Let us start with something honest: almost every student who opens this chapter for the first time feels a small wave of panic. There are so many arrows. Nitro compounds turning into amines, amines turning into diazonium salts, diazonium salts turning into what feels like half the periodic table. It looks like a chapter you have to memorise by brute force. It is not. Amines is one of the most logical chapters in Class 12 Organic Chemistry once you see the single idea hiding underneath it, and that is exactly what we are going to build here, slowly, from zero.
Here is the idea. A nitrogen atom in an amine has one lone pair of electrons that it has not used for bonding. That lone pair is the whole chapter. When the lone pair grabs a proton, we call it basicity. When it grabs a carbon, we call it alkylation. When it grabs an acyl group, we call it acylation. When it pushes electrons into a benzene ring, we get activation towards electrophilic substitution. And when we deliberately trap that nitrogen into a −N2+ group and let it leave, we get diazonium chemistry. One lone pair, five behaviours. That is the map.
This page is written as a full teaching chapter, not a summary sheet. We will cover structure and classification, both naming systems, all six preparation methods, physical properties, and the complete reaction set — including a proper, patient treatment of the basic strength order of amines explained from first principles, because that single topic is where most marks are quietly lost. You will find a diazonium salt reactions chart drawn as a hub diagram, a worked set of amines class 12 important questions, and a practice worksheet with answers you can check yourself. Take it one section at a time. There is no prize for rushing.
Meet Your Tutor
Amines become easier when structure decides the reaction instead of memory deciding it. I will help you compare basic strength, preparations, diazonium chemistry and conversions through electron availability, medium and reaction conditions, with a product check after every step.
What You’ll Learn
- What Amines Are And How Nitrogen Holds Them Together
- Classification: Primary, Secondary, Tertiary, Aliphatic, Aromatic
- Nomenclature: Common Names And IUPAC Names
- Preparation 1: Reduction Of Nitro Compounds
- Preparation 2: Ammonolysis Of Alkyl Halides
- Preparation 3: Reduction Of Nitriles And Amides
- Preparation 4: Gabriel Phthalimide Synthesis
- Preparation 5: Hoffmann Bromamide Degradation
- Physical Properties: Boiling Point, Solubility And Hydrogen Bonding
- Basic Character Of Amines And What Controls It
- Alkylation And Acylation Of Amines
- The Carbylamine Reaction: A Test For Primary Amines
- Reaction With Nitrous Acid
- Hinsberg’s Test With Arylsulphonyl Chloride
- Electrophilic Substitution On Aniline
- Preparation Of Diazonium Salts: Diazotisation
- Diazonium Salts: Reactions That Displace Nitrogen
- Diazonium Salts: Coupling Reactions That Keep The Diazo Group
- Why Diazonium Salts Matter In Synthetic Organic Chemistry
Your Game Plan
- Get the structure and classification straight first. Ten minutes here saves an hour later.
- Learn nomenclature by writing ten structures yourself, not by reading a table.
- Group the six preparations by what happens to the carbon count — that one question sorts them instantly.
- Give basicity a full study session on its own. Understand it; never memorise it.
- Learn the four identification tests as one connected flowchart, not four separate facts.
- Finish with diazonium salts, and treat the hub diagram as your revision poster.
- Close the notes and attempt the worksheet cold. Only then check the answers.
Study Notes
What Amines Are And How Nitrogen Holds Them Together
Take ammonia, NH3. Now replace one, two or three of those hydrogen atoms with carbon-containing groups. Whatever you get is an amine. That is genuinely the entire definition. Amines are ammonia wearing different outfits.
The nitrogen atom is sp3 hybridised. It forms three sigma bonds and keeps one lone pair in the fourth sp3 orbital. Because a lone pair takes up more room than a bonding pair, it squeezes the three bonds slightly closer together. So the shape is pyramidal, not flat and not perfectly tetrahedral. In trimethylamine the C–N–C angle is about 108°, a little under the ideal 109.5°. In ammonia itself the H–N–H angle is about 107°.
Think of the lone pair as an outstretched hand. It is always available, always looking for something electron-poor to hold onto. Anything that makes that hand more crowded, or ties it up elsewhere, will change how the molecule behaves. Hold onto that picture — we will cash it in repeatedly.
Classification: Primary, Secondary, Tertiary, Aliphatic, Aromatic
Amines get sorted in two independent ways at the same time, and mixing these two systems up is the single most common early mistake in this chapter.
Sorting rule one — count the carbon groups on nitrogen. One carbon group attached to N means primary (1°), written RNH2. Two means secondary (2°), written R2NH. Three means tertiary (3°), written R3N. Notice we are counting groups on the nitrogen.
Sorting rule two — look at what kind of carbon touches nitrogen. If nitrogen is attached only to alkyl groups, the amine is aliphatic. If nitrogen is attached directly to a benzene ring, it is aromatic (aniline is the parent). If a compound has both an aryl and an alkyl group on nitrogen, such as N-methylaniline, it is called a mixed or aryl-alkyl amine.
| Type | General form | Aliphatic example | Aromatic example | N–H bonds left |
|---|---|---|---|---|
| Primary (1°) | R–NH2 | CH3NH2 | C6H5NH2 | 2 |
| Secondary (2°) | R2NH | (CH3)2NH | C6H5NHCH3 | 1 |
| Tertiary (3°) | R3N | (CH3)3N | C6H5N(CH3)2 | 0 |
(b) C6H5–NH–C6H5 — two carbons on N, both aryl → 2° aromatic (diphenylamine).
