (NaOH), migration of the alkyl or aryl group occurs from carbonyl carbon to nitrogen. The resulting primary amine contains one carbon less than the parent amide (descent of series).

R–CONH₂ + Br₂ + 4 NaOH ──▶ R–NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

NCERT Example 9.3:
(i) Which amide gives Propanamine (3 C) by Hoffmann bromamide reaction? → Must start with a 4-carbon amide: Butanamide (CH₃CH₂CH₂CONH₂).
(ii) Which amine is produced by Hoffmann degradation of Benzamide (C₆H₅CONH₂, 7 C)? → Aniline / Benzenamine (C₆H₅NH₂, 6 C)!

Memorise this golden contrast for 2-mark board questions: Gabriel phthalimide synthesis produces only aliphatic (and benzylic) 1° amines—it CANNOT produce Aniline! In contrast, Hoffmann bromamide degradation prepares both aliphatic AND aromatic 1° amines (Benzamide → Aniline), always with one carbon atom less than the starting amide!

Topic 9.5: Physical Properties of Amines: Odour, Solubility & Boiling Point Trends

Why does a forgotten fish stall smell unmistakable—and why do three isomeric amines weighing the exact same 73 g mol^-1 boil 40 K apart? Imagine molecules holding hands in a liquid crowd via intermolecular hydrogen bonds. Primary n-C₄H₉NH₂ has two N–H hands to grip neighbours; secondary (C₂H₅)₂NH has only one N–H hand; and tertiary C₂H₅N(CH₃)₂ has zero N–H hands—floating free and escaping into vapour 40 K earlier!

1. State, Colour & Odour

Lower aliphatic amines are gases with a characteristic fishy odour; 1° amines with 3 or more carbons are liquids; higher ones are solids.
Aniline and other arylamines are colourless when freshly prepared, but become coloured on storage due to atmospheric oxidation.

2. Solubility in Water

Lower aliphatic amines are soluble in water because they form hydrogen bonds with water molecules.
Solubility decreases with increase in molar mass as the size of the hydrophobic alkyl part grows (C₂H₅NH₂ > (C₂H₅)₂NH > C₆H₅NH₂; aniline is essentially insoluble in water).
Alcohols vs Amines: Oxygen (electronegativity 3.5) is more electronegative than Nitrogen (3.0), so alcohols are more polar and form stronger intermolecular H-bonds than amines. Hence butan-1-ol is more soluble and boils higher than butan-1-amine.

3. Boiling Points of Isomeric Amines (Table 9.2)

Primary and secondary amines undergo intermolecular association via N–H···N hydrogen bonding (stronger in 1° with 2 H atoms than in 2° with 1 H atom). Tertiary amines lack N–H hydrogens, so they have no intermolecular H-bonding! Thus, for isomeric amines of comparable molar mass:

Boiling Point Order: Alcohol > 1° Amine > 2° Amine > 3° Amine > Alkane

Compound (Table 9.2)	Class	Molar Mass	Boiling Point (K)	Intermolecular H-Bonds
`n-C₄H₉OH`	1° Alcohol	74	390.3 K	Strong O–H···O
`n-C₄H₉NH₂`	1° Amine	73	350.8 K	Extensive (2 N–H per molecule)
`(C₂H₅)₂NH`	2° Amine	73	329.3 K	Moderate (1 N–H per molecule)
`C₂H₅N(CH₃)₂`	3° Amine	73	310.5 K	None (0 N–H; dipole only)
`C₂H₅CH(CH₃)₂`	Alkane	72	300.8 K	None (weak van der Waals)

Two classic traps: (1) Even though tertiary amines have NO intermolecular H-bonding among themselves (giving them the lowest b.p. among isomeric amines), lower 3° amines like (CH₃)₃N still dissolve in water because nitrogen's lone pair accepts H-bonds from water's O–H! (2) Always check electronegativity: O (3.5) > N (3.0) makes C₂H₅OH boil higher than C₂H₅NH₂.

