Structure and nomenclature of carbonyl compounds
Link to Std 11 : In Std 11 you learnt functional groups, the sp² carbon of alkenes and IUPAC naming of simple chains.
The smell of cinnamon, the sharp taste of vinegar and the nail-polish remover in a dressing table are all carbonyl compounds.
A carbonyl group is a carbon doubly bonded to oxygen (C=O). In aldehydes it is bonded to at least one hydrogen, in ketones to two carbon groups, and in carboxylic acids to an –OH group. Names end in -al, -one and -oic acid.
In the figure : The carbon of C=O is sp² with three sigma bonds in one plane; because oxygen is more electronegative, the carbon carries a partial positive charge.
Rule 1: Naming aldehydes and ketones — Aldehyde: longest chain containing –CHO, which is carbon 1, named alkanal (butanal). Ketone: number from the end nearest to C=O, named alkanone (pentan-2-one). Common names: formaldehyde, acetaldehyde, acetone.
- Easy : CH₃CH₂CHO is propanal.
- Medium : CH₃COCH₂CH₃ is butanone.
- Exam-type : CH₃CH(CH₃)CH₂CHO is 3-methylbutanal; CH₃COCH₂CH(CH₃)₂ is 4-methylpentan-2-one.
Rule 2: Naming carboxylic acids — Chain containing –COOH, which is carbon 1, named alkanoic acid. Dicarboxylic acids end in -dioic acid (ethanedioic = oxalic acid). Aromatic: benzoic acid. Common names: formic (HCOOH), acetic (CH₃COOH).
- Easy : CH₃CH₂COOH is propanoic acid.
- Medium : HOOC–COOH is ethanedioic acid (oxalic acid).
- Exam-type : CH₃CH(Cl)COOH is 2-chloropropanoic acid.
Rule 3: Structure and polarity — In C=O, carbon is sp² and the geometry is trigonal planar. The π bond is polarised towards oxygen, so the molecule has a dipole moment and the carbon is electrophilic. Carbonyl compounds are more polar than ethers and alkenes of similar mass, which is why their boiling points are higher than those of hydrocarbons.
- Easy : Propanone (bp 329 K) boils higher than butane (bp 273 K) because of dipole–dipole attraction.
- Medium : A carboxylic acid forms a cyclic dimer through two hydrogen bonds.
- Exam-type : Why does the C=O carbon react with nucleophiles but not with electrophiles? Its carbon is δ⁺, so nucleophiles attack it.
Carbonyl carbon is electrophilic (δ⁺); nucleophiles attack it.
Counting the –CHO carbon wrongly: in aldehydes and acids the carbonyl carbon is always carbon 1.
Memory clue: "al-one-oic" — aldehyde, ketone, acid.
Spot the difference
| Point | Aldehyde | Ketone |
|---|---|---|
| Group | –CHO (H on C=O) | –CO– (two carbons on C=O) |
| Suffix | -al | -one |
| Oxidised easily? | Yes | No |
Practice questions
- 1Name CH₃CH₂CH₂CHO.
- 2What is the hybridisation of the carbonyl carbon?
- 3Give the IUPAC name of acetone.
- 4Write the structure of 3-methylpentan-2-one.
- 5Why is the carbonyl carbon electrophilic?
- 6Name CH₃CH₂CH(CH₃)COOH.
- 7Explain why propanone boils at a higher temperature than butane although their masses are similar.
- 8Name HOOCCH₂COOH and C₆H₅CHO.
Show answers (try them yourself first)
- 1Butanal.
- 2sp².
- 3Propanone.
- 4CH₃COCH(CH₃)CH₂CH₃.
- 5Oxygen is more electronegative and pulls the π electrons, leaving carbon δ⁺.
- 62-Methylbutanoic acid.
- 7Propanone is polar and has dipole–dipole attraction; butane is non-polar and has only weak London forces.
- 8Propanedioic acid (malonic acid) and benzaldehyde.
Preparation of aldehydes, ketones and carboxylic acids
Link to Std 11 : In Std 11 you learnt oxidation of alcohols and the Friedel–Crafts reaction; both appear again here.
Vanilla flavour (vanillin) and cherry flavour (benzaldehyde) are aldehydes; acetone is the commonest ketone in the home.