(c) C6H5–CH2–NH2 — careful. The ring is there, but nitrogen is attached to a CH2, not to the ring itself. So this is 1° aliphatic (benzylamine), and it behaves chemically much more like ethylamine than like aniline.
(d) C6H5–N(CH3)2 — three carbons on N, one of them aryl → 3°, aryl-alkyl (mixed).
Why it works: in every case you put your finger on the nitrogen and asked two questions — how many carbons touch me, and what kind are they.
Nomenclature: Common Names And IUPAC Names
Two naming systems run side by side, and CBSE expects you to recognise both.
Common system. Name the alkyl group or groups attached to nitrogen, in alphabetical order, then add the word “amine” as one word. CH3NH2 is methylamine. (CH3)2NH is dimethylamine. CH3NHC2H5 is ethylmethylamine. Simple and quick.
IUPAC system. Treat –NH2 as a suffix. Take the parent alkane, drop the final “e”, add “amine”, and number the chain to give nitrogen the lowest locant. So CH3NH2 is methanamine and CH3CH2CH2NH2 is propan-1-amine. When extra groups sit on the nitrogen itself, mark them with a capital N– prefix: (CH3)2NH becomes N-methylmethanamine, and (CH3)3N becomes N,N-dimethylmethanamine. For the IUPAC name of a 2° or 3° amine, the largest alkyl group is chosen as the parent chain and the rest become N-substituents.
For aromatic amines, IUPAC keeps the name aniline as an accepted name (the fully systematic name benzenamine also exists). Derivatives are then named from aniline: C6H5NHCH3 is N-methylaniline, and 4-CH3–C6H4–NH2 is 4-methylaniline (commonly p-toluidine).
| Structure | Common name | IUPAC name | Class |
|---|---|---|---|
| CH3NH2 | Methylamine | Methanamine | 1° |
| CH3CH2CH2NH2 | n-Propylamine | Propan-1-amine | 1° |
| (CH3)2CHNH2 | Isopropylamine | Propan-2-amine | 1° |
| (CH3)2NH | Dimethylamine | N-Methylmethanamine | 2° |
| CH3NHC2H5 | Ethylmethylamine | N-Methylethanamine | 2° |
| (CH3)3N | Trimethylamine | N,N-Dimethylmethanamine | 3° |
| H2NCH2CH2NH2 | Ethylenediamine | Ethane-1,2-diamine | 1° (di) |
| C6H5NH2 | Aniline | Aniline (benzenamine) | 1° aromatic |
| C6H5NHCH3 | N-Methylaniline | N-Methylaniline | 2° mixed |
Step 2 — number from the end that gives nitrogen the lower locant. Numbering from the right puts –NH2 at C-2; from the left it would be C-3. Choose C-2.
Step 3 — drop the “e” from butane, add “amine” with the locant: butan-2-amine. Common name: sec-butylamine.
Why it works: the amine suffix takes numbering priority here because it is the principal functional group in the molecule.
Preparation 1: Reduction Of Nitro Compounds
This is the industrial route to aniline, and the easiest preparation to understand. A nitro group, –NO2, is nitrogen already attached to the carbon skeleton but loaded with oxygen. Strip the oxygen off and add hydrogen, and you are left with –NH2. Nothing about the carbon skeleton changes.
Three reagent sets are used, and CBSE expects you to know when each is chosen:
- H2 gas with a Pd, Pt or Ni catalyst — clean and complete, but it will also reduce other reducible groups such as C=C or C≡N in the same molecule. Not selective.
- Sn (or Fe) with concentrated HCl — the classic laboratory method. It is selective: it reduces –NO2 without touching a carbon–carbon double bond. The amine first forms as its hydrochloride salt, so you must add NaOH at the end to liberate the free amine.
- Fe scrap with dilute HCl — the industrial choice. The FeCl2 formed is hydrolysed by water and regenerates HCl, so only a small catalytic quantity of acid is actually consumed. Cheaper, and that is exactly why industry uses it.
Use Sn/conc. HCl, then NaOH. Dissolving-metal reduction in acid attacks the nitro group but leaves an isolated carbon–carbon double bond untouched. The NaOH at the end frees the amine from its hydrochloride salt.
Why it works: selectivity is the whole point. Whenever a question puts a second reducible group in the molecule, the answer is almost always the metal/acid method, never H2/catalyst.
Preparation 2: Ammonolysis Of Alkyl Halides
Ammonia has a lone pair, so it is a nucleophile. An alkyl halide has a carbon carrying a partial positive charge. Put them together and ammonia attacks that carbon, kicking out the halide. This is ordinary nucleophilic substitution — the same mechanism you met when studying alcohols, phenols and ethers in Class 12 Chemistry, just with nitrogen doing the attacking instead of oxygen.
And there is the problem. The primary amine you just made is also a nucleophile — in fact a better one than ammonia, because the alkyl group pushes electron density onto nitrogen. So it attacks another molecule of RX, and the reaction runs onwards to the secondary amine, then the tertiary amine, then finally a quaternary ammonium salt. You end up with a mixture of all four.
Preparation 3: Reduction Of Nitriles And Amides
Two closely related reductions. What separates them in an exam is the carbon count, so let us make that the headline.
Nitriles. A nitrile R–C≡N is reduced by H2 over Ni, or by LiAlH4, or by sodium in ethanol (the Mendius reaction) to give a primary amine. The nitrile carbon becomes a CH2 group, so the product has one more carbon than the alkyl group you started with.
