ChemistryISC Class 12 · Chapter 9

Organic Compounds Containing Nitrogen ISC Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Organic Compounds Containing Nitrogen is Chapter 9 of ISC Class 12 Chemistry. It covers amines (classification, preparation including Hofmann degradation and Gabriel synthesis, basicity, alkylation and acylation with mechanism, nitrous acid, carbylamine and Hinsberg tests), aniline and its reactions, cyanides and isocyanides, and diazonium salts with Sandmeyer, Gattermann, Balz-Schiemann and coupling reactions.

What Organic Compounds Containing Nitrogen is about

Amines are derivatives of ammonia in which one, two or three hydrogens are replaced by alkyl or aryl groups: primary RNH₂, secondary R₂NH, tertiary R₃N (general formula of saturated aliphatic amines $C_{n} H_{2 n + 3} N$ ). The nitrogen is sp³ with a lone pair, giving a pyramidal shape and making amines bases and nucleophiles. This chapter belongs to the Organic Chemistry group (31 of the 70 theory marks) and is popular for 'identify A, B, C' chains because each reaction changes the carbon count in a predictable way.

You need six preparations: from alcohols, from alkyl halides (ammonolysis gives a mixture), from cyanides (reduction adds the carbon of CN), from amides (Hofmann degradation removes one carbon), from nitro compounds, and Gabriel phthalimide synthesis (pure primary amines). Physical properties depend on hydrogen bonding: 1° > 2° > 3° in boiling point, all lower than alcohols. Basic strength in water follows a balance of inductive effect, solvation and steric hindrance, so for ethylamines 2° > 3° > 1° > NH₃; aniline is much weaker because its lone pair is delocalised into the ring.

Reactions include alkylation and acylation (with mechanism), reaction with nitrous acid (different for 1°, 2°, 3°), the carbylamine reaction (1° only) and Hinsberg's test with benzenesulphonyl chloride. Aniline is made by reducing nitrobenzene; it forms salts, is acetylated, alkylated and benzoylated, undergoes diazotisation, and reacts so readily with electrophiles that bromine water gives 2,4,6-tribromoaniline at once (acetylation tames it). It does not give Friedel-Crafts reactions.

Finally, cyanides and isocyanides (from alkyl halides with KCN/AgCN, amides and amines) and diazonium salts, the key intermediates of aromatic synthesis: replacement of -N₂⁺ by Cl, Br, CN (Sandmeyer, Gattermann), F (Balz-Schiemann), H, OH and NO₂, and coupling with phenol and aniline to form azo dyes.