Topic 9.6.1: Basicity of Amines: Gas Phase vs Aqueous Phase (+I, Solvation & Steric Effects) & Aniline

Suppose nitrogen's lone pair is an open hand offering an electron pair to catch an incoming proton (H⁺). In empty gas phase, alkyl groups simply push more electrons onto N (+I effect), so 3° > 2° > 1° > NH₃ wins smoothly! Now drop the reaction into water: water molecules swarm around the ammonium cation to stabilise it by H-bond solvation (RNH₃⁺ gets 3 water hugs, R₂NH₂⁺ gets 2, R₃NH⁺ gets just 1!), while bulky alkyl arms block water (**steric hindrance**). In this three-way tug-of-war, 2° amine emerges champion in water!

Amines carry a lone pair on nitrogen and behave as Lewis bases, reacting with mineral acids to form water-soluble ammonium salts (regenerated upon adding NaOH).

R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH^- | pK_b = −log K_b
Rule: Larger K_b or Smaller pK_b = Stronger Base! (NH₃ has pK_b = 4.75)

1. Gas Phase vs Aqueous Phase Order (Alkanamines vs NH₃)

Gas Phase: Only the +I (electron-releasing) effect acts to stabilise the ammonium cation → 3° > 2° > 1° > NH₃.
Aqueous Phase: Basic strength depends on a delicate interplay of three factors: (i) +I inductive effect (favours 3°), (ii) Solvation / H-bonding of substituted ammonium ion with water (favours 1°: RNH₃⁺ > R₂NH₂⁺ > R₃NH⁺), and (iii) Steric hindrance of bulky alkyl groups blocking H-bonding (disfavours 3°).

Aqueous Ethyl Series ("231"): (C₂H₅)₂NH (3.00) > (C₂H₅)₃N (3.25) > C₂H₅NH₂ (3.29) > NH₃ (4.75)
Aqueous Methyl Series ("213"): (CH₃)₂NH (3.27) > CH₃NH₂ (3.38) > (CH₃)₃N (4.22) > NH₃ (4.75)

Amine (Table 9.3, Aqueous)	pK_b	Amine	pK_b
`(C₂H₅)₂NH` (N-Ethylethanamine)	3.00	`(CH₃)₂NH` (N-Methylmethanamine)	3.27
`(C₂H₅)₃N` (N,N-Diethylethanamine)	3.25	`CH₃NH₂` (Methanamine)	3.38
`C₂H₅NH₂` (Ethanamine)	3.29	`(CH₃)₃N` (N,N-Dimethylmethanamine)	4.22
`C₆H₅CH₂NH₂` (Phenylmethanamine)	4.70	`NH₃` (Ammonia)	4.75
`C₆H₅N(CH₃)₂` (N,N-Dimethylaniline)	8.92	`C₆H₅NH₂` (Benzenamine / Aniline)	9.38

2. Why Aniline (pK_b = 9.38) is Far Weaker than NH₃ & Aliphatic Amines

In aniline, the –NH₂ group is attached directly to the benzene ring, so the lone pair on N enters conjugation (resonance) with the ring (5 resonating structures), making the lone pair less available for protonation.
The anilinium ion (C₆H₅NH₃⁺) formed after accepting a proton has only 2 Kekulé resonating structures—so aniline is far more stabilised than anilinium ion, lowering basicity!
Benzylamine (C₆H₅CH₂NH₂, pK_b = 4.70): The –CH₂– spacer blocks conjugation with the ring, so it is a much stronger base than aniline!
Substituted Anilines: Electron-releasing groups (–OCH₃, –CH₃) increase basicity; electron-withdrawing groups (–NO₂, –SO₃H, –COOH, –X) decrease basicity (p-nitroaniline < aniline < p-toluidine).

NCERT Exercise 9.3(iii): Why does aqueous methylamine precipitate hydrated ferric oxide with FeCl₃?
Methylamine (more basic than water) generates OH^- ions in water (CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH^-), which react with Fe³⁺ to precipitate reddish-brown Fe(OH)₃ / Fe₂O₃·xH₂O!

