Amines & Diazonium Salts — Basicity, Tests & Synthesis

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Every reaction of an amine starts from the lone pair on nitrogen. It makes amines bases and nucleophiles, it activates the benzene ring of aniline, and when nitrous acid turns that nitrogen into N₂⁺ you get the diazonium salt — the most versatile intermediate in aromatic synthesis.

What you will be able to do

• Classify amines and prepare them by reduction, ammonolysis, Gabriel synthesis and Hofmann bromamide degradation
• Explain basicity orders of aliphatic and aromatic amines in water and in the gas phase
• Distinguish 1°, 2° and 3° amines by the carbylamine, Hinsberg and nitrous acid tests
• Predict ring substitution of aniline and use acetylation to control it
• Use diazonium salts in Sandmeyer, Gattermann, replacement and coupling reactions, and plan conversions

The big picture

One lone pair runs the chapter.

In one line: The nitrogen lone pair makes amines basic and nucleophilic and makes aniline’s ring highly reactive; converting –NH₂ to –N₂⁺ (diazotisation) turns it into an excellent leaving group that can be replaced by almost anything.

Class	Structure	Example
Primary (1°)	R–NH₂	CH₃NH₂, aniline C₆H₅NH₂
Secondary (2°)	R₂NH	(CH₃)₂NH, N-methylaniline
Tertiary (3°)	R₃N	(CH₃)₃N, N,N-dimethylaniline
Quaternary salt	R₄N⁺ X⁻	(CH₃)₄N⁺ I⁻

Think of it like this: Think of the nitrogen lone pair as a hand held out. In an alkylamine the hand is free and eager to grab a proton (strong base). In aniline the benzene ring holds that hand (resonance), so it is less available for protons — but the ring itself gets stronger and grabs electrophiles easily.

Structure and physical properties

Pyramidal nitrogen and hydrogen bonding.

• Nitrogen is sp³ with one lone pair: pyramidal. The C–N–C angle in trimethylamine is 108°.
• In aniline the lone pair is partly delocalised into the ring, so N is slightly flattened toward sp².
• Hydrogen bonding: 1° amines (two N–H) > 2° (one N–H) > 3° (none). So boiling points: 1° > 2° > 3° for isomers, e.g. butan-1-amine (350.8 K) > N-ethylethanamine (329.3 K) > N,N-dimethylethanamine (310.5 K).
• Amines boil lower than alcohols of similar mass, because N is less electronegative than O and forms weaker H bonds.
• Lower aliphatic amines are water-soluble and smell fishy; aniline is a colourless liquid that darkens in air by oxidation.

Preparing amines

Seven routes and what each is good for.

Route	Reagent	Product	Note
Reduction of nitro compounds	Fe/HCl (preferred), Sn/HCl, H₂/Pd	ArNH₂ / RNH₂	Fe/HCl: FeCl₂ formed is hydrolysed, releasing HCl, so little acid is needed
Ammonolysis of halides	NH₃ (alcoholic), heat in sealed tube	Mixture of 1°, 2°, 3° amines and R₄N⁺X⁻	Excess NH₃ favours 1°; aryl halides do not react easily
Reduction of nitriles	LiAlH₄ or H₂/Ni; Na/ethanol (Mendius)	R–CH₂NH₂	Adds one carbon to the halide R–X used to make R–CN
Reduction of amides	LiAlH₄, then H₂O	R–CH₂NH₂	Same carbon count as the amide
Gabriel phthalimide synthesis	Phthalimide + KOH; then R–X; then NaOH (hydrolysis)	Pure 1° aliphatic amine	Not for aryl amines (ArX does not undergo SN2)
Hofmann bromamide degradation	RCONH₂ + Br₂ + 4NaOH	R–NH₂ + Na₂CO₃ + 2NaBr + 2H₂O	One carbon fewer than the amide; via isocyanate
Reductive amination	Aldehyde/ketone + NH₃, then H₂/Ni	1° amine	Via the imine

Key idea: Count the carbons. Nitrile reduction adds one carbon (relative to RX); amide reduction keeps the count; Hofmann bromamide loses one. Questions on “ascending” and “descending” a series test exactly this.

Solved example 01 — Hofmann bromamide

What amine is obtained from benzamide with Br₂ and NaOH? How many carbons does it lose?