Weightage: Organic Compounds Containing Nitrogen is part of the Organic Chemistry group, which carries 31 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Classification, structure and nomenclature of amines: Primary (1°): one alkyl/aryl group, RNH₂ (ethanamine, aniline). Secondary (2°): R₂NH (N-methylethanamine). Tertiary (3°): R₃N (N,N-dimethylmethanamine). Quaternary ammonium salts: R₄N⁺X⁻. Simple amines have identical groups, mixed amines different ones. Nomenclature: common names write alkyl + amine (ethylamine, dimethylamine, ethylmethylamine); IUPAC names use alkanamine with N- locants for groups on nitrogen: CH₃CH₂NH₂ ethanamine; CH₃NHC₂H₅ N-methylethanamine; (CH₃)₂NC₂H₅ N,N-dimethylethanamine; C₆H₅NH₂ aniline (benzenamine). Structure: N is sp³, three σ bonds and a lone pair in the fourth orbital; pyramidal, C-N-H angle about 108° (C-N-C in trimethylamine about 108°). In aniline the N is nearly planar because the lone pair conjugates with the ring. Physical properties: lower amines are gases or volatile liquids with a fishy smell; lower members are water-soluble (hydrogen bonding), solubility falling with size. Boiling points: 1° > 2° > 3° for isomers (1° have two N-H, 3° none), but lower than alcohols of similar mass because N is less electronegative than O (weaker hydrogen bonds). Aniline is a colourless oily liquid (bp 457 K) that darkens in air and is only slightly soluble in water.; Worked example
Arrange in increasing order of boiling point: C₂H₅NHCH₃, C₂H₅OH, (CH₃)₃N, CH₃CH₂CH₂NH₂ (the three amines are isomers of C₃H₉N).
Try it first, then show the solutionSolution
(CH₃)₃N has no N-H, so no hydrogen bonding: lowest (about 276 K).
The secondary amine has one N-H per molecule (about 310 K); the primary amine has two and associates more strongly (about 321 K).
Ethanol, though lighter, forms stronger O-H…O hydrogen bonds (O is more electronegative than N) and boils highest (351 K).
Answer: (CH₃)₃N < C₂H₅NHCH₃ < CH₃CH₂CH₂NH₂ < C₂H₅OH
Preparation of amines: From alcohols: vapour with NH₃ over alumina at about 633 K: $C_{2} H_{5} O H + N H_{3} \to C_{2} H_{5} N H_{2} + H_{2} O$ (mixture of amines). From alkyl halides (ammonolysis): heat with alcoholic ammonia in a sealed tube: $C_{2} H_{5} B r + N H_{3} \to C_{2} H_{5} N H_{2} + H B r$ ; the product amine reacts further, giving a mixture of 1°, 2°, 3° amines and the quaternary salt (excess NH₃ favours 1°). From cyanides: reduction with LiAlH₄ or Na/ethanol (Mendius reduction) gives a 1° amine with the same number of carbons as the cyanide (one more than the alkyl halide): $C_{2} H_{5} C N + 4 [H] \to C_{2} H_{5} C H_{2} N H_{2}$ . Isocyanides give 2° amines: $C_{2} H_{5} N C + 4 [H] \to C_{2} H_{5} N H C H_{3}$ . From amides (Hofmann bromamide degradation): amide + Br₂ + KOH gives a 1° amine with one carbon fewer: $C H_{3} C O N H_{2} + B r_{2} + 4 K O H \to C H_{3} N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$ . From nitro compounds: Sn/HCl, Fe/HCl or H₂/Ni: $C H_{3} N O_{2} + 6 [H] \to C H_{3} N H_{2} + 2 H_{2} O$ ; $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ (aniline). Gabriel phthalimide synthesis (pure 1° aliphatic amines): phthalimide with ethanolic KOH gives potassium phthalimide ( $C_{8} H_{5} N O_{2} + K O H \to C_{8} H_{4} K N O_{2} + H_{2} O$ ); this with an alkyl halide gives N-alkylphthalimide ( $C_{8} H_{4} K N O_{2} + C_{2} H_{5} B r \to C_{8} H_{4} N O_{2} C_{2} H_{5} + K B r$ ), hydrolysed to the amine and phthalic acid: $C_{8} H_{4} N O_{2} C_{2} H_{5} + 2 H_{2} O \to C_{2} H_{5} N H_{2} + C_{8} H_{6} O_{4}$ . Aryl amines cannot be made this way (aryl halides do not undergo SN2 with the phthalimide anion).; Worked example
Identify A, B and C: CH₃CH₂CH₂Br + KCN → A; A + Na/C₂H₅OH → B; B + CH₃COCl → C.
Try it first, then show the solutionSolution
A = CH₃CH₂CH₂CN, butanenitrile (one carbon added).
Mendius reduction: $C_{3} H_{7} C N + 4 [H] \to C_{3} H_{7} C H_{2} N H_{2}$ , so B = butan-1-amine.
Acylation: $C_{4} H_{9} N H_{2} + C H_{3} C O C l \to C_{4} H_{9} N H C O C H_{3} + H C l$ , so C = N-butylethanamide.
Answer: A butanenitrile, B butan-1-amine, C N-butylethanamide
Basic character of amines: Amines accept a proton through the lone pair: $R N H_{2} + H_{2} O ⇌ R N H_{3}^{+} + O H^{-}$ ; larger $K_{b}$ (smaller pK_b) means stronger base. They form salts: $C_{2} H_{5} N H_{2} + H C l \to C_{2} H_{5} N H_{3} C l$ . Aliphatic amines are stronger bases than NH₃: alkyl groups (+I) push electron density onto N. In the gas phase: 3° > 2° > 1° > NH₃ (pure inductive order). In water: the order is also affected by solvation of the ammonium ion by hydrogen bonding (more N-H, better solvation; 1° best) and steric hindrance to solvation and protonation (worst for 3°). Net: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃ and (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. Aniline is a weaker base than ammonia (pK_b 9.38 vs 4.75): the lone pair is delocalised into the benzene ring by resonance and is less available, and the anilinium ion has no such resonance stabilisation. Substituents in aniline: electron-releasing groups (-CH₃, -OCH₃) at para increase basicity (p-toluidine > aniline); electron-withdrawing groups (-NO₂, -Cl, -CN) decrease it (p-nitroaniline is far weaker). Order: p-methoxyaniline > p-toluidine > aniline > p-chloroaniline > p-nitroaniline. Benzylamine (C₆H₅CH₂NH₂), whose N is not attached to the ring, behaves like an aliphatic amine.; Worked example
Arrange in increasing order of basic strength in aqueous solution: C₆H₅NH₂, NH₃, C₂H₅NH₂, (C₂H₅)₂NH.
Try it first, then show the solutionSolution
Aniline is weakest (lone pair delocalised by resonance).
NH₃ has no +I group; ethylamines are stronger.
Among ethylamines in water, 2° > 1° (+I effect, with adequate solvation).
Answer: C₆H₅NH₂ < NH₃ < C₂H₅NH₂ < (C₂H₅)₂NH
Alkylation and acylation (with mechanism): Alkylation: the amine's lone pair attacks the carbon of an alkyl halide by SN2, displacing X⁻; loss of H⁺ gives the next amine. Each product is also a nucleophile, so the reaction continues to the quaternary salt: $C_{2} H_{5} N H_{2} + C_{2} H_{5} B r \to (C_{2} H_{5})_{2} N H + H B r$ ; then (C₂H₅)₃N and (C₂H₅)₄N⁺Br⁻ (tetraethylammonium bromide). Acylation: 1° and 2° amines react with acid chlorides, anhydrides or esters by nucleophilic acyl substitution, replacing an N-H hydrogen by an acyl group to give amides: $C_{2} H_{5} N H_{2} + C H_{3} C O C l \to C_{2} H_{5} N H C O C H_{3} + H C l$ (N-ethylethanamide); with ethanoic anhydride: $C_{2} H_{5} N H_{2} + (C H_{3} C O)_{2} O \to C_{2} H_{5} N H C O C H_{3} + C H_{3} C O O H$ . Mechanism: (1) N lone pair attacks the carbonyl carbon of CH₃COCl, giving a tetrahedral intermediate; (2) C=O re-forms, expelling Cl⁻; (3) loss of H⁺ (taken by a base such as pyridine, which also drives the equilibrium by removing HCl) gives the amide. Tertiary amines (no N-H) are not acylated. Benzoylation uses benzoyl chloride (C₆H₅COCl) in NaOH (Schotten-Baumann method).; Worked example
Write the acylation of N-methylethanamine with ethanoic anhydride. Why does N,N-dimethylethanamine not react?
Try it first, then show the solutionSolution
$C_{2} H_{5} N H C H_{3} + (C H_{3} C O)_{2} O \to C_{2} H_{5} N (C H_{3}) C O C H_{3} + C H_{3} C O O H$ (N-ethyl-N-methylethanamide).
Acylation replaces an N-H hydrogen by an acyl group; the tertiary amine has no N-H, so no amide can form.
Answer: N-ethyl-N-methylethanamide; a 3° amine has no N-H to replace