Two giant exam pitfalls: (1) Remember 231 vs 213 in water—for ethyl amines, 2° > 3° > 1°, whereas for methyl amines, 2° > 1° > 3°! (2) Read the wording carefully: "decreasing order of pK_b" means weakest base first → strongest base last, because higher pK_b = weaker base!

Topic 9.6.2 – 9.6.6: Acylation, Carbylamine Test, Nitrous Acid & Hinsberg's Distinction of 1°, 2° & 3° Amines

Three clear liquid bottles sit unlabeled on your exam practical desk—one contains a 1° amine, one a 2° amine, and one a 3° amine. How do you unmask all three in two quick reactions? First, heat a drop with chloroform and alcoholic KOH—only a 1° amine fires off the unmistakable foul smell of an isocyanide (Carbylamine Test)! Next, shake them with Hinsberg's Reagent (C₆H₅SO₂Cl) and KOH: 1° dissolves to a clear solution, 2° forms an alkali-insoluble precipitate, and 3° refuses to react at all!

1. Acylation & Benzoylation (9.6.3)

Aliphatic and aromatic 1° and 2° amines react with acid chlorides (CH₃COCl), acid anhydrides ((CH₃CO)₂O), or benzoyl chloride (C₆H₅COCl) in the presence of a stronger base like pyridine (to remove HCl and shift equilibrium forward) to form substituted amides. 3° amines lack replaceable N–H hydrogen, so they do not undergo acylation!

2. Carbylamine Reaction / Isocyanide Test (1° Amines ONLY — 9.6.4)

Aliphatic AND aromatic primary amines on heating with chloroform (CHCl₃) and ethanolic KOH form foul-smelling isocyanides (carbylamines). Secondary and tertiary amines do NOT give this reaction.

R–NH₂ + CHCl₃ + 3 KOH (alc.) ──(Δ)──▶ R–NC (Foul smell) + 3 KCl + 3 H₂O

3. Reaction with Nitrous Acid (HNO₂ from NaNO₂ + HCl, 9.6.5)

1° Aliphatic amines: Form unstable alkyldiazonium salts that decompose immediately to liberate nitrogen gas (N₂↑) quantitatively and form alcohols (used to estimate amino acids/proteins).
1° Aromatic amines (Aniline): At 273–278 K, form stable arenediazonium chloride (C₆H₅N₂⁺Cl^-), which couples with alkaline β-naphthol to give a scarlet-orange dye!

4. Hinsberg's Test (C₆H₅SO₂Cl — Differentiates 1°, 2°, 3°, 9.6.6)

Benzenesulphonyl chloride (Hinsberg's reagent; modern lab substitute: p-toluenesulphonyl chloride) reacts differently with all three classes:

Amine Class	Reaction with C₆H₅SO₂Cl	Product	Behaviour in aq. KOH
1° Amine (`RNH₂`)	Reacts readily	`C₆H₅SO₂NHR` (N-Alkylbenzenesulphonamide)	SOLUBLE (N–H hydrogen is strongly acidic due to –SO₂–)
2° Amine (`R₂NH`)	Reacts readily	`C₆H₅SO₂NR₂` (N,N-Dialkylbenzenesulphonamide)	INSOLUBLE (No H left on N atom!)
3° Amine (`R₃N`)	Does NOT react	No sulphonamide formed	Insoluble in KOH (dissolves on adding HCl)

NCERT Exercise 9.2 Quick Distinction Pairs:
• Methylamine (1°) vs Dimethylamine (2°): Carbylamine test (1° gives foul R–NC; 2° negative) OR Hinsberg test.
• Secondary vs Tertiary amine: Hinsberg test (2° reacts to give KOH-insoluble sulphonamide; 3° does not react).
• Ethylamine vs Aniline (or Benzylamine vs Aniline): Azo-dye test / HNO₂ at 273–278 K (Aniline gives orange/red dye with β-naphthol; Ethylamine/Benzylamine gives brisk N₂ gas bubbles!).
• Aniline vs N-Methylaniline: Carbylamine test (Aniline is 1°, positive; N-methylaniline is 2°, negative).

