ChemistryISC Class 12 · Chapter 8

Aldehydes, Ketones and Carboxylic Acids ISC Class 12 Chemistry

Q: Which mechanisms are required in this ISC chapter?

Nucleophilic addition (with ammonia derivatives, HCN, NaHSO₃ and Grignard reagent), and the base-catalysed aldol condensation, cross aldol condensation and Cannizzaro reaction.

Q: Which benzaldehyde reactions are listed?

Oxidation and reduction, addition of HCN and NaHSO₃, reactions with hydroxylamine, hydrazine and phenylhydrazine, PCl₅, Cannizzaro, benzoin condensation, Perkin's reaction and electrophilic substitution (halogenation, nitration, sulphonation).

Q: How much detail is needed on acidity of carboxylic acids?

The syllabus says the effect of substituents on aliphatic and aromatic acids is to be dealt with in detail: inductive and resonance effects, number and position of substituents.

Q: What tests for acids are needed?

Tests for formic acid (Tollens', Fehling's), acetic acid (neutral FeCl₃, ester smell) and benzoic acid (neutral FeCl₃ buff precipitate), plus NaHCO₃ for all.

Q: Which decarboxylation methods should I know?

Chemical decarboxylation by heating the sodium salt with soda lime, and Kolbe's electrolytic method.

Q: 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.

Updated 15 October 2026 · Fact-checked

Aldehydes, Ketones and Carboxylic Acids is Chapter 8 of ISC Class 12 Chemistry. It covers preparation of carbonyl compounds, nucleophilic addition with mechanism, oxidation, iodoform and reduction reactions, aldol and Cannizzaro reactions with mechanism, benzaldehyde (benzoin, Perkin), and carboxylic acids: acidity, derivatives, decarboxylation, HVZ, ring substitution and tests.

What Aldehydes, Ketones and Carboxylic Acids is about

Aldehydes (R-CHO) and ketones (R-CO-R') contain the carbonyl group >C=O; carboxylic acids contain the carboxyl group -COOH. This is the longest chapter in the Organic Chemistry group (31 of the 70 theory marks) and the most common source of 'identify A, B, C' and conversion questions in Sections B to D.

The C=O bond is polar (C δ+, O δ−) and planar, so the carbonyl carbon is attacked by nucleophiles. Aldehydes are more reactive than ketones (less steric hindrance, only one electron-releasing group), and aromatic carbonyls less reactive than aliphatic ones (resonance with the ring). You need the preparations (from alcohols, ozonolysis, hydration of alkynes, Rosenmund reduction, dialkyl cadmium, calcium salts, nitriles by Stephen reaction or Grignard, esters), nucleophilic addition (HCN, NaHSO₃, Grignard, ammonia and its derivatives, with mechanism), oxidation (Tollens', Fehling's; the iodoform reaction), reduction (to alcohols; to alkanes by Clemmensen, Wolff-Kishner or red P/HI) and base-catalysed reactions with mechanism: aldol, cross aldol and Cannizzaro.

Benzaldehyde is made from toluene with chromyl chloride (Étard reaction) and also shows benzoin condensation, Perkin's reaction, reaction with PCl₅ and meta-substitution of the ring.

Carboxylic acids are made by oxidation, hydrolysis of nitriles and Grignard reagents with CO₂. They are stronger acids than phenol because the carboxylate ion is stabilised by two equivalent resonance structures; substituents change acidity by inductive and resonance effects (dealt with in detail). They react with metals, alkalis, carbonates and hydrogencarbonates, form acid chlorides, anhydrides, esters and amides, lose CO₂ with soda lime or by Kolbe's electrolysis, undergo HVZ α-halogenation, and benzoic acid is substituted at the meta position. Tests distinguish formic, acetic and benzoic acids.

Weightage: Aldehydes, Ketones and Carboxylic Acids 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