Nitrous acid, carbylamine reaction and Hinsberg's test: Nitrous acid (NaNO₂ + HCl, made in situ, cold): 1° aliphatic amines liberate N₂ quantitatively (brisk effervescence), giving alcohols via unstable diazonium ions: $C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$ . 1° aromatic amines at 273-278 K give stable diazonium salts. 2° amines give yellow oily N-nitrosoamines: $(C_{2} H_{5})_{2} N H + H N O_{2} \to (C_{2} H_{5})_{2} N N O + H_{2} O$ (Liebermann's nitroso test). 3° aliphatic amines form soluble nitrite salts: $(C_{2} H_{5})_{3} N + H N O_{2} \to (C_{2} H_{5})_{3} N H N O_{2}$ . Carbylamine (isocyanide) reaction: only primary amines (aliphatic and aromatic) heated with chloroform and alcoholic KOH give isocyanides with a very offensive smell: $C H_{3} N H_{2} + C H C l_{3} + 3 K O H \to C H_{3} N C + 3 K C l + 3 H_{2} O$ . Hinsberg's test with benzenesulphonyl chloride (C₆H₅SO₂Cl): 1° gives an N-alkylbenzenesulphonamide soluble in KOH (its N-H is acidic due to the strongly electron-withdrawing sulphonyl group): $C_{6} H_{5} S O_{2} C l + C_{2} H_{5} N H_{2} \to C_{6} H_{5} S O_{2} N H C_{2} H_{5} + H C l$ . 2° gives an N,N-dialkylbenzenesulphonamide insoluble in KOH (no acidic H): $C_{6} H_{5} S O_{2} C l + (C_{2} H_{5})_{2} N H \to C_{6} H_{5} S O_{2} N (C_{2} H_{5})_{2} + H C l$ . 3° does not react (no N-H).; Worked example
Write the reactions of ethanamine, N-ethylethanamine and N,N-diethylethanamine with cold nitrous acid. Which gives effervescence?
Try it first, then show the solutionSolution
1°: $C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$ : brisk effervescence of N₂.
2°: $(C_{2} H_{5})_{2} N H + H N O_{2} \to (C_{2} H_{5})_{2} N N O + H_{2} O$ : yellow oily N-nitrosamine.
3°: $(C_{2} H_{5})_{3} N + H N O_{2} \to (C_{2} H_{5})_{3} N H N O_{2}$ : a soluble nitrite salt, no gas.
Answer: Only the primary amine (ethanamine) liberates N₂
Aniline: preparation and reactions: Preparation: reduction of nitrobenzene with Sn (or Fe) and conc. HCl, then NaOH to free the base: $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ . Salts: $C_{6} H_{5} N H_{2} + H C l \to C_{6} H_{5} N H_{3} C l$ (anilinium chloride); $2 C_{6} H_{5} N H_{2} + H_{2} S O_{4} \to (C_{6} H_{5} N H_{3})_{2} S O_{4}$ . Acetylation: $C_{6} H_{5} N H_{2} + C H_{3} C O C l \to C_{6} H_{5} N H C O C H_{3} + H C l$ (acetanilide). Alkylation: $C_{6} H_{5} N H_{2} + C H_{3} I \to C_{6} H_{5} N H C H_{3} + H I$ . Benzoylation: $C_{6} H_{5} N H_{2} + C_{6} H_{5} C O C l \to C_{6} H_{5} N H C O C_{6} H_{5} + H C l$ (benzanilide, NaOH). Carbylamine: $C_{6} H_{5} N H_{2} + C H C l_{3} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$ . Diazotisation (273-278 K): $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ . Electrophilic substitution: -NH₂ is strongly activating and ortho-para directing (+R). Bromination: bromine water gives a white precipitate of 2,4,6-tribromoaniline at once: $C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$ . For monobromination, first acetylate (the -NHCOCH₃ group is less activating), brominate to 4-bromoacetanilide, then hydrolyse. Nitration: direct nitration gives tarry oxidation products and a large amount of the meta isomer (about 47%), because in strong acid much of the aniline becomes anilinium ion (-NH₃⁺, meta-directing). So acetylate, nitrate ( $C_{6} H_{5} N H C O C H_{3} + H N O_{3} \to C_{6} H_{4} (N O_{2}) N H C O C H_{3} + H_{2} O$ ) and hydrolyse ( $C_{6} H_{4} (N O_{2}) N H C O C H_{3} + H_{2} O \to C_{6} H_{4} (N O_{2}) N H_{2} + C H_{3} C O O H$ ) to obtain mainly 4-nitroaniline. Sulphonation: anilinium hydrogensulphate heated at 453-473 K gives sulphanilic acid (4-aminobenzenesulphonic acid), which exists as a zwitterion: overall $C_{6} H_{5} N H_{2} + H_{2} S O_{4} \to C_{6} H_{4} (N H_{2}) S O_{3} H + H_{2} O$ . No Friedel-Crafts: aniline is a Lewis base; it forms a salt with the Lewis acid catalyst AlCl₃, placing a positive charge on N that strongly deactivates the ring. Tests: carbylamine test; azo dye test (diazotise, add alkaline β-naphthol: orange-red dye); bleaching powder solution gives a violet colour. Uses: dyes (azo dyes), drugs (sulpha drugs, paracetamol precursors), rubber chemicals, explosives.; Worked example
Give a reason: aniline is a much weaker base than cyclohexylamine.
Try it first, then show the solutionSolution
In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is less available to accept a proton.
Cyclohexylamine has N on an sp³ carbon with no resonance, and the alkyl ring releases electrons (+I), so its lone pair is fully available.
Answer: Resonance delocalises aniline's lone pair; cyclohexylamine has no such delocalisation
Cyanides and isocyanides: Alkyl cyanides (nitriles, R-C≡N): from alkyl halides with alcoholic KCN or NaCN: $C_{2} H_{5} B r + K C N \to C_{2} H_{5} C N + K B r$ ; from amides by dehydration with P₂O₅ (or SOCl₂): $C H_{3} C O N H_{2} \to C H_{3} C N + H_{2} O$ . Properties: pleasant smell, polar, hydrolysed to acids, reduced to 1° amines. Alkyl isocyanides (carbylamines, R-N≡C): from alkyl halides with alcoholic AgCN: $C_{2} H_{5} B r + A g C N \to C_{2} H_{5} N C + A g B r$ ; from primary amines by the carbylamine reaction: $C_{2} H_{5} N H_{2} + C H C l_{3} + 3 K O H \to C_{2} H_{5} N C + 3 K C l + 3 H_{2} O$ . Properties: very unpleasant smell, poisonous; hydrolysed by acid to a 1° amine and methanoic acid; reduced to 2° amines. KCN vs AgCN: KCN is ionic and gives free CN⁻, which attacks through carbon (the more nucleophilic end); AgCN is largely covalent (Ag-C bond), so only the nitrogen lone pair is available.; Worked example
Identify A, B and C: CH₃CH₂Br + AgCN → A; A + LiAlH₄ → B; B + HNO₂ (cold) → C.
Try it first, then show the solutionSolution
AgCN attacks through N: A = C₂H₅NC, ethyl isocyanide.
Isocyanide reduction gives a secondary amine: B = C₂H₅NHCH₃, N-methylethanamine.
A secondary amine with nitrous acid gives a nitrosamine: C = C₂H₅N(CH₃)NO, a yellow oil.
Answer: A ethyl isocyanide, B N-methylethanamine, C N-nitroso-N-methylethanamine
Diazonium salts: preparation, reactions and importance: Preparation (diazotisation): aniline in excess HCl with NaNO₂ at 273-278 K: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ . Arenediazonium ions are stabilised by resonance with the ring, but decompose above about 278 K, so they are used in solution at once. Aliphatic diazonium ions are not stable at all. Replacement of -N₂⁺ (with loss of N₂): Sandmeyer: CuCl/HCl, CuBr/HBr or CuCN/KCN: $C_{6} H_{5} N_{2} C l C u C l / H C l C_{6} H_{5} C l + N_{2}$ ; $C_{6} H_{5} N_{2} C l + K C N \to C_{6} H_{5} C N + N_{2} + K C l$ (CuCN). Gattermann: copper powder with HCl or HBr: $C_{6} H_{5} N_{2} C l + H B r \to C_{6} H_{5} B r + N_{2} + H C l$ (Cu). Iodide: with KI: $C_{6} H_{5} N_{2} C l + K I \to C_{6} H_{5} I + N_{2} + K C l$ . Balz-Schiemann (fluoride): $C_{6} H_{5} N_{2} C l + H B F_{4} \to C_{6} H_{5} N_{2} B F_{4} + H C l$ ; heat: $C_{6} H_{5} N_{2} B F_{4} \to C_{6} H_{5} F + B F_{3} + N_{2}$ . By -H: hypophosphorous acid (or ethanol): $C_{6} H_{5} N_{2} C l + H_{3} P O_{2} + H_{2} O \to C_{6} H_{6} + N_{2} + H_{3} P O_{3} + H C l$ . By -OH: warm with water: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ . By -NO₂: the fluoroborate heated with NaNO₂ and Cu: $C_{6} H_{5} N_{2} B F_{4} + N a N O_{2} \to C_{6} H_{5} N O_{2} + N_{2} + N a B F_{4}$ . Coupling (N₂ retained): the diazonium ion is a weak electrophile that attacks activated rings at the para position. With phenol in mildly alkaline medium (pH 9-10): $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N_{2} C_{6} H_{4} O H + H C l$ , orange p-hydroxyazobenzene. With aniline in mildly acidic medium (pH 4-5): $C_{6} H_{5} N_{2} C l + C_{6} H_{5} N H_{2} \to C_{6} H_{5} N_{2} C_{6} H_{4} N H_{2} + H C l$ , yellow p-aminoazobenzene. Importance: routes to fluoro, iodo, cyano and nitro benzenes and to compounds that cannot be made by direct substitution (e.g. 1,3,5-tribromobenzene from tribromoaniline); azo dyes.; Worked example
Illustrate the coupling of benzenediazonium chloride with aniline, stating the medium and the colour of the product.
Try it first, then show the solutionSolution
Mildly acidic medium (pH 4-5), cold.
$C_{6} H_{5} N_{2} C l + C_{6} H_{5} N H_{2} \to C_{6} H_{5} N_{2} C_{6} H_{4} N H_{2} + H C l$ : yellow p-aminoazobenzene; the -N=N- link is retained and attack is at the para position.
Answer: Aniline + benzenediazonium chloride at pH 4-5: yellow p-aminoazobenzene