Classic exam mistake: Students try using the Carbylamine test to distinguish Ethylamine from Aniline! BOTH are 1° amines, so BOTH give foul-smelling isocyanides! Always use the Azo-Dye / Nitrous Acid test (273–278 K + β-naphthol) to separate aromatic 1° amines from aliphatic or benzylic 1° amines!

Topic 9.6.7: Electrophilic Substitution in Aniline: Bromination, The Nitration Trap, Sulphonation & Friedel-Crafts

Drop bromine water into aniline and instantly a snowy white precipitate of 2,4,6-tribromoaniline crashes out—the –NH₂ lone pair pumps so much electron density into the ortho and para corners that substitution happens at all three positions at once! Try direct nitration, and worse happens: strong acid protonates –NH₂ into meta-directing –NH₃⁺, spitting out an astonishing 47% meta-nitroaniline alongside tar! How do chemists tame this hyperactive ring? Put a temporary acetyl shield (–COCH₃) on nitrogen, substitute cleanly at para, then hydrolyse the shield away!

In aniline, the –NH₂ group is a powerful activating and ortho/para-directing group because its nitrogen lone pair resonates strongly into the ortho and para positions of the benzene ring.

1. Bromination & How to Control Reactivity

Direct Bromination: Aniline + 3 Br₂(aq) at room temp → white precipitate of 2,4,6-tribromoaniline.
Monosubstitution via Acetylation Protection: Treat aniline with acetic anhydride ((CH₃CO)₂O / pyridine) to form Acetanilide (C₆H₅NHCOCH₃). The lone pair on N now resonates with the carbonyl oxygen (C=O) too, making it less available to the benzene ring and moderating activation! Reacting acetanilide with Br₂ / CH₃COOH followed by hydrolysis (OH^- or H⁺) yields p-bromoaniline as major product.

2. Nitration — The Famous 47% Meta Trap!

Direct nitration (conc. HNO₃ + conc. H₂SO₄, 288 K) yields tarry oxidation products AND a surprising mixture: 51% p-nitroaniline, 47% m-nitroaniline, and 2% o-nitroaniline!
Why 47% meta? In strongly acidic medium, aniline gets protonated to form the anilinium ion (C₆H₅NH₃⁺), which is strongly electron-withdrawing and meta-directing!
Clean Fix: Protect as acetanilide → nitrate (HNO₃/H₂SO₄, 288 K) → p-nitroacetanilide → hydrolyse → p-nitroaniline cleanly!

3. Sulphonation & Zwitterion Formation

Aniline reacts with conc. H₂SO₄ to form anilinium hydrogensulphate, which on heating at 453–473 K forms p-aminobenzenesulphonic acid (Sulphanilic acid) as major product—existing as a dipolar zwitterion (⁺H₃N–C₆H₄–SO₃^-).

4. Why Aniline Does NOT Undergo Friedel-Crafts Reaction

Aniline (a Lewis base) reacts with anhydrous AlCl₃ (the Lewis acid catalyst used in Friedel-Crafts alkylation/acetylation) to form a salt (C₆H₅NH₂⁺AlCl₃^-). Nitrogen acquires a positive charge and acts as a strong deactivating group, blocking further reaction!

Master the 3 most-repeated "Account for" board questions in Chapter 9: (1) 47% m-nitroaniline → protonation to meta-directing –NH₃⁺; (2) No Friedel-Crafts → salt formation with catalyst AlCl₃ puts + charge on N; (3) Why acetylation lowers reactivity → N lone pair participates in resonance with C=O of –NHCOCH₃!

Topic 9.7 – 9.10: Diazonium Salts: Diazotisation, Sandmeyer/Gattermann Reactions & Azo Dye Coupling

Keep aniline chilled in an ice bath at 273–278 K (0–5 °C), stir in sodium nitrite and HCl, and you create organic chemistry's Swiss Army knife: Benzenediazonium Chloride (C₆H₅N₂⁺Cl^-)! With a ready-to-escape N₂ gas molecule perched on the ring, it effortlessly swaps in –Cl, –Br, –I, –F, –CN, –OH, –NO₂, or –H! Even better, keep the two nitrogens locked together as an –N=N– bridge with phenol or aniline, and vibrant orange and yellow azo dyes bloom before your eyes!