1. C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
2. The carbonyl carbon leaves as CO₃²⁻.

Answer: Aniline; one carbon fewer (C₇ → C₆).

Solved example 02 — Why Gabriel fails for aniline

Why can aniline not be prepared by the Gabriel phthalimide synthesis?

1. The key step is SN2 attack of the phthalimide anion on an alkyl halide.
2. Aryl halides do not undergo SN2 (partial double-bond character of C–X, sp² carbon, ring repels the nucleophile).

Answer: The phthalimide ion cannot substitute the halogen of an aryl halide.

Check yourself

• C1. Which method gives a pure primary amine free of 2° and 3° amines?
• C2. Starting from CH₃Br, how would you get ethanamine?

Answers at the end.

Basicity

The most-asked topic in this chapter.

Amines are bases because the nitrogen lone pair accepts H⁺: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻. A smaller pKb means a stronger base.

Amine	pKb (in water)
Diethylamine (C₂H₅)₂NH	3.00
Triethylamine (C₂H₅)₃N	3.25
Dimethylamine (CH₃)₂NH	3.27
Ethanamine C₂H₅NH₂	3.29
Methanamine CH₃NH₂	3.38
Trimethylamine (CH₃)₃N	4.22
Benzylamine C₆H₅CH₂NH₂	4.70
Ammonia	4.75
N,N-Dimethylaniline	8.92
N-Methylaniline	9.30
Aniline	9.38

• Aliphatic amines are stronger bases than NH₃: alkyl groups push electron density onto N (+I).
• In the gas phase only the +I effect matters: 3° > 2° > 1° > NH₃.
• In water three effects compete — +I, solvation of the ammonium ion by H bonding (best for 1°, worst for 3°), and steric hindrance. Results: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ and (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃.
• Aryl amines are much weaker bases: the lone pair is delocalised into the ring, and the anilinium ion has fewer resonance structures than aniline. Electron-donating groups (–CH₃, –OCH₃) on the ring raise basicity; electron-withdrawing groups (–NO₂, –Cl) lower it — p-nitroaniline is a very weak base.
• Benzylamine (CH₂ between ring and N) behaves like an aliphatic amine — the lone pair is not conjugated.

Solved example 03 — Ordering basicity in water

Arrange CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, NH₃ and C₆H₅NH₂ in decreasing basicity in aqueous solution.

1. Aniline is weakest: lone pair delocalised.
2. NH₃ is weaker than all methylamines.
3. Among methylamines in water, 2° > 1° > 3° (solvation and steric effects).

Answer: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃ > C₆H₅NH₂.

Check yourself

• C3. Which is the stronger base: aniline or benzylamine?
• C4. Why is (CH₃)₃N a weaker base than CH₃NH₂ in water but stronger in the gas phase?

Answers at the end.

Reactions of amines

At nitrogen.

Reaction	Reagent	1°	2°	3°
Alkylation	R–X	→ 2°, 3°, R₄N⁺X⁻	→ 3°, R₄N⁺X⁻	→ R₄N⁺X⁻
Acylation	RCOCl or (RCO)₂O, pyridine	Amide RNHCOR	Amide	No reaction (no N–H)
Benzoylation (Schotten–Baumann)	C₆H₅COCl, aq. NaOH	N-substituted benzamide	Benzamide	No reaction
Carbylamine	CHCl₃ + alc. KOH, heat	Isocyanide (foul smell)	No reaction	No reaction
Hinsberg	C₆H₅SO₂Cl	Sulphonamide, soluble in alkali	Sulphonamide, insoluble in alkali	No reaction
Nitrous acid (NaNO₂ + HCl)	HNO₂	Aliphatic: alcohol + N₂ (bubbles); aromatic (273–278 K): diazonium salt	Yellow oily N-nitrosamine	Aliphatic: soluble salt

R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O

Carbylamine (isocyanide) test — given only by primary amines, aliphatic and aromatic

Key idea: Hinsberg logic: the sulphonamide from a 1° amine still has an N–H, made acidic by the strongly electron-withdrawing SO₂ group, so it dissolves in KOH. The 2° sulphonamide has no N–H left, so it does not. A 3° amine has no N–H to react at all.

Solved example 04 — Identifying the class

Three isomeric amines C₃H₉N are tested. A gives a foul smell with CHCl₃/KOH; B forms a sulphonamide insoluble in alkali with Hinsberg’s reagent; C does not react with Hinsberg’s reagent. Identify them.