Chemistry in real life: Organic Compounds Containing Nitrogen

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Azo dyes

Suppose a textile chemist wants an orange dye and has aniline, sodium nitrite, hydrochloric acid, phenol and sodium hydroxide.

Outline how to make the dye and why the temperature must be kept near 0 °C.

Try it first, then show the solutionSolution

Diazotise: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ at 273-278 K.
Couple with phenol in alkaline solution: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} O H \to C_{6} H_{5} N_{2} C_{6} H_{4} O H + H C l$ (orange). Above about 278 K the diazonium salt decomposes to phenol and N₂.

Answer: Diazotise cold, then couple with alkaline phenol; warmth destroys the diazonium salt.

Memory hook: Ice keeps the diazo alive.

Fishy smell and amines

Imagine a cook who rubs lemon juice on her hands after cleaning fish to remove the smell.

Explain using the basic character of amines.

Try it first, then show the solutionSolution

The smell comes from volatile amines such as trimethylamine.
Citric acid converts them into non-volatile ammonium salts: $(C H_{3})_{3} N + H^{+} \to (C H_{3})_{3} N H^{+}$ , which wash away and do not smell.

Answer: Acid turns volatile amines into odourless salts.

Memory hook: Acid + amine = salt, no smell.

Protecting group in synthesis

Suppose a pharmaceutical lab needs 4-bromoaniline, but adding bromine water to aniline gives only the tribromo compound.

How can they get the monobromo product?

Try it first, then show the solutionSolution

Acetylate: $C_{6} H_{5} N H_{2} + C H_{3} C O C l \to C_{6} H_{5} N H C O C H_{3} + H C l$ ; the lone pair on N is now shared with C=O, so the ring is less activated.
Brominate (Br₂ in ethanoic acid) to 4-bromoacetanilide (para, bulky group), then hydrolyse with acid or alkali to 4-bromoaniline.

Answer: Acetylate, brominate, hydrolyse.

Memory hook: Put a hat on -NH₂ to calm it down.

Topics in this chapter

Organic Compounds Containing Nitrogen formula sheet

Ammonolysis: $C_{2} H_{5} B r + N H_{3} \to C_{2} H_{5} N H_{2} + H B r$; Alcoholic NH₃, sealed tube; mixture of amines.
Reduction of nitrile: $C_{2} H_{5} C N + 4 [H] \to C_{2} H_{5} C H_{2} N H_{2}$; LiAlH₄ or Na/C₂H₅OH; one carbon more than RX.
Reduction of isocyanide: $C_{2} H_{5} N C + 4 [H] \to C_{2} H_{5} N H C H_{3}$; Secondary amine.
Hofmann bromamide degradation: $C H_{3} C O N H_{2} + B r_{2} + 4 K O H \to C H_{3} N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$; One carbon fewer.
Reduction of nitrobenzene: $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$; Sn/HCl or Fe/HCl.
Gabriel synthesis (last step): $C_{8} H_{4} N O_{2} C_{2} H_{5} + 2 H_{2} O \to C_{2} H_{5} N H_{2} + C_{8} H_{6} O_{4}$; Pure primary amine and phthalic acid.
Acylation: $C_{2} H_{5} N H_{2} + C H_{3} C O C l \to C_{2} H_{5} N H C O C H_{3} + H C l$; Pyridine; 1° and 2° only.
Nitrous acid with 1° amine: $C_{2} H_{5} N H_{2} + H N O_{2} \to C_{2} H_{5} O H + N_{2} + H_{2} O$; Effervescence of N₂.
Nitrous acid with 2° amine: $(C_{2} H_{5})_{2} N H + H N O_{2} \to (C_{2} H_{5})_{2} N N O + H_{2} O$; Yellow oily nitrosamine.
Carbylamine reaction: $C H_{3} N H_{2} + C H C l_{3} + 3 K O H \to C H_{3} N C + 3 K C l + 3 H_{2} O$; Primary amines only.
Hinsberg (1° amine): $C_{6} H_{5} S O_{2} C l + C_{2} H_{5} N H_{2} \to C_{6} H_{5} S O_{2} N H C_{2} H_{5} + H C l$; Product soluble in KOH.
Hinsberg (2° amine): $C_{6} H_{5} S O_{2} C l + (C_{2} H_{5})_{2} N H \to C_{6} H_{5} S O_{2} N (C_{2} H_{5})_{2} + H C l$; Insoluble in KOH; 3° no reaction.
Diazotisation: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$; 273-278 K.
Bromination of aniline: $C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$; White precipitate.
Sandmeyer (cyanide): $C_{6} H_{5} N_{2} C l + K C N \to C_{6} H_{5} C N + N_{2} + K C l$; CuCN catalyst.
Balz-Schiemann: $C_{6} H_{5} N_{2} B F_{4} \to C_{6} H_{5} F + B F_{3} + N_{2}$; Heat the fluoroborate.
Replacement by H: $C_{6} H_{5} N_{2} C l + H_{3} P O_{2} + H_{2} O \to C_{6} H_{6} + N_{2} + H_{3} P O_{3} + H C l$; Hypophosphorous acid.
Replacement by OH: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$; Warm.
Coupling with aniline: $C_{6} H_{5} N_{2} C l + C_{6} H_{5} N H_{2} \to C_{6} H_{5} N_{2} C_{6} H_{4} N H_{2} + H C l$; pH 4-5; yellow p-aminoazobenzene.
Nitrile from amide: $C H_{3} C O N H_{2} \to C H_{3} C N + H_{2} O$; P₂O₅, heat.