1. Diazotisation & Stability (9.7 – 9.8)

C₆H₅NH₂ + NaNO₂ + 2 HCl ──(273–278 K)──▶ C₆H₅N₂⁺Cl^- + NaCl + 2 H₂O

Why Arenediazonium salts are stable in cold while Alkyldiazonium salts are unstable: Dispersal of positive charge over the benzene ring by resonance stabilises Ar–N₂⁺ at 273–278 K; aliphatic R–N₂⁺ lacks resonance and decomposes immediately to R–OH + N₂↑.
Benzenediazonium chloride is water-soluble and decomposes if warmed (≥ 283 K) or dried; C₆H₅N₂⁺BF₄^- is water-insoluble and stable at room temperature.

2. Reactions Involving Displacement of N₂ (Group A, 9.9A)

Reaction / Name	Reagent on C₆H₅N₂⁺Cl^-	Product (+ N₂↑)
Sandmeyer Reaction (Better yield)	`CuCl/HCl`, `CuBr/HBr`, or `CuCN/KCN`	`C₆H₅Cl`, `C₆H₅Br`, or `C₆H₅CN`
Gattermann Reaction	`Cu powder + HCl` or `Cu powder + HBr`	`C₆H₅Cl` or `C₆H₅Br`
Replacement by –I	`KI` (aq, warm)	Iodobenzene (`C₆H₅I`)
Replacement by –F (Balz-Schiemann)	(i) `HBF₄` → ppt, (ii) Heat (`Δ`)	Fluorobenzene (`C₆H₅F`) + BF₃
Replacement by –H (Deamination)	`H₃PO₂ + H₂O` (→ H₃PO₃) or `C₂H₅OH` (→ CH₃CHO)	Benzene (`C₆H₆`)
Replacement by –OH	Warm `H₂O` at 283 K	Phenol (`C₆H₅OH`)
Replacement by –NO₂	(i) `HBF₄`, (ii) `NaNO₂ / Cu, Δ`	Nitrobenzene (`C₆H₅NO₂`)

3. Coupling Reactions — Retention of Diazo Group (Group B, 9.9B)

Benzenediazonium chloride undergoes electrophilic substitution at the para position of phenol or aniline, forming extended conjugated azo dyes linked by –N=N–:

C₆H₅N₂⁺Cl^- + Phenol (mildly alkaline, OH^-) → p-Hydroxyazobenzene (Orange dye)
C₆H₅N₂⁺Cl^- + Aniline (mildly acidic, H⁺) → p-Aminoazobenzene (Yellow dye)

Importance in Synthesis (9.10 & NCERT Example 9.5): Diazonium salts unlock syntheses impossible by direct ring substitution, such as Ar–F, Ar–I, Ar–CN, 1,3,5-tribromobenzene (Aniline → 2,4,6-tribromoaniline → diazotise → H₃PO₂), and 4-nitrotoluene → 2-bromobenzoic acid:
4-Nitrotoluene ──(Br₂/FeBr₃)──▶ 2-Bromo-4-nitrotoluene ──(Sn/HCl)──▶ 3-Bromo-4-methylaniline ──(NaNO₂/HCl, then H₃PO₂)──▶ 2-Bromotoluene ──(KMnO₄/OH^-, H₃O⁺)──▶ 2-Bromobenzoic acid!

Watch three high-frequency exam traps: (1) Sandmeyer uses Cu(I) salts (CuCl/CuBr/CuCN) and gives better yield, while Gattermann uses Cu powder + HCl/HBr. (2) Keep diazotisation at 273–278 K; warming to 283 K turns the salt into phenol! (3) Phenol couples in alkaline medium (OH^-, orange dye), whereas Aniline couples in acidic medium (H⁺, yellow dye)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 3 Marks): (a) Arrange the following in increasing order of basic strength in aqueous solution: NH₃, C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂.
(b) Why is pK_b of aniline higher than that of methylamine?