1. A: carbylamine positive → 1°: propan-1-amine (or propan-2-amine).
2. B: 2° → N-methylethanamine.
3. C: 3° → N,N-dimethylmethanamine (trimethylamine).

Answer: A 1° (propan-1-amine), B 2° (N-methylethanamine), C 3° (trimethylamine).

Ring substitution of aniline

A very reactive ring — and how to tame it.

Reaction	Conditions	Product	Note
Bromination	Bromine water, room temperature	2,4,6-Tribromoaniline (white ppt)	Too fast to stop at one Br
Controlled bromination	Acetylate → Br₂/CH₃COOH → hydrolyse	p-Bromoaniline	–NHCOCH₃ is a weaker activator and bulky
Direct nitration	conc. HNO₃ + H₂SO₄	p- (51%), m- (47%), o- (2%) nitroaniline + tar	Anilinium ion (–NH₃⁺) is a meta director; ring also oxidised
Controlled nitration	Acetylate → nitrate → hydrolyse	p-Nitroaniline (major)	Protection again
Sulphonation	conc. H₂SO₄, heat at 453–473 K	Sulphanilic acid (p-aminobenzenesulphonic acid)	Exists as a zwitterion
Friedel–Crafts	RCl / AlCl₃	No reaction	–NH₂ (Lewis base) complexes AlCl₃ and deactivates the ring

Key idea: Why acetylation works: in acetanilide the N lone pair is shared with the C=O of the acetyl group, so it donates less to the ring (moderate activation, monosubstitution); the bulky –NHCOCH₃ group also favours the para position. Hydrolysis afterwards regenerates –NH₂.

Solved example 05 — Why m-nitroaniline appears

Direct nitration of aniline gives 47% of the meta product, although –NH₂ is an o/p director. Explain.

1. The nitrating mixture is strongly acidic; most aniline is protonated to the anilinium ion C₆H₅NH₃⁺.
2. –NH₃⁺ has no lone pair and withdraws electrons: it is a meta director.

Answer: Substitution on the anilinium ion gives the meta product.

Check yourself

• C5. How would you prepare p-bromoaniline from aniline?
• C6. Why does aniline not undergo Friedel–Crafts reactions?

Answers at the end.

Diazonium salts

Making them and why aryl ones survive.

C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

Diazotisation at 273–278 K; the salt is used at once in solution

• Aryl diazonium salts are stable at 0–5 °C because the positive charge is delocalised over the ring; aliphatic diazonium ions decompose at once to carbocations and N₂.
• Benzenediazonium fluoroborate is stable enough to be isolated as a solid; the chloride is used in solution.
• The –N₂⁺ group is an excellent leaving group (N₂ gas is very stable), so it can be replaced by many nucleophiles.

Reactions of diazonium salts

Replacement and coupling.

Replace –N₂⁺ by	Reagent	Name
–Cl	CuCl / HCl	Sandmeyer
–Br	CuBr / HBr	Sandmeyer
–CN	CuCN / KCN	Sandmeyer
–Cl or –Br	Cu powder + HCl or HBr	Gattermann
–I	KI (no catalyst)	—
–F	HBF₄, then heat the fluoroborate	Balz–Schiemann
–H	H₃PO₂ (or ethanol)	Reduction / deamination
–OH	Warm water	Hydrolysis
–NO₂	HBF₄, then NaNO₂/Cu, heat	—

Coupling reactions — azo dyes

• With phenol in mildly alkaline solution: p-hydroxyazobenzene (orange dye).
• With aniline in mildly acidic solution: p-aminoazobenzene (yellow dye).
• Coupling happens at the para position of the activated ring; the –N=N– group is retained (the electrophile is the diazonium ion).

Solved example 06 — Benzene to iodobenzene

Convert benzene into iodobenzene.

1. Nitrate: benzene + conc. HNO₃/H₂SO₄ → nitrobenzene.
2. Reduce: Sn/HCl (or Fe/HCl) → aniline.
3. Diazotise: NaNO₂ + HCl, 273–278 K → benzenediazonium chloride.
4. Add KI → iodobenzene + N₂.

Answer: Benzene → nitrobenzene → aniline → diazonium salt → iodobenzene.

