Amines are derivatives of ammonia (NH₃) obtained by replacement of one, two or all three hydrogen atoms by alkyl and/or aryl groups. They occur widely in nature — in proteins, vitamins, alkaloids, and hormones — and are used in synthesis of dyes, drugs, and polymers.
Definition
Amines: Organic compounds derived from ammonia by replacing one or more H atoms with alkyl/aryl groups. The nitrogen atom retains an unshared pair of electrons, which makes amines basic and nucleophilic.
Hybridisation & Geometry
Like ammonia, the nitrogen atom in amines is sp³ hybridised. Three of its sp³ orbitals overlap with orbitals of C or H, while the fourth sp³ orbital holds the lone pair. This lone pair causes bond-angle compression below the tetrahedral ideal of 109.5°.
Pyramidal geometry of trimethylamine — N is sp³ hybridised; lone pair occupies the 4th orbital, compressing the C–N–C bond angle to 108°
Key Rules
Nitrogen in amines: trivalent, sp³ hybridised, pyramidal geometry
Bond angle C–N–C or C–N–H is less than 109.5° due to lone pair repulsion
Trimethylamine: C–N–C = 108°
The lone pair makes amines Lewis bases and nucleophiles
🧪
Fig. 9.1 — Pyramidal Shape of Trimethylamine
NCERT p.260
3D orbital diagram showing the four sp³ hybrid orbitals of nitrogen in trimethylamine — three overlapping with CH₃ groups and one holding the unshared electron pair pointing upward.
📌 Image Source
NCERT Chemistry Class 12, Chapter 9, Fig. 9.1, Page 260. Replace this placeholder with the actual NCERT figure once image files are available.
9.2 Classification of Amines
Amines are classified as primary (1°), secondary (2°), and tertiary (3°) based on the number of H atoms in NH₃ replaced by alkyl/aryl groups.
Type
H Atoms Replaced
General Formula
Example
Name
Primary (1°)
1
RNH₂ or ArNH₂
CH₃NH₂
Methylamine
Secondary (2°)
2
R₂NH or RNHR′
CH₃NHCH₃
Dimethylamine
Tertiary (3°)
3
R₃N
(CH₃)₃N
Trimethylamine
Quaternary Ammonium Salt
—
R₄N⁺ X⁻
(CH₃)₄N⁺Cl⁻
Tetramethylammonium chloride
Stepwise replacement of H atoms in NH₃ by alkyl/aryl groups yields 1°, 2°, 3° amines, and finally quaternary ammonium salt
⚠️ Common Confusion
In amines, 1°, 2°, 3° refers to the number of carbon groups attached to N, NOT to the type of carbon. This is the opposite of how we classify alcohols (where it refers to the carbon bearing –OH). A tertiary amine has three alkyl groups on N.
Amines are called 'simple' when all alkyl/aryl groups are the same and 'mixed' when groups are different.
9.3 Nomenclature of Amines
Common System
Primary amines: alkyl group name + "amine" as one word
E.g., CH₃NH₂ = methylamine
For 2° and 3° with same groups: di/tri prefix
E.g., (CH₃)₂NH = dimethylamine
Simplest arylamine: C₆H₅NH₂ = aniline
IUPAC System
Primary amines: replace 'e' of alkane by 'amine' → alkanamines
E.g., CH₃NH₂ = methanamine
Multiple –NH₂ groups: di, tri prefix + retain 'e' of alkane
E.g., H₂N–CH₂–CH₂–NH₂ = ethane-1,2-diamine
For 2°/3°: use locant N for substituent on N
Arylamines: suffix 'e' of arene → 'amine' → benzenamine
IUPAC Rule for 2° and 3° Amines
Identify the longest chain containing N as parent. Other groups on N are named as N-substituents.
