Classification and Nomenclature
Amines are derivatives of ammonia (NH₃) in which one or more hydrogen atoms are replaced by alkyl or aryl groups. The degree of substitution on nitrogen defines the class.
Classification
| Class | Formula | N-H bonds | Example |
|---|---|---|---|
| Primary (1°) | RNH₂ | 2 | CH₃NH₂ (methylamine) |
| Secondary (2°) | R₂NH | 1 | (CH₃)₂NH (dimethylamine) |
| Tertiary (3°) | R₃N | 0 | (CH₃)₃N (trimethylamine) |
| Quaternary ammonium | R₄N⁺ X⁻ | 0 | (CH₃)₄N⁺Cl⁻ |
Aromatic vs aliphatic: if an aryl group (benzene ring) is directly attached to nitrogen, the amine is aromatic. Aniline (C₆H₅NH₂) is the parent aromatic amine.
IUPAC Nomenclature
For simple amines: name the alkyl/aryl group(s) and add the suffix -amine. Examples: CH₃NH₂ = methanamine; C₂H₅NH₂ = ethanamine; (CH₃)₂NH = N-methylmethanamine. For substituted aryl amines, use aniline as the parent: CH₃C₆H₄NH₂ = methylaniline (specify o/m/p or 2/3/4 position). Common names (methylamine, aniline, ethylamine) are widely accepted in NEET contexts.
Preparation of Amines
1. Reduction of Nitro Compounds
Aromatic primary amines are almost always made by reducing nitro compounds. Reagents: Fe/HCl, Sn/HCl, or catalytic hydrogenation (H₂, Ni/Pd).
C₆H₅NO₂ + 6[H] Fe/HCl C₆H₅NH₂ + 2H₂O
This is the industrial route to aniline. Nitrobenzene (prepared by nitration) is reduced in acidic medium. The primary amine salt precipitates and is then made free by adding NaOH.
2. Reduction of Amides, Nitriles, and Oximes
| Starting material | Reagent | Product | Note |
|---|---|---|---|
| Amide RCONH₂ | LiAlH₄ / H₂O | RCH₂NH₂ (1°) | Same carbon count |
| Nitrile RCN | LiAlH₄ or H₂/Ni | RCH₂NH₂ (1°) | One extra CH₂ |
| Oxime R₂C=NOH | H₂/Ni or Na/C₂H₅OH | R₂CHNH₂ (1°) | Secondary carbon |
3. Hoffmann Bromamide Degradation
This converts a primary amide RCONH₂ to a primary amine RNH₂ with one fewer carbon. Reagents: Br₂ + NaOH (aqueous).
RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Mechanism: NaOBr brominates the amide nitrogen. The N-bromo intermediate undergoes a 1,2-migration of the R group (Curtius-type) to form an isocyanate (RNCO). Hydrolysis of RNCO gives RNH₂ + CO₂. The key point: carbon count decreases by 1 (the C=O of the amide is lost as CO₂).
4. Gabriel Phthalimide Synthesis
Produces pure primary aliphatic amines free from secondary or tertiary contamination.
Steps: (1) Phthalimide + KOH → potassium phthalimide (K-salt). (2) Potassium phthalimide + RX (1° alkyl halide) → N-alkylphthalimide (via SN2). (3) Hydrolysis with NaOH (aq) or NH₂NH₂ (hydrazinolysis) → RNH₂ + phthalic acid.
Limitation: ArX cannot undergo SN2, so aromatic primary amines cannot be made by this method.
Track Your NEET Score Across All 90 Chapters
Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.
Physical Properties
Boiling Points
Primary and secondary amines can form hydrogen bonds (N-H...N). Tertiary amines cannot form N-H hydrogen bonds (no N-H present) and have the lowest boiling points among amines of similar MW. Among primary and secondary amines of the same formula weight, primary amines generally have higher b.p. (more N-H bonds).
Amines have lower boiling points than corresponding alcohols because N-H...N hydrogen bonds are weaker than O-H...O bonds (N is less electronegative and O-H is more polar).