(a) Increasing order of basic strength in aqueous solution:
C₆H₅NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH

(b) Reason why pK_b of aniline is higher (weaker base):
In aniline, the unshared lone pair on nitrogen is delocalised into the benzene ring by resonance across 5 resonating structures, reducing its availability for protonation (whereas anilinium ion has only 2 Kekulé structures). In methylamine, the +I effect of the –CH₃ group increases electron density on nitrogen, making it a much stronger base (lower pK_b).

Q2 (Kerala HSE March (Very Important) • 2 Marks): Describe Hinsberg's test to distinguish between primary, secondary, and tertiary amines with chemical equations.

Treat the amine with benzenesulphonyl chloride (C₆H₅SO₂Cl, Hinsberg's reagent) followed by aqueous KOH:

Primary (1°) amine: Forms N-alkylbenzenesulphonamide (C₆H₅SO₂NHR), which has an acidic N–H hydrogen and is soluble in aqueous KOH.
Secondary (2°) amine: Forms N,N-dialkylbenzenesulphonamide (C₆H₅SO₂NR₂), which has no N–H hydrogen and is insoluble in aqueous KOH.
Tertiary (3°) amine: Does not react with benzenesulphonyl chloride.

Q3 (Kerala HSE SAY / Model • 2 Marks): Explain (i) Carbylamine reaction and (ii) Hoffmann bromamide degradation reaction with balanced equations.

(i) Carbylamine reaction: Aliphatic or aromatic 1° amines heated with chloroform and ethanolic KOH form foul-smelling isocyanides (carbylamines):
R–NH₂ + CHCl₃ + 3 KOH (alc.) ──(Δ)──▶ R–NC + 3 KCl + 3 H₂O
(ii) Hoffmann bromamide degradation: Treating a primary amide with Br₂ in aqueous/ethanolic NaOH yields a 1° amine with one carbon atom less:
R–CONH₂ + Br₂ + 4 NaOH ──▶ R–NH₂ + Na₂CO₃ + 2 NaBr + 2 H₂O

Q4 (Kerala HSE March (Repeated) • 2 Marks): Give reasons:
(i) Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.
(ii) Aniline does not undergo Friedel-Crafts reaction.

(i) Gabriel limitation: Aryl halides do not undergo nucleophilic substitution (S_N2) with the potassium phthalimide anion (due to partial double-bond character of the C–X bond by resonance).
(ii) No Friedel-Crafts: Aniline (Lewis base) reacts with anhydrous AlCl₃ (Lewis acid catalyst) to form a salt (C₆H₅NH₂⁺AlCl₃^-), giving nitrogen a positive charge that strongly deactivates the benzene ring.

Q5 (Kerala HSE March • 2 Marks): Although the –NH₂ group is ortho and para directing, direct nitration of aniline gives a substantial amount (47%) of m-nitroaniline. Why? How can pure p-nitroaniline be obtained?

Reason: In the strongly acidic nitrating mixture (conc. HNO₃ + conc. H₂SO₄), aniline is protonated to form the anilinium ion (C₆H₅NH₃⁺), which is strongly meta-directing, producing 47% m-nitroaniline.

Preparation of pure p-nitroaniline: Protect –NH₂ by acetylation with acetic anhydride to form acetanilide, nitrate at 288 K to get p-nitroacetanilide (major), and hydrolyse to yield pure p-nitroaniline.

Q6 (Kerala HSE SAY • 2 Marks): What is diazotisation? Write equations for the conversion of benzenediazonium chloride into (a) chlorobenzene (Sandmeyer) and (b) p-hydroxyazobenzene (Orange dye).

Diazotisation: Reaction of aniline with nitrous acid (NaNO₂ + 2 HCl) at 273–278 K to form benzenediazonium chloride (C₆H₅N₂⁺Cl^-).