Organic Compounds Containing Nitrogen: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Resonance in aniline and the anilinium ion

Explaining why aniline is a weaker base than ammonia, or why -NH₂ is o/p directing.

Draw

Aniline with lone pair on N; three resonance structures with C=N⁺ and negative charge at the two ortho carbons and the para carbon.
Anilinium ion C₆H₅NH₃⁺ with only the two Kekulé structures (no lone pair to delocalise).

Label

lone pair delocalised
δ− at o and p
anilinium ion: no extra resonance

Watch out: Conclude: aniline is stabilised more than its conjugate acid, so it accepts H⁺ less readily.

Mechanism of acylation of an amine

'Explain acylation of ethylamine with mechanism'.

Draw

Curved arrow from N lone pair of C₂H₅NH₂ to the carbonyl C of CH₃COCl; arrow from C=O π bond to O.
Tetrahedral intermediate with O⁻ and N⁺H₂.
C=O re-forms, Cl⁻ leaves; base removes H⁺ from N, giving C₂H₅NHCOCH₃.

Label

nucleophile
tetrahedral intermediate
Cl⁻ leaves
pyridine removes H⁺
amide

Watch out: For alkylation, draw a single SN2 step instead.

How the board asks Organic Compounds Containing Nitrogen

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Identify A, B, C	Chains with KCN, LiAlH₄, HNO₂, Br₂/KOH, CHCl₃/KOH, Na/C₂H₅OH.	Track carbon count: CN adds one, Hofmann removes one; isocyanide reduction gives a 2° amine.
Basicity order	Order of NH₃, aliphatic amines and substituted anilines in water or gas phase.	Balance +I effect, solvation and steric effect; resonance for aniline.
Distinguishing tests	1°/2°/3° amines; aniline vs ethylamine; methylamine vs dimethylamine.	Hinsberg (solubility in KOH), carbylamine, HNO₂, azo dye test; observation for each.
Named reactions	Hofmann, Gabriel, carbylamine, Sandmeyer, Gattermann, Balz-Schiemann, coupling.	Balanced equation with conditions (temperature, medium).
Aniline reasoning	No Friedel-Crafts; why acetylate before nitration; meta product in nitration; tribromo product.	Lewis base/salt with AlCl₃; anilinium ion; activation control by acetylation.
Diazonium conversions	Aniline to fluorobenzene, benzonitrile, phenol, benzene, iodobenzene.	Diazotise at 273-278 K, then the right replacement reagent.

Tricks to remember Organic Compounds Containing Nitrogen

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hofmann takes one, cyanide gives one
Amide → amine loses 1 C; RX → RCN → RCH₂NH₂ gains 1 C.
Use it when: Identification chains.
Hinsberg: 1 Soluble, 2 Insoluble, 3 No reaction
1-2-3: S-I-N.
Use it when: Distinguishing amines.
Basicity in water: 2 > 3 > 1 > NH₃ > aniline (ethyl)
Secondary wins in water; tertiary wins in gas.
Use it when: Ordering bases.
Sandmeyer uses Cu(I) salts; Gattermann uses Cu powder
S for Salts, G for Granules of copper.
Use it when: Named diazonium reactions.
Phenol couples in Alkali, Aniline in Acid
Phenol-Alkaline (pH 9-10), Aniline-Acidic (pH 4-5).
Use it when: Coupling conditions.

Organic Compounds Containing Nitrogen: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Sandmeyer reagent · 1 mark · MCQBoard level

Benzenediazonium chloride is converted into chlorobenzene by:
(a) HBF₄ followed by heat
(b) CuCl and HCl
(c) H₃PO₂ and water
(d) KI

Try it first, then show the solutionSolution

Sandmeyer reaction: CuCl/HCl replaces -N₂⁺ by Cl; HBF₄ gives F, H₃PO₂ gives H and KI gives I.

Answer: (b)

Example 2: Why Gabriel gives pure amines · 1 mark · Assertion-reasonBoard level

Assertion: Gabriel phthalimide synthesis gives primary amines free from secondary and tertiary amines.
Reason: Once an alkyl group is attached, N-alkylphthalimide has no hydrogen on nitrogen, so it cannot be alkylated again.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Try it first, then show the solutionSolution

Unlike ammonolysis, where each amine formed is itself a nucleophile, the single N-H of phthalimide is used up in one alkylation; hydrolysis then releases only RNH₂.

Answer: (a)

Example 3: Balanced equations · 2 marks · Short answerBoard level

Write balanced equations: (a) nitrobenzene reduced with Sn and HCl; (b) aniline with ethanoic anhydride. [2]

Try it first, then show the solutionSolution

(a) $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ (Sn/HCl; NaOH then frees aniline from its salt).
(b) $C_{6} H_{5} N H_{2} + (C H_{3} C O)_{2} O \to C_{6} H_{5} N H C O C H_{3} + C H_{3} C O O H$ (acetanilide).

Answer: Aniline; acetanilide + ethanoic acid

Example 4: Basicity of substituted anilines · 2 marks · Short answerHarder board level

Arrange in increasing order of basic strength and explain: aniline, p-nitroaniline, p-toluidine, p-methoxyaniline. [2]

Try it first, then show the solutionSolution

-NO₂ withdraws electrons (−I, −R), making the N lone pair less available; -CH₃ (+I, hyperconjugation) and -OCH₃ (+R) donate electrons.
p-nitroaniline < aniline < p-toluidine < p-methoxyaniline.