Aldehydes: oxidise 1° alcohols with PCC, reduce acid chlorides (Rosenmund), nitriles (Stephen) or esters (DIBAL-H), and oxidise toluene (Etard). Ketones: oxidise 2° alcohols, use acid chlorides with R₂Cd, or Friedel–Crafts acylation. Acids: oxidise 1° alcohols, hydrolyse nitriles, or use Grignard + CO₂.
In the figure : Each box shows a starting material and the reagent on the arrow. Rosenmund uses a poisoned catalyst so the reduction stops at the aldehyde.
Rule 1: Aldehydes — Rosenmund: RCOCl + H₂ (Pd–BaSO₄) → RCHO + HCl. Stephen: RCN + SnCl₂ + HCl, then H₂O → RCHO. Etard: C₆H₅CH₃ + CrO₂Cl₂ → C₆H₅CHO. Gattermann–Koch: benzene + CO + HCl (AlCl₃/CuCl) → benzaldehyde.
- Easy : CH₃COCl + H₂ (Pd–BaSO₄) → CH₃CHO + HCl.
- Medium : CH₃CN + SnCl₂/HCl, then H₂O → CH₃CHO.
- Exam-type : Toluene → benzaldehyde by Etard; chromyl chloride avoids further oxidation.
Rule 2: Ketones — Oxidation of 2° alcohols: R₂CHOH → R₂C=O (CrO₃, K₂Cr₂O₇). Acid chloride + dialkylcadmium: 2RCOCl + R′₂Cd → 2RCOR′. Nitrile + Grignard: RCN + R′MgX, then H₃O⁺ → RCOR′. Friedel–Crafts acylation: benzene + CH₃COCl (AlCl₃) → acetophenone.
- Easy : CH₃CH(OH)CH₃ → CH₃COCH₃ with CrO₃.
- Medium : C₆H₆ + CH₃COCl (AlCl₃) → C₆H₅COCH₃ + HCl.
- Exam-type : 5.0 g propan-2-ol (60.1 g mol⁻¹) fully oxidised gives at most 5.0 ÷ 60.1 × 58.08 = 4.83 g propanone (58.08 g mol⁻¹).
Rule 3: Carboxylic acids — (i) Oxidise 1° alcohols or aldehydes with KMnO₄ or K₂Cr₂O₇. (ii) Hydrolyse nitriles: RCN + 2H₂O (H⁺ or OH⁻) → RCOOH + NH₃ (this adds one carbon compared with R–X). (iii) Grignard + CO₂: RMgX + CO₂ → RCOOMgX → RCOOH (acid has one more carbon). (iv) Side-chain oxidation of alkylbenzenes (toluene → benzoic acid).
- Easy : CH₃CH₂CN + 2H₂O → CH₃CH₂COOH + NH₃.
- Medium : CH₃MgBr + CO₂, then H₃O⁺ → CH₃COOH.
- Exam-type : Toluene + hot alkaline KMnO₄, then H⁺ → benzoic acid (any alkyl side chain gives COOH).
Nitrile hydrolysis and Grignard + CO₂ both give an acid with one more carbon than the halide.
Thinking Rosenmund gives an acid; the poisoned catalyst stops the reduction at the aldehyde.
Memory clue: "Rosenmund: Pd poisoned by BaSO₄; Stephen: tin" — Rosenmund uses H₂, Stephen uses SnCl₂.
Spot the difference
| Point | Rosenmund | Stephen |
|---|---|---|
| Starting material | Acid chloride | Nitrile |
| Reagent | H₂, Pd–BaSO₄ | SnCl₂/HCl, then H₂O |
| Product | Aldehyde | Aldehyde |
Practice questions
- 1Which reaction converts toluene into benzaldehyde?
- 2What is formed from CH₃MgBr and CO₂ (after acid)?
- 3Name the catalyst poison in Rosenmund reduction.
- 4Write the Stephen reaction for ethanenitrile.
- 5How would you convert benzene to acetophenone?
- 6Write the oxidation of propan-2-ol to a ketone.
- 7How can ethyl bromide be converted into propanoic acid by two different routes?
- 8Find the mass of propanone formed from 5.0 g propan-2-ol (60.1 g mol⁻¹) at 100% yield (58.08 g mol⁻¹).
Show answers (try them yourself first)
- 1Etard reaction (CrO₂Cl₂).
- 2Ethanoic acid (acetic acid).
- 3BaSO₄ (also quinoline/sulphur).
- 4CH₃CN + SnCl₂ + HCl, then H₂O → CH₃CHO.
- 5Friedel–Crafts acylation with CH₃COCl and anhydrous AlCl₃.