This is genuinely useful, because you can make the nitrile from an alkyl halide (R–X + KCN → R–CN), so the two steps together are a reliable way to lengthen a carbon chain by exactly one carbon while installing an amine.
Amides. An amide R–CONH2 is reduced by LiAlH4 to a primary amine. Here the carbonyl carbon becomes CH2 but stays in the molecule, so the product has the same number of carbons as the amide.
Step 1: CH3CH2Br + KCN (alcoholic) → CH3CH2CN (propanenitrile). The cyanide carbon is the new third carbon.
Step 2: CH3CH2CN + 4[H], H2/Ni → CH3CH2CH2NH2 (propan-1-amine).
Why it works: the carbon count told us immediately that ammonolysis or amide reduction could never get there, because neither changes the number of carbons.
Preparation 4: Gabriel Phthalimide Synthesis
Ammonolysis failed us because the primary amine kept reacting again. Gabriel synthesis is the clever fix: it keeps nitrogen locked in a cage while the alkyl group is attached, and only releases it at the very end, when there is no alkyl halide left to over-react with.
- Phthalimide + KOH → potassium phthalimide. The N–H of phthalimide is unusually acidic because it sits between two carbonyl groups, which stabilise the resulting anion by resonance. KOH removes it easily.
- Potassium phthalimide + R–X → N-alkylphthalimide. The phthalimide nitrogen attacks the alkyl halide. Crucially, this nitrogen now has no free N–H and is flanked by two electron-withdrawing carbonyls, so it cannot attack a second alkyl halide. The over-alkylation problem is designed out.
- Hydrolysis with aqueous NaOH (or acid), or treatment with hydrazine, cleaves both amide bonds and releases the pure primary amine along with phthalic acid (as its salt).
Why it works: the same reasoning explains why aryl halides are inert in almost every substitution you have met so far.
Preparation 5: Hoffmann Bromamide Degradation
The name contains a promise: degradation. Something gets smaller. An amide with n carbons gives an amine with n − 1 carbons. The carbonyl carbon leaves the molecule entirely, ending up as carbonate.
The reagent is bromine in aqueous or ethanolic sodium hydroxide (equivalently, NaOBr). The heart of the reaction is a migration: the alkyl or aryl group R moves from the carbonyl carbon directly to the nitrogen atom. That migration is intramolecular — R never becomes free, it slides across within the same molecule. This is why the migrating group keeps its stereochemistry, and it is a favourite one-mark question.
Equation: C6H5CONH2 + Br2 + 4NaOH → C6H5NH2 + Na2CO3 + 2NaBr + 2H2O.
Why it works: the phenyl group migrates intact from the carbonyl carbon to nitrogen; the carbonyl carbon is expelled as carbonate.
• Starting from butanamide (CH3CH2CH2CONH2, 4 C) → use Hoffmann, since 4 C must drop to 3 C.
• Starting from 1-bromopropane (3 C) → use Gabriel, since the carbon count must stay at 3 and we need a pure product.
• Starting from bromoethane (2 C) → use the nitrile route, since 2 C must rise to 3 C.
Why it works: one arithmetic step — compare carbon counts — picks the method every time.
Physical Properties: Boiling Point, Solubility And Hydrogen Bonding
All the physical behaviour of amines comes from one question: how many N–H bonds does the molecule have? An N–H bond can donate a hydrogen bond. No N–H, no donating.
Compare three isomers, all with molar mass 59.11 g mol−1:
| Isomer (M = 59.11) | Class | N–H bonds | Boiling point |
|---|---|---|---|
| CH3CH2CH2NH2 (propan-1-amine) | 1° | 2 | about 49 °C |
| CH3NHCH2CH3 (N-methylethanamine) | 2° | 1 | about 37 °C |
| (CH3)3N (trimethylamine) | 3° | 0 | about 3 °C |
Same mass, same shape roughly, and yet a 46-degree spread in boiling point. The only variable is the number of N–H bonds available for intermolecular hydrogen bonding. So among isomeric amines the boiling point order is always 1° > 2° > 3°.
Amines versus alcohols. Ethanol boils at 78 °C but propan-1-amine, which is heavier, boils at only 49 °C. Nitrogen is less electronegative than oxygen, so the N–H bond is less polar than O–H and the hydrogen bonds it forms are weaker. Hence alcohol > amine > alkane in boiling point at comparable mass.
Solubility. Lower amines dissolve well in water because they form hydrogen bonds with it. As the alkyl chain grows, the greasy hydrocarbon part dominates and solubility falls away — the same trend you saw with alcohols. Aniline, with its bulky benzene ring, is only slightly soluble in water but dissolves readily in organic solvents.
(CH3)3N has no N–H at all, so no intermolecular hydrogen bonding — lowest, about 3 °C.
CH3CH2CH2NH2 has two N–H bonds and hydrogen bonds, but N–H is only moderately polar — about 49 °C.
CH3CH2CH2OH has a strongly polar O–H, giving the strongest hydrogen bonds — about 97 °C.
Answer: (CH3)3N < CH3CH2CH2NH2 < CH3CH2CH2OH.
Why it works: rank by hydrogen-bonding strength, which means first counting donor H atoms and then comparing electronegativity of the atom they sit on.
Basic Character Of Amines And What Controls It
This is the heart of the chapter, so we will go slowly. An amine is basic because its nitrogen lone pair can grab a proton. The more freely available that lone pair is, and the more stable the resulting cation, the stronger the base.
