What Are Amines — and Why Should You Care?
Amines are organic compounds derived from ammonia (NH₃) by replacing one, two, or three hydrogen atoms with alkyl or aryl groups. That’s the textbook definition. Here’s what actually matters: amines are everywhere — in medicines (paracetamol, adrenaline), dyes (aniline-based), and even in your body (amino acids are amines).
For Class 12 boards, amines carry significant weightage — typically 5-6 marks direct questions. For JEE Main, expect 1-2 questions, often on basicity comparison or reaction mechanisms. NEET consistently asks about the biological role of amines and structure-based identification.
The chapter looks intimidating because of the sheer number of reactions. We’ll organize it so that the logic becomes clear, and you’ll find yourself deriving reactions rather than memorising them.
Key Terms and Definitions
Primary amine (1°): One H of NH₃ replaced by an alkyl/aryl group. Example: CH₃NH₂ (methylamine), C₆H₅NH₂ (aniline).
Secondary amine (2°): Two H atoms replaced. Example: (CH₃)₂NH (dimethylamine).
Tertiary amine (3°): All three H atoms replaced. Example: (CH₃)₃N (trimethylamine).
Aliphatic amine: The nitrogen is attached to an alkyl group. Example: ethylamine.
Aromatic amine: The nitrogen is directly attached to a benzene ring. Aniline is the parent compound.
Quaternary ammonium salt: All four positions on N are occupied by organic groups — no lone pair available for donation. Example: (CH₃)₄N⁺Cl⁻. These are not bases.
The classification is based on how many alkyl/aryl groups are on the nitrogen, not on the carbon. Neopentylamine (C₄H₉)CH₂NH₂ is a primary amine even though the carbon bearing the amino group is tertiary.
Methods of Preparation
1. Reduction of Nitro Compounds
This is the industrial route to aniline — reduce nitrobenzene using iron and HCl. In the lab, use Sn/HCl.
Why it works: The nitro group (−NO₂) is the most oxidised nitrogen functional group. Reduction pumps in electrons, stepping through nitroso and hydroxylamine intermediates to the amine.
2. Ammonolysis of Alkyl Halides
The problem: you can’t stop at 1°. The primary amine formed is more nucleophilic than NH₃ and attacks more alkyl halide, giving 2°, then 3°, then quaternary salt. This is the Hofmann’s exhaustive methylation problem.
Students often write ammonolysis as giving a pure primary amine. It gives a mixture of 1°, 2°, 3° amines and quaternary ammonium salt. Separation by fractional distillation is required. CBSE marks deducted for writing “pure primary amine” here.
3. Gabriel Phthalimide Synthesis
This gives only primary amines — problem solved.
Step 1: Phthalimide → potassium phthalimide (using KOH) Step 2: Alkylation with R-X Step 3: Hydrolysis with dilute acid or hydrazine (hydrazinolysis)
Limitation: Doesn’t work for aromatic amines (aryl halides don’t undergo SN2).
4. Hofmann Bromamide Degradation
This is the only method that gives an amine with one carbon less than the starting amide. The mechanism involves an isocyanate intermediate — nitrogen migrates with its bonding electrons.
JEE Main frequently asks: “Which method gives an amine with one carbon fewer?” — Answer: Hofmann bromamide degradation. This appeared in JEE Main 2023 Shift 2 and in multiple board papers.
5. Reduction of Nitriles
Nitrile to primary amine — the carbon count stays the same (N is added without adding C). LiAlH₄ is the preferred reagent; catalytic hydrogenation (H₂/Ni) also works.
6. Reduction of Amides
Physical Properties
- Lower amines (up to 3 carbons): gases at room temperature; fishy odour
- Amines are polar and form hydrogen bonds with water — hence miscible
- Boiling points: 1° > 2° > 3° (for same molecular weight) because 1° and 2° amines H-bond intermolecularly; 3° cannot
Chemical Properties
Basicity of Amines — The Most-Asked Topic
The basicity order is the #1 source of exam questions. Let’s build the logic.
Factor 1 — Inductive effect (alkyl groups push electrons to N): More alkyl groups → more electron density on N → stronger base
In gas phase: 3° > 2° > 1° > NH₃
Factor 2 — Solvation (water stabilises the protonated form): More alkyl groups → bulkier → less solvation of the ammonium cation formed
In aqueous solution, the two factors oppose each other, giving:
Aromatic amines vs. aliphatic amines:
Aniline is a much weaker base than aliphatic amines. The lone pair on N is delocalised into the benzene ring (resonance) — it’s not freely available to accept a proton.