Solubility
Lower amines (C₁-C₃) are freely miscible with water because they can hydrogen bond with water molecules. As chain length increases, the hydrophobic alkyl part dominates and solubility decreases. Aniline is slightly soluble in water (only 3.4 g/100 mL at 25°C) because the aromatic ring is hydrophobic.
Smell
Simple aliphatic amines smell like fish (trimethylamine, (CH₃)₃N, is the odour of rotting fish). Higher amines and aromatic amines have unpleasant odours. Isocyanides (RNC), formed in the carbylamine test, are famously foul-smelling.
Basicity of Amines
Amines are Lewis and Brønsted bases. They accept protons via the lone pair on nitrogen:
RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻ (Kb, pKb)
Higher Kb = stronger base = lower pKb.
Key Rule 1: Aliphatic Amines are Stronger Bases than Aromatic Amines
In aniline (C₆H₅NH₂), the lone pair on N is delocalised into the benzene ring by resonance. There are five resonance structures where the lone pair moves into the ring. This delocalisation reduces lone pair availability, making aniline a much weaker base than aliphatic amines.
Methylamine pKb = 3.36. Aniline pKb = 9.40. Difference of 6 units means aniline is 10⁶ times weaker as a base.
Key Rule 2: In Aqueous Solution: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N
Two opposing effects determine basicity of aliphatic amines in water:
| Effect | Predicts | Reason |
|---|---|---|
| +I inductive effect (alkyl groups) | 3° > 2° > 1° | More alkyl groups push more electrons onto N |
| Solvation of conjugate acid in water | 1° > 2° > 3° | More N-H bonds → more H-bonding → more stable RNH₃⁺ |
Net result: dimethylamine has the best balance (2 alkyl groups for +I, still 2 N-H for solvation). Trimethylamine has maximum +I but minimal solvation (only 1 N-H in conjugate acid), so it is less basic than dimethylamine and sometimes even methyl amine. Actual pKb: (CH₃)₂NH 3.27, CH₃NH₂ 3.36, (CH₃)₃N 4.22.
Effect of Substituents on Aniline Basicity
Electron-withdrawing groups (NO₂, Cl, CHO) on the benzene ring further withdraw electron density from N via the ring, reducing basicity. Electron-donating groups (CH₃, OCH₃) increase basicity slightly.
Most important: o-nitroaniline < m-nitroaniline < p-nitroaniline (all weaker than aniline). Para-nitro lowers basicity most for resonance-withdrawal reasons. p-Methoxyaniline (anisidine) is slightly more basic than aniline.
Amine Basicity Comparator
Compare pKb values across aliphatic and aromatic amines. Click any compound to see the inductive, solvation, and resonance effects that determine its basicity. The key NEET fact: aliphatic amines (pKb 3-4) are 10⁶× more basic than aniline (pKb 9.4) because of resonance delocalisation.
Dimethylamine
(CH₃)₂NH
Chemical Reactions of Amines
1. Alkylation
Amines react with alkyl halides by nucleophilic substitution. The primary amine attacks the alkyl halide and displaces halide. Stepwise alkylation can produce 2°, 3°, and quaternary ammonium salts. This is why direct alkylation gives a mixture and Gabriel synthesis is preferred for pure primary amines.
2. Acylation (Acetylation and Benzoylation)
Primary and secondary amines react with acyl chlorides (RCOCl) or acid anhydrides ((RCO)₂O) to give amides. Tertiary amines do not react with acylating agents under normal conditions because they have no N-H.
Aniline + (CH₃CO)₂O → CH₃CONHC₆H₅ (acetanilide) + CH₃COOH
Acetylation protects the -NH₂ group (converts it to amide) during synthesis. The amide can be hydrolysed back to the amine under acidic or basic conditions.
Benzoylation (Schotten-Baumann reaction): amine + C₆H₅COCl + NaOH → N-benzoyl amide. Tertiary amines do not react.
3. Carbylamine Reaction (Isocyanide Test)
Primary amines (both aliphatic and aromatic) react with CHCl₃ + alcoholic KOH to form isocyanides (RNC), which have a distinctive foul smell.
RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O
Secondary and tertiary amines do not give this test. The carbylamine reaction is used as a confirmatory test for primary amines.