Strength is measured by Kb, or more conveniently by pKb = −log Kb. Because of that minus sign, a smaller pKb means a stronger base. Write that on your hand if you must; half the errors in this topic are students reading the ladder upside down.
Look hard at that picture before reading on. Two things should jump out. First, the three methylamines are packed into a narrow band between pKb 3.27 and 4.22 — they are all similar, and all stronger bases than ammonia. Second, aniline is in a different world, more than five pKb units away. Those are two separate stories, so let us tell them separately.
Story one: the aliphatic amines and their three competing factors
Factor 1 — the inductive (+I) effect. An alkyl group is electron-releasing. Attach one to nitrogen and it pushes electron density towards the lone pair, making it more eager to grab a proton. It also stabilises the positive charge on the ammonium ion formed. On this factor alone, more alkyl groups means a stronger base, predicting 3° > 2° > 1° > NH3.
Factor 2 — solvation of the cation. Basicity in water is not decided by the amine alone; it is decided by how happy the product ion is. When an amine takes a proton it becomes an ammonium ion, and water molecules stabilise that ion by hydrogen bonding to its N–H bonds. Count them: RNH3+ has three N–H bonds, R2NH2+ has two, R3NH+ has only one. So solvation stabilises the primary ammonium ion best and the tertiary one worst — exactly the opposite ranking to the inductive effect. (If the idea of solvent molecules stabilising a dissolved ion feels shaky, it is worth revisiting how solvation is treated in the Solutions chapter for Class 12 Chemistry.)
Factor 3 — steric hindrance. Three bulky alkyl groups physically crowd the nitrogen. A proton, and the water molecules that would solvate the resulting ion, find it harder to get close. This also works against tertiary amines.
• Methyl series PIN = 2130 → (CH3)2NH > CH3NH2 > (CH3)3N > NH3
• Ethyl series PIN = 2310 → (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
Both PINs start with 2 and end with 0 — the secondary amine always wins in water, and ammonia always loses. Only the middle two digits swap. The ethyl groups are bigger, so their extra +I push is strong enough to lift the tertiary amine above the primary one. Remember “2130 for methyl, 2310 for ethyl” and you have both orders, plus the reason they differ.
On inductive grounds alone trimethylamine should have been stronger, since it has three electron-releasing methyl groups instead of one. It is not, for two reasons acting together. Its conjugate acid (CH3)3NH+ has only one N–H bond to hydrogen bond with water, against three in CH3NH3+, so it is far less stabilised by solvation. And the three methyl groups sterically crowd the nitrogen, hindering the approach of both the proton and the solvent.
Why it works: the answer that earns full marks names all three factors and states which ones dominate — not just “because of steric hindrance.”
Story two: why aromatic amines are dramatically weaker
Aniline has pKb 9.38 against ammonia’s 4.75. That gap of 4.63 pKb units means aniline is about forty thousand times weaker a base than ammonia. Something dramatic is going on, and it is resonance.
In aniline, the nitrogen lone pair is not sitting quietly in an sp3 orbital. It overlaps with the π system of the benzene ring and is delocalised into the ring, spreading onto the ortho and para carbons. A lone pair that is shared across five atoms is simply not as available to grab a proton as one that belongs to nitrogen alone.
There is a second half to the argument, and it is the half students forget. Once aniline does accept a proton, the anilinium ion C6H5NH3+ has its lone pair used up in the new N–H bond, so that delocalisation is no longer possible. The neutral amine was resonance-stabilised; the cation is not. Protonation therefore costs aniline a stabilisation it was enjoying, and the equilibrium sits well to the left.
Substituents on the ring then fine-tune this. An electron-releasing group pushes more density towards nitrogen and raises basicity: 4-methylaniline has pKb 8.92, making it a stronger base than aniline itself. An electron-withdrawing group pulls density away and crushes basicity: 4-nitroaniline has pKb about 13.0, which is a very weak base indeed. So the order runs 4-methylaniline > aniline > 4-nitroaniline. Piling on more aryl groups makes things worse still — diphenylamine, with two rings competing for one lone pair, is barely basic at all.
Within the aromatic pair, N-methylaniline has an extra electron-releasing methyl group on nitrogen, which partly compensates for the delocalisation, so it is slightly the stronger of the two.
Answer: C2H5NH2 > NH3 > C6H5NHCH3 > C6H5NH2.
Why it works: resonance is a far bigger effect than induction, so the aliphatic/aromatic split always dominates the ordering. Sort into families first, then rank within each family.
Alkylation And Acylation Of Amines
The lone pair that grabbed a proton can equally well grab a carbon. That is all alkylation and acylation are.
Alkylation. Treat an amine with an alkyl halide and nitrogen attacks the electron-poor carbon. This is the same reaction as ammonolysis viewed from the other side, and it suffers the same flaw: the product is a better nucleophile than the starting material, so the reaction keeps going, up to the quaternary ammonium salt. Useful for climbing the ladder deliberately; useless for making one clean product.
Acylation. Now treat the amine with an acid chloride (such as CH3COCl) or an acid anhydride, usually with a base like pyridine present to mop up the HCl produced. Nitrogen attacks the carbonyl carbon and you get an N-substituted amide.
aniline → acetanilide (N-phenylethanamide)
Crucially, acylation stops after one step. The acyl group is strongly electron-withdrawing, so the nitrogen in the amide is far less nucleophilic than it was in the amine. It has no appetite for a second acyl chloride. That single fact makes acylation genuinely useful.