Structure, nomenclature and preparation of aldehydes and ketones: The carbonyl carbon is sp², trigonal planar (bond angles about 120°), with a σ and a π bond to oxygen. IUPAC: aldehydes -al (ethanal, CH₃CHO), ketones -one (propanone, CH₃COCH₃; butan-2-one). Preparation: (1) From alcohols: oxidation of 1° alcohols ( $C H_{3} C H_{2} O H + [O] \to C H_{3} C H O + H_{2} O$ , acidified K₂Cr₂O₇; ethanal is distilled off at once because it would be oxidised further to ethanoic acid) and 2° alcohols (ketones); or dehydrogenation over Cu at 573 K. (2) Ozonolysis of alkenes: ozonide then Zn/H₂O: $C_{2} H_{4} O_{3} + Z n \to 2 H C H O + Z n O$ (the ozonide of ethene). (3) Hydration of alkynes (HgSO₄, dil. H₂SO₄, 333 K): $C_{2} H_{2} + H_{2} O \to C H_{3} C H O$ ; propyne gives propanone. (4) Rosenmund reduction: acid chloride + H₂ over Pd poisoned with BaSO₄ in boiling xylene: $C H_{3} C O C l + H_{2} \to C H_{3} C H O + H C l$ (the poison stops reduction to alcohol). (5) Dialkyl cadmium (ketones): $2 C H_{3} C O C l + (C H_{3})_{2} C d \to 2 C H_{3} C O C H_{3} + C d C l_{2}$ . (6) Dry distillation of calcium salts: $(C H_{3} C O O)_{2} C a \to C H_{3} C O C H_{3} + C a C O_{3}$ ; with calcium formate, an aldehyde: $(H C O O)_{2} C a \to H C H O + C a C O_{3}$ . (7) From nitriles: Stephen reaction: SnCl₂/HCl reduces RCN to the imine, hydrolysed to RCHO; with Grignard reagent a ketone forms: $C H_{3} C N + C H_{3} M g B r \to (C H_{3})_{2} C N M g B r$ , then $(C H_{3})_{2} C N M g B r + 2 H_{2} O \to C H_{3} C O C H_{3} + N H_{3} + M g (O H) B r$ . (8) From esters: DIBAL-H reduces esters to aldehydes: $C H_{3} C O O C_{2} H_{5} + 2 [H] \to C H_{3} C H O + C_{2} H_{5} O H$ . Physical properties: methanal is a gas; others liquids or solids. Boiling points lie between those of alkanes and alcohols of similar mass (dipole-dipole forces, no intermolecular H-bonds); lower members are water-soluble (H-bonds with water).; Worked example
Identify A, B, C, D: CH₃CH₂COOH + PCl₅ → A; A + H₂ (Pd/BaSO₄, boiling xylene) → B; B with dilute NaOH → C; C heated → D.
Try it first, then show the solutionSolution
A = CH₃CH₂COCl (propanoyl chloride); B = CH₃CH₂CHO (propanal, Rosenmund reduction).
Propanal has α-hydrogens, so dilute alkali gives the aldol C = CH₃CH₂CH(OH)CH(CH₃)CHO (3-hydroxy-2-methylpentanal).
Heating removes water: D = CH₃CH₂CH=C(CH₃)CHO (2-methylpent-2-enal).
Answer: A propanoyl chloride, B propanal, C 3-hydroxy-2-methylpentanal, D 2-methylpent-2-enal
Nucleophilic addition (mechanism) and reactivity: Mechanism: (1) a nucleophile attacks the electrophilic carbonyl carbon from above or below the plane; the π electrons move to oxygen, giving a tetrahedral alkoxide intermediate (sp³); (2) the alkoxide takes up H⁺ (from the solvent or reagent) to give the product. Reactivity: HCHO > CH₃CHO > CH₃COCH₃: alkyl groups (+I) reduce the positive charge on carbon and add steric crowding. Benzaldehyde is less reactive than propanal because the ring's resonance (+R) reduces the electrophilicity of the carbonyl carbon. Additions: HCN (base catalyst generates CN⁻): $C H_{3} C H O + H C N \to C H_{3} C H (O H) C N$ (cyanohydrin; hydrolysis gives 2-hydroxy acids). Sodium hydrogensulphite: $C H_{3} C H O + N a H S O_{3} \to C H_{3} C H (O H) S O_{3} N a$ , crystalline adduct used to separate and purify aldehydes and methyl ketones (regenerated by dilute acid or alkali). Grignard reagent: gives alcohols after hydrolysis. Alcohols (dry HCl): hemiacetals and acetals (ketals with ethylene glycol), used to protect C=O.; Worked example
Write the mechanism of addition of HCN to propanone.
Try it first, then show the solutionSolution
Base generates the nucleophile: $H C N + O H^{-} \to C N^{-} + H_{2} O$ .
CN⁻ attacks the carbonyl carbon; π electrons move to O, giving the tetrahedral alkoxide (CH₃)₂C(O⁻)CN (slow step).
The alkoxide is protonated by HCN (regenerating CN⁻): (CH₃)₂C(OH)CN, acetone cyanohydrin.
Answer: 2-hydroxy-2-methylpropanenitrile
Reactions with ammonia and its derivatives: Derivatives H₂N-Z add to C=O and then lose water (addition-elimination), giving >C=N-Z, in weakly acidic medium (pH about 3.5). Hydroxylamine → oxime: $C H_{3} C H O + N H_{2} O H \to C H_{3} C H N O H + H_{2} O$ (acetaldoxime, CH₃CH=NOH); propanone gives acetoxime, (CH₃)₂C=NOH. Hydrazine → hydrazone: $C H_{3} C H O + N H_{2} N H_{2} \to C H_{3} C H N N H_{2} + H_{2} O$ . Phenylhydrazine → phenylhydrazone: $C H_{3} C H O + C_{6} H_{5} N H N H_{2} \to C H_{3} C H N N H C_{6} H_{5} + H_{2} O$ . 2,4-Dinitrophenylhydrazine (Brady's reagent) gives yellow-orange 2,4-DNP derivatives: a general test for C=O. Semicarbazide → semicarbazone: $C H_{3} C O C H_{3} + N H_{2} N H C O N H_{2} \to (C H_{3})_{2} C N N H C O N H_{2} + H_{2} O$ . These crystalline derivatives have sharp melting points, used to identify carbonyl compounds. With ammonia: methanal forms urotropine (hexamethylenetetramine): $6 H C H O + 4 N H_{3} \to (C H_{2})_{6} N_{4} + 6 H_{2} O$ (used as a urinary antiseptic and to make the explosive RDX); ethanal gives aldehyde ammonia: $C H_{3} C H O + N H_{3} \to C H_{3} C H (O H) N H_{2}$ .; Worked example
Ethanal is treated with hydroxylamine. Give the product's name and structure and say why the reaction is done at pH about 4.
Try it first, then show the solutionSolution
$C H_{3} C H O + N H_{2} O H \to C H_{3} C H = N O H + H_{2} O$ : ethanal oxime (acetaldoxime).
Mild acid protonates the C=O oxygen, increasing electrophilicity; too much acid would protonate the -NH₂ of hydroxylamine and destroy its nucleophilicity.
Answer: Acetaldoxime, CH₃CH=NOH; pH balance between activating C=O and keeping a free nucleophile
Oxidation, iodoform reaction and reduction: Oxidation: aldehydes are oxidised easily (H on the carbonyl carbon) to acids, so they are reducing agents; ketones resist mild oxidants. Tollens' reagent (ammoniacal silver nitrate) gives a silver mirror: $C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 4 N H_{3} + 2 H_{2} O$ . Fehling's solution (alkaline Cu²⁺ tartrate) gives a red precipitate of Cu₂O: $C H_{3} C H O + 2 C u^{2 +} + 5 O H^{-} \to C H_{3} C O O^{-} + C u_{2} O + 3 H_{2} O$ . Aromatic aldehydes (benzaldehyde) do not reduce Fehling's solution. Schiff's reagent turns pink with aldehydes. Iodoform reaction: compounds with CH₃CO- (ethanal and methyl ketones) or CH₃CH(OH)- with I₂/NaOH give a yellow precipitate of CHI₃: $C H_{3} C O C H_{3} + 3 I_{2} + 4 N a O H \to C H I_{3} + C H_{3} C O O N a + 3 N a I + 3 H_{2} O$ . Methanal does not respond. Reduction to alcohols: H₂/Ni, NaBH₄ or LiAlH₄ (aldehydes → 1°, ketones → 2°). Reduction to alkanes (C=O → CH₂): Clemmensen (Zn-Hg/conc. HCl): $C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$ . Wolff-Kishner: the hydrazone is heated with KOH in ethylene glycol: $C H_{3} C H O + N H_{2} N H_{2} \to C H_{3} C H N N H_{2} + H_{2} O$ , then $C H_{3} C H N N H_{2} \to C_{2} H_{6} + N_{2}$ . Red phosphorus and HI (heat): $C H_{3} C H O + 4 H I \to C_{2} H_{6} + H_{2} O + 2 I_{2}$ .; Worked example
Give two methods of converting propanone into propane, with equations.
Try it first, then show the solutionSolution
Clemmensen reduction (Zn-Hg, conc. HCl): $C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$
Wolff-Kishner reduction: $C H_{3} C O C H_{3} + N H_{2} N H_{2} \to (C H_{3})_{2} C N N H_{2} + H_{2} O$ , then the hydrazone is heated with KOH in ethylene glycol: $(C H_{3})_{2} C N N H_{2} \to C H_{3} C H_{2} C H_{3} + N_{2}$
Answer: Clemmensen (acidic) or Wolff-Kishner (basic) reduction; the carbon count stays at three
Aldol and cross aldol condensation (mechanism): Aldehydes and ketones with at least one α-hydrogen react with dilute alkali. α-Hydrogens are acidic because the carbanion (enolate) formed is resonance-stabilised by the C=O group. Aldol: $2 C H_{3} C H O \to C H_{3} C H (O H) C H_{2} C H O$ (3-hydroxybutanal, 'aldol'); on heating it dehydrates to but-2-enal: $C H_{3} C H (O H) C H_{2} C H O \to C H_{3} C H = C H C H O + H_{2} O$ . Propanone gives diacetone alcohol, $(C H_{3})_{2} C (O H) C H_{2} C O C H_{3}$ , with Ba(OH)₂, and then 4-methylpent-3-en-2-one. Mechanism: (1) OH⁻ removes an α-H: $C H_{3} C H O + O H^{-} \to^{-} C H_{2} C H O + H_{2} O$ ; (2) the carbanion attacks the carbonyl carbon of a second molecule, forming an alkoxide; (3) the alkoxide takes H⁺ from water, regenerating OH⁻. Cross aldol: two different carbonyl compounds with α-H give four products (two self, two crossed); it is useful when one partner has no α-H. Claisen-Schmidt: benzaldehyde + ethanal (dil. NaOH) gives cinnamaldehyde: $C_{6} H_{5} C H O + C H_{3} C H O \to C_{6} H_{5} C H = C H C H O + H_{2} O$ .; Worked example