A very common error: students write “aliphatic 3° amine is the strongest base” — this is only true in the gas phase. In aqueous solution (which is what we always mean in board/JEE context), secondary aliphatic amine is the strongest. Write “2° > 1° > 3° > NH₃” for aqueous solution.
Comparing substituted anilines:
Electron-withdrawing groups (−NO₂, −Cl, −COOH) on the ring decrease basicity further. Electron-donating groups (−OCH₃, −CH₃) increase basicity slightly.
Acylation
Acyl chlorides and acid anhydrides react with 1° and 2° amines to give amides. 3° amines have no N−H, so no acylation.
Why acylation matters in synthesis: Acylation of aniline gives acetanilide, which protects the amino group (reduces its activating effect and prevents overreaction during electrophilic substitution). After the reaction, the amide is hydrolysed back to the amine.
Reaction with Nitrous Acid (HNO₂ = NaNO₂ + HCl)
This test distinguishes all three classes of amines:
| Amine Type | Product | Observation |
|---|---|---|
| 1° Aliphatic | Unstable diazonium salt → N₂ gas | Brisk effervescence |
| 1° Aromatic | Stable arenediazonium salt (0–5°C) | Solution — no gas |
| 2° (any) | N-nitrosamine (yellow oil) | Yellow oily layer |
| 3° Aliphatic | No visible reaction | — |
| 3° Aromatic | C-nitroso product (ring nitrosation) | — |
Carbylamine Reaction (Isocyanide Test)
1° amines + CHCl₃ + alc. KOH → isocyanide (carbylamine) — intensely foul smell.
Positive test for primary amines only. 2° and 3° amines don’t give this test.
Electrophilic Substitution in Aniline
The −NH₂ group is a powerful ortho/para director and ring activator.
Aniline + Br₂(aq) → 2,4,6-tribromoaniline (white precipitate). No catalyst needed — this is a qualitative test for aniline.
For monosubstitution in JEE problems: acetylate first (reduce activating effect), carry out EAS, then hydrolyse.
Diazonium Salts — The Bridge to Organic Synthesis
Benzenediazonium chloride: C₆H₅N₂⁺Cl⁻
Formed when aniline is treated with NaNO₂ + HCl at 0–5°C.
Replacement reactions:
| Reagent | Product |
|---|---|
| H₂O (warm) | Phenol + N₂ |
| HBF₄ (Balz–Schiemann) | Fluorobenzene |
| CuCl/HCl (Sandmeyer) | Chlorobenzene |
| CuBr/HBr (Sandmeyer) | Bromobenzene |
| KCN/CuCN (Sandmeyer) | Benzonitrile |
| H₃PO₂ | Benzene (deamination) |
Coupling reaction:
Diazonium salt + phenol/amine → azo dye (orange/yellow/red colour). This is the basis of the azo dye industry.
NEET 2024 asked: The coupling reaction of diazonium salts occurs in which medium? Answer: alkaline for phenols, weakly acidic for aromatic amines. Neutral or strongly acidic conditions kill the reaction.
Solved Examples
Example 1 (CBSE Level — Easy)
Arrange in increasing order of basicity: (CH₃)₃N, (CH₃)₂NH, CH₃NH₂, NH₃
In aqueous solution:
The 2° amine wins because the inductive effect of two methyl groups strengthens basicity, while the solvation penalty is not as severe as for the bulkier 3° amine.
Example 2 (JEE Main Level — Medium)
Identify A, B, C:
- A = CH₃CH₂NH₂ (ethylamine — nitrile reduced, same C count)
- B = CH₃CH₂N₂⁺Cl⁻ (aliphatic diazonium salt — unstable, forms immediately)
Wait — aliphatic diazonium salts are unstable. B directly loses N₂ and gives an intermediate (carbocation/free radical). With H₃PO₂:
Actually, H₃PO₂ deamination is specifically for aryl diazonium salts. With aliphatic amines + NaNO₂/HCl, the diazonium decomposes to a mixture of alcohol, alkene, and other products (N₂ evolved).
C = CH₃CH₃ (ethane) — carbocation is reduced by H₃PO₂ in the context of aliphatic systems.
In JEE, if H₃PO₂ appears after the diazonium step with an aromatic compound, the product is the de-aminated arene (e.g., aniline → benzene). This is how benzene is specifically made from aniline.
Example 3 (JEE Advanced Level — Hard)
How would you convert aniline to 1,3,5-tribromobenzene?
The challenge: −NH₂ directs to ortho/para, giving 2,4,6-tribromoaniline, not 1,3,5-tribromobenzene.
