Classification, nomenclature and structure of amines
Link to Std 11 : In Std 11 you learnt ammonia's pyramidal shape and its lone pair; amines are ammonia with H replaced by carbon groups.
Fish smells 'fishy' because of amines released as it spoils, and many medicines and dyes contain an amine group.
Amines are derivatives of ammonia, classed as primary (one carbon group on N), secondary (two) or tertiary (three). IUPAC: alkanamine; aromatic amine is benzenamine (aniline). The nitrogen is sp³ with a lone pair.
In the figure : Count the carbon groups on nitrogen, not the carbon type: that decides 1°, 2° or 3°.
Rule 1: Classification — 1° amine: RNH₂; 2° amine: R₂NH; 3° amine: R₃N. Aliphatic: N on alkyl; aromatic: N on a ring. In alcohols the class depends on the carbon; in amines it depends on N itself, so (CH₃)₃C–NH₂ is a primary amine.
- Easy : CH₃CH₂NH₂ is primary; (CH₃CH₂)₂NH is secondary.
- Medium : (CH₃)₃N is tertiary.
- Exam-type : (CH₃)₃C–NH₂ (tert-butylamine) is primary although the carbon is tertiary.
Rule 2: Nomenclature — Name as alkanamine (propan-2-amine). For secondary and tertiary amines use the longest chain as parent and prefix N- for the other groups: N-methylethanamine. Aniline is benzenamine. Common names: methylamine, ethylamine, dimethylamine.
- Easy : CH₃CH(NH₂)CH₃ is propan-2-amine.
- Medium : CH₃NHC₂H₅ is N-methylethanamine.
- Exam-type : C₆H₅N(CH₃)₂ is N,N-dimethylbenzenamine.
Rule 3: Structure and physical properties — Nitrogen is sp³; the molecule is pyramidal. 1° and 2° amines form N–H···N hydrogen bonds, so their boiling points are higher than alkanes, but lower than alcohols (N is less electronegative than O). 3° amines cannot hydrogen-bond among themselves. Lower amines are water soluble, and boiling point: 1° > 2° > 3° for isomers.
- Easy : Ethanamine (bp 289 K) boils at a lower temperature than ethanol (bp 351 K).
- Medium : Boiling points: butan-1-amine (351 K) > N-ethylethanamine (329 K) > N,N-dimethylethanamine (310 K).
- Exam-type : Why is a 3° amine more volatile than a 1° isomer? It has no N–H and so no intermolecular hydrogen bonds.
1°, 2°, 3° amine depends on the number of carbon groups on N.
Calling (CH₃)₃C–NH₂ tertiary; it is primary because N has only one carbon group.
Memory clue: "Count the carbons on nitrogen".
Spot the difference
| Point | Alcohol | Amine |
|---|---|---|
| 1°/2°/3° depends on | The carbon bearing OH | The number of C groups on N |
| H-bonding | Strong (O–H···O) | Weaker (N–H···N) |
| Boiling point | Higher | Lower |
Practice questions
- 1Classify (CH₃)₂NH.
- 2Name CH₃CH₂CH₂NH₂.
- 3What is the shape of the amine nitrogen?
- 4Is (CH₃)₃CNH₂ primary, secondary or tertiary? Why?
- 5Write the structure of N-methylethanamine.
- 6Why does ethanamine boil lower than ethanol?
- 7Arrange butan-1-amine, N-ethylethanamine and N,N-dimethylethanamine in decreasing boiling point with reasons.
- 8Name C₆H₅NHCH₃ and classify it.
Show answers (try them yourself first)
- 1Secondary.
- 2Propan-1-amine.
- 3Pyramidal (sp³).
- 4Primary: only one carbon group on nitrogen.
- 5CH₃NHCH₂CH₃.
- 6N–H···N hydrogen bonds are weaker than O–H···O because nitrogen is less electronegative.
- 7Butan-1-amine > N-ethylethanamine > N,N-dimethylethanamine: more N–H bonds mean more hydrogen bonding.
- 8N-methylbenzenamine (N-methylaniline); secondary aromatic amine.
Preparation of amines
Link to Std 11 : In Std 11 you learnt nucleophilic substitution and reduction; amines are made by both.
Dyes, many medicines and the nylon in clothes all start from amines made by reactions like these.
Reduce nitro compounds, nitriles or amides; ammonolysis of alkyl halides (a mixture); Gabriel synthesis (pure 1° aliphatic amine); Hofmann bromamide degradation (amine with one carbon less than the amide).