Propanone is shaken with Ba(OH)₂. Write the aldol product and the product formed on heating it. Why does 2,2-dimethylpropanal not undergo aldol condensation?
Try it first, then show the solutionSolution
$2 C H_{3} C O C H_{3} \to (C H_{3})_{2} C (O H) C H_{2} C O C H_{3}$ : 4-hydroxy-4-methylpentan-2-one (diacetone alcohol).
On heating it loses water: $(C H_{3})_{2} C (O H) C H_{2} C O C H_{3} \to (C H_{3})_{2} C = C H C O C H_{3} + H_{2} O$ (4-methylpent-3-en-2-one).
(CH₃)₃CCHO has no α-hydrogen, so no enolate can form; with concentrated alkali it gives the Cannizzaro reaction instead.
Answer: Diacetone alcohol, then 4-methylpent-3-en-2-one; 2,2-dimethylpropanal has no α-H
Cannizzaro reaction (mechanism) and benzaldehyde: Cannizzaro reaction: aldehydes with no α-hydrogen (HCHO, C₆H₅CHO) disproportionate with concentrated alkali: one molecule is oxidised to the acid salt, the other reduced to the alcohol. $2 H C H O + N a O H \to C H_{3} O H + H C O O N a$ ; $2 C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$ . Mechanism: (1) OH⁻ adds to C=O; (2) the tetrahedral anion transfers a hydride ion to the carbonyl carbon of a second molecule (slow step); (3) proton exchange gives alcohol and carboxylate. Crossed Cannizzaro: with HCHO and another aldehyde, HCHO is oxidised (to formate) and the other reduced. Benzaldehyde: prepared from toluene with chromyl chloride in CS₂, then hydrolysis (Étard reaction): overall $C_{6} H_{5} C H_{3} + 2 [O] \to C_{6} H_{5} C H O + H_{2} O$ . Colourless liquid smelling of bitter almonds; slowly oxidised in air to benzoic acid. Reactions: oxidation to benzoic acid, reduction to benzyl alcohol; HCN ( $C_{6} H_{5} C H O + H C N \to C_{6} H_{5} C H (O H) C N$ , mandelonitrile) and NaHSO₃; hydroxylamine ( $C_{6} H_{5} C H O + N H_{2} O H \to C_{6} H_{5} C H N O H + H_{2} O$ ), hydrazine and phenylhydrazine; PCl₅ → benzal chloride: $C_{6} H_{5} C H O + P C l_{5} \to C_{6} H_{5} C H C l_{2} + P O C l_{3}$ ; benzoin condensation with alcoholic KCN: $2 C_{6} H_{5} C H O \to C_{6} H_{5} C H (O H) C O C_{6} H_{5}$ ; Perkin's reaction with ethanoic anhydride and sodium ethanoate at about 453 K gives cinnamic acid: $C_{6} H_{5} C H O + (C H_{3} C O)_{2} O \to C_{6} H_{5} C H = C H C O O H + C H_{3} C O O H$ . Electrophilic substitution: -CHO is deactivating and meta-directing: nitration gives 3-nitrobenzaldehyde ( $C_{6} H_{5} C H O + H N O_{3} \to C_{6} H_{4} (N O_{2}) C H O + H_{2} O$ ), sulphonation gives 3-formylbenzenesulphonic acid, chlorination with FeCl₃ gives 3-chlorobenzaldehyde (without a catalyst Cl₂ gives benzoyl chloride: $C_{6} H_{5} C H O + C l_{2} \to C_{6} H_{5} C O C l + H C l$ ). Test: aromatic aldehydes give Tollens' test but not Fehling's test.; Worked example
A mixture of methanal and benzaldehyde is heated with concentrated NaOH. Predict the main products and explain.
Try it first, then show the solutionSolution
Neither aldehyde has an α-hydrogen, so a crossed Cannizzaro reaction occurs.
Methanal, the more reactive carbonyl, is attacked by OH⁻ and transfers hydride, so it is oxidised: $H C H O + C_{6} H_{5} C H O + N a O H \to H C O O N a + C_{6} H_{5} C H_{2} O H$ .
Answer: Sodium methanoate and benzyl alcohol (methanal oxidised, benzaldehyde reduced)
Carboxylic acids: preparation and acidity: Classification: monocarboxylic (HCOOH, CH₃COOH, C₆H₅COOH) and dicarboxylic (oxalic acid, HOOC-COOH; malonic; succinic). The -COOH carbon is sp²; the group is planar. Preparation: from primary alcohols and aldehydes by oxidation (KMnO₄, K₂Cr₂O₇); alkylbenzenes give benzoic acid ( $C_{6} H_{5} C H_{3} + 3 [O] \to C_{6} H_{5} C O O H + H_{2} O$ , alkaline KMnO₄ then acid); nitriles by acid or base hydrolysis: $C H_{3} C N + 2 H_{2} O \to C H_{3} C O O H + N H_{3}$ (via the amide); Grignard reagents with dry ice: $C H_{3} M g B r + C O_{2} \to C H_{3} C O O M g B r$ , then $C H_{3} C O O M g B r + H_{2} O \to C H_{3} C O O H + M g (O H) B r$ . Physical: higher boiling points than alcohols of similar mass because they form hydrogen-bonded dimers; lower acids are water-soluble. Acidity (in detail): RCOOH ⇌ RCOO⁻ + H⁺. The carboxylate ion is stabilised by two equivalent resonance structures (both C-O bonds equal, 127 pm), so carboxylic acids are stronger than phenols and alcohols. Electron-withdrawing groups (-NO₂, -CN, -F, -Cl) increase acidity by dispersing the negative charge; the effect is larger with more and more electronegative groups and closer to -COOH: F₃CCOOH > Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH; FCH₂COOH > ClCH₂COOH > BrCH₂COOH; 2-chlorobutanoic > 3-chlorobutanoic > 4-chlorobutanoic acid. Electron-releasing groups (alkyl) decrease acidity: HCOOH (pKa 3.75) > CH₃COOH (4.76) > CH₃CH₂COOH. Aromatic: benzoic acid (pKa 4.19) is stronger than ethanoic acid (sp² carbon is electron-withdrawing) but weaker than methanoic acid; -NO₂ at para increases acidity (4-nitrobenzoic acid pKa 3.41), -CH₃ or -OCH₃ at para decrease it.; Worked example
Arrange in increasing order of acidic strength and explain: CH₃COOH, ClCH₂COOH, HCOOH, C₆H₅COOH.
Try it first, then show the solutionSolution
CH₃ (+I) destabilises the anion: ethanoic acid is weakest.
Benzoic acid: sp² phenyl is mildly electron-withdrawing; methanoic acid has no electron-releasing group; Cl (−I) in chloroethanoic acid strongly stabilises the anion (pKa 2.86).
CH₃COOH < C₆H₅COOH < HCOOH < ClCH₂COOH
Answer: CH₃COOH < C₆H₅COOH < HCOOH < ClCH₂COOH
Reactions of carboxylic acids and tests: Acidic reactions: active metals: $2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}$ ; alkalis: $C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$ ; carbonates: $2 C H_{3} C O O H + N a_{2} C O_{3} \to 2 C H_{3} C O O N a + H_{2} O + C O_{2}$ ; hydrogencarbonates: $C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$ (brisk effervescence: test for -COOH; phenols do not respond). Acid derivatives: acid chloride: $C H_{3} C O O H + P C l_{5} \to C H_{3} C O C l + P O C l_{3} + H C l$ (or SOCl₂: $C H_{3} C O O H + S O C l_{2} \to C H_{3} C O C l + S O_{2} + H C l$ ); anhydride with P₂O₅ or heat: $2 C H_{3} C O O H \to (C H_{3} C O)_{2} O + H_{2} O$ ; ester (esterification); amide via the ammonium salt: $C H_{3} C O O H + N H_{3} \to C H_{3} C O O N H_{4}$ , then $C H_{3} C O O N H_{4} \to C H_{3} C O N H_{2} + H_{2} O$ on heating. Decarboxylation: sodium salt heated with soda lime (NaOH + CaO): $C H_{3} C O O N a + N a O H \to C H_{4} + N a_{2} C O_{3}$ ; $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$ . Kolbe's electrolysis: aqueous potassium ethanoate gives ethane at the anode: $2 C H_{3} C O O K + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}$ . HVZ (Hell-Volhard-Zelinsky) reaction: acids with α-H react with Cl₂ or Br₂ in the presence of red phosphorus: $C H_{3} C O O H + C l_{2} \to C l C H_{2} C O O H + H C l$ (further to di- and trichloroethanoic acid). Reduction: LiAlH₄ reduces -COOH to -CH₂OH. Ring substitution in benzoic acid: -COOH deactivates and directs meta: nitration gives 3-nitrobenzoic acid ( $C_{6} H_{5} C O O H + H N O_{3} \to C_{6} H_{4} (N O_{2}) C O O H + H_{2} O$ ); sulphonation gives 3-sulphobenzoic acid. Benzoic acid does not undergo Friedel-Crafts reaction (ring deactivated and AlCl₃ binds the -COOH). Tests: *methanoic acid* has an aldehyde-like H: it reduces Tollens' reagent (silver mirror) and Fehling's solution, and decolourises acidified KMnO₄; *ethanoic acid* with neutral FeCl₃ gives a deep red colour (brown-red precipitate on boiling) and with ethanol/conc. H₂SO₄ a fruity smell; *benzoic acid* with neutral FeCl₃ gives a buff (pale brown) precipitate. All give effervescence with NaHCO₃. Uses: methanoic acid in leather tanning and coagulating rubber latex; ethanoic acid as vinegar (food), solvent, for cellulose acetate and aspirin; benzoic acid and sodium benzoate as food preservatives, in medicines and dyes.; Worked example
Give a chemical test to distinguish ethanoic acid from ethyl ethanoate.
Try it first, then show the solutionSolution
Add NaHCO₃ solution: ethanoic acid gives brisk effervescence of CO₂ ( $C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$ ).
Ethyl ethanoate has no acidic -COOH hydrogen and gives no effervescence.
Answer: NaHCO₃: effervescence with the acid only