- 6CH₃CH(OH)CH₃ + [O] (CrO₃) → CH₃COCH₃ + H₂O.
- 7(i) C₂H₅Br + KCN → C₂H₅CN, then hydrolysis → C₂H₅COOH. (ii) C₂H₅Br + Mg → C₂H₅MgBr, then CO₂ and H₃O⁺ → C₂H₅COOH.
- 84.83 g.
Nucleophilic addition reactions
Link to Std 11 : In Std 11 you learnt that electron-rich species attack electron-poor carbon (nucleophiles and electrophiles) and the inductive effect.
Cold-water solubility of small aldehydes and the orange-red test paper used to detect ketones both depend on this one reaction.
A nucleophile attacks the δ⁺ carbon of C=O; the π bond breaks and the oxygen picks up the extra electron pair, then a proton is added. Aldehydes are more reactive than ketones because they have less crowding and less electron donation to the carbon.
In the figure : The nucleophile adds to carbon; the oxygen becomes an alkoxide and then takes H⁺ to form an alcohol-type product.
Rule 1: Why aldehydes beat ketones — Ketones have two alkyl groups that push electrons towards C=O (+I) and crowd the carbon, so the carbon is less positive and harder to attack. Aromatic aldehydes are less reactive than aliphatic ones because the ring donates electrons by resonance.
- Easy : Methanal reacts fastest because it has no alkyl group.
- Medium : Order: HCHO > CH₃CHO > CH₃COCH₃.
- Exam-type : Why is benzaldehyde less reactive than ethanal? Resonance with the ring reduces the positive charge on carbon.
Rule 2: Typical additions — HCN: RCHO + HCN → RCH(OH)CN (cyanohydrin; acid-catalysed hydrolysis gives a hydroxy acid). NaHSO₃: bisulphite adduct (crystalline, used to purify aldehydes). Alcohol/dry HCl: hemiacetal then acetal (protects the C=O group). Ammonia derivatives: NH₂OH gives an oxime, NH₂NH₂ a hydrazone, 2,4-dinitrophenylhydrazine an orange-red crystalline 2,4-DNP derivative.
- Easy : CH₃CHO + HCN → CH₃CH(OH)CN.
- Medium : CH₃COCH₃ + NH₂OH → (CH₃)₂C=NOH + H₂O.
- Exam-type : Why is the 2,4-DNP test used? A ketone or aldehyde gives a coloured precipitate; the other functional groups do not.
Rule 3: Grignard reaction and reduction — Grignard: RMgX adds to C=O; H₃O⁺ gives 1° (from HCHO), 2° (other aldehydes) or 3° (ketones) alcohols. Reduction to alcohols: NaBH₄, LiAlH₄ or H₂/Ni. To hydrocarbons: Clemmensen (Zn–Hg/conc. HCl) or Wolff–Kishner (NH₂NH₂ then KOH/ethylene glycol, heat) turn C=O into CH₂.
- Easy : CH₃COCH₃ + Zn–Hg/HCl → CH₃CH₂CH₃.
- Medium : Wolff–Kishner works under basic conditions and Clemmensen under acidic ones: choose according to the other groups in the molecule.
- Exam-type : Acetophenone (C₆H₅COCH₃) + Zn–Hg/HCl → ethylbenzene.
Aldehyde > ketone in reactivity: less +I, less steric hindrance.
Saying ketones are more reactive because they have more alkyl groups; more alkyl groups mean less positive carbon and more crowding.
Memory clue: "H is small and pulls; R is big and pushes" — H on C=O makes it more reactive.
Spot the difference
| Point | Clemmensen | Wolff–Kishner |
|---|---|---|
| Reagent | Zn–Hg/conc. HCl | NH₂NH₂, then KOH/heat |
| Medium | Acidic | Basic |
| Product | CH₂ | CH₂ |
Practice questions
- 1Which is more reactive towards nucleophiles: HCHO or CH₃COCH₃?
- 2What is the product of CH₃CHO with HCN?
- 3What colour is the 2,4-DNP derivative?
- 4Why are aldehydes more reactive than ketones?
- 5Write the reaction of propanone with hydroxylamine.
- 6Which reaction converts C=O into CH₂ in a base-stable compound?
- 7Explain why benzaldehyde is less reactive than propanal.
- 8A compound C₃H₆O gives an orange-red precipitate with 2,4-DNP and reacts with NaBH₄ to give a secondary alcohol. Identify it.