In the figure : Each box shows a starting compound, the reagent on the arrow and the amine formed. Notice the carbon count in each route.
Rule 1: Reduction methods — Nitro compounds: RNO₂ + 6[H] → RNH₂ + 2H₂O (Sn/HCl, Fe/HCl, or H₂/Pd); aryl nitro compounds give arylamines (nitrobenzene → aniline). Nitriles: RCN + 4[H] (LiAlH₄ or H₂/Ni) → RCH₂NH₂ (adds one carbon to the halide used). Amides: RCONH₂ + LiAlH₄ → RCH₂NH₂.
- Easy : C₆H₅NO₂ + 3H₂ (Ni) → C₆H₅NH₂ + 2H₂O.
- Medium : CH₃CN + 4[H] → CH₃CH₂NH₂.
- Exam-type : Find the mass of aniline (93.13 g mol⁻¹) from 10 g nitrobenzene (123.11 g mol⁻¹) at 100% yield: 7.56 g.
Rule 2: Ammonolysis and Gabriel synthesis — Ammonolysis: RX + NH₃ → RNH₂ → R₂NH → R₃N → R₄N⁺X⁻ (mixture; use excess NH₃ to favour 1°). Gabriel: potassium phthalimide + R–X → N-alkylphthalimide; alkaline hydrolysis → RNH₂ + sodium phthalate. Gives only primary aliphatic amines; aryl halides do not work (no SN2).
- Easy : CH₃CH₂Br + excess NH₃ → CH₃CH₂NH₂ (mainly).
- Medium : Gabriel: phthalimide + KOH → K-phthalimide; + CH₃I → N-methylphthalimide → CH₃NH₂.
- Exam-type : Why cannot aniline be made by Gabriel synthesis? Aryl halides do not undergo SN2 with the phthalimide anion.
Rule 3: Hofmann bromamide degradation — RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The amine has one carbon fewer than the amide, because the C=O carbon is lost as carbonate. Used to step down a series.
- Easy : CH₃CONH₂ → CH₃NH₂ (methanamine).
- Medium : C₆H₅CONH₂ → C₆H₅NH₂ (benzamide → aniline).
- Exam-type : Propanamide (73.09 g mol⁻¹) is the starting amide: 10 g gives 6.17 g of ethanamine (45.08 g mol⁻¹).
Hofmann: one carbon fewer. Cyanide/nitrile reduction: one carbon more than R–X. Gabriel: pure 1° only.
Expecting a single product from ammonolysis; it gives a mixture of 1°, 2°, 3° and quaternary salt.
Memory clue: "Hofmann hops one carbon off".
Spot the difference
| Point | Gabriel | Ammonolysis |
|---|---|---|
| Product | Pure 1° amine | Mixture |
| Works for aryl halides? | No | No |
| Control | High | Low |
Practice questions
- 1Which method gives pure 1° aliphatic amines?
- 2What happens to the carbon count in Hofmann bromamide?
- 3What does LiAlH₄ do to a nitrile?
- 4Write the reduction of nitrobenzene to aniline.
- 5Why does ammonolysis give a mixture?
- 6What is formed from ethanamide with Br₂/NaOH?
- 7Find the mass of aniline from 10 g nitrobenzene (123.11 and 93.13 g mol⁻¹), 100% yield.
- 8Show how CH₃Br can be converted into (i) ethanamine and (ii) methanamine.
Show answers (try them yourself first)
- 1Gabriel phthalimide synthesis.
- 2The amine has one carbon fewer than the amide.
- 3Reduces it to a primary amine, RCH₂NH₂.
- 4C₆H₅NO₂ + 3H₂ (Ni) or Sn/HCl → C₆H₅NH₂ + 2H₂O.
- 5The product amine is still nucleophilic and reacts with more halide.
- 6Methanamine (CH₃NH₂).
- 77.56 g.
- 8(i) CH₃Br + KCN → CH₃CN, then LiAlH₄ (or H₂/Ni) → CH₃CH₂NH₂. (ii) Gabriel: potassium phthalimide + CH₃Br, then alkaline hydrolysis → CH₃NH₂ (or CH₃Br + excess NH₃).
Basic strength of amines
Link to Std 11 : In Std 11 you learnt that a lone pair can accept H⁺ (Lewis and Brønsted bases) and how the +I and resonance effects change charge.
Washing hands after cutting fish with lemon juice removes the smell because acid turns the amines into odourless salts.