Chemistry in real life: Aldehydes, Ketones and Carboxylic Acids

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.

Acidity of vinegar

Suppose a cook uses vinegar that is 0.10 M ethanoic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ).

Estimate its degree of ionisation and pH.

Try it first, then show the solutionSolution

$α = K_{a} / C = 1.8 \times 1 0^{- 5} /0.10 = 0.0134$
$[H^{+}] = C α = 1.34 \times 1 0^{- 3}$ M; pH = 2.87

Answer: About 1.3% ionised; pH about 2.9.

Memory hook: Weak acid: only about one molecule in seventy-five gives up its proton.

Formalin and urotropine

Imagine a biology museum keeping specimens in 40% aqueous methanal (formalin).

Why does methanal preserve tissue, and what happens if ammonia leaks into a jar?

Try it first, then show the solutionSolution

Methanal reacts with -NH₂ groups of proteins (nucleophilic addition), cross-linking and hardening them so microbes cannot digest them.
With ammonia it forms urotropine: $6 H C H O + 4 N H_{3} \to (C H_{2})_{6} N_{4} + 6 H_{2} O$ , using up the preservative.

Answer: It cross-links proteins; ammonia would consume it as urotropine.

Memory hook: Formalin fixes by adding to amines.

Sodium benzoate preservative

Suppose a juice label lists sodium benzoate (E211), and a student wonders how it is made from benzoic acid and whether heating the salt with soda lime would be dangerous.

Write the equations for making the salt and for heating it with soda lime.

Try it first, then show the solutionSolution

$C_{6} H_{5} C O O H + N a H C O_{3} \to C_{6} H_{5} C O O N a + H_{2} O + C O_{2}$
With soda lime and strong heat: $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$ , giving benzene, a toxic compound: so this is never done to food.

Answer: Neutralisation gives the preservative; soda lime would decarboxylate it to benzene.

Memory hook: Soda lime strips CO₂ off a carboxylate.