Answer: p-nitroaniline < aniline < p-toluidine < p-methoxyaniline

Example 5: Hinsberg's test · 3 marks · Short answerHarder board level

How will you distinguish methylamine, dimethylamine and trimethylamine using Hinsberg's reagent? Write equations. [3]

Try it first, then show the solutionSolution

Methylamine: $C_{6} H_{5} S O_{2} C l + C H_{3} N H_{2} \to C_{6} H_{5} S O_{2} N H C H_{3} + H C l$ ; the sulphonamide dissolves in KOH (acidic N-H).
Dimethylamine: $C_{6} H_{5} S O_{2} C l + (C H_{3})_{2} N H \to C_{6} H_{5} S O_{2} N (C H_{3})_{2} + H C l$ ; insoluble in KOH.
Trimethylamine: no N-H, no reaction.

Answer: 1° soluble product, 2° insoluble product, 3° no reaction

Example 6: From aniline by diazonium salts · 3 marks · Short answerHarder board level

Starting from aniline, how would you prepare (a) fluorobenzene, (b) benzonitrile, (c) phenol? Give equations and conditions. [3]

Try it first, then show the solutionSolution

Diazotise first: $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ (273-278 K).
(a) $C_{6} H_{5} N_{2} C l + H B F_{4} \to C_{6} H_{5} N_{2} B F_{4} + H C l$ ; heat: $C_{6} H_{5} N_{2} B F_{4} \to C_{6} H_{5} F + B F_{3} + N_{2}$ . (b) $C_{6} H_{5} N_{2} C l + K C N \to C_{6} H_{5} C N + N_{2} + K C l$ (CuCN).
(c) Warm with water: $C_{6} H_{5} N_{2} C l + H_{2} O \to C_{6} H_{5} O H + N_{2} + H C l$ .

Answer: Balz-Schiemann; Sandmeyer; hydrolysis

Example 7: Amines in full · 5 marks · Long answerHarder board level

(a) Describe Gabriel phthalimide synthesis with equations. Why can't aniline be made this way? [3]
(b) Explain why the order of basicity of ethylamines in water is (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂. [2]

Try it first, then show the solutionSolution

(a) $C_{8} H_{5} N O_{2} + K O H \to C_{8} H_{4} K N O_{2} + H_{2} O$ (potassium phthalimide).
$C_{8} H_{4} K N O_{2} + C_{2} H_{5} B r \to C_{8} H_{4} N O_{2} C_{2} H_{5} + K B r$ ; hydrolysis: $C_{8} H_{4} N O_{2} C_{2} H_{5} + 2 H_{2} O \to C_{2} H_{5} N H_{2} + C_{8} H_{6} O_{4}$ .
Aryl halides do not undergo nucleophilic substitution with the phthalimide anion (partial double-bond character of C-X), so aniline cannot be made.
(b) +I effect increases basicity with more ethyl groups, but in water the protonated amine must be stabilised by hydrogen bonding (solvation), best for C₂H₅NH₃⁺ (three N-H) and worst for (C₂H₅)₃NH⁺.
Steric crowding in the 3° amine also hinders protonation. The balance gives 2° > 3° > 1°.

Answer: Gabriel for pure 1° aliphatic amines; basicity balance of inductive, solvation and steric effects

Example 8: Why nitrate acetanilide · 3 marks · Short answerHOTS / competency

Direct nitration of aniline gives a large amount of m-nitroaniline and tarry matter. Explain both observations and describe how p-nitroaniline is made. [3]

Try it first, then show the solutionSolution

In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion; -NH₃⁺ is deactivating and meta-directing, so about 47% m-nitroaniline forms.
Nitric acid is also an oxidising agent and oxidises the highly activated ring, giving tar.
Acetylate to acetanilide, nitrate (mainly para), then hydrolyse: $C_{6} H_{4} (N O_{2}) N H C O C H_{3} + H_{2} O \to C_{6} H_{4} (N O_{2}) N H_{2} + C H_{3} C O O H$ .

Answer: Anilinium ion is m-directing; oxidation gives tar; protect by acetylation

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Three unlabelled amines

Suppose a student has three bottles of C₃H₉N isomers: propan-1-amine, N-methylethanamine and N,N-dimethylmethanamine. She has nitrous acid, chloroform with alcoholic KOH and benzenesulphonyl chloride with KOH.

(i) · 1 mark
Which isomer gives an offensive smell with chloroform and alcoholic KOH?
Answer
Answer: Propan-1-amine.
Only primary amines give isocyanides: $C_{3} H_{7} N H_{2} + C H C l_{3} + 3 K O H \to C_{3} H_{7} N C + 3 K C l + 3 H_{2} O$ .
(ii) · 1 mark
Which gives brisk effervescence with nitrous acid, and what is the gas?
Answer
Answer: Propan-1-amine; nitrogen.
1° aliphatic amines give an alcohol and N₂.
(iii) · 1 mark
What does N-methylethanamine give with nitrous acid?
Answer
Answer: A yellow oily N-nitrosamine.
$C_{2} H_{5} N H C H_{3} + H N O_{2} \to C_{2} H_{5} N (C H_{3}) N O + H_{2} O$
(iv) · 2 marks
Predict the result of Hinsberg's test for each isomer.
Answer
Answer: 1°: sulphonamide soluble in KOH; 2°: insoluble sulphonamide; 3°: no reaction.
The 1° product has an acidic N-H made acidic by the sulphonyl group; the 2° product has no N-H; the 3° amine has no N-H to replace.

Making a dye intermediate

Imagine a dye factory reducing nitrobenzene to aniline, diazotising it, and coupling the product with aniline itself to make a yellow dye.