Solved example 07 — 1,3,5-Tribromobenzene

Make 1,3,5-tribromobenzene from aniline — a product that direct bromination of benzene cannot give.

1. Bromine water → 2,4,6-tribromoaniline.
2. Diazotise (NaNO₂/HCl, 273–278 K).
3. H₃PO₂ replaces –N₂⁺ by H.

Answer: The –NH₂ is used to place three Br atoms meta to each other, then removed.

Conversions

The problems examiners love.

Solved example 08 — Ethanamine to methanamine

Convert ethanamine into methanamine (descending the series).

1. C₂H₅NH₂ + HNO₂ → C₂H₅OH + N₂ + H₂O.
2. Oxidise: C₂H₅OH → CH₃COOH (K₂Cr₂O₇/H⁺).
3. CH₃COOH → CH₃COCl (SOCl₂) → CH₃CONH₂ (NH₃).
4. Hofmann bromamide: CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂.

Answer: Via ethanol, ethanoic acid and ethanamide; Hofmann removes one carbon.

Solved example 09 — Methanamine to ethanamine

Convert methanamine into ethanamine (ascending the series).

1. CH₃NH₂ + HNO₂ → CH₃OH.
2. CH₃OH → CH₃Br (PBr₃) → CH₃CN (KCN).
3. Reduce: LiAlH₄ → CH₃CH₂NH₂.

Answer: Via methanol, bromomethane and ethanenitrile; nitrile reduction adds one carbon.

Solved example 10 — Aniline to benzoic acid

Convert aniline into benzoic acid.

1. Diazotise: NaNO₂ + HCl, 273–278 K.
2. Sandmeyer: CuCN/KCN → benzonitrile.
3. Hydrolyse: H₃O⁺ → benzoic acid.

Answer: Aniline → diazonium salt → C₆H₅CN → C₆H₅COOH.

Solved example 11 — Nitrobenzene to phenol

Convert nitrobenzene into phenol.

1. Reduce: Sn/HCl → aniline.
2. Diazotise at 273–278 K.
3. Warm with water → phenol + N₂.

Answer: Nitrobenzene → aniline → diazonium salt → phenol.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	Aniline is a stronger base than methylamine.	Methylamine is far stronger (pKb 3.38 vs 9.38).	Aniline’s lone pair is delocalised into the ring.
2	In water, 3° amines are the strongest bases.	In water 2° is usually strongest; 3° can be weaker than 1°.	Solvation and steric effects oppose +I.
3	Gabriel synthesis gives aniline.	It gives only 1° aliphatic amines.	Aryl halides do not undergo SN2.
4	Hofmann bromamide gives an amine with the same number of carbons.	One carbon fewer.	The C=O carbon leaves as carbonate.
5	Secondary amines give the carbylamine test.	Only primary amines.	Isocyanide needs two N–H.
6	The 2° sulphonamide in the Hinsberg test dissolves in alkali.	It is insoluble; the 1° sulphonamide dissolves.	No acidic N–H on the 2° product.
7	Aniline + bromine water gives p-bromoaniline.	It gives 2,4,6-tribromoaniline.	Protect by acetylation for monobromination.
8	Direct nitration of aniline gives only o- and p-products.	About 47% meta forms.	Anilinium ion is a meta director.
9	Aniline undergoes Friedel–Crafts alkylation.	It does not.	–NH₂ forms a complex with AlCl₃.
10	Aliphatic diazonium salts are stable at 0–5 °C.	Only aryl diazonium salts are.	Resonance with the ring stabilises the aryl ion.
11	Sandmeyer converts diazonium salt to iodobenzene with CuI.	KI alone is enough for –I.	I⁻ is a good enough nucleophile.
12	Benzylamine is as weak a base as aniline.	Benzylamine (pKb 4.70) is much stronger.	Its lone pair is not conjugated with the ring.