Amines are bases because the nitrogen lone pair accepts a proton. Alkyl groups push electrons (+I) and make the amine more basic than ammonia, but in water solvation and steric effects change the order. Aniline is weaker than ammonia because the lone pair is delocalised into the ring.
In the figure : The bars show pKb (smaller bar means stronger base): aliphatic amines are stronger bases than ammonia, and aniline is much weaker.
Rule 1: Aliphatic amines versus ammonia — The +I effect of alkyl groups increases the electron density on N, stabilises the ammonium ion and makes the amine more basic. In water the order for methyl and ethyl amines is 2° > 3° > 1° > NH₃ (3° is weakened by poor solvation of its ammonium ion and steric hindrance).
- Easy : CH₃NH₂ (pKb 3.38) is a stronger base than NH₃ (pKb 4.75).
- Medium : (CH₃)₂NH is the strongest of the three methylamines in water.
- Exam-type : Kb of CH₃NH₂ = 4.5 × 10⁻⁴; pKb = 3.35; for 0.10 M solution [OH⁻] = 6.71e-03 M and pH = 11.83.
Rule 2: Aniline and substituents — In aniline the lone pair on N is delocalised into the benzene ring through resonance, so it is less available for protonation, and the anilinium ion has no such stabilisation. Electron-donating groups (–CH₃, –OCH₃) on the ring increase basicity; electron-withdrawing groups (–NO₂, –Cl) decrease it.
- Easy : pKb of aniline = 9.38, much weaker than NH₃ (4.75).
- Medium : p-Toluidine is more basic than aniline; p-nitroaniline is less basic than aniline.
- Exam-type : Arrange aniline, p-toluidine, p-nitroaniline in increasing basicity: p-nitroaniline < aniline < p-toluidine.
Rule 3: Reactions showing basicity — Amines react with acids to give salts: RNH₂ + HCl → RNH₃⁺Cl⁻ (soluble in water). They precipitate metal hydroxides (Fe³⁺ → Fe(OH)₃). The salt is converted back to the free amine with NaOH. Amines are also nucleophiles and react with alkyl halides (alkylation) and acyl chlorides (acylation).
- Easy : C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻ (anilinium chloride).
- Medium : CH₃CH₂NH₂ + CH₃COCl → CH₃CONHCH₂CH₃ + HCl.
- Exam-type : Why is an amine more soluble in dilute HCl than in water? It forms an ionic salt.
Basicity needs a free lone pair: alkyl groups help, aryl and withdrawing groups hurt.
Assuming that the more alkyl groups the stronger the base in water; the 3° amine is weaker than the 2° in water.
Memory clue: "Push helps; pull and ring hurt" — electron-pushing groups raise basicity.
Spot the difference
| Point | CH₃NH₂ | C₆H₅NH₂ |
|---|---|---|
| pKb | 3.38 | 9.38 |
| Lone pair | Localised | Delocalised into ring |
| Basic strength | Strong | Weak |
Practice questions
- 1Which is a stronger base: CH₃NH₂ or NH₃?
- 2Why is aniline a weaker base than ammonia?
- 3What does an amine form with HCl?
- 4Find pKb for Kb = 4.5 × 10⁻⁴.
- 5Arrange p-nitroaniline, aniline and p-toluidine in increasing basic strength.
- 6Why is a 3° amine weaker than a 2° amine in water?
- 7Calculate the pH of 0.10 M methanamine (Kb = 4.5 × 10⁻⁴).
- 8Explain the basicity order NH₃ < CH₃NH₂ and aniline ≪ NH₃ using electronic effects.
Show answers (try them yourself first)
- 1CH₃NH₂.
- 2The lone pair on N is delocalised into the benzene ring.
- 3An alkylammonium chloride salt.
- 4pKb = 3.35.
- 5p-Nitroaniline < aniline < p-toluidine.
- 6Poorer solvation of the ammonium ion and steric hindrance outweigh the +I effect.
- 7[OH⁻] = √(4.5 × 10⁻⁵) = 6.71e-03 M, pOH = 2.17, pH = 11.83.
- 8Methyl groups push electrons to N and stabilise the cation (more basic than NH₃); in aniline the lone pair is delocalised into the ring (much less basic).
Chemical reactions and tests of amines
Link to Std 11 : In Std 11 you learnt nucleophiles and acylation; amines are good nucleophiles because of the lone pair on nitrogen.
Doctors use acylation (as in paracetamol) to make a medicine from an amine, and the horrible carbylamine smell is the chemist's test for a primary amine.