(i) · 1 mark
Write the reduction of nitrobenzene and name the reagent.
Answer
Answer: $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ ; Sn (or Fe) and conc. HCl.
NaOH is added afterwards to release free aniline from its salt.
(ii) · 1 mark
Why must diazotisation be carried out at 273-278 K?
Answer
Answer: Benzenediazonium chloride decomposes at higher temperatures.
It would form phenol and nitrogen.
(iii) · 1 mark
Name the dye and the pH used for coupling with aniline.
Answer
Answer: p-Aminoazobenzene; mildly acidic (pH 4-5).
$C_{6} H_{5} N_{2} C l + C_{6} H_{5} N H_{2} \to C_{6} H_{5} N_{2} C_{6} H_{4} N H_{2} + H C l$
(iv) · 2 marks
Why does the diazonium ion attack the para position of aniline rather than the ring of benzene?
Answer
Answer: It is a weak electrophile and needs a strongly activated ring.
-NH₂ strongly activates the ring and directs ortho-para; para is favoured for steric reasons. Benzene is not activated enough.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Illustrate with one example each: (a) benzoylation of aniline, (b) the Gattermann reaction. (2 marks)

Step	Marks
(a) $C_{6} H_{5} N H_{2} + C_{6} H_{5} C O C l \to C_{6} H_{5} N H C O C_{6} H_{5} + H C l$ in aqueous NaOH (benzanilide)	1
(b) $C_{6} H_{5} N_{2} C l + H B r \to C_{6} H_{5} B r + N_{2} + H C l$ with copper powder	1

Identify A, B, C: C₆H₅NO₂ + Sn/HCl → A; A + NaNO₂/HCl (273-278 K) → B; B + CuCN/KCN → C. Write the equations. (3 marks)

Step	Marks
A: aniline, $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$	1
B: benzenediazonium chloride, $C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$	1
C: benzonitrile, $C_{6} H_{5} N_{2} C l + K C N \to C_{6} H_{5} C N + N_{2} + K C l$ (CuCN, Sandmeyer)	1

(a) Prepare benzenediazonium chloride. (b) Give the Sandmeyer, Gattermann and Balz-Schiemann reactions. (c) How is the diazo group replaced by H? (5 marks)

Step	Marks
$C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l \to C_{6} H_{5} N_{2} C l + N a C l + 2 H_{2} O$ at 273-278 K	1
Sandmeyer: CuCl/HCl gives C₆H₅Cl + N₂	1
Gattermann: Cu/HBr gives C₆H₅Br + N₂ + HCl	1
Balz-Schiemann: HBF₄ then heat, C₆H₅F + BF₃ + N₂	1
H₃PO₂ (or ethanol): $C_{6} H_{5} N_{2} C l + H_{3} P O_{2} + H_{2} O \to C_{6} H_{6} + N_{2} + H_{3} P O_{3} + H C l$	1

Where students lose marks in Organic Compounds Containing Nitrogen

The slips examiners see most often, and what to write instead.

Mistake: Forgetting that Hofmann degradation removes a carbon
CISCE examiners' comments, 2024: for ethanamide with Br₂ and NaOH, candidates chose ethylamine, forgetting that Hofmann bromamide degradation removes one carbon atom.
Wrong working
CH₃CONH₂ → C₂H₅NH₂
Right working
CH₃CONH₂ → CH₃NH₂ (carbonyl carbon lost as K₂CO₃)
Do this: Count carbons before and after every step.
Usually costs: 1 mark.
Mistake: Losing the cyanide carbon or misreading isocyanide reduction
CISCE examiners' comments, 2025: in C₂H₅Br → KCN → LiAlH₄ → HNO₂, candidates wrote ethylamine and ethanol instead of propylamine and n-propyl alcohol; in the amide-isocyanide sequence they wrote diethylamine instead of ethyl methyl amine; and in reductions they wrote molecular hydrogen instead of nascent hydrogen.
Wrong working
B = C₂H₅NH₂; C (from C₂H₅NC) = (C₂H₅)₂NH
Right working
B = CH₃CH₂CH₂NH₂; C = C₂H₅NHCH₃ (the isocyanide carbon becomes the N-methyl group).
Do this: Remember R-N≡C reduces to R-NH-CH₃.
Usually costs: 1 mark each.
Mistake: Missing medium, by-products or key words
CISCE examiners' comments, 2024 and 2025: candidates did not mention the medium for the coupling reaction, showed alkylation instead of acetylation or omitted the by-product, gave resonance structures instead of 'Lewis base forms a salt with the Lewis acid' for aniline's failure in Friedel-Crafts, omitted HBr from the bromination of aniline, and wrote unbalanced equations.
Wrong working
$C_{6} H_{5} N H_{2} + B r_{2} \to C_{6} H_{2} B r_{3} N H_{2}$
Right working
$C_{6} H_{5} N H_{2} + 3 B r_{2} \to C_{6} H_{2} B r_{3} N H_{2} + 3 H B r$
Do this: Write conditions above arrows and every by-product.
Usually costs: 1 mark.
Mistake: Choosing a non-primary amine for the carbylamine test
CISCE examiners' comments, 2025: for the pair that gives methyl isocyanide, candidates chose chloroform with aniline or with dimethylamine instead of methylamine; in 2024 some judged a true assertion about alcoholic ammonia to be false.
Wrong working
Chloroform + dimethylamine gives methyl isocyanide.
Right working
Only a primary amine with the same R gives R-NC: methylamine → methyl isocyanide.
Do this: Isocyanide = primary amine's R + NC.
Usually costs: 1 mark.