4. Reaction with Nitrous Acid (HNO₂ = NaNO₂ + HCl)
The reaction with HNO₂ at 0–5°C is different for each class:
| Amine class | Product | Observation |
|---|---|---|
| 1° aliphatic | Alkanol + N₂ (unstable diazonium) | N₂ gas evolved; diazonium decomposes instantly |
| 1° aromatic | Arene diazonium salt (stable at 0–5°C) | No immediate gas; diazonium used in reactions |
| 2° aliphatic or aromatic | N-nitrosamine (yellow oil) | Yellow N-nitrosodimethylamine type product |
| 3° aliphatic | N-nitroso salt (reversible) | Dissolves as nitrous acid salt; on basification, free amine |
Distinction Tests: Hinsberg and HNO₂
Two key tests distinguish primary, secondary, and tertiary amines. Know both for NEET.
Hinsberg Test
Reagent: benzenesulfonyl chloride (C₆H₅SO₂Cl) + aqueous KOH.
| Amine | Reaction | Product in KOH | On acidification |
|---|---|---|---|
| 1° (RNH₂) | Reacts | Soluble (N-H is acidic, forms K-salt) | Precipitate forms |
| 2° (R₂NH) | Reacts | Insoluble precipitate (no N-H to ionise) | Precipitate remains |
| 3° (R₃N) | No reaction | No change | No change |
Memory tip: 1° is soluble in alkali (has one acidic N-H on sulfonamide), 2° is insoluble (no N-H on sulfonamide), 3° does not react at all.
Amine Distinction Tests Simulator
Select an amine class (primary, secondary, or tertiary) and a distinction test (Hinsberg, carbylamine, or HNO₂). See what happens, why it happens, and what conclusion you can draw. Covers all three tests that NEET uses to identify amine type.
Hinsberg Test
onSoluble in excess KOH (alkali-soluble)
| Test | 1° Amine | 2° Amine | 3° Amine |
|---|---|---|---|
| Hinsberg (PhSO₂Cl + KOH) | Alkali-soluble sulfonamide | Alkali-insoluble ppt. | No reaction |
| Carbylamine (CHCl₃ + KOH) | Foul smell (isocyanide) | No reaction | No reaction |
| HNO₂ (NaNO₂ + HCl, 0°C) | Ar: diazonium salt; Alkyl: N₂ gas | Yellow N-nitrosamine | Salt (reversible) |
Diazonium Salts
Preparation (Diazotisation)
Aromatic primary amines react with NaNO₂ + HCl at 0–5°C to form arene diazonium chlorides.
C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
The temperature must be kept at 0–5°C. Above this temperature, the diazonium salt decomposes (N₂ is lost and phenol forms). Aliphatic primary amine diazonium salts are too unstable to isolate.
Reactions of Diazonium Salts
| Reaction | Reagent | Product | Named reaction |
|---|---|---|---|
| ArN₂⁺ → ArCl | CuCl + HCl | Chlorobenzene | Sandmeyer |
| ArN₂⁺ → ArBr | CuBr + HBr | Bromobenzene | Sandmeyer |
| ArN₂⁺ → ArCN | CuCN + KCN | Benzonitrile | Sandmeyer |
| ArN₂⁺ → ArI | KI | Iodobenzene | Finkelstein variant |
| ArN₂⁺ → ArF | HBF₄ then heat | Fluorobenzene | Balz-Schiemann |
| ArN₂⁺ → ArH | H₃PO₂ | Benzene | Deamination |
| ArN₂⁺ → ArOH | H₂O + heat | Phenol | Hydrolysis |
| ArN₂⁺ → Ar-N=N-Ar' | Activated ArH (phenol/aniline) | Azo compound | Coupling reaction |
Coupling Reaction (Azo Dye Formation)
The diazonium ion is an electrophile. It attacks activated aromatic rings (phenol in alkaline medium, aniline in acidic/neutral medium) at the para position.
C₆H₅N₂⁺Cl⁻ + C₆H₅OH (alkaline) → C₆H₅-N=N-C₆H₄-OH (p-hydroxyazobenzene) + HCl
The product contains the azo group (-N=N-), which gives characteristic orange/red colour. These azo compounds are dyes used in textiles, food colouring, and pH indicators (methyl orange, Congo red).