Show answers (try them yourself first)
- 1HCHO.
- 2CH₃CH(OH)CN (a cyanohydrin).
- 3Orange-red.
- 4Aldehydes have only one electron-donating alkyl group and less crowding, so carbon is more positive and easier to reach.
- 5(CH₃)₂C=O + NH₂OH → (CH₃)₂C=NOH + H₂O.
- 6Wolff–Kishner reduction.
- 7In benzaldehyde the ring donates electrons by resonance and lowers the positive charge on carbon.
- 8Propanone (CH₃COCH₃).
Oxidation tests, Cannizzaro and aldol reactions
Link to Std 11 : In Std 11 you learnt oxidation and reduction as loss and gain of electrons; here the aldehyde is a reducing agent.
Old mirrors were silvered with a silver-mirror-type reaction of an aldehyde, and the yellow iodoform smell was once used to detect ethanol.
Aldehydes are easily oxidised, so they reduce Tollens' and Fehling's reagents; ketones do not. Aldehydes with no α-hydrogen undergo Cannizzaro in strong base, while those with α-hydrogen undergo aldol condensation in dilute base.
In the figure : The figure lists the tests: silver mirror (Tollens), red-brown precipitate (Fehling), yellow precipitate (iodoform), and the two base-driven reactions of aldehydes.
Rule 1: Tollens' and Fehling's tests — Tollens': RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 4NH₃ + 2H₂O (silver mirror). Fehling's: RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O (red-brown) + 3H₂O; aromatic aldehydes do not react. Ketones react with neither.
- Easy : CH₃CHO gives a silver mirror with Tollens' reagent.
- Medium : C₆H₅CHO gives Tollens' test but not Fehling's.
- Exam-type : 10 g of ethanal (44.05 g mol⁻¹) can give at most 2 × 107.87 × 10 ÷ 44.05 = 49.0 g of silver.
Rule 2: Haloform (iodoform) test — Methyl ketones (CH₃CO–R) and CH₃CH(OH)–R compounds give a yellow precipitate of CHI₃ (iodoform) with I₂/NaOH. Ethanol, ethanal, propanone and propan-2-ol give it; methanol, propanal and pentan-3-one do not.
- Easy : CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O.
- Medium : Pentan-2-one gives iodoform; pentan-3-one does not.
- Exam-type : Distinguish ethanol from methanol: only ethanol gives yellow iodoform.
Rule 3: Aldol and Cannizzaro reactions — Aldol: aldehydes/ketones with α-H, in dilute base, add to each other to give a β-hydroxy aldehyde (aldol), which dehydrates on heating to an α,β-unsaturated carbonyl compound. Cannizzaro: aldehydes without α-H (HCHO, C₆H₅CHO) in conc. NaOH give an alcohol and a carboxylate salt (one molecule is oxidised, one reduced).
- Easy : 2CH₃CHO ⇌ CH₃CH(OH)CH₂CHO (aldol) in dil. NaOH.
- Medium : 2HCHO + NaOH(conc.) → CH₃OH + HCOONa.
- Exam-type : 2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa. Why no aldol for benzaldehyde? It has no α-hydrogen.
α-Hydrogen present → aldol. α-Hydrogen absent → Cannizzaro. Methyl ketone/ethanol → iodoform.
Predicting Cannizzaro for ethanal; ethanal has α-H, so it undergoes aldol under dilute base.
Memory clue: "Alpha present: aldol. Alpha absent: Cannizzaro".
Spot the difference
| Point | Aldol | Cannizzaro |
|---|---|---|
| α-H needed? | Yes | No |
| Base | Dilute | Concentrated |
| Products | β-hydroxy carbonyl | Alcohol + acid salt |
Practice questions
- 1Which reagent gives a silver mirror with an aldehyde?
- 2Does propanone give Tollens' test?
- 3Name a compound that gives yellow iodoform.
- 4Write the Cannizzaro reaction of methanal.
- 5Why does benzaldehyde not give an aldol condensation?
- 6How do you distinguish pentan-2-one from pentan-3-one?
- 7Find the mass of silver formed from 10 g of ethanal (44.05 g mol⁻¹, Ag = 107.87) by Tollens' test.
- 8Explain the difference between the aldol reaction and Cannizzaro reaction using ethanal and benzaldehyde.
Show answers (try them yourself first)
- 1Tollens' reagent.
- 2No.
- 3Ethanol (or propanone, ethanal, propan-2-ol).