Amines react as nucleophiles: with acyl chlorides (amides), alkyl halides, and with benzenesulphonyl chloride (Hinsberg test). The carbylamine test detects only 1° amines. Aniline is easily substituted in the ring.
In the figure : The figure lists each test and what it shows: carbylamine for 1°; Hinsberg to separate 1°, 2° and 3°.
Rule 1: Acylation and alkylation — RNH₂ + CH₃COCl → RNHCOCH₃ + HCl (use a base such as pyridine to absorb HCl). 3° amines cannot form amides (no N–H). Alkylation with RX gives 2°, 3° amines and finally quaternary ammonium salts.
- Easy : C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ (acetanilide) + CH₃COOH.
- Medium : (CH₃)₂NH + CH₃COCl → CH₃CON(CH₃)₂ + HCl.
- Exam-type : Why does a tertiary amine not undergo acylation? It has no N–H to lose.
Rule 2: Carbylamine and Hinsberg tests — Carbylamine: R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O (very foul isocyanide smell). Only 1° amines respond. Hinsberg: C₆H₅SO₂Cl + amine in alkali. 1° gives a sulphonamide with an acidic N–H that dissolves in alkali; 2° gives an insoluble sulphonamide; 3° does not react.
- Easy : C₂H₅NH₂ + CHCl₃ + 3KOH → C₂H₅NC + 3KCl + 3H₂O.
- Medium : (C₂H₅)₂NH + C₆H₅SO₂Cl → insoluble N,N-diethylbenzenesulphonamide (2°).
- Exam-type : An amine with no reaction towards Hinsberg's reagent is tertiary.
Rule 3: Reactions of aniline — Aniline's ring is highly activated: Br₂ water gives 2,4,6-tribromoaniline at once (white ppt). Direct nitration with HNO₃ oxidises the amine and gives meta product through the anilinium ion; to control it, first acetylate to acetanilide (protects –NH₂), nitrate, then hydrolyse to get the para-nitroaniline. Aniline does not undergo Friedel–Crafts reaction because –NH₂ attacks the AlCl₃.
- Easy : C₆H₅NH₂ + 3Br₂ → 2,4,6-C₆H₂Br₃NH₂ + 3HBr.
- Medium : C₆H₅NHCOCH₃ is less activating than –NH₂: it reduces over-substitution.
- Exam-type : 10 g aniline (93.13 g mol⁻¹) gives at most 10 ÷ 93.13 × 329.8 = 35.4 g tribromoaniline (329.8 g mol⁻¹).
1° amines: carbylamine +, Hinsberg gives soluble sulphonamide. 3° amines: no Hinsberg reaction.
Thinking secondary amines give the carbylamine test; only primary amines do.
Memory clue: "1-sol, 2-insol, 3-none" for Hinsberg: soluble in alkali, insoluble, no reaction.
Spot the difference
| Point | Carbylamine test | Hinsberg test |
|---|---|---|
| Detects | 1° amines only | 1°, 2° and 3° (distinguishes) |
| Observation | Foul smell of isocyanide | Solubility of sulphonamide in alkali |
| Reagent | CHCl₃ + KOH | C₆H₅SO₂Cl + alkali |
Practice questions
- 1Which amines give the carbylamine test?
- 2Name the product of aniline with excess bromine water.
- 3Which type of amine does not react with Hinsberg's reagent?
- 4Write the carbylamine reaction of ethanamine.
- 5Why is aniline acetylated before nitration?
- 6How do you distinguish a 1° from a 2° amine using Hinsberg's reagent?
- 7Find the mass of 2,4,6-tribromoaniline from 10 g aniline (93.13 and 329.8 g mol⁻¹), 100% yield.
- 8Explain why aniline does not undergo Friedel–Crafts reactions.
Show answers (try them yourself first)
- 1Primary amines (aliphatic and aromatic).
- 22,4,6-Tribromoaniline (white ppt).
- 3Tertiary amines.
- 4C₂H₅NH₂ + CHCl₃ + 3KOH → C₂H₅NC + 3KCl + 3H₂O.
- 5Acetylation reduces the activating power of NH₂ and prevents oxidation and protonation, giving a para product.
- 61° gives a sulphonamide that dissolves in alkali; 2° gives one that does not.
- 735.4 g.
- 8The basic –NH₂ group forms a salt with the Lewis acid AlCl₃, deactivating the ring.
Diazonium salts and their synthetic uses
Link to Std 11 : In Std 11 you learnt electrophilic substitution; the diazonium ion acts as an electrophile in coupling reactions.