Sn + HCl or Fe + HCl — Fe/HCl preferred industrially because FeCl₂ formed gets hydrolysed to release HCl, so only a small amount is needed to initiate the reaction
(ii) Aromatic — Chemical:
C₆H₅–NO₂ —Sn+HCl or Fe+HCl→ C₆H₅–NH₂
(iii) Aliphatic:
RNO₂ + 6[H] —Ni/H₂→ RNH₂ + 2H₂O
2. Ammonolysis of Alkyl Halides
An alkyl or benzyl halide reacts with ethanolic NH₃ in a sealed tube at 373 K — the C–X bond is cleaved nucleophilically by NH₃ (ammonolysis). The primary amine formed can react further to give 2°, 3° amines and finally quaternary ammonium salts.
Ammonolysis gives a mixture of 1°, 2°, 3° amines and quaternary salt. To obtain 1° amine as major product, use large excess of NH₃.
Reactivity order of halides: RI > RBr > RCl
3. Reduction of Nitriles
Nitriles (R–C≡N) are reduced by LiAlH₄ or catalytic hydrogenation to give primary amines. This is used for ascent of amine series — the product has one more carbon than the starting amine.
Step-by-step reaction scheme: Phthalimide → K-phthalimide (KOH) → N-alkylphthalimide (R–X) → primary amine (NaOH hydrolysis), showing all intermediates and products.
📌 Image Source
NCERT Chemistry Class 12, Chapter 9, Page 264. Replace with the actual NCERT diagram when image files are available.
⚠️ Why Gabriel can't prepare ArNH₂
Aryl halides (Ar–X) do not undergo nucleophilic substitution with the phthalimide anion because the C–X bond in aryl halides is very strong (resonance stabilisation). Hence, aromatic primary amines cannot be prepared by this method.
Amides react with Br₂ in aqueous/ethanolic NaOH to give primary amines. The amine formed has one carbon less than the starting amide — a migration of the alkyl/aryl group from carbonyl C to N occurs.
1Reduction of nitro compounds → RNH₂ (1°) via H₂/Ni or Fe+HCl
2Ammonolysis of alkyl halides → mixture; 1° major with excess NH₃
3Reduction of nitriles → RCH₂NH₂ (one carbon more) via LiAlH₄
4Reduction of amides → RCH₂NH₂ (same carbons) via LiAlH₄
5Gabriel synthesis → pure 1° alkyl amines only (not ArNH₂)
6Hoffmann rearrangement → pure 1° amine with one carbon less
9.5 Physical Properties of Amines
State, Odour & Colour
Lower aliphatic amines (C1–C2): gases with fishy/ammoniacal odour
3–6 carbons: liquids
Higher amines: solids
Aniline and arylamines: colourless liquids but turn brown on storage due to atmospheric oxidation
Hydrogen Bonding & Boiling Points
Primary and secondary amines can form intermolecular N–H···N hydrogen bonds. The N–H bond is less polar than O–H (N electronegativity = 3.0 vs O = 3.5), so H-bonds in amines are weaker than those in alcohols.
Boiling Point Order
For isomeric amines of the same molecular formula:
1° Aminemost H-bonds
>
2° Aminefewer H-bonds
>
3° Amineno N–H, no H-bond
But all amines boil lower than alcohols of similar molar mass because N–H···N bonds are weaker than O–H···O bonds.
Solubility in Water
Lower amines are soluble in water — they can form H-bonds with water.
Solubility decreases with increase in molar mass — the hydrophobic alkyl part increases.
Higher amines are essentially insoluble in water.
All amines are soluble in organic solvents (alcohol, ether, benzene).