The most important use of acylation is protection. Convert aniline into acetanilide, do a reaction that raw aniline would have handled badly, then hydrolyse the amide back to the amine with acid or alkali. We will use this trick properly in the electrophilic substitution section — it is the standard route to para-substituted anilines.
The ring is therefore still activated — acetanilide is more reactive than benzene — but much less violently so than aniline. That controlled reactivity is precisely why we acetylate before halogenating or nitrating.
Why it works: two electron-withdrawing destinations competing for one lone pair means each gets a smaller share.
The Carbylamine Reaction: A Test For Primary Amines
Heat a primary amine with chloroform and alcoholic potassium hydroxide and you get an isocyanide (also called a carbylamine), a compound with an extraordinarily foul smell. The smell is the test.
Secondary and tertiary amines do not give this reaction. The test needs two hydrogen atoms on the same nitrogen, and only a primary amine has them. Both aliphatic and aromatic primary amines respond, so aniline gives a positive carbylamine test just as methylamine does.
Reaction With Nitrous Acid
Nitrous acid, HNO2, is unstable, so it is always generated in the flask from sodium nitrite and a mineral acid (NaNO2 + HCl) at low temperature. It gives a different visible result with each class of amine, which makes it one of the most useful diagnostic reagents in the chapter.
- Aliphatic 1° amine → an unstable diazonium salt forms and immediately falls apart, releasing nitrogen gas. You see brisk effervescence and get an alcohol. R–NH2 + HNO2 → R–OH + N2 + H2O. Because the nitrogen evolved is quantitative, this reaction was historically used to estimate amino groups.
- Aromatic 1° amine at 273–278 K → the diazonium salt survives instead of decomposing. This is diazotisation, and it gets its own section below.
- 2° amine (aliphatic or aromatic) → an N-nitrosamine, seen as a yellow oily liquid. R2NH + HNO2 → R2N–N=O + H2O.
- 3° aliphatic amine → no N–H to substitute, so it simply acts as a base and forms a soluble trialkylammonium nitrite salt.
- 3° aromatic amine such as N,N-dimethylaniline → the ring is so activated that the nitrosonium ion attacks the ring instead, giving a p-nitroso compound (a green solid).
Hinsberg’s Test With Arylsulphonyl Chloride
Hinsberg’s reagent is benzenesulphonyl chloride, C6H5SO2Cl. It separates all three classes of amine in one test tube, which is why it is the classic exam question.
- Primary amine → C6H5SO2NHR, an N-alkylbenzenesulphonamide. This product still has one hydrogen on nitrogen, and that hydrogen is acidic because the strongly electron-withdrawing sulphonyl group stabilises the anion left behind. So the product dissolves in aqueous KOH.
- Secondary amine → C6H5SO2NR2, an N,N-dialkylbenzenesulphonamide. There is no hydrogen left on nitrogen, so nothing acidic, so the product is insoluble in alkali.
- Tertiary amine → no reaction at all. It has no N–H for the sulphonyl group to replace, so the amine is recovered unchanged.
Bottle that gives a product dissolving in KOH → CH3NH2. Its sulphonamide C6H5SO2NHCH3 keeps one acidic N–H.
Bottle that gives a product not dissolving in KOH → (CH3)2NH. Its sulphonamide has no N–H.
Bottle that gives no reaction → (CH3)3N.
Cross-check with chloroform and alcoholic KOH: only the first bottle produces the revolting isocyanide smell, confirming it is primary.
Why it works: the test converts an invisible structural difference (number of N–H bonds) into a visible one (dissolves, does not dissolve, does not react).
Electrophilic Substitution On Aniline
We established that aniline’s lone pair is delocalised into the ring. That was bad news for basicity. It is spectacular news for electrophilic substitution, because it dumps extra electron density onto the ortho and para carbons. So –NH2 is a strongly activating, ortho/para-directing group. Aniline is enormously more reactive than benzene.
That reactivity is the problem. Aniline is too eager, and reactions run away from us.
Bromination
Add bromine water to aniline at room temperature and, instantly, a white precipitate of 2,4,6-tribromoaniline drops out. No catalyst, no heating, and no stopping at one bromine — all three activated positions react at once.
To get just the para product, we use the protection trick. Acetylate aniline to acetanilide, which is only moderately activated; brominate to get p-bromoacetanilide; then hydrolyse the amide back with acid or alkali to give p-bromoaniline. Three steps, but complete control.
Nitration
Here something genuinely surprising happens. Nitrate aniline directly with concentrated nitric and sulphuric acids and you get a substantial amount of the meta product — roughly 47% meta alongside about 51% para and only 2% ortho.
Once again, protection solves it. Acetanilide is not basic enough to be protonated significantly, so nitrating acetanilide gives predominantly the para product; hydrolysis then yields p-nitroaniline cleanly.
Sulphonation
Aniline dissolves in concentrated sulphuric acid to give anilinium hydrogensulphate. Heat this at about 453–473 K and it rearranges to sulphanilic acid (4-aminobenzenesulphonic acid).
Sulphanilic acid has a lovely quirk worth knowing: it contains a basic –NH2 group and a strongly acidic –SO3H group in the same molecule, so the acid simply protonates its own amine. The molecule exists as an internal salt, a zwitterion, with a −SO3− and an –NH3+ at opposite ends. That is why it has an unusually high melting point and low solubility in organic solvents — it behaves like an ionic solid.
Why aniline does not undergo Friedel–Crafts reactions
Friedel–Crafts alkylation and acylation need a Lewis acid catalyst, usually anhydrous AlCl3. Aluminium chloride is electron-deficient — it is looking for an electron pair. And aniline is holding out exactly that: a nitrogen lone pair.