- 42HCHO + NaOH → CH₃OH + HCOONa.
- 5Benzaldehyde has no α-hydrogen, so no enolate can form.
- 6Pentan-2-one has a CH₃CO– group and gives yellow CHI₃ with I₂/NaOH; pentan-3-one does not.
- 749.0 g.
- 8Ethanal has α-H, so in dilute NaOH it gives aldol; benzaldehyde has none, so in conc. NaOH one molecule is oxidised to benzoate and another reduced to benzyl alcohol.
Acidity and reactions of carboxylic acids
Link to Std 11 : In Std 11 you learnt resonance and the inductive effect; both decide how strong a carboxylic acid is.
Vinegar is 5–8% ethanoic acid; it dissolves limescale and releases bubbles of CO₂ when poured on baking soda.
Carboxylic acids are weak acids but stronger than phenols and alcohols because the carboxylate ion is stabilised by resonance. They react with carbonates to give CO₂ (a test) and form esters, acid chlorides, amides and anhydrides.
In the figure : pKa values: smaller = stronger. Electron-withdrawing groups (Cl, F) increase strength; electron-donating alkyl groups decrease it.
Rule 1: Acid strength — Resonance spreads the negative charge equally over two oxygens in RCOO⁻. Electron-withdrawing groups (–Cl, –F, –NO₂) stabilise the anion and increase acidity (CF₃COOH is far stronger than CH₃COOH); more of them and nearer to –COOH have a larger effect. Electron-donating groups (–CH₃) reduce acidity.
- Easy : Order: Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH.
- Medium : Ka of ethanoic acid = 1.74 × 10⁻⁵, pKa = 4.76.
- Exam-type : For 0.10 M ethanoic acid, [H⁺] = √(Ka × c) = 1.32e-03 M, pH = 2.88, degree of dissociation = 1.32 %.
Rule 2: Reactions of the –COOH group — With NaHCO₃: RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂ (phenols do not do this — a test). Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O. With PCl₅, PCl₃ or SOCl₂: acid chloride. With NH₃, heat: amide. Reduction with LiAlH₄ gives 1° alcohol. Decarboxylation: sodium salt + soda lime → alkane with one carbon less.
- Easy : CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑.
- Medium : CH₃COONa + NaOH (CaO, heat) → CH₄ + Na₂CO₃.
- Exam-type : Differentiate benzoic acid from phenol: only benzoic acid releases CO₂ with NaHCO₃.
Rule 3: α-Substitution and uses — Hell–Volhard–Zelinsky (HVZ): RCH₂COOH + Cl₂/red P, then H₂O → RCHClCOOH (α-halo acid). Aromatic ring: –COOH is meta-directing and deactivating (nitration of benzoic acid gives m-nitrobenzoic acid). Uses: ethanoic acid (vinegar, solvent), benzoic acid (food preservative), long-chain acids (soaps).
- Easy : CH₃CH₂COOH + Cl₂/P → CH₃CHClCOOH.
- Medium : Benzoic acid + HNO₃/H₂SO₄ → m-nitrobenzoic acid.
- Exam-type : A good test for α-hydrogen acids: HVZ reaction needs at least one α-H.
Electron-withdrawing group: stronger acid. Electron-donating group: weaker acid.
Saying chloroacetic acid is weaker than acetic acid; chlorine pulls electrons and makes it stronger.
Memory clue: "Pull more, pKa lower" — electron-withdrawing groups lower pKa.
Spot the difference
| Point | Phenol | Carboxylic acid |
|---|---|---|
| pKa (approx.) | 10 | 4–5 |
| Reacts with NaHCO₃? | No | Yes (CO₂ released) |
| Anion stabilised by | Ring resonance | Two equivalent resonance forms |
Practice questions
- 1Which is the stronger acid: CH₃COOH or ClCH₂COOH?
- 2What gas is released when RCOOH reacts with NaHCO₃?
- 3What reagent converts RCOOH to RCOCl?
- 4Find the pKa of ethanoic acid (Ka = 1.74 × 10⁻⁵).
- 5Write the HVZ reaction of propanoic acid.
- 6How can benzoic acid be distinguished from phenol?
- 7Calculate the pH of 0.10 M ethanoic acid (Ka = 1.74 × 10⁻⁵).
- 8Arrange CH₃COOH, HCOOH and Cl₂CHCOOH in decreasing acid strength, with reasons.
Show answers (try them yourself first)
- 1ClCH₂COOH.
- 2CO₂.