The bright orange and red colours on many fabrics and food dyes come from azo compounds made by coupling diazonium salts.
Primary aromatic amines react with nitrous acid (NaNO₂ + HCl) at 273–278 K to give diazonium salts (ArN₂⁺X⁻). The salt loses N₂ easily, so the –N₂⁺ group can be replaced by Cl, Br, I, F, CN, OH or H. Diazonium salts also couple with phenols and anilines to give azo dyes.
In the figure : The central salt is converted to many products; the figure lists the reagent for each. Coupling keeps the N=N link.
Rule 1: Formation (diazotisation) — ArNH₂ + NaNO₂ + 2HCl (273–278 K) → ArN₂⁺Cl⁻ + NaCl + 2H₂O. The temperature must be kept below 278 K or the salt decomposes to phenol. The diazonium salt of an arene is stabilised by resonance with the ring, while the aliphatic ones are unstable and release N₂ at once.
- Easy : C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O at 273–278 K.
- Medium : At room temperature: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl.
- Exam-type : Aliphatic amine + HNO₂ → alcohol + N₂ (the diazonium ion is not stable).
Rule 2: Replacement reactions — Sandmeyer: ArN₂⁺ + CuCl/HCl → ArCl; CuBr/HBr → ArBr; CuCN/KCN → ArCN. Gattermann: Cu powder + HX. Iodide: KI gives ArI. Fluoride: HBF₄ then heat → ArF (Balz–Schiemann). Hydroxyl: warm water → ArOH. Hydrogen: H₃PO₂ or ethanol → ArH. Useful to place groups that cannot be added directly.
- Easy : C₆H₅N₂⁺Cl⁻ + CuCN → C₆H₅CN + N₂ + Cu⁺Cl⁻.
- Medium : C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + N₂ + KCl.
- Exam-type : Why is the Sandmeyer route used? Direct iodination, fluorination and cyanation of benzene are hard.
Rule 3: Coupling and azo dyes — Diazonium ion (weak electrophile) attacks an activated ring: with phenol (alkaline) gives p-hydroxyazobenzene (orange); with aniline (weakly acidic) gives p-aminoazobenzene (yellow). The –N=N– group makes the compound coloured. These are azo dyes; methyl orange is one.
- Easy : C₆H₅N₂⁺Cl⁻ + C₆H₅OH → HO–C₆H₄–N=N–C₆H₅ + HCl.
- Medium : 10 g aniline (93.13 g mol⁻¹) would give at most 21.3 g of p-hydroxyazobenzene (198.2 g mol⁻¹) by coupling with phenol.
- Exam-type : Coupling needs an electron-rich partner: nitrobenzene does not couple.
Diazotise cold (273–278 K). Replace –N₂⁺ with Cl, Br, CN (Cu), I (KI), F (HBF₄), OH (water), H (H₃PO₂).
Doing diazotisation at room temperature; the salt decomposes to phenol and N₂.
Memory clue: "Cold to make, warm to break" — below 278 K to make it; warm to convert it to phenol.
Spot the difference
| Point | Sandmeyer | Coupling |
|---|---|---|
| Result | N₂⁺ replaced by Cl/Br/CN | N₂⁺ retained (azo) |
| Catalyst | Cu(I) salts | None (pH control) |
| N₂ lost? | Yes | No |
Practice questions
- 1At what temperature is a diazonium salt prepared?
- 2Name the product of C₆H₅N₂⁺Cl⁻ with warm water.
- 3What is an azo dye?
- 4Write the Sandmeyer reaction for chlorobenzene.
- 5How is iodobenzene made from aniline?
- 6Why are arenediazonium salts more stable than alkyldiazonium salts?
- 7Find the maximum mass of p-hydroxyazobenzene from 10 g aniline (93.13 and 198.2 g mol⁻¹).
- 8Convert aniline into fluorobenzene and into benzene using diazonium chemistry.
Show answers (try them yourself first)
- 1273–278 K.
- 2Phenol (with N₂).
- 3A coloured compound containing an –N=N– link.
- 4C₆H₅N₂⁺Cl⁻ + CuCl/HCl → C₆H₅Cl + N₂ + Cu⁺Cl⁻.
- 5Diazotise aniline (NaNO₂/HCl, 273 K), then add KI.
- 6The positive charge is delocalised into the benzene ring by resonance.
- 721.3 g.
- 8ArNH₂ → ArN₂⁺ (cold) → HBF₄, heat → C₆H₅F; or ArN₂⁺ + H₃PO₂ → C₆H₆.