So the two react with each other first. The nitrogen donates its lone pair to aluminium, forming an acid–base complex, C6H5NH2→AlCl3. This is the same lone-pair donation that makes a ligand bind to a metal centre in Coordination Compounds for Class 12 Chemistry. Two consequences follow, and a full-mark answer names both:
- The nitrogen now carries a positive charge, which strongly deactivates the ring by pulling electron density out of it — the ring becomes poor at attacking electrophiles.
- The catalyst has been consumed by complexation rather than being available to generate the electrophile.
Step 2 — nitrate. Conc. HNO3 / conc. H2SO4 gives predominantly p-nitroacetanilide. Acetanilide is far less basic than aniline, so it is not protonated to a meta-director, and the bulky acetamido group blocks the ortho positions, steering the electrophile to para.
Step 3 — deprotect. Hydrolyse with dilute acid (or alkali) to cleave the amide, giving p-nitroaniline.
Why it works: direct nitration of aniline would waste roughly half the material as the meta isomer. Protection removes the basicity that caused the problem, then we take it back off.
Preparation Of Diazonium Salts: Diazotisation
We now arrive at the most powerful idea in the chapter. Treat a primary aromatic amine with nitrous acid (NaNO2 plus HCl, generated in the flask) at 273–278 K and you get an arenediazonium salt. This process is called diazotisation.
Two details in that equation are marks. The temperature range 273–278 K (0–5 °C) must be stated, and there must be two moles of HCl — one to make the nitrous acid, one to supply the chloride counter-ion.
Why does aniline give a stable salt when ethylamine does not? Because the positive charge on the –N2+ group is delocalised into the benzene ring by resonance. An alkyl group cannot do that, so an alkanediazonium salt falls apart the instant it forms, fizzing out nitrogen gas. The benzene ring is what buys us the stability, and that stability is what makes the whole of diazonium chemistry possible.
Physical properties. Benzenediazonium chloride is a colourless crystalline solid, readily soluble in water. It is stable in cold solution but decomposes when warmed. In the dry state it is unstable and can decompose explosively, so it is essentially always prepared and used in solution without isolation. Benzenediazonium fluoroborate, by contrast, is water-insoluble and stable at room temperature — which is exactly why it can be filtered off and heated in the Balz–Schiemann route to aryl fluorides.
Aniline: the benzenediazonium ion delocalises its positive charge over the ring, so it survives in the cold solution. No gas is evolved, and you are left with a solution of C6H5N2+Cl− ready for the next step.
Why it works: resonance stabilisation of a cation is the deciding factor, exactly as it was when we compared the stability of carbocations.
Diazonium Salts: Reactions That Displace Nitrogen
Here is the distinctive way to think about this section. The –N2+ group is the best leaving group in all of organic chemistry, because when it leaves it becomes N2 gas — an extremely stable molecule that bubbles out of the flask and never comes back. Every departure is irreversible.
So picture the diazonium salt as a vending machine. You have paid your money by putting nitrogen onto the ring. Now you press a button — each button is a different reagent — and out comes a different substituent in that position. One starting material, eight products.
• Cl and Br sit in the middle — both need copper. Salt means Sandmeyer (Cu2Cl2, a cuprous salt); Grains mean Gatterman (copper powder). S for salt, G for grains — both start with the right letter, so you can never swap them.
• Iodine, at the bottom, is easy: plain KI and a little warmth, no catalyst.
• Fluorine, at the top, is awkward and needs its own trick entirely — boron, as HBF4, and heat.
• CN tags along with the copper crowd (CuCN/KCN), because it behaves like a pseudohalide.
Then the two non-halogens: Hypophosphorous acid gives Hydrogen, and plain warm water gives the –OH. Both are alliterative on purpose.
A note on the copper chemistry: the Sandmeyer reaction relies on copper in the +1 oxidation state, which transfers a single electron to the diazonium ion and triggers loss of nitrogen. If the ability of transition metals to shuttle between oxidation states feels unfamiliar, it is worth reading the variable-oxidation-state discussion in the d- and f-Block Elements for Class 12 Chemistry — it is exactly the property being exploited here.
Step 1: benzene + conc. HNO3/conc. H2SO4, 323–333 K → nitrobenzene.
Step 2: Sn/conc. HCl, then NaOH → aniline.
Step 3: NaNO2/HCl at 273–278 K → benzenediazonium chloride.
Step 4: warm with KI → iodobenzene. C6H5N2+Cl− + KI → C6H5I + KCl + N2.
Why it works: this four-step sequence is the single most examinable synthesis in the chapter. Nitrogen is installed as a temporary handle, then traded for the group we actually wanted.
Diazonium Salts: Coupling Reactions That Keep The Diazo Group
Every reaction so far threw the nitrogen away. Coupling is the one family that keeps both nitrogen atoms, building them into the product as an azo bridge, –N=N–. These reactions are why your clothes have colour.
Mechanically, coupling is an electrophilic substitution on a second aromatic ring, with the diazonium ion acting as the electrophile. But the diazonium ion is a fairly weak electrophile, so it cannot attack an ordinary benzene ring. It only works on a ring that is strongly activated — in practice a phenol or an amine. Attack occurs at the para position, or ortho if para is already occupied.
- Coupling with phenol, in mildly alkaline medium (about pH 9–10), gives p-hydroxyazobenzene, an orange dye.
- Coupling with aniline, in mildly acidic medium (about pH 4–5), gives p-aminoazobenzene, a yellow dye.