Compound
Molar Mass
b.p. (K)
Remarks
n-C₄H₉NH₂ (1° amine)
73
350.8
Highest b.p. — 2 N–H available
(C₂H₅)₂NH (2° amine)
73
329.3
1 N–H available
C₂H₅N(CH₃)₂ (3° amine)
73
310.5
No N–H; no H-bonding
C₂H₅CH(CH₃)₂ (alkane)
72
300.8
No H-bonding at all
n-C₄H₉OH (alcohol)
74
390.3
Highest — strong O–H···O bonds
🔗
Fig. 9.2 — Intermolecular H-bonding in Primary Amines
NCERT p.266
Diagram showing intermolecular N–H···N hydrogen bonds between primary amine molecules (R–NH₂), illustrating the zigzag chain of H-bonds and explaining why 1° amines have higher boiling points than 2° and 3° amines.
📌 Image Source
NCERT Chemistry Class 12, Chapter 9, Fig. 9.2, Page 266. Replace with actual NCERT figure when image files are available.
9.6 Chemical Reactions of Amines
Amines are reactive due to: (a) the lone pair on N → Lewis base / nucleophile behaviour, and (b) the N–H bonds → allowing acylation, sulphonylation, etc.
Alkyl groups have a +I (inductive) effect — they push electrons towards N, increasing availability of the lone pair for protonation. Moreover, the substituted ammonium ion formed is stabilised by +I dispersal of the positive charge.
In gas phase: Basicity order follows inductive effect perfectly:
3° > 2° > 1° > NH₃
In aqueous phase: The trend is disturbed by solvation effect and steric hindrance:
1° cation has 3 N–H bonds → best H-bonded/solvated by water → most stabilised
3° cation has only 1 N–H → least solvated → less stabilised than expected
Net result: secondary amines are usually the strongest bases in aqueous solution
In aniline, the –NH₂ group is directly attached to benzene ring. The lone pair on N is in conjugation with the π system of the ring → delocalised → less available for protonation.
Aniline has 5 resonating structures (delocalized lone pair). Anilinium ion (protonated) has only 2 Kekulé structures. More resonating structures = more stability → aniline is more stable → resists protonation → weaker base.
∴ Arylamines are much weaker bases than alkylamines or NH₃ (pKb aniline ≈ 9.38 vs NH₃ ≈ 4.75)
🔄
Resonance Structures of Aniline (5 structures) & Anilinium Ion (2 structures)
NCERT p.269
Five resonating structures of aniline showing electron delocalisation of N lone pair into the benzene ring (ortho and para positions carry negative charge). Two Kekulé structures of anilinium ion (only ring resonance). Demonstrates why aniline is a weaker base.
📌 Image Source
NCERT Chemistry Class 12, Chapter 9, Page 269. Resonance structures of aniline and anilinium ion.
Amine
pKb
Reason
Methanamine (CH₃NH₂)
3.38
+I of methyl → strong base
N-Methylmethanamine ((CH₃)₂NH)
3.27
More +I → stronger (aqueous)
N,N-Dimethylmethanamine ((CH₃)₃N)
4.22
Steric + poor solvation → weaker
Ethanamine (C₂H₅NH₂)
3.29
+I of ethyl
N-Ethylethanamine ((C₂H₅)₂NH)
3.00
Strongest in ethyl series (aqueous)
N,N-Diethylethanamine ((C₂H₅)₃N)
3.25
Steric hindrance effect
Benzenamine (C₆H₅NH₂)
9.38
Resonance delocalisation → very weak
Phenylmethanamine (C₆H₅CH₂NH₂)
4.70
–CH₂– insulates N from ring
N-Methylaniline
9.30
Still aryl amine; weak base
N,N-Dimethylaniline
8.92
Slightly stronger than aniline
2. Alkylation
Amines react with alkyl halides (R–X) in nucleophilic substitution to give successively higher amines and finally quaternary ammonium salts (same mechanism as ammonolysis — see Section 9.4).
3. Acylation
Primary and secondary amines react with acid chlorides, acid anhydrides, or esters to give amides. This is a nucleophilic substitution. A stronger base (pyridine) is added to remove the HCl formed and shift equilibrium to the right.
Primary amines only (both aliphatic and aromatic) react with CHCl₃ and ethanolic KOH on heating to give isocyanides (carbylamines) — foul smelling substances.