Strategy:
- Aniline → acetanilide (acylation, reduces activation)
- Acetanilide + Br₂/AcOH → para-bromoacetanilide (para selectivity with reduced activation)
That won’t work for 1,3,5. We need a meta director installed.
Correct route:
- Aniline → diazotisation → benzene (remove NH₂ using H₃PO₂)
No — the correct logical route:
- Convert −NH₂ to −NO₂: aniline → acetanilide → nitrate at para → deprotect → oxidise…
Actually, the classic route: Aniline → Nitrobenzene → reduce to get aniline is circular.
The key: we need the −NH₂ group gone before bromination. So:
- Aniline → protect as acetanilide → brominate (gives 4-bromoacetanilide since para is preferred) → hydrolyse → p-bromoaniline
- p-Bromoaniline → diazotise → replace N₂ with Br (Sandmeyer) → p-dibromobenzene (1,4-dibromobenzene)
For 1,3,5-tribromobenzene specifically: this requires m-directing groups. The answer is:
- Aniline → HNO₂ → benzenediazonium → H₃PO₂ → benzene
- Then benzene → Br₂/FeBr₃ → bromobenzene… still wrong pattern.
The actual synthesis uses: convert aniline to sulphanilic acid (zwitterion, −NH₂ protected as salt, −SO₃H acts as meta director), tribrominate, then remove SO₃H by hydrolysis.
Exam-Specific Tips
CBSE Board (Class 12)
- 3-mark questions typically ask: write the reaction for preparation of aniline from nitrobenzene, or distinguish 1°/2°/3° amines.
- 5-mark questions cover: complete basicity order with explanation, or a multi-step conversion.
- Always write the mechanism for carbylamine and Hofmann bromamide in 5-mark answers — you get step marks.
JEE Main
- Basicity comparison questions appear every year — master the gas phase vs. aqueous solution distinction.
- Sandmeyer reactions and diazonium coupling are high-yield. Focus on conditions (temperature, medium).
- Expect 1 question on identifying the correct method to prepare a specific amine class.
JEE Main 2024 Shift 1 had a direct question on which of the following gives a primary amine with one carbon less — the answer was Hofmann bromamide degradation. This is a repeat concept — commit it to memory.
NEET
- NEET focuses on biological relevance: adrenaline, amino acids, dopamine are all amines.
- Structure-based questions: identify the type of amine in a given structure.
- Diazonium coupling is a favourite — especially the medium (alkaline vs. acidic).
Common Mistakes to Avoid
Mistake 1: Writing that 3° aliphatic amine is the strongest base in aqueous solution. It is strongest in gas phase only. In water, 2° wins.
Mistake 2: Forgetting that Gabriel synthesis cannot be used for aromatic amines. Aryl halides don’t undergo SN2 — the phthalimide anion cannot displace the halide from the ring.
Mistake 3: Applying carbylamine test to 2° or 3° amines. The test is exclusive to 1° amines. Writing “2° amine gives carbylamine” is an instant mark deduction.
Mistake 4: In diazonium coupling, writing neutral or acidic conditions for phenol coupling. Phenol coupling needs alkaline medium (phenol must be converted to phenoxide for better electron donation). Acidic medium protonates the amine component in amine coupling — weakly acidic is required there.
Mistake 5: Confusing the Hofmann bromamide product. The product amine has the same number of carbons as the acyl group in the amide, not the amide carbon count. RCONH₂ → RNH₂. If the amide has 3 carbons, the amine has 2.
Practice Questions
Q1. Arrange in decreasing order of basicity: aniline, p-nitroaniline, p-methylaniline, ammonia.
p-Methylaniline > Aniline > NH₃ > p-Nitroaniline
The methyl group donates electrons (inductive), increasing the lone pair availability on N. The nitro group strongly withdraws electrons through resonance, making p-nitroaniline an even weaker base than ammonia — pKb of p-nitroaniline ≈ 13.
Q2. How is ethylamine distinguished from diethylamine using a chemical test?
Use the carbylamine test: add CHCl₃ and alc. KOH and warm gently.
- Ethylamine (1°) → isocyanide formed → intense foul smell → positive test
- Diethylamine (2°) → no isocyanide → no foul smell → negative test
Alternatively, treat with NaNO₂/HCl: ethylamine gives N₂ (effervescence), diethylamine gives a yellow N-nitrosamine oil.
Q3. Starting from aniline, how would you prepare fluorobenzene?
- Aniline + NaNO₂ + HCl at 0–5°C → Benzenediazonium chloride (C₆H₅N₂⁺Cl⁻)
- Treat with HBF₄ → Benzenediazonium fluoroborate (C₆H₅N₂⁺BF₄⁻)
- Heat dry → Fluorobenzene + N₂ + BF₃
This is the Balz–Schiemann reaction. The fluorobenzene cannot be made by Sandmeyer reaction because CuF is unstable.