Topics in this chapter

Aldehydes, Ketones and Carboxylic Acids formula sheet

Rosenmund reduction: $C H_{3} C O C l + H_{2} \to C H_{3} C H O + H C l$; Pd/BaSO₄, boiling xylene.
Dialkyl cadmium: $2 C H_{3} C O C l + (C H_{3})_{2} C d \to 2 C H_{3} C O C H_{3} + C d C l_{2}$; Ketones only.
Calcium salt distillation: $(C H_{3} C O O)_{2} C a \to C H_{3} C O C H_{3} + C a C O_{3}$; With calcium formate: aldehydes.
Hydration of ethyne: $C_{2} H_{2} + H_{2} O \to C H_{3} C H O$; HgSO₄, dil. H₂SO₄, 333 K.
Nitrile + Grignard: $C H_{3} C N + C H_{3} M g B r \to (C H_{3})_{2} C N M g B r$; Hydrolysis gives propanone.
Cyanohydrin: $C H_{3} C H O + H C N \to C H_{3} C H (O H) C N$; Base catalysed nucleophilic addition.
Bisulphite adduct: $C H_{3} C H O + N a H S O_{3} \to C H_{3} C H (O H) S O_{3} N a$; Purification of aldehydes.
Oxime: $C H_{3} C H O + N H_{2} O H \to C H_{3} C H N O H + H_{2} O$; CH₃CH=NOH.
Urotropine: $6 H C H O + 4 N H_{3} \to (C H_{2})_{6} N_{4} + 6 H_{2} O$; Hexamethylenetetramine.
Tollens' test: $C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 4 N H_{3} + 2 H_{2} O$; Silver mirror.
Fehling's test: $C H_{3} C H O + 2 C u^{2 +} + 5 O H^{-} \to C H_{3} C O O^{-} + C u_{2} O + 3 H_{2} O$; Red precipitate; not given by benzaldehyde.
Iodoform reaction: $C H_{3} C O C H_{3} + 3 I_{2} + 4 N a O H \to C H I_{3} + C H_{3} C O O N a + 3 N a I + 3 H_{2} O$; Yellow precipitate.
Clemmensen reduction: $C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$; Zn-Hg, conc. HCl.
Red P + HI reduction: $C H_{3} C H O + 4 H I \to C_{2} H_{6} + H_{2} O + 2 I_{2}$; Heat.
Aldol condensation: $2 C H_{3} C H O \to C H_{3} C H (O H) C H_{2} C H O$; Dil. NaOH; heat gives but-2-enal.
Cannizzaro reaction: $2 C_{6} H_{5} C H O + N a O H \to C_{6} H_{5} C H_{2} O H + C_{6} H_{5} C O O N a$; Conc. alkali; no α-H.
Benzoin condensation: $2 C_{6} H_{5} C H O \to C_{6} H_{5} C H (O H) C O C_{6} H_{5}$; Alcoholic KCN.
Benzaldehyde + PCl₅: $C_{6} H_{5} C H O + P C l_{5} \to C_{6} H_{5} C H C l_{2} + P O C l_{3}$; Benzal chloride.
Grignard + CO₂: $C H_{3} C O O M g B r + H_{2} O \to C H_{3} C O O H + M g (O H) B r$; From $C H_{3} M g B r + C O_{2}$ .
Hydrolysis of nitrile: $C H_{3} C N + 2 H_{2} O \to C H_{3} C O O H + N H_{3}$; Acid or alkali, heat.
Hydrogencarbonate test: $C H_{3} C O O H + N a H C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}$; Effervescence.
Soda-lime decarboxylation: $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$; Heat with soda lime.
Kolbe's electrolysis: $2 C H_{3} C O O K + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}$; Ethane and CO₂ at anode, H₂ at cathode.
HVZ reaction: $C H_{3} C O O H + C l_{2} \to C l C H_{2} C O O H + H C l$; Red phosphorus.
Amide from acid: $C H_{3} C O O N H_{4} \to C H_{3} C O N H_{2} + H_{2} O$; Heat the ammonium salt.

Aldehydes, Ketones and Carboxylic Acids: diagrams to practise

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

Mechanism of nucleophilic addition to C=O

'Explain nucleophilic addition with mechanism' (HCN, NaHSO₃, Grignard).

Draw

Planar carbonyl with δ+ on C and δ− on O; Nu⁻ approaching C with a curved arrow; a second curved arrow from the C=O π bond onto O.
Tetrahedral alkoxide intermediate with O⁻.
Curved arrow from O⁻ to H⁺, giving the product.

Label

δ+
δ−
Nu⁻
tetrahedral intermediate (sp³)
protonation

Watch out: Write 'slow' over step 1 and state the reactivity order HCHO > RCHO > R₂CO.

Mechanism of aldol condensation

'Write the mechanism of aldol condensation of ethanal'.

Draw

OH⁻ removing an α-H from CH₃CHO, giving the resonance-stabilised carbanion ⁻CH₂CHO ↔ CH₂=CHO⁻.
Carbanion attacking the carbonyl C of a second CH₃CHO, giving CH₃CH(O⁻)CH₂CHO.
Alkoxide taking H⁺ from water: CH₃CH(OH)CH₂CHO + OH⁻.

Label

α-hydrogen
enolate (carbanion)
nucleophilic attack
aldol
OH⁻ regenerated

Watch out: Add the heating step (−H₂O) if the question asks for the condensation product.

Mechanism of Cannizzaro reaction

'Explain Cannizzaro reaction with mechanism'.

Draw

OH⁻ adding to the carbonyl carbon of HCHO (or C₆H₅CHO) to give a tetrahedral anion.
Curved arrow showing hydride (H⁻) transfer from this anion to the carbonyl carbon of a second molecule.
Proton exchange giving HCOO⁻ and CH₃OH.

Label

OH⁻ addition
hydride transfer (slow)
carboxylate
alcohol

Watch out: State 'concentrated alkali' and 'no α-hydrogen' as the conditions.

How the board asks Aldehydes, Ketones and Carboxylic Acids

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

Question type	How it is asked	Method
Identify A, B, C, D	Sequences with PCl₅, Rosenmund, oxidation, Ca(OH)₂; Cannizzaro; molecular formulas C₇H₆O₂, C₂H₄O.	Use molecular formulas, tests (Tollens', iodoform, NaHCO₃) and each reagent's role; give names and structures.
Conversions	Benzaldehyde → benzene, benzoic acid → benzaldehyde, ethanol → propanoic acid.	Reagents and conditions above arrows; balanced equations.
Named reactions	Rosenmund, Stephen, Clemmensen, Wolff-Kishner, aldol, Cannizzaro, benzoin, Perkin, HVZ, Kolbe electrolysis.	One example equation each with conditions.
Mechanisms	Nucleophilic addition, aldol, Cannizzaro.	Curved arrows, intermediates, slow step.
Distinguishing tests	Aldehyde/ketone, HCHO/CH₃CHO, aromatic/aliphatic aldehyde, formic/acetic/benzoic acid, phenol/benzoic acid, ethanol/propanal.	Reagent, positive observation, negative for the other.
Acidity reasoning	Order of acids with substituents; why RCOOH > phenol; effect of position of Cl.	Resonance of carboxylate; −I/+I effects; distance from -COOH.

Tricks to remember Aldehydes, Ketones and Carboxylic Acids

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

No α-H? Cannizzaro. α-H? Aldol.
Concentrated alkali + no α-H = Cannizzaro; dilute alkali + α-H = aldol.
Use it when: Choosing the base-catalysed reaction.
Clemmensen in acid, Wolff-Kishner in base
C for Clemmensen and Conc. HCl; W-K uses hydrazine and KOH.
Use it when: Reduction to alkanes.
Rosenmund is poisoned
Pd poisoned by BaSO₄ stops at the aldehyde.
Use it when: Acid chloride to aldehyde.
BAP: Benzoin, Alkaline KCN; Perkin, anhydride
Benzoin needs alcoholic KCN; Perkin needs (CH₃CO)₂O + CH₃COONa.
Use it when: Benzaldehyde reactions.
Fehling fails for phenyl
Benzaldehyde gives Tollens' but not Fehling's.
Use it when: Aromatic vs aliphatic aldehyde test.