Track Your NEET Score Across All 90 Chapters
Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.
Worked NEET Problems
NEET-style problem · Basicity
Question
Solution
Correct order: (B) (CH₃)₂NH > (A) CH₃NH₂ > (C) (CH₃)₃N > (D) C₆H₅NH₂.
Two factors compete for aliphatic amines in water. +I inductive effect (alkyl groups) favours 3° > 2° > 1°. Solvation of the conjugate acid (N-H bonds available for H-bonding with water) favours 1° > 2° > 3°. The balance gives 2° as the most basic, then 1°, then 3°. Aniline is far less basic (pKb 9.4 vs 3-4 for aliphatic) because the lone pair is delocalised into the ring.
NEET-style problem · Preparation
Question
Solution
Product: methylamine (CH₃NH₂). Reaction type: Hoffmann bromamide degradation.
Reason for carbon loss: in the mechanism, the R group migrates from the carbonyl carbon to nitrogen (Curtius-type 1,2-shift) via an isocyanate intermediate (CH₃NCO). NaOH then hydrolyses the isocyanate: CH₃NCO + H₂O → CH₃NH₂ + CO₂. The CO₂ carries away the original carbonyl carbon. So acetamide (2C) gives methylamine (1C) + CO₂. Carbon count decreases by 1.
NEET-style problem · Diazonium Salts
Question
Solution
Step 1 (common): diazotise aniline with NaNO₂ + HCl at 0–5°C to get benzenediazonium chloride (C₆H₅N₂⁺Cl⁻).
(a) Fluorobenzene: treat diazonium salt with HBF₄ to precipitate diazonium tetrafluoroborate (ArN₂⁺BF₄⁻). Heat it: ArN₂⁺BF₄⁻ → ArF + N₂ + BF₃. This is the Balz-Schiemann reaction.
(b) Iodobenzene: treat diazonium salt with KI directly: C₆H₅N₂⁺ + KI → C₆H₅I + N₂ + KCl. No copper catalyst needed for iodine.
(c) p-hydroxyazobenzene: treat diazonium salt with phenol in alkaline medium (NaOH solution). The diazonium ion couples at the para position of phenol to give C₆H₅-N=N-C₆H₄-OH (orange azo compound). This is the coupling reaction used in azo dye synthesis.
NEET-style problem · Distinction Tests
Question
Solution
A is a primary amine (1°): Hinsberg test gives alkali-soluble sulfonamide (has N-H); carbylamine test positive (CHCl₃ + KOH gives foul isocyanide smell). Only primary amines give both results.
B is a secondary amine (2°): Hinsberg test gives alkali-insoluble sulfonamide (no N-H to ionise); carbylamine test negative (secondary amines do not react with CHCl₃/KOH).
C is a tertiary amine (3°): no reaction with benzenesulfonyl chloride (no N-H); no carbylamine test (no N-H for isocyanide formation).
Summary Cheat Sheet
Basicity Quick Reference
| Compound | pKb | Key reason |
|---|---|---|
| (CH₃)₂NH (dimethylamine) | 3.27 | Best inductive + solvation balance |
| CH₃NH₂ (methylamine) | 3.36 | +I effect, good solvation |
| (CH₃)₃N (trimethylamine) | 4.22 | Max +I but poor solvation |
| NH₃ (ammonia) | 4.74 | No +I, no resonance |
| C₆H₅NH₂ (aniline) | 9.40 | Lone pair delocalised into ring |
| p-NO₂-C₆H₄-NH₂ (p-nitroaniline) | 13.0 | NO₂ withdraws further by resonance |
Named Reactions Summary
| Reaction | Reagents | Product | Key point |
|---|---|---|---|
| Hoffmann bromamide | Br₂ + NaOH | RNH₂ (1° amine, 1 fewer C) | Only primary amide → primary amine |
| Gabriel phthalimide | Phthalimide + KOH + RX then hydrolysis | Pure RNH₂ | No ArX; only 1° aliphatic amines |
| Carbylamine | CHCl₃ + alc. KOH | RNC (foul smell) | Only 1° amines react |
| Diazotisation | NaNO₂ + HCl, 0–5°C | ArN₂⁺Cl⁻ | Only aromatic 1° amine is stable |
| Sandmeyer (Cl/Br/CN) | CuX/HX | ArX or ArCN | Copper catalyst needed for Cl, Br, CN |
| Balz-Schiemann (F) | HBF₄ then heat | ArF | Only route to ArF from ArNH₂ |
| Azo coupling | ArN₂⁺ + phenol/aniline | Ar-N=N-Ar' (azo dye) | Electrophilic; para attack; colour |
Distinction Tests at a Glance
| Test | 1° Amine | 2° Amine | 3° Amine |
|---|---|---|---|
| Hinsberg (PhSO₂Cl + KOH) | Alkali-soluble sulfonamide | Alkali-insoluble precipitate | No reaction |
| Carbylamine (CHCl₃ + alc. KOH) | Foul smell (positive) | No reaction | No reaction |
| HNO₂ test at 0°C | Aromatic: stable diazonium; aliphatic: N₂ gas | Yellow N-nitrosamine | Salt formation (reversible) |
Frequently asked questions
Why are aliphatic amines more basic than aniline?