(a) Sandmeyer: C₆H₅N₂⁺Cl^- ──(CuCl / HCl)──▶ C₆H₅Cl + N₂↑
(b) Coupling: C₆H₅N₂⁺Cl^- + C₆H₅OH ──(OH^-)──▶ p-HO–C₆H₄–N=N–C₆H₅ (Orange dye) + Cl^- + H₂O

Q7 (Kerala HSE March / NCERT Ex 9.8 • 4 Marks): An aromatic compound 'A' (C₇H₆O₂) on treatment with aqueous ammonia and heating forms compound 'B', which on heating with Br₂ and KOH yields compound 'C' (C₆H₇N). Identify A, B, and C with equations, and write how 'C' is converted into benzene.

Identification (NCERT Ex 9.10):

A = Benzoic acid (C₆H₅COOH)
B = Benzamide (C₆H₅CONH₂) — formed via ammonium benzoate upon heating with NH₃
C = Aniline / Benzenamine (C₆H₅NH₂) — formed by Hoffmann bromamide degradation of B with Br₂/KOH

Conversion of Aniline (C) to Benzene: Diazotise with NaNO₂ + HCl at 273–278 K to form C₆H₅N₂⁺Cl^-, then reduce with H₃PO₂ + H₂O (or C₂H₅OH) to obtain Benzene (C₆H₆).

Q8 (Kerala HSE Model / NCERT Ex 9.9 • 3 Marks): How will you accomplish the following conversions?
(i) Aniline into p-bromoaniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into 1,3,5-tribromobenzene

Aniline → p-bromoaniline: Acetylate with (CH₃CO)₂O / pyridine → Acetanilide → Br₂ / CH₃COOH → p-bromoacetanilide → OH^- or H⁺ hydrolysis → p-bromoaniline.
Ethanoic acid → methanamine: CH₃COOH + NH₃, Δ → CH₃CONH₂ (Ethanamide) ──(Br₂ / NaOH)──▶ CH₃NH₂ (Methanamine).
Aniline → 1,3,5-tribromobenzene: C₆H₅NH₂ + 3 Br₂(aq) → 2,4,6-tribromoaniline → diazotise (NaNO₂/HCl, 273–278 K) → reduce with H₃PO₂ + H₂O → 1,3,5-tribromobenzene.

Q9 (Kerala HSE Improvement • 3 Marks): Account for the following:
(a) Primary amines have higher boiling points than tertiary amines.
(b) Ethylamine is soluble in water whereas aniline is not.
(c) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(a) Boiling point: Primary amines have two N–H hydrogens and form extensive intermolecular H-bonds, whereas tertiary amines have no N–H hydrogens and thus no intermolecular H-bonding.
(b) Solubility: Ethylamine readily forms H-bonds with water; in aniline, the large hydrophobic phenyl ring prevents H-bonding with water.
(c) Diazonium stability: Arenediazonium ion (C₆H₅N₂⁺) is stabilised by resonance dispersal of the positive charge over the benzene ring; alkyldiazonium ions lack resonance and decompose immediately.

Q10 (Kerala HSE March • 1 Marks): Give one chemical test to distinguish between Aniline and N-methylaniline.

Carbylamine test: Heating with chloroform and ethanolic KOH gives a foul-smelling phenyl isocyanide with Aniline (1° amine), whereas N-methylaniline (2° amine) gives no reaction.

Q11 (Kerala HSE SAY • 1 Marks): Why is reduction of nitro compounds using iron scrap and hydrochloric acid preferred over tin and hydrochloric acid?

Because FeCl₂ formed during the reaction gets hydrolysed to release HCl, so only a very small amount of HCl is needed to initiate the reaction.

Q12 (Kerala HSE March • 1 Marks): Why does methylamine in water react with ferric chloride to precipitate hydrated ferric oxide?

Methylamine is more basic than water and generates OH^- ions (CH₃NH₂ + H₂O ⇌ CH₃NH₃⁺ + OH^-), which react with aqueous FeCl₃ to precipitate reddish-brown hydrated ferric oxide, Fe₂O₃·xH₂O / Fe(OH)₃.