Secondary and tertiary amines do NOT show this reaction. Used as a test for primary amines.
Very important for synthesis — covered in Section 9.7–9.9.
Secondary & Tertiary Amines with HNO₂
2° amines: form N-nitrosamines (R₂N–N=O) — yellow oily substances, many carcinogenic.
3° aliphatic amines: form soluble salts only (no N–H to react).
3° aromatic amines (like N,N-dimethylaniline): undergo ring nitrosation at the para position.
6. Reaction with Benzenesulphonyl Chloride (Hinsberg's Test)
Benzenesulphonyl chloride (C₆H₅SO₂Cl, Hinsberg's reagent) distinguishes between 1°, 2°, and 3° amines.
Amine Type
Reaction with C₆H₅SO₂Cl
With NaOH
1° Amine
Forms N-ethylbenzenesulphonamide C₆H₅SO₂–NHR (has N–H)
Soluble — N–H is acidic due to electron-withdrawing –SO₂– group
2° Amine
Forms N,N-disubstituted sulphonamide C₆H₅SO₂–NR₂ (no N–H)
Insoluble — no acidic H on N
3° Amine
Does NOT react with C₆H₅SO₂Cl
Remains as such
7. Electrophilic Substitution of Aromatic Amines
The –NH₂ group is a powerful ortho-para director and activating group for electrophilic aromatic substitution. It increases electron density at ortho and para positions through resonance (+M effect).
(a) Bromination
Aniline reacts with bromine water (without catalyst) at room temperature to give a white precipitate of 2,4,6-tribromoaniline (all three positions activated simultaneously).
C₆H₅–NH₂ + 3Br₂ —Br₂/H₂O→ 2,4,6-Br₃–C₆H₂–NH₂↓ + 3HBr White precipitate — confirms aniline; no FeBr₃ catalyst needed
To get monosubstituted product: protect –NH₂ by acetylation → reduce reactivity → brominate → hydrolyse the amide.
Direct nitration gives tarry products + meta derivative (because in strongly acidic HNO₃/H₂SO₄ medium, aniline is protonated → anilinium ion → –NH₃⁺ is a meta director). Products: o- (51%) + p- (47%) + m- (2%).
For pure para-nitroaniline: acetylate first (–NHCOCH₃ is weaker o/p director) → nitrate → hydrolyse.
(c) Sulphonation of Aniline
C₆H₅–NH₂ —H₂SO₄→ C₆H₅–NH₃⁺ HSO₄⁻ —453–473 K, H₂SO₄→ p-H₂N–C₆H₄–SO₃H Sulphanilic acid — a zwitterion (H₃N⁺–C₆H₄–SO₃⁻); used in azo dye synthesis
Why Aniline Doesn't Undergo Friedel-Crafts?
AlCl₃ (Lewis acid catalyst) forms a complex with N of aniline: C₆H₅NH₂·AlCl₃. This gives N a positive charge, making it a strong deactivating group — the ring is now deactivated for electrophilic substitution.
9.7 Diazonium Salts — Preparation
Definition
Diazonium Salts: Compounds of general formula Ar–N₂⁺ X⁻, where Ar is an aryl group and X⁻ = Cl⁻, Br⁻, HSO₄⁻, BF₄⁻, etc. They are named by suffixing "diazonium" to the parent hydrocarbon name + anion name.
E.g., C₆H₅–N₂⁺ Cl⁻ = benzenediazonium chloride
Diazotisation
The conversion of primary aromatic amines into diazonium salts is called diazotisation.
C₆H₅–NH₂ + NaNO₂ + 2HCl —273–278 K→ C₆H₅–N₂⁺ Cl⁻ + NaCl + 2H₂O • Temperature MUST be maintained at 273–278 K (0–5°C) to prevent decomposition of the salt • NaNO₂ + HCl → HNO₂ (generated in situ)
Why Aryl vs Alkyl Diazonium Salts?