Aldehydes, Ketones and Carboxylic Acids: board-level solved questions

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

Example 1: Named reductions · 1 mark · Very short answerBoard level

Fill in the blanks: Propanone is reduced to propane by zinc amalgam and concentrated HCl in the ______ reduction, while a nitrile reduced by SnCl₂/HCl and then hydrolysed gives an aldehyde in the ______ reaction.

Try it first, then show the solutionSolution

Clemmensen: C=O → CH₂ in acid. Stephen: RCN → imine (RCH=NH) → RCHO on hydrolysis.

Answer: Clemmensen; Stephen

Example 2: Reagent for Rosenmund reduction · 1 mark · MCQBoard level

Which converts benzoyl chloride to benzaldehyde?
(a) LiAlH₄
(b) H₂, Pd/BaSO₄
(c) Zn-Hg/HCl
(d) NaBH₄

Try it first, then show the solutionSolution

Poisoned catalyst stops at the aldehyde; the others reduce further.

Answer: (b)

Example 3: Reactivity and solubility · 2 marks · Short answerBoard level

Give reasons: (a) Propanone is less reactive than ethanal towards HCN. (b) Ethanal mixes with water in all proportions but benzaldehyde is only slightly soluble. [2]

Try it first, then show the solutionSolution

(a) Propanone has two electron-releasing methyl groups (+I), which reduce the positive charge on the carbonyl carbon and crowd the approaching CN⁻; ethanal has only one.
(b) Both can accept hydrogen bonds from water at the carbonyl oxygen, but benzaldehyde's large hydrophobic phenyl group outweighs this.

Answer: (a) +I effect and steric hindrance of two CH₃ groups (b) large hydrophobic phenyl group

Example 4: Two conversions · 2 marks · Short answerHarder board level

Convert: (a) ethanal to but-2-enal; (b) ethanoic acid to methane. [2]

Try it first, then show the solutionSolution

(a) Aldol with dilute NaOH: $2 C H_{3} C H O \to C H_{3} C H (O H) C H_{2} C H O$ ; heat: $C H_{3} C H (O H) C H_{2} C H O \to C H_{3} C H = C H C H O + H_{2} O$ .
(b) $C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$ ; heat with soda lime: $C H_{3} C O O N a + N a O H \to C H_{4} + N a_{2} C O_{3}$ .

Answer: Aldol then dehydration; sodium salt then soda-lime decarboxylation

Example 5: Identify from a nitrile · 3 marks · Short answerHarder board level

A compound A (C₇H₅N) is hydrolysed by dilute acid to B (C₇H₆O₂), which gives effervescence with NaHCO₃. The sodium salt of B heated with soda lime gives C (C₆H₆). B with a mixture of conc. HNO₃ and conc. H₂SO₄ gives D. Identify A to D. [3]

Try it first, then show the solutionSolution

A = benzonitrile, C₆H₅CN: $C_{6} H_{5} C N + 2 H_{2} O H^{+} C_{6} H_{5} C O O H + N H_{3}$ (as NH₄⁺), so B = benzoic acid.
Decarboxylation: $C_{6} H_{5} C O O N a + N a O H \to C_{6} H_{6} + N a_{2} C O_{3}$ , so C = benzene.
-COOH is meta-directing: D = 3-nitrobenzoic acid, $C_{6} H_{5} C O O H + H N O_{3} \to C_{6} H_{4} (N O_{2}) C O O H + H_{2} O$ .

Answer: A benzonitrile, B benzoic acid, C benzene, D 3-nitrobenzoic acid

Example 6: Base-catalysed reactions · 5 marks · Long answerHarder board level

(a) Explain the mechanism of aldol condensation of ethanal. [3]
(b) What happens when methanal is heated with concentrated NaOH? Give the equation and name the reaction. [1]
(c) Why does methanal not undergo aldol condensation? [1]

Try it first, then show the solutionSolution

(a) OH⁻ removes an α-H: $C H_{3} C H O + O H^{-} \to^{-} C H_{2} C H O + H_{2} O$ (enolate, resonance-stabilised).
The carbanion attacks the carbonyl carbon of another CH₃CHO, giving CH₃CH(O⁻)CH₂CHO.
The alkoxide is protonated by water: CH₃CH(OH)CH₂CHO (aldol), regenerating OH⁻; on heating, but-2-enal forms.
(b) Cannizzaro: $2 H C H O + N a O H \to C H_{3} O H + H C O O N a$ .
(c) Methanal has no α-hydrogen, so no enolate can form.

Answer: Aldol via enolate; Cannizzaro gives methanol and sodium methanoate

Example 7: Acidity and substituent position · 2 marks · Short answerHOTS / competency

Arrange in decreasing order of acidity and justify: 2-chlorobutanoic acid, 3-chlorobutanoic acid, butanoic acid, 2,2-dichlorobutanoic acid. [2]

Try it first, then show the solutionSolution

−I effect of Cl stabilises the carboxylate; more Cl atoms and closer to -COOH means a larger effect; the effect fades with distance.
2,2-dichlorobutanoic > 2-chlorobutanoic > 3-chlorobutanoic > butanoic acid.

Answer: 2,2-dichloro > 2-chloro > 3-chloro > butanoic

Example 8: Methyl ketone identification · 3 marks · Short answerHOTS / competency

A compound A (C₈H₈O) gives an orange precipitate with 2,4-DNP and a yellow precipitate with I₂/NaOH, but does not reduce Tollens' reagent. On vigorous oxidation it gives benzoic acid. Identify A and write the iodoform reaction. [3]

Try it first, then show the solutionSolution

2,4-DNP positive: carbonyl; no Tollens': ketone; iodoform positive: CH₃CO- group; oxidation to benzoic acid: one side chain on benzene.
A = acetophenone, C₆H₅COCH₃ (C₈H₈O).
$C_{6} H_{5} C O C H_{3} + 3 I_{2} + 4 N a O H \to C H I_{3} + C_{6} H_{5} C O O N a + 3 N a I + 3 H_{2} O$

Answer: Acetophenone (1-phenylethanone)

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.

Unknown in the practical exam

Suppose a student receives three unlabelled liquids: ethanal, propanone and benzaldehyde. She has Tollens' reagent, Fehling's solution and iodine with NaOH.

(i) · 1 mark
Which reagent identifies propanone among the three by giving no reaction?
Answer
Answer: Tollens' reagent.
Both aldehydes give a silver mirror; the ketone does not.
(ii) · 1 mark
How will she distinguish ethanal from benzaldehyde?
Answer
Answer: Fehling's solution: red precipitate with ethanal only.
Aromatic aldehydes do not reduce Fehling's solution (iodoform also works: ethanal positive).
(iii) · 1 mark
Which of the three give a yellow precipitate with I₂/NaOH?
Answer
Answer: Ethanal and propanone.
Both contain the CH₃CO- group.
(iv) · 2 marks
Write the equation for the Tollens' test with ethanal.
Answer
Answer: Silver mirror.
$C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 4 N H_{3} + 2 H_{2} O$

From toluene to cinnamic acid

Imagine a perfume company making benzaldehyde from toluene and then converting it to cinnamic acid and benzoin.