Memory hook: “Nitrile adds one, amide keeps, Hofmann takes one.” · “Carbylamine for 1° only — follow your nose.” · “Hinsberg: 1° dissolves, 2° stays, 3° ignores.” · “Protect, substitute, deprotect” — acetanilide route.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. Carbylamine test is given by: (Easy)

• (a) 1° amines
• (b) 2° amines
• (c) 3° amines
• (d) all amines

Q02. Hinsberg’s reagent is: (Easy)

• (a) C₆H₅COCl
• (b) C₆H₅SO₂Cl
• (c) CHCl₃
• (d) NaNO₂ + HCl

Q03. Aniline is less basic than ammonia because: (Easy)

• (a) it has a higher molar mass
• (b) its lone pair is delocalised into the ring
• (c) N is sp
• (d) it is a liquid

Q04. Aniline + bromine water gives: (Easy)

• (a) o-bromoaniline
• (b) p-bromoaniline
• (c) 2,4,6-tribromoaniline
• (d) bromobenzene

Q05. Diazotisation is carried out at: (Easy)

• (a) 273–278 K
• (b) 373 K
• (c) 298 K
• (d) 573 K

Q06. Benzenediazonium chloride + KI gives: (Easy)

• (a) chlorobenzene
• (b) iodobenzene
• (c) phenol
• (d) benzene

Q07. Gabriel synthesis is used to prepare: (Easy)

• (a) 1° aliphatic amines
• (b) 2° amines
• (c) 3° amines
• (d) aryl amines

Q08. Hofmann bromamide reaction of propanamide gives: (Moderate)

• (a) propanamine
• (b) ethanamine
• (c) methanamine
• (d) butanamine

Q09. Strongest base in water: (Moderate)

• (a) CH₃NH₂
• (b) (CH₃)₂NH
• (c) (CH₃)₃N
• (d) NH₃

Q10. Strongest base in the gas phase: (Moderate)

• (a) NH₃
• (b) CH₃NH₂
• (c) (CH₃)₂NH
• (d) (CH₃)₃N

Q11. Reduction of CH₃CN with LiAlH₄ gives: (Moderate)

• (a) CH₃NH₂
• (b) CH₃CH₂NH₂
• (c) CH₃CONH₂
• (d) CH₃CH₂OH

Q12. Sandmeyer reaction converts ArN₂⁺ to ArCl using: (Moderate)

• (a) Cu/HCl
• (b) CuCl/HCl
• (c) KCl
• (d) Cl₂

Q13. Benzenediazonium chloride + H₃PO₂ gives: (Moderate)

• (a) phenol
• (b) benzene
• (c) aniline
• (d) chlorobenzene

Q14. A 2° amine with HNO₂ gives: (Moderate)

• (a) alcohol
• (b) diazonium salt
• (c) N-nitrosamine
• (d) no reaction

Q15. Coupling of benzenediazonium chloride with phenol gives: (Moderate)

• (a) p-aminoazobenzene
• (b) p-hydroxyazobenzene
• (c) azobenzene
• (d) biphenyl

Q16. Acetylation of aniline before nitration is done to: (Moderate)

• (a) increase activation
• (b) protect –NH₂ and get the p-product
• (c) make it soluble
• (d) remove the ring

Q17. Sulphanilic acid exists as: (Moderate)

• (a) a neutral molecule
• (b) a zwitterion
• (c) a cation only
• (d) an anion only

Q18. Which amine does not react with Hinsberg’s reagent? (Moderate)

• (a) CH₃NH₂
• (b) (CH₃)₂NH
• (c) (CH₃)₃N
• (d) C₆H₅NH₂

Q19. Weakest base: (Tough)

• (a) aniline
• (b) p-toluidine
• (c) p-nitroaniline
• (d) benzylamine

Q20. Direct nitration of aniline gives a large amount of m-nitroaniline because of: (Tough)

• (a) steric hindrance
• (b) formation of the anilinium ion
• (c) oxidation
• (d) hydrolysis

Q21. Fluorobenzene is made from benzenediazonium chloride by: (Tough)

• (a) KF
• (b) HBF₄ then heating
• (c) CuF
• (d) F₂

Q22. Number of carbons in the amine from Hofmann degradation of hexanamide: (Tough)

• (a) 4
• (b) 5
• (c) 6
• (d) 7

Q23. Correct basicity order in water: (Tough)

• (a) (C₂H₅)₃N > (C₂H₅)₂NH > C₂H₅NH₂
• (b) (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂
• (c) C₂H₅NH₂ > (C₂H₅)₂NH > (C₂H₅)₃N
• (d) all equal

Q24. 1,3,5-Tribromobenzene can be made from aniline by: (Tough)

• (a) Br₂/FeBr₃
• (b) Br₂ water, then diazotisation and H₃PO₂
• (c) NBS
• (d) HBr