Primary aliphatic amines form highly unstable alkyldiazonium salts → immediately decompose to N₂ + alcohol
Primary aromatic amines form relatively stable arenediazonium salts at low T (273–278 K) due to resonance stabilisation of Ar–N≡N⁺
Arenediazonium salts are not stored — used immediately after preparation
🔄
Resonance Structures of Arenediazonium Ion (Ar–N₂⁺)
NCERT p.274
Four resonating structures of benzenediazonium ion showing delocalisation of the positive charge from the diazo group into the benzene ring — ortho and para positions become electron-deficient. This explains the relative stability of the arenediazonium ion compared to the aliphatic counterpart.
📌 Image Source
NCERT Chemistry Class 12, Chapter 9, Page 274. Resonance structures of arenediazonium ion.
9.8–9.9 Properties & Chemical Reactions of Diazonium Salts
B. Reactions Involving Retention of Diazo Group — Coupling Reactions
Diazonium salts react with electron-rich aromatic compounds (phenols, arylamines) in electrophilic aromatic substitution at the para position. The –N=N– group is retained. Products are azo compounds used as azo dyes.
Diazonium salts are extremely versatile synthetic intermediates. They allow introduction of substituents that cannot be introduced by direct substitution:
Key Synthetic Advantages
Aryl fluorides: cannot be made by direct halogenation; made via Balz-Schiemann
Aryl iodides: I₂ is too weak an electrophile for direct substitution; made via KI reaction
Cyanobenzene: CN⁻ cannot do nucleophilic substitution on chlorobenzene; but easy via diazonium + CuCN
Phenol: can be made without direct substitution from aniline
Azo dyes: extended conjugated –Ar–N=N–Ar– chromophore; many commercial dyes
🎯
Mnemonic: "F I Cl Br CN OH NO₂ Azo"
Remember all substituents that can be introduced via diazonium salts using: Fluoride (Balz-Schiemann), Iodide (KI), Chloride/Bromide (Sandmeyer/Gattermann), Cyanide (CuCN), Hydroxyl (hydrolysis), Nitro (NaNO₂/Cu), Azo (coupling)
✏️ Practice Questions
Q1
Classify the following as 1°, 2°, or 3° amines and write their IUPAC names:
(i) (CH₃)₂CHNH₂ (ii) CH₃(CH₂)₂NH₂ (iii) CH₃NHCH(CH₃)₂ (iv) (CH₃CH₂)₂NCH₃
Arrange the following in decreasing order of basic strength:
C₆H₅NH₂, C₂H₅NH₂, (C₂H₅)₂NH, NH₃
Decreasing order: (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂
Reason: (C₂H₅)₂NH — secondary, +I of two ethyl groups + solvation; C₂H₅NH₂ — primary alkylamine; NH₃ — no alkyl groups; C₆H₅NH₂ — lone pair delocalised into ring → weakest base.
Q3
Explain why aniline is a weaker base than methylamine even though both have an unshared pair on N.
In aniline, the –NH₂ group is directly attached to the benzene ring. The lone pair on N conjugates with the π system of the ring (resonance), making it less available for protonation. Aniline has 5 resonating structures while anilinium ion has only 2 — so aniline is more stabilised than its conjugate acid → it is a weaker base. In methylamine, the +I effect of CH₃ increases electron density on N, making the lone pair more available → stronger base. pKb: C₆H₅NH₂ = 9.38 vs CH₃NH₂ = 3.38.
Q4
How do you distinguish between primary, secondary and tertiary amines using (a) Hinsberg's test and (b) Carbylamine test?