Aniline needs mild acid to keep the solution from going alkaline, but again only mild: in strongly acidic solution aniline is protonated to the anilinium ion, whose lone pair is tied up, so the ring is deactivated and coupling stops.
Both partners are therefore squeezed into a narrow pH window. “Just enough, not too much” is the whole idea — and stating why is what separates a two-mark answer from a one-mark answer. If phenol’s acidity and the strength of the phenoxide ion are hazy, the treatment in Alcohols, Phenols and Ethers for Class 12 is the place to refresh it.
Azo compounds are intensely coloured because the –N=N– bridge joins the two aromatic rings into one long conjugated system. Extended conjugation lowers the energy gap between the ground and excited states, so the molecule absorbs visible rather than ultraviolet light — and we see the complementary colour.
Why Diazonium Salts Matter In Synthetic Organic Chemistry
It is fair to ask why chemists bother with a four-step sequence when a one-step reaction would be nicer. Three reasons, and CBSE asks for them directly.
- It installs groups that cannot be installed directly. You cannot put –F or –I onto benzene by direct halogenation in any useful way. Through a diazonium salt, both become routine. The same goes for –CN and –OH on a simple ring.
- It gives clean, single products. Direct substitution on benzene often produces an ortho/para mixture that is painful to separate. The diazonium route places the new group exactly where the amino group was, because the reaction happens at that carbon and nowhere else. Position is guaranteed.
- It reaches substitution patterns that are otherwise impossible. The amino group is a strong ortho/para director, and it can be converted afterwards into groups that are meta directors, or removed altogether with H3PO2. So –NH2 can be used purely as a temporary blocking group or director and then discarded. That is how chemists build tri-substituted rings with patterns no direct route could ever give.
And of course the coupling reactions gave the world synthetic azo dyes, which is where a large part of the modern chemical industry began.
Step 1: aniline + Br2/H2O → 2,4,6-tribromoaniline. The powerfully activating –NH2 group puts bromines at exactly the two ortho and one para positions — and those three positions are mutually 1,3,5 to each other.
Step 2: NaNO2/HCl at 273–278 K → the corresponding diazonium salt.
Step 3: warm with H3PO2 and water → the diazo group is replaced by –H, giving 1,3,5-tribromobenzene.
Why it works: the amino group was never wanted in the final product. It was hired to do a job — direct three bromines into place — and then dismissed. That is the deepest idea in this chapter.
Amines Class 12 Important Questions with Solved Examples
These are the question shapes that recur year after year. Read the reasoning, not just the answer.
C3H9N has two primary possibilities: propan-1-amine, CH3CH2CH2NH2, and propan-2-amine, (CH3)2CHNH2.
Now use the optical inactivity. Neither of these actually has a chiral carbon — in propan-2-amine the C-2 carbon carries two identical methyl groups — so both are optically inactive and this clue does not separate them. What does separate them in a full question is usually an additional detail. If the question instead stated that A is optically active, the formula would have to be C4H11N and the answer butan-2-amine, whose C-2 bears H, NH2, CH3 and C2H5 — four different groups.
Why it works: always test a chirality claim by actually counting the four groups on the candidate carbon rather than assuming. This is a trap question, and the counting is the defence.
In aniline the nitrogen lone pair is delocalised into the ring. The resonance structures that describe this have a C=N double bond between the ring carbon and nitrogen. The real molecule is a hybrid, so the actual C–N bond has partial double-bond character. It is therefore shorter and stronger. Additionally, the ring carbon is sp2 hybridised, and sp2 carbon forms shorter bonds than sp3 carbon because of its greater s character.
Why it works: the same delocalisation that weakened aniline’s basicity shortens its C–N bond. One cause, two observable consequences — examiners love asking you to connect them.
Step 1: C6H5NO2 + Sn/conc. HCl, then NaOH → A = C6H5NH2.
Step 2: NaNO2/HCl, 273–278 K → B = C6H5N2+Cl−.
Step 3: warm with H2O → C6H5OH + N2 + HCl.
Why it works: in conversion chains, identify the step nearest the product first. There are usually fewer ways to make the final compound than to consume the starting one.
Shake the mixture with benzenesulphonyl chloride and aqueous KOH. Aniline forms C6H5SO2NHC6H5, which has an acidic N–H and therefore dissolves in the alkali. N,N-dimethylaniline has no N–H, does not react, and can be separated off as an insoluble layer.
Acidify the alkaline layer to precipitate the sulphonamide, then hydrolyse it with hot mineral acid to recover pure aniline.
Why it works: a separation question is always a solubility question in disguise. Convert one component into something with different solubility, remove the other, then convert back.
Practice Worksheet
Close the notes. Attempt all ten with a pen and paper before you open a single answer — recognising an answer is not the same as producing one, and only one of those two skills earns marks.
Q1. Classify (CH3)3C–NH2 as 1°, 2° or 3°, and give its IUPAC name.
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IUPAC name: the longest chain through the carbon bearing –NH2 is three carbons (propane), with the amino group and a methyl group both on C-2. So the name is 2-methylpropan-2-amine. The common name is tert-butylamine. Molecular formula C4H11N, molar mass 73.14 g mol−1.
Q2. Arrange in increasing order of basic strength in aqueous solution: NH3, CH3NH2, (CH3)2NH, (CH3)3N. Quote the pKb values.
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pKb: ammonia 4.75, trimethylamine 4.22, methylamine 3.38, dimethylamine 3.27. Remember that lower pKb means stronger base, so increasing basic strength means decreasing pKb.