The key factor is availability of the lone pair on nitrogen. In aliphatic amines (like methylamine, CH₃NH₂), the alkyl group donates electrons by the +I (inductive) effect, increasing electron density on nitrogen. This makes the lone pair more available to accept a proton, so aliphatic amines are stronger bases (pKb 3–4). In aniline (C₆H₅NH₂), the lone pair on nitrogen is delocalised into the benzene ring by conjugation (resonance). The lone pair is spread over the ring and is less available to accept a proton. This lowers the basicity significantly (pKb 9.4). Quantitatively: Kb of methylamine = 5 × 10⁻⁴, Kb of aniline = 4 × 10⁻¹⁰. The resonance withdrawal in aniline reduces basicity by six orders of magnitude.
Why is the basicity order 2° > 1° > 3° for aliphatic amines in water?
Purely by inductive effect, you would expect 3° > 2° > 1° (more alkyl groups donating more electrons). But the actual order in aqueous solution is 2° > 1° > 3°. Two opposing factors are at work: (1) Inductive effect: more alkyl groups push more electrons onto N, increasing lone pair availability. This predicts 3° > 2° > 1°. (2) Solvation (hydration): the ammonium ion (conjugate acid) is stabilised by hydrogen bonding with water. Primary ammonium R-NH₃⁺ has 3 N-H bonds and is heavily solvated. Secondary R₂NH₂⁺ has 2 N-H bonds — moderate solvation. Tertiary R₃NH⁺ has only 1 N-H bond — least solvated. Less solvation means the conjugate acid is less stable, which reduces apparent basicity of the tertiary amine. The balance of these two opposing effects gives 2° (best balance) > 1° > 3°.
What is the Hoffmann bromamide degradation? What product forms?
Hoffmann bromamide degradation converts a primary amide (RCONH₂) to a primary amine (RNH₂) with one fewer carbon. Reagents: Br₂ + NaOH (aqueous). Mechanism: (1) Br₂ and NaOH form NaOBr (or OBr⁻). (2) OBr⁻ bromaminates the amide nitrogen: RCONH₂ → RCONHBr. (3) A strongly basic N-bromo intermediate forms and the acyl-N bond rearranges in a Curtius-type 1,2-shift: R migrates from C to N with the bonding pair. An isocyanate (RNCO) is the intermediate. (4) RNCO is hydrolysed by NaOH to give RNH₂ + CO₂. Product: a primary amine (RNH₂) with one carbon fewer than the starting amide. This is important in NEET because it distinguishes primary amides and yields primary amines specifically. Example: CH₃CONH₂ + Br₂/NaOH → CH₃NH₂.
How does the Hinsberg test distinguish primary, secondary, and tertiary amines?
Reagent: benzenesulfonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. Primary amine (RNH₂): reacts to give a sulfonamide (RNH-SO₂C₆H₅). This sulfonamide has one acidic N-H and dissolves in excess KOH (alkali soluble). On acidification, it precipitates. Secondary amine (R₂NH): reacts to give a disubstituted sulfonamide (R₂N-SO₂C₆H₅). This has NO acidic N-H and is insoluble in KOH — precipitate forms in alkaline solution and remains on acidification. Tertiary amine (R₃N): does NOT react with benzenesulfonyl chloride because the nitrogen has no N-H. No reaction. Summary: 1° gives alkali-soluble sulfonamide; 2° gives alkali-insoluble precipitate; 3° gives no reaction.