- 3SOCl₂ (or PCl₅, PCl₃).
- 4pKa = 4.76.
- 5CH₃CH₂COOH + Cl₂ (red P) → CH₃CHClCOOH + HCl.
- 6Add NaHCO₃: benzoic acid gives effervescence of CO₂; phenol does not.
- 7[H⁺] = √(1.74 × 10⁻⁶) = 1.32e-03 M; pH = 2.88.
- 8Cl₂CHCOOH > HCOOH > CH₃COOH: chlorine withdraws electrons, and the methyl group in CH₃COOH donates electrons.
Derivatives of carboxylic acids and uses
Link to Std 11 : In Std 11 you learnt esterification as an equilibrium; here you use it to make fragrances and soaps.
Fruity smells like banana and pineapple come from esters; soap is made by hydrolysing fats with alkali.
Acid derivatives replace the –OH of –COOH with –Cl (acid chloride), –OR′ (ester), –NH₂ (amide) or –O–COR (anhydride). Acid chlorides are the most reactive and amides the least; the order of reactivity guides how they are interconverted.
In the figure : Each derivative is made from the acid or its chloride; reactivity decreases in the order shown.
Rule 1: Acid chlorides and anhydrides — RCOCl + R′OH → RCOOR′ + HCl; RCOCl + NH₃ → RCONH₂ + HCl. Anhydride: 2RCOOH → (RCO)₂O + H₂O (P₂O₅, heat). Both are strong acylating agents (used to make aspirin, paracetamol).
- Easy : CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl.
- Medium : (CH₃CO)₂O + salicylic acid → aspirin + CH₃COOH.
- Exam-type : Why are acid chlorides stored away from moisture? They hydrolyse rapidly to give acids and HCl.
Rule 2: Esters — Fischer esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O (conc. H₂SO₄, heat); the equilibrium is pushed forward by excess alcohol or removal of water. Hydrolysis: acid-catalysed (reversible) or alkaline (saponification): RCOOR′ + NaOH → RCOONa + R′OH (irreversible).
- Easy : CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O.
- Medium : Soap = sodium salt of long-chain fatty acid, formed by saponification of fat.
- Exam-type : If 1.0 mol each of acid and alcohol reach equilibrium with K = 4, the ester formed is x where x² ÷ (1 − x)² = 4, so x = 0.67 mol.
Rule 3: Amides and uses — Amides: RCOOH + NH₃ → RCOONH₄ → RCONH₂ (heat); hydrolysis gives acid + amine. Uses: acids in food (vinegar, citric), polymers (nylon-6,6 is a polyamide), medicines, flavours.
- Easy : CH₃COCl + NH₃ → CH₃CONH₂ + HCl.
- Medium : Amide hydrolysis in aqueous acid gives the acid and an ammonium salt.
- Exam-type : Hofmann bromamide degradation uses an amide to give an amine with one carbon less (see Amines).
Reactivity: acid chloride > anhydride > ester > amide.
Writing saponification as reversible; with excess alkali the carboxylate salt is formed and the reaction does not go back.
Memory clue: "Ester + alkali = soap and alcohol".
Spot the difference
| Point | Acid hydrolysis of ester | Alkaline hydrolysis |
|---|---|---|
| Reversible? | Yes | No |
| Product | Acid + alcohol | Carboxylate salt + alcohol |
| Name | — | Saponification |
Practice questions
- 1Which acid derivative is the most reactive?
- 2What is saponification?
- 3Name the product of CH₃COCl with NH₃.
- 4Write the esterification of propanoic acid with methanol.
- 5How can an ester be made irreversibly from an acid?
- 6What do nylon-6,6 and proteins have in common?
- 7For 1.0 mol each of ethanoic acid and ethanol, if K = 4, find the moles of ester at equilibrium.
- 8Explain why aspirin is made with ethanoic anhydride rather than ethanoic acid.
Show answers (try them yourself first)
- 1Acid chloride.
- 2Alkaline hydrolysis of an ester to give a carboxylate salt and alcohol.
- 3Ethanamide (acetamide).
- 4CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O (H⁺).
- 5Convert the acid to an acid chloride (SOCl₂) and add the alcohol.
- 6Both have amide (peptide) linkages.
- 7x²/(1−x)² = 4 gives x/(1−x) = 2 and x = 0.67 mol.
- 8The anhydride is a much stronger acylating agent and the reaction goes to completion; the acid alone gives a poor equilibrium yield.