Q25. Aniline + CHCl₃ + alc. KOH gives: (Tough)

• (a) phenyl isocyanide
• (b) benzonitrile
• (c) chlorobenzene
• (d) benzaldehyde

Answer key with explanations

1. (a) 1° amines — Two N–H needed.
2. (b) C₆H₅SO₂Cl — Benzenesulphonyl chloride.
3. (b) its lone pair is delocalised into the ring — Resonance.
4. (c) 2,4,6-tribromoaniline — Highly activated ring.
5. (a) 273–278 K — Diazonium salt is unstable when warm.
6. (b) iodobenzene — Replacement by I.
7. (a) 1° aliphatic amines — Pure 1°.
8. (b) ethanamine — One carbon fewer.
9. (b) (CH₃)₂NH — 2° in water.
10. (d) (CH₃)₃N — +I only.
11. (b) CH₃CH₂NH₂ — Adds one carbon.
12. (b) CuCl/HCl — Cu(I) salt.
13. (b) benzene — –N₂⁺ → H.
14. (c) N-nitrosamine — Yellow oil.
15. (b) p-hydroxyazobenzene — Orange dye.
16. (b) protect –NH₂ and get the p-product — Controlled substitution.
17. (b) a zwitterion — –SO₃⁻ and –NH₃⁺.
18. (c) (CH₃)₃N — No N–H.
19. (c) p-nitroaniline — –NO₂ withdraws.
20. (b) formation of the anilinium ion — –NH₃⁺ meta director.
21. (b) HBF₄ then heating — Balz–Schiemann.
22. (b) 5 — 6 − 1.
23. (b) (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ — pKb 3.00, 3.25, 3.29.
24. (b) Br₂ water, then diazotisation and H₃PO₂ — Use and remove –NH₂.
25. (a) phenyl isocyanide — Carbylamine test.

Check-yourself answers

• C1. Gabriel phthalimide synthesis (or Hofmann bromamide degradation).
• C2. CH₃Br + KCN → CH₃CN; reduce with LiAlH₄ (or Na/C₂H₅OH) → CH₃CH₂NH₂.
• C3. Benzylamine — its lone pair is not delocalised into the ring.
• C4. In water, the (CH₃)₃NH⁺ ion has only one N–H for H bonding with water and is poorly solvated, and the three methyl groups crowd the nitrogen; in the gas phase only the +I effect counts, so 3° is strongest.
• C5. Acetylate (acetic anhydride) → acetanilide; brominate (Br₂ in ethanoic acid) → p-bromoacetanilide; hydrolyse (acid or base) → p-bromoaniline.
• C6. The –NH₂ group (a Lewis base) forms a salt with AlCl₃, making nitrogen positive and deactivating the ring.

One-page revision

Read this page the night before the exam.

Preparation

• Nitro + Fe/HCl → amine
• RCN + LiAlH₄ → RCH₂NH₂ (+1 C)
• Gabriel: pure 1° aliphatic only
• Hofmann: RCONH₂ + Br₂ + 4NaOH → RNH₂ (−1 C)

Basicity

• Aliphatic > NH₃ > aryl
• Water: 2° > 1° ≈ 3° (Et: 2° > 3° > 1°)
• Gas: 3° > 2° > 1°
• EDG ↑, EWG ↓ basicity of anilines

Tests

• Carbylamine: 1° only, foul RNC
• Hinsberg: 1° soluble in KOH; 2° insoluble; 3° none
• HNO₂: 1° aliphatic → ROH + N₂
• 2° → yellow N-nitrosamine

Aniline ring

• Br₂ water → 2,4,6-tribromoaniline
• Acetylate to get mono / para
• Direct nitration: 47% meta (anilinium)
• No Friedel–Crafts; sulphanilic acid zwitterion

Diazonium

• ArNH₂ + NaNO₂ + HCl at 273–278 K
• CuCl, CuBr, CuCN: Sandmeyer
• KI → ArI; HBF₄ → ArF; H₃PO₂ → ArH
• Warm H₂O → phenol

Coupling and conversions

• Phenol → p-hydroxyazobenzene (orange)
• Aniline → p-aminoazobenzene (yellow)
• Ascend: via nitrile; descend: via Hofmann
• –NH₂ as a removable director