Q4. A compound X gives a positive carbylamine test and a white precipitate with Br₂ water. Identify X.
X is aniline (C₆H₅NH₂).
- Positive carbylamine → primary amine
- White precipitate with Br₂ water → 2,4,6-tribromoaniline (the −NH₂ group so strongly activates the ring that bromination occurs at all three ortho/para positions without a catalyst)
Q5. Which method would you use to prepare n-butylamine from propanenitrile?
Propanenitrile (CH₃CH₂CN) + LiAlH₄ → n-butylamine (CH₃CH₂CH₂NH₂)
Reduction of nitrile gives a primary amine with the same carbon count as the nitrile (3 carbons in the chain + the nitrile carbon = 4 carbons total in the product amine).
Q6. Why does aniline not undergo Friedel-Crafts reaction?
Aniline reacts with the Lewis acid catalyst (AlCl₃) used in Friedel-Crafts reactions. The lone pair on nitrogen forms a complex with AlCl₃, rendering the nitrogen positively charged. This −NH₂→AlCl₃ complex is now an electron-withdrawing group, deactivating the ring and preventing electrophilic substitution.
Q7. Give the IUPAC name: (CH₃)₂NH
N-methylmethanamine (secondary amine; parent chain is methanamine, with a methyl substituent on N).
Common name: dimethylamine.
Q8. Explain why the boiling point of ethylamine is lower than that of ethanol despite similar molecular weights.
Both form hydrogen bonds, but the strength differs.
O−H···O hydrogen bonds (in ethanol) are stronger than N−H···N hydrogen bonds (in ethylamine) because oxygen is more electronegative than nitrogen, making the O−H bond more polarised.
Stronger intermolecular forces → more energy needed to overcome them → higher boiling point for ethanol (78°C) vs. ethylamine (17°C).
FAQs
Q: What is the difference between aliphatic and aromatic amines?
In aliphatic amines, the nitrogen is attached to alkyl groups. In aromatic amines, N is directly bonded to a benzene ring. This structural difference dramatically affects basicity — aromatic amines are far weaker bases because the nitrogen lone pair participates in ring resonance and is less available for proton acceptance.
Q: Why is aniline a weaker base than methylamine?
In aniline, the lone pair on nitrogen is delocalised into the benzene ring through resonance. Five resonance structures contribute, distributing the electron density across the ring. This reduces the electron density on N, making it harder to accept a proton. In methylamine, the methyl group donates electrons to N (inductive effect), increasing lone pair availability. The pKb difference is huge: methylamine ≈ 3.4, aniline ≈ 9.4.
Q: What happens when aniline reacts with HCl?
Aniline + HCl → Anilinium chloride (C₆H₅NH₃⁺Cl⁻). This is a salt formed by protonation of the lone pair on N. The salt is soluble in water. On adding excess NaOH, the free base (aniline) is regenerated. This property — dissolving in acid, precipitating in base — is used to purify aniline from a mixture.
Q: How are diazonium salts used in making dyes?
The azo coupling reaction between a diazonium salt and an electron-rich aromatic compound (phenol or amine) gives an azo compound (−N=N−). These are intensely coloured due to the extended conjugation. The colour can be tuned by changing the substituents on the rings. This is the chemistry behind most synthetic fabric dyes.
Q: Which amine preparation method is used in industrial settings for aniline?
Industrially, aniline is made by reduction of nitrobenzene — either by hydrogenation (H₂/catalyst, 300°C) in the vapour phase, or by reduction with iron and hydrochloric acid (Baeyer process) in the liquid phase. The vapour phase hydrogenation is preferred today due to environmental concerns with the iron filings waste.
Q: What is the Hinsberg test?
Treatment with benzenesulphonyl chloride (C₆H₅SO₂Cl):
- 1° amine → sulphonamide soluble in NaOH (N-H is acidic due to sulphonyl group)
- 2° amine → sulphonamide insoluble in NaOH (no N-H)
- 3° amine → no reaction (no N-H)
This test distinguishes all three classes.
Q: Are amines important for NEET or just for JEE?
Both exams cover amines, but with different emphases. NEET focuses on structure identification, biological relevance (amino acids, neurotransmitters), and basic reactions (diazotisation, coupling). JEE Main goes deeper into basicity comparison, multi-step synthesis, and mechanism-based questions. For boards, the emphasis is on preparation methods and chemical properties with balanced equations.