(a) Hinsberg's test (C₆H₅SO₂Cl + NaOH):
1° amine → sulphonamide with N–H → soluble in NaOH
2° amine → sulphonamide with no N–H → insoluble in NaOH
3° amine → no reaction
(b) Carbylamine test (CHCl₃ + alc. KOH + heat):
1° amine → foul smelling isocyanide (R–N≡C) — POSITIVE
2° and 3° amines → no isocyanide formed — NEGATIVE
Q5
Write the reactions for: (i) preparation of aniline from nitrobenzene (ii) Gabriel synthesis for propan-1-amine
(ii) Gabriel synthesis for CH₃CH₂CH₂NH₂:
Step 1: Phthalimide + KOH → K-phthalimide
Step 2: K-phthalimide + CH₃CH₂CH₂Br → N-propylphthalimide
Step 3: N-propylphthalimide + NaOH(aq) → CH₃CH₂CH₂NH₂ + sodium phthalate
Q6
Why do primary amines have higher boiling points than tertiary amines of similar molecular mass?
Primary amines have 2 N–H bonds → can form more extensive intermolecular hydrogen bonds (N–H···N). Tertiary amines have no N–H bond → cannot form N–H···N hydrogen bonds, only weak van der Waals forces. More H-bonding → higher energy needed to separate molecules → higher boiling point. E.g., n-C₄H₉NH₂ (1°, bp 350.8 K) vs C₂H₅N(CH₃)₂ (3°, bp 310.5 K) for same molar mass (73 g/mol).
Q7
What is diazotisation? Write the conditions required and the reaction for preparation of benzenediazonium chloride.
Diazotisation: Conversion of primary aromatic amines into diazonium salts using NaNO₂ + HCl at 273–278 K (0–5°C).
Conditions: low temperature (0–5°C) is critical — at higher T, the diazonium salt decomposes to phenol. NaNO₂ + HCl generate HNO₂ in situ. The salt is used immediately after preparation.
Q8
Convert aniline to (i) chlorobenzene (ii) iodobenzene (iii) fluorobenzene (iv) cyanobenzene using diazonium salt reactions.
All start with: C₆H₅NH₂ —(NaNO₂/HCl, 273–278 K)→ C₆H₅N₂⁺Cl⁻
Why cannot aryl fluorides and aryl iodides be prepared by direct halogenation, and how does the diazonium salt route solve this problem?
Aryl fluorides: F₂ is too reactive; it causes multiple substitutions and side reactions. Via diazonium salt (Balz-Schiemann): controlled mono-fluorination by mild thermal decomposition of the diazonium fluoroborate salt.
Aryl iodides: I₂ is a weak electrophile — the benzene ring does not get activated enough for electrophilic substitution with I₂ alone (unless a very activated ring). Via diazonium salt + KI: the I⁻ nucleophile directly replaces the good leaving group N₂, giving clean conversion.
The diazonium route allows selective mono-substitution under mild, controlled conditions.
Q10
An aromatic compound A on treatment with aqueous NH₃ and heating forms compound B which on heating with Br₂ and KOH forms compound C (mol. formula C₆H₇N). Identify A, B, C with IUPAC names.
C has mol. formula C₆H₇N → C₆H₅NH₂ (aniline, MW = 93 = 6×12 + 7×1 + 14 = 93 ✓)
Working backwards using Hoffmann degradation (Br₂ + KOH gives amine with one less C):
C = C₆H₅NH₂ (aniline) ← formed by Hoffmann from B
B must have 7 carbons: C₆H₅–CO–NH₂ = benzamide (B)
A + aqueous NH₃ → B (benzamide): A must be C₆H₅COOH = benzoic acid (A)
A = C₆H₅COOH (Benzoic acid / benzenecarboxylic acid)
B = C₆H₅CONH₂ (Benzamide / benzenecarbamide)
C = C₆H₅NH₂ (Aniline / benzenamine)
Q11
Arrange in increasing order of basic strength:
(i) C₆H₅NH₂, C₆H₅CH₂NH₂, NH₃, C₂H₅NH₂, (C₂H₅)₂NH
(ii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