What is the carbylamine reaction and who does it identify?
Carbylamine reaction (also called isocyanide test): a primary amine reacts with chloroform (CHCl₃) and alcoholic KOH to give a foul-smelling isocyanide (RNC, also called carbylamine). Reagents: CHCl₃ + alc. KOH + primary amine. Observation: characteristic intensely unpleasant smell. Primary aliphatic and aromatic amines both give a positive carbylamine test. Secondary and tertiary amines do NOT react. This makes it a specific test for primary amines (both aliphatic and aromatic). In NEET questions, the carbylamine test is used to confirm whether a compound is a primary amine. The isocyanide is toxic, which is why the characteristic smell is diagnostic.
What are diazonium salts and what reactions do they undergo?
Diazonium salts (ArN₂⁺ X⁻) are formed when aromatic primary amines react with NaNO₂ + HCl at 0–5°C (diazotisation). The benzene diazonium chloride (C₆H₅N₂⁺Cl⁻) is relatively stable at low temperature and undergoes many useful reactions: (1) Sandmeyer reactions: replacement of N₂⁺ by Cl, Br (with CuCl/CuBr + HCl/HBr), CN (with CuCN), or I (with KI). (2) Balz-Schiemann reaction: N₂⁺ replaced by F via ArN₂⁺BF₄⁻ → ArF + N₂ + BF₃. (3) Reduction: N₂⁺ → ArH (deamination) using H₃PO₂. (4) Coupling reactions: the diazonium ion acts as an electrophile and couples with activated aromatic rings (phenol in alkaline medium, aniline) at the para position to give azo compounds (ArN=NAr'). These orange/red azo dyes are used in NEET synthesis questions. Diazonium salts are only stable at 0–5°C; above this temperature they decompose.
What is Gabriel phthalimide synthesis and what is it used for?
Gabriel phthalimide synthesis makes pure primary aliphatic amines without contamination by secondary or tertiary amines. Steps: (1) Phthalimide (a cyclic imide) is treated with KOH to form potassium phthalimide (the N-K salt). (2) Potassium phthalimide reacts with a primary alkyl halide (RX, usually 1°) by SN2 displacement to give N-alkylphthalimide. (3) N-alkylphthalimide is hydrolysed with NaOH (aq.) OR hydrazinolysis (NH₂NH₂) to give the primary amine (RNH₂) + phthalic acid (or phthalhydrazide with hydrazine). Why it is useful: since phthalimide has only one N-H, only one alkyl group can be introduced, guaranteeing a primary amine product. Phenyl halides (ArX) do not work because SN2 on Ar is not possible. NEET often asks which synthesis gives a pure primary amine — answer: Gabriel phthalimide.
How many NEET questions come from Amines and what are the most important topics?
Amines contributes 3–5 NEET questions per year from Class 12 Chemistry. The highest-frequency topics are: (1) Basicity order — aliphatic vs aromatic, 2° > 1° > 3° in water. Know the reason (inductive vs solvation). (2) Hoffmann bromamide degradation — identify that the product has one fewer carbon and is a primary amine. (3) Distinction tests — Hinsberg test (1°: alkali-soluble; 2°: alkali-insoluble; 3°: no reaction), carbylamine test (only 1°). (4) Diazonium chemistry — Sandmeyer reaction sequence (Cl, Br, CN, I replacement), Balz-Schiemann (F), coupling reactions (azo dyes). (5) Gabriel synthesis — makes pure 1° aliphatic amine. Learn the reagent–product pairs precisely, as NEET frequently tests "which reagent gives which product" in the form of match-the-following or single correct MCQs.
Continue with the next chapter notes
Stay in NCERT order — the next chapter's notes are one click away.
Track Your NEET Score Across All 90 Chapters
Free 14-day trial. AI tutor, full mock tests and chapter analytics — built for NEET 2027.
Free 14-day trial · No credit card required