Reason: the +I effect of methyl groups makes all three amines stronger bases than ammonia. But trimethylamine falls back below the other two because its conjugate acid (CH3)3NH+ has only one N–H bond available for hydrogen bonding with water (poor solvation), and because three methyl groups sterically crowd the nitrogen. This is the PIN 2130 read backwards.
Q3. Give the product: CH3CH2CN, treated with LiAlH4 followed by water. Name it and state how the carbon count changes.
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CH3CH2CN is propanenitrile and already contains three carbons (the nitrile carbon counts). Reduction converts the C≡N into CH2–NH2, so the product also has three carbons — the carbon count is unchanged in this step.
Watch the wording of such questions carefully: the chain grew only if you count from the alkyl halide CH3CH2Br that would have been used to make the nitrile. Reduction of the nitrile itself adds no carbon.
Q4. Aniline cannot be prepared by the Gabriel phthalimide synthesis. Explain fully.
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Two reasons: (i) the lone pairs on the halogen overlap with the ring π system, giving the C–Br bond partial double-bond character, so it is shorter and much harder to break than an ordinary C–Br single bond; (ii) the electron-rich benzene ring repels the approaching phthalimide anion. The reaction therefore never gets past step two.
Note also the general limitation: even where it works, Gabriel synthesis gives only primary amines.
Q5. Write the balanced equation for the Hoffmann bromamide degradation of propanamide and name the amine formed.
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The amine formed is ethanamine (ethylamine). Propanamide has three carbons; the product has two, because the carbonyl carbon leaves the molecule as carbonate. That loss of exactly one carbon is the signature of this reaction.
Check the coefficients: sodium 4 = 2 + 2. Bromine 2 = 2. Oxygen 1 + 4 = 3 + 2. Hydrogen 2 + 4 = 2 + 4. Balanced. The ethyl group migrates intramolecularly from carbon to nitrogen.
Q6. Give two chemical tests to distinguish between ethylamine and aniline.
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Test 2 — nitrous acid at 273–278 K, then alkaline phenol. Ethylamine gives brisk effervescence of N2 at once, because the alkanediazonium ion is unstable, and leaves ethanol behind. Aniline gives no gas; instead a stable benzenediazonium salt forms, and adding it to alkaline phenol produces a bright orange azo dye (p-hydroxyazobenzene). The dye test is the more decisive of the two.
A third option: the azo dye test alone is often accepted, since only aromatic primary amines give it.
Q7. Direct nitration of aniline gives a large proportion of the meta product even though –NH2 is an ortho/para director. Explain.
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The nitrating mixture is concentrated HNO3 with concentrated H2SO4 — a strongly acidic medium. Aniline is a base, so it is largely protonated to the anilinium ion, C6H5NH3+. That ion has no lone pair left to donate into the ring and carries a full positive charge, making it deactivating and meta-directing.
So the meta product (about 47%) comes from the protonated form, while the para product (about 51%) comes from the small amount of free aniline still present. Only about 2% ortho is formed, since the ortho positions are sterically crowded.
The fix is to acetylate aniline to acetanilide first. Acetanilide is far less basic, so it is not protonated, and nitration then gives predominantly the para product; hydrolysis afterwards gives clean p-nitroaniline.
Q8. How would you obtain 4-bromoaniline from aniline? Give reagents for each step and explain why the direct route fails.
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Step 1 (protect): aniline + acetic anhydride, (CH3CO)2O → acetanilide, C6H5NHCOCH3. The acetyl group competes for the nitrogen lone pair, so the ring is now only moderately activated.
Step 2 (brominate): Br2 in acetic acid → p-bromoacetanilide. Only one bromine enters, and it goes para because the bulky acetamido group blocks the ortho positions.
Step 3 (deprotect): hydrolyse with dilute HCl (or NaOH) and warm → 4-bromoaniline plus acetic acid.
Q9. Name the reagent used to convert benzenediazonium chloride into each of: (a) fluorobenzene, (b) iodobenzene, (c) benzene, (d) phenol, (e) benzonitrile.
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(b) Iodobenzene: simply potassium iodide, KI, with warming. No copper catalyst is needed — this is the one halogen that is easy.
(c) Benzene: hypophosphorous acid, H3PO2, with water (ethanol also works). The diazo group is replaced by hydrogen, removing it from the molecule altogether.
(d) Phenol: just warm with water. The salt hydrolyses, evolving N2.
(e) Benzonitrile: CuCN in the presence of KCN — a Sandmeyer reaction. Copper(I) is essential here.
Sorting rule: copper is needed for the middle of the halogen column (Cl, Br) and for the pseudohalide CN; the ends (I and F) and the non-halogens (H, OH) each have their own trick.
Q10. Dimethylamine has pKb 3.27 and ammonia has pKb 4.75. Calculate Kb for each and state how many times stronger a base dimethylamine is.
Show Answer
For dimethylamine: Kb = 10−3.27 = 5.37 × 10−4.
For ammonia: Kb = 10−4.75 = 1.78 × 10−5.
Ratio = (5.37 × 10−4) ÷ (1.78 × 10−5) = 10(4.75 − 3.27) = 101.48 ≈ 30.
So dimethylamine is roughly 30 times stronger a base than ammonia. Note the useful shortcut: you never need the individual Kb values to get the ratio — a difference of one pKb unit is a factor of ten, so a difference of 1.48 units is 101.48.
Now score yourself honestly, and note down only the questions you got wrong. Those are your revision list — not the whole chapter.