(i) · 1 mark
Name the reagent that converts toluene to benzaldehyde and the reaction.
Answer
Answer: Chromyl chloride (CrO₂Cl₂) in CS₂, then hydrolysis; Étard reaction.
The methyl group is oxidised only as far as -CHO.
(ii) · 1 mark
What reagents convert benzaldehyde into cinnamic acid? Name the reaction.
Answer
Answer: Ethanoic anhydride and sodium ethanoate, about 453 K; Perkin's reaction.
$C_{6} H_{5} C H O + (C H_{3} C O)_{2} O \to C_{6} H_{5} C H = C H C O O H + C H_{3} C O O H$
(iii) · 1 mark
Write the benzoin condensation.
Answer
Answer: $2 C_{6} H_{5} C H O \to C_{6} H_{5} C H (O H) C O C_{6} H_{5}$
Alcoholic KCN catalyst.
(iv) · 2 marks
Why does nitration of benzaldehyde give mainly the meta product?
Answer
Answer: -CHO withdraws electrons by −I and −R effects.
Ortho and para positions become more electron-poor than meta, so NO₂⁺ attacks meta: 3-nitrobenzaldehyde.

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.

Write equations for: (a) Rosenmund reduction, (b) Clemmensen reduction. (2 marks)

Step	Marks
$C H_{3} C O C l + H_{2} \to C H_{3} C H O + H C l$ (Pd/BaSO₄, xylene)	1
$C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$ (Zn-Hg, conc. HCl)	1

An organic compound C₃H₆O does not reduce Tollens' reagent but gives a yellow precipitate with I₂/NaOH. (i) Identify it with its IUPAC name. (ii) Write its product with semicarbazide. (iii) Write its product with HCN. (3 marks)

Step	Marks
(i) CH₃COCH₃, propanone (a methyl ketone)	1
(ii) $(C H_{3})_{2} C = N N H C O N H_{2}$ , propanone semicarbazone, with H₂O	1
(iii) $(C H_{3})_{2} C (O H) C N$ , 2-hydroxy-2-methylpropanenitrile (acetone cyanohydrin)	1

(a) Explain why carboxylic acids are more acidic than phenols. (b) Write equations for HVZ reaction and Kolbe's electrolysis. (c) How is formic acid distinguished from acetic acid? (5 marks)

Step	Marks
Carboxylate has two equivalent resonance structures with charge on two O atoms	1
Phenoxide charge partly on less electronegative C atoms: less stabilised	1
HVZ: $C H_{3} C O O H + C l_{2} \to C l C H_{2} C O O H + H C l$ (red P)	1
Kolbe: $2 C H_{3} C O O K + 2 H_{2} O \to C_{2} H_{6} + 2 C O_{2} + 2 K O H + H_{2}$	1
Formic acid gives silver mirror with Tollens' (acetic does not)	1

Where students lose marks in Aldehydes, Ketones and Carboxylic Acids

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

Mistake: Vague reasons in 'why' questions
CISCE examiners' comments, 2024: for benzaldehyde being less reactive than propionaldehyde, candidates missed key terms such as resonance or electron-releasing effect; for removing ethanal as it forms, they said it is poisonous or explosive instead of saying it would be oxidised further to ethanoic acid.
Wrong working
Ethanal is removed because it is poisonous.
Right working
Ethanal is removed to prevent its further oxidation to ethanoic acid.
Do this: Use the exact chemical reason, not a safety guess.
Usually costs: 1 mark.
Mistake: Wrong final product in sequences
CISCE examiners' comments, 2024 and 2025: candidates wrote C₂H₅COONH₄ instead of the amide C₂H₅CONH₂, C₆H₅CHO instead of C₆H₅COOH after dichromate oxidation, C₆H₅COOH instead of sodium benzoate C₆H₅COONa in a Cannizzaro problem, calcium acetate instead of acetone after heating with Ca(OH)₂, ignored the molecular formula C₇H₆O₂ and wrote acetic acid, and failed to identify acetone oxime.
Wrong working
D = (CH₃COO)₂Ca
Right working
Dry distillation of calcium ethanoate gives D = CH₃COCH₃ (+ CaCO₃).
Do this: Follow each reagent to its final product (heat, hydrolysis, soda lime) and check the molecular formula.
Usually costs: 1 mark per compound.
Mistake: Wrong observation or reagent in tests
CISCE examiners' comments, 2024 and 2025: candidates gave a white precipitate instead of a silver mirror for Tollens' test, gave common names or formulas when the IUPAC name was asked, used the Lucas test for ethanol vs propanal, gave the FeCl₃ observation for phenol but not for benzoic acid, and omitted soda lime for decarboxylation or used wrong reagents for benzoic acid to benzaldehyde.
Wrong working
Tollens' reagent gives a white precipitate.
Right working
Tollens' reagent gives a silver mirror (or grey-black precipitate of Ag).
Do this: Learn reagent → observation pairs and write both compounds' results.
Usually costs: 1 mark each.
Mistake: Mixing up named reductions
CISCE examiners' comments, 2025: for ethanal with hydrazine and KOH in glycol, candidates wrote propane and Stephen instead of ethane and Wolff-Kishner.
Wrong working
Propane; Stephen reduction.
Right working
Ethane (same carbon count, C=O → CH₂); Wolff-Kishner reduction.
Do this: Reduction to alkane never changes the number of carbons.
Usually costs: 1 mark.