Practice questions on Organic Compounds Containing Nitrogen

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which amine does not react with benzenesulphonyl chloride?
(a) ethylamine
(b) diethylamine
(c) triethylamine
(d) aniline
Hint
No N-H.
Answer
(c)
Q2 · 1 mark · MCQBoard level
The most basic compound in aqueous solution is:
(a) NH₃
(b) CH₃NH₂
(c) (CH₃)₂NH
(d) C₆H₅NH₂
Hint
Secondary methylamine.
Answer
(c)
Q3 · 1 mark · Very short answerBoard level
Fill in the blanks: Benzenediazonium chloride with HBF₄ followed by heating gives ______; the reaction is called ______ reaction.
Hint
F.
Answer
fluorobenzene; Balz-Schiemann
Q4 · 1 mark · Very short answerBoard level
Why are amines less acidic than alcohols of comparable mass?
Hint
Electronegativity.
Answer
N is less electronegative than O, so the N-H bond is less polar and the amide ion RNH⁻ is less stable than RO⁻.
Q5 · 2 marks · Short answerHarder board level
Convert ethanoic acid into methanamine.
Hint
Amide then Hofmann.
Answer
$C H_{3} C O O H + N H_{3} \to C H_{3} C O O N H_{4}$ ; heat: $C H_{3} C O O N H_{4} \to C H_{3} C O N H_{2} + H_{2} O$ ; Hofmann: $C H_{3} C O N H_{2} + B r_{2} + 4 K O H \to C H_{3} N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$ .
Q6 · 3 marks · Short answerHarder board level
Convert benzene into aniline and then aniline into iodobenzene.
Hint
Nitration, reduction, diazotisation, KI.
Answer
$C_{6} H_{6} + H N O_{3} \to C_{6} H_{5} N O_{2} + H_{2} O$ (conc. H₂SO₄); $C_{6} H_{5} N O_{2} + 6 [H] \to C_{6} H_{5} N H_{2} + 2 H_{2} O$ (Sn/HCl); diazotise at 273-278 K; $C_{6} H_{5} N_{2} C l + K I \to C_{6} H_{5} I + N_{2} + K C l$ .
Q7 · 2 marks · Short answerHarder board level
Why does aniline dissolve in hydrochloric acid but not in water to any great extent?
Hint
Salt formation.
Answer
Its large hydrophobic phenyl group limits hydrogen bonding with water. In HCl it forms the ionic salt anilinium chloride, $C_{6} H_{5} N H_{3}^{+} C l^{-}$ , which is water-soluble.
Q8 · 3 marks · Short answerHarder board level
Write the mechanism of alkylation of ammonia by bromoethane and explain why a mixture of amines forms.
Hint
SN2, and products are nucleophiles too.
Answer
N lone pair of NH₃ attacks the α-carbon of C₂H₅Br from the back (SN2), Br⁻ leaves, giving C₂H₅NH₃⁺Br⁻; NH₃ removes H⁺ to give C₂H₅NH₂. The primary amine is a stronger nucleophile than NH₃ and reacts with more C₂H₅Br, giving (C₂H₅)₂NH, (C₂H₅)₃N and (C₂H₅)₄N⁺Br⁻.
Q9 · 2 marks · Short answerHOTS / competency
An aromatic compound A (C₇H₉N) gives an offensive smell with CHCl₃/KOH and dissolves in HCl. With NaNO₂/HCl at 278 K and then warm water it gives B (C₇H₈O), which gives a violet colour with FeCl₃. A has its substituents para. Identify A and B.
Hint
Aryl primary amine.
Answer
A is a primary aromatic amine with a methyl group: 4-methylaniline (p-toluidine). B is 4-methylphenol (p-cresol), formed via the diazonium salt.
Q10 · 1 mark · Very short answerBoard level
Distinguish between aniline and ethylamine by one chemical test.
Hint
Azo dye.
Answer
Treat with NaNO₂/HCl at 273-278 K then alkaline β-naphthol: aniline gives an orange-red azo dye; ethylamine gives N₂ gas and no dye.
Q11 · 5 marks · Long answerHOTS / competency
(a) Compare the basicity of aniline and benzylamine with reasons. [2]
(b) Explain why -NH₂ is ortho-para directing but aniline gives m-nitroaniline on direct nitration. [2]
(c) Write the formation of sulphanilic acid. [1]
Hint
Anilinium ion.
Answer
(a) Benzylamine is more basic: its N is attached to CH₂, so the lone pair is not delocalised into the ring; in aniline it is. (b) The lone pair donates electron density to o and p positions (+R). In the nitrating acid, aniline is protonated; -NH₃⁺ withdraws electrons and is meta-directing, giving much meta product. (c) $C_{6} H_{5} N H_{2} + H_{2} S O_{4} \to C_{6} H_{4} (N H_{2}) S O_{3} H + H_{2} O$ (via anilinium hydrogensulphate heated to 453-473 K; product exists as a zwitterion).
Q12 · 3 marks · Short answerHOTS / competency
1,3,5-Tribromobenzene cannot be made by brominating benzene. Show how it is made from aniline.
Hint
Use -NH₂ to direct, then remove it.
Answer
Brominate aniline with bromine water: 2,4,6-tribromoaniline. Diazotise (NaNO₂/HCl, 273-278 K), then replace -N₂⁺ by H with H₃PO₂: 1,3,5-tribromobenzene.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.General formula of saturated aliphatic amines?Show answer
$C_{n} H_{2 n + 3} N$
2.Which method gives pure primary aliphatic amines?Show answer
Gabriel phthalimide synthesis.
3.What does Hofmann bromamide degradation do?Show answer
Converts an amide to a primary amine with one carbon fewer.
4.Which amines give the carbylamine reaction?Show answer
Primary amines only.
5.Temperature for diazotisation?Show answer
273-278 K.
6.Reagent for Hinsberg's test?Show answer
Benzenesulphonyl chloride.
7.Why is aniline a weaker base than NH₃?Show answer
Its lone pair is delocalised into the benzene ring.
8.Colour of dye from benzenediazonium chloride and phenol?Show answer
Orange (p-hydroxyazobenzene).

Best order to study it

Learn classification, naming, structure and physical properties of amines.
Learn the six preparations with their carbon-count effects.
Learn basicity orders with the three factors and substituent effects in aniline.
Learn alkylation and acylation mechanisms, nitrous acid, carbylamine and Hinsberg tests.
Learn aniline's preparation and reactions, including why acetylation is used.
Learn cyanides/isocyanides and all diazonium reactions; practise identification chains.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name 1°, 2°, 3° amines by IUPAC rules?
Can you prepare amines by Hofmann, Gabriel and nitrile reduction with equations?
Can you order ethylamines, methylamines and substituted anilines by basicity?
Can you write acylation with mechanism?
Can you distinguish 1°, 2°, 3° amines by HNO₂ and Hinsberg's test?
Can you explain why aniline gives no Friedel-Crafts reaction?
Can you prepare cyanides and isocyanides two ways each?
Can you convert benzenediazonium chloride to C₆H₅X (X = Cl, Br, I, F, CN, H, OH, NO₂) and azo dyes?

Organic Compounds Containing Nitrogen: frequently asked questions

Which mechanisms are in this chapter for ISC 2027?

Alkylation and acylation of amines are listed with mechanism.

Is Hinsberg's test in the syllabus?

Yes. Distinction between primary, secondary and tertiary amines by Hinsberg's test is listed.

Which diazonium reactions should I learn?

Preparation from aniline; Sandmeyer, Gattermann and Balz-Schiemann reactions; replacement of the diazo group by -H, -OH and -NO₂; and coupling with phenol and aniline.

Do I need cyanides and isocyanides in detail?

Their methods of preparation: cyanides from alkyl halides and amides; isocyanides from alkyl halides and primary amines. Their reduction products also appear in identification questions.

What tests for aniline are expected?

The carbylamine test, the azo dye test (diazotise, couple with β-naphthol) and bromine water (white precipitate). The carbylamine test is also part of the practical syllabus.

How many marks does this chapter carry?

It is part of the Organic Chemistry group, 31 of the 70 theory marks, shared with four other organic chapters.

Based on: CISCE ISC 2027 Chemistry syllabus; CISCE ISC 2027 Chemistry specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Chemistry, 2025; CISCE Analysis of Pupil Performance, ISC Chemistry, 2024