Practice questions on Aldehydes, Ketones and Carboxylic Acids

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 gives the Cannizzaro reaction?
(a) CH₃CHO
(b) CH₃COCH₃
(c) C₆H₅CHO
(d) CH₃CH₂CHO
Hint
No α-H.
Answer
(c)
Q2 · 1 mark · MCQBoard level
The reagent for converting an acid into its α-chloro acid is:
(a) Cl₂/FeCl₃
(b) Cl₂/red P
(c) PCl₅
(d) SOCl₂
Hint
HVZ.
Answer
(b)
Q3 · 1 mark · Assertion-reasonBoard level
Assertion: Benzoic acid does not undergo Friedel-Crafts alkylation.
Reason: The -COOH group deactivates the ring and the catalyst AlCl₃ bonds to the carboxyl group.
(a) Both true, Reason explains Assertion
(b) Both true, Reason does not explain
(c) Assertion true, Reason false
(d) Both false
Hint
Two effects together.
Answer
(a)
Q4 · 1 mark · Very short answerBoard level
Fill in the blanks: Benzaldehyde with alcoholic KCN gives ______; with ethanoic anhydride and sodium ethanoate it gives ______.
Hint
Two named reactions.
Answer
benzoin; cinnamic acid
Q5 · 1 mark · Very short answerBoard level
Give a test to distinguish methanal from ethanal.
Hint
CH₃CO-.
Answer
Iodoform test: ethanal gives a yellow precipitate of CHI₃ with I₂/NaOH; methanal does not.
Q6 · 2 marks · Short answerHarder board level
How is propanone obtained from (a) propyne, (b) ethanoyl chloride?
Hint
Hydration; dialkyl cadmium.
Answer
(a) $C_{3} H_{4} + H_{2} O \to C H_{3} C O C H_{3}$ (HgSO₄, dil. H₂SO₄). (b) $2 C H_{3} C O C l + (C H_{3})_{2} C d \to 2 C H_{3} C O C H_{3} + C d C l_{2}$ .
Q7 · 3 marks · Short answerHarder board level
Write the mechanism of the Cannizzaro reaction of benzaldehyde.
Hint
Hydride transfer.
Answer
(1) OH⁻ adds to C=O of C₆H₅CHO giving C₆H₅CH(OH)O⁻. (2) This anion transfers H⁻ to the carbonyl carbon of a second benzaldehyde, forming C₆H₅COOH and C₆H₅CH₂O⁻ (slow). (3) Proton exchange gives C₆H₅COO⁻ and C₆H₅CH₂OH.
Q8 · 2 marks · Short answerHarder board level
Convert ethanol into propanoic acid.
Hint
Add a carbon via nitrile.
Answer
$C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$ ; $C_{2} H_{5} C l + K C N \to C_{2} H_{5} C N + K C l$ ; $C_{2} H_{5} C N + 2 H_{2} O \to C_{2} H_{5} C O O H + N H_{3}$ (acid hydrolysis).
Q9 · 2 marks · Short answerHarder board level
Arrange in increasing order of boiling point: propanal, propan-1-ol, butane, propanoic acid. Explain.
Hint
Dimers.
Answer
Butane < propanal < propan-1-ol < propanoic acid: dispersion only; dipole-dipole; hydrogen bonding; hydrogen-bonded dimers (strongest).
Q10 · 2 marks · Short answerHarder board level
An aldehyde A (C₃H₆O) undergoes aldol condensation; a ketone B with the same formula does not reduce Tollens' reagent. Name both and write the aldol product of A.
Hint
Isomers of C₃H₆O.
Answer
A propanal, B propanone. $2 C H_{3} C H_{2} C H O \to C H_{3} C H_{2} C H (O H) C H (C H_{3}) C H O$ (3-hydroxy-2-methylpentanal).
Q11 · 5 marks · Long answerHarder board level
(a) Write the mechanism of nucleophilic addition of NaHSO₃ to ethanal and state its use. [2]
(b) How will you distinguish formic acid, acetic acid and benzoic acid? [2]
(c) Why is HCOOH a stronger acid than CH₃COOH? [1]
Hint
+I of CH₃.
Answer
(a) The S of HSO₃⁻ attacks the carbonyl carbon forming an alkoxide; proton transfer gives CH₃CH(OH)SO₃Na, a crystalline adduct; used to separate and purify aldehydes. (b) Formic acid: silver mirror with Tollens'. Acetic acid: neutral FeCl₃ gives red colour (fruity ester with ethanol/H₂SO₄). Benzoic acid: neutral FeCl₃ gives buff precipitate. (c) The CH₃ group's +I effect destabilises ethanoate; methanoate has no such group.
Q12 · 2 marks · Short answerHOTS / competency
Why does the reaction of a carbonyl compound with 2,4-DNP need a weakly acidic medium? What would happen at very low pH?
Hint
Nucleophile protonation.
Answer
Mild acid protonates the carbonyl oxygen, making carbon more electrophilic. At very low pH the -NH₂ of the reagent is protonated (-NH₃⁺) and loses its lone pair, so it cannot act as a nucleophile; the reaction stops.
Q13 · 2 marks · Short answerHOTS / competency
Explain why 4-nitrobenzoic acid is stronger, and 4-methoxybenzoic acid weaker, than benzoic acid.
Hint
−R vs +R.
Answer
-NO₂ withdraws electrons by −I and −R, dispersing the negative charge of the carboxylate (pKa 3.41 < 4.19). -OCH₃ donates electrons by +R, intensifying the negative charge and destabilising the anion (pKa about 4.5).

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.Catalyst for Rosenmund reduction?Show answer
Pd on BaSO₄ (poisoned), in boiling xylene.
2.Which aldehydes give Cannizzaro reaction?Show answer
Those without α-hydrogen, e.g. HCHO, C₆H₅CHO.
3.Product of aldol condensation of ethanal?Show answer
3-Hydroxybutanal (aldol); on heating, but-2-enal.
4.Reagent for Clemmensen reduction?Show answer
Zinc amalgam and concentrated HCl.
5.What is Tollens' reagent?Show answer
Ammoniacal silver nitrate, [Ag(NH₃)₂]⁺.
6.How is benzaldehyde made from toluene?Show answer
Chromyl chloride (Étard reaction), then hydrolysis.
7.What is the HVZ reaction?Show answer
α-Halogenation of carboxylic acids with Cl₂ or Br₂ and red phosphorus.
8.Test for carboxylic acids?Show answer
Brisk effervescence of CO₂ with NaHCO₃.

Best order to study it

Learn structure, naming and all preparations of aldehydes and ketones.
Master nucleophilic addition with mechanism and ammonia-derivative reactions.
Learn oxidation tests, iodoform, and the three reductions to alkanes.
Learn aldol, cross aldol and Cannizzaro with mechanisms.
Learn benzaldehyde: preparation, benzoin, Perkin, PCl₅, meta substitution, tests.
Learn carboxylic acids: preparation, acidity with substituents, reactions, tests; practise identification.

Quick revision checklist

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

Can you prepare an aldehyde by Rosenmund, Stephen and from calcium salts?
Can you write the nucleophilic addition mechanism?
Can you give products with NH₂OH, NH₂NH₂, C₆H₅NHNH₂ and NH₃?
Can you distinguish aldehydes and ketones, and aliphatic and aromatic aldehydes?
Can you write Clemmensen, Wolff-Kishner and red P/HI reductions?
Can you write aldol and Cannizzaro mechanisms?
Can you order acids by strength with reasons?
Can you write decarboxylation, Kolbe electrolysis and HVZ?

Aldehydes, Ketones and Carboxylic Acids: frequently asked questions

Which mechanisms are required in this ISC chapter?

Nucleophilic addition (with ammonia derivatives, HCN, NaHSO₃ and Grignard reagent), and the base-catalysed aldol condensation, cross aldol condensation and Cannizzaro reaction.

Which benzaldehyde reactions are listed?

Oxidation and reduction, addition of HCN and NaHSO₃, reactions with hydroxylamine, hydrazine and phenylhydrazine, PCl₅, Cannizzaro, benzoin condensation, Perkin's reaction and electrophilic substitution (halogenation, nitration, sulphonation).

How much detail is needed on acidity of carboxylic acids?

The syllabus says the effect of substituents on aliphatic and aromatic acids is to be dealt with in detail: inductive and resonance effects, number and position of substituents.

What tests for acids are needed?

Tests for formic acid (Tollens', Fehling's), acetic acid (neutral FeCl₃, ester smell) and benzoic acid (neutral FeCl₃ buff precipitate), plus NaHCO₃ for all.

Which decarboxylation methods should I know?

Chemical decarboxylation by heating the sodium salt with soda lime, and Kolbe's electrolytic method.

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