Aldehydes, Ketones and Carboxylic Acids Class 12: Named Reactions and Board Answers

What Aldehydes, Ketones and Carboxylic Acids is about

Weightage: This chapter is Unit VIII, Aldehydes, Ketones and Carboxylic Acids, which carries 8 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit VIII Aldehydes, Ketones and Carboxylic Acids

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.9 marks per paper. See the questions

Key concepts

Naming, structure and physical properties of carbonyl compounds

NCERT section 8.1 Nomenclature and Structure of Carbonyl Group; 8.3 Physical Properties

IUPAC names: aldehydes end in -al (CH₃CHO is ethanal); ketones end in -one with a locant (CH₃COCH₂CH₃ is butan-2-one); acids end in -oic acid (CH₃COOH is ethanoic acid). Common names: formaldehyde (methanal), acetaldehyde (ethanal), acetone (propanone), benzaldehyde, acetophenone (1-phenylethanone), benzophenone. When the –CHO is attached to a ring, the suffix is -carbaldehyde.

Structure: the carbonyl carbon is sp² hybridised and forms three σ bonds in a plane (about 120°) and one π bond with oxygen. The C=O is strongly polar; it can be shown as a resonance hybrid of a neutral structure and a dipolar one ($C^{+}$–$O^{-}$). This polarity makes the carbon an electrophilic (Lewis acid) centre and the oxygen a nucleophilic (Lewis base) centre.

Physical properties:

• Methanal is a gas; ethanal is a volatile liquid; other aldehydes and ketones are liquids or solids.
• Boiling points are higher than those of hydrocarbons and ethers of similar mass, because of dipole–dipole interactions, but lower than those of alcohols, because they cannot hydrogen-bond with each other.
• Lower members are miscible with water because they form hydrogen bonds with water molecules; solubility falls as the alkyl chain grows.
• Lower aldehydes have sharp, pungent smells; larger ones and ketones often smell pleasant (used in perfumes and flavourings).

Worked example

Arrange in increasing order of boiling point and explain: CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, CH₃CH₂OCH₂CH₃, CH₃CH₂CH₂CH₃. (Molar masses are close: 72, 74, 74, 58.) (2 marks)

Try it first, then show the solutionSolution

1. Butane has only weak van der Waals forces; ethoxyethane has weak dipole–dipole forces; butanal has stronger dipole–dipole forces (polar C=O); butan-1-ol has hydrogen bonding. (1)
2. Order: n-butane < ethoxyethane < butanal < butan-1-ol. (1)

Answer: Butane < ethoxyethane < butanal < butan-1-ol

Preparation of aldehydes and ketones

NCERT section 8.2 Preparation of Aldehydes and Ketones

General methods:

• Oxidation of alcohols: 1° alcohols give aldehydes, 2° alcohols give ketones (Chapter 7); dehydrogenation over heated Cu or Ag does the same and is useful for volatile alcohols.
• Ozonolysis of alkenes, followed by Zn dust and water, gives aldehydes, ketones or both, depending on the alkene.
• Hydration of alkynes with H₂SO₄ and HgSO₄: ethyne gives ethanal; other alkynes give ketones.

Aldehydes only:

• Rosenmund reduction: an acyl chloride is hydrogenated over Pd on barium sulphate (the poisoned catalyst stops reduction at the aldehyde): $C_6{H}_{5}COCl+H_2\stackrel{Pd-BaS{O}_4}{\to }{C}_6{H}_{5}CHO+HCl$.
• Stephen reaction: a nitrile is reduced by SnCl₂ + HCl to an imine, which is hydrolysed to the aldehyde.
• Nitriles and esters can also be reduced to aldehydes by DIBAL-H (diisobutylaluminium hydride) followed by hydrolysis.
• Etard reaction: toluene is oxidised by chromyl chloride (CrO₂Cl₂) to a chromium complex, which is hydrolysed to benzaldehyde.
• Toluene with chromic oxide in acetic anhydride gives benzylidene diacetate, which hydrolyses to benzaldehyde; side-chain chlorination of toluene to benzal chloride, followed by hydrolysis, also gives benzaldehyde.
• Gatterman–Koch reaction: benzene with CO and HCl in the presence of anhydrous AlCl₃ or CuCl gives benzaldehyde.

Ketones only:

• Acyl chlorides react with dialkylcadmium (made from CdCl₂ and a Grignard reagent) to give ketones.
• A nitrile with a Grignard reagent, followed by hydrolysis, gives a ketone.
• Friedel–Crafts acylation of benzene with an acyl chloride and anhydrous AlCl₃ gives aryl ketones: $C_6{H}_6+C{H}_{3}COCl\stackrel{AlC{l}_3}{\to }{C}_6{H}_{5}COC{H}_3+HCl$.

Worked example

Name the reactions and write equations to prepare benzaldehyde from (a) benzoyl chloride, (b) toluene, (c) benzene. (3 marks)

Try it first, then show the solutionSolution

1. (a) Rosenmund reduction: $C_6{H}_{5}COCl+H_2\stackrel{Pd-BaS{O}_4}{\to }{C}_6{H}_{5}CHO+HCl$. (1)
2. (b) Etard reaction: toluene + CrO₂Cl₂ in CS₂ gives a chromium complex; hydrolysis (H₃O⁺) gives C₆H₅CHO. (1)
3. (c) Gatterman–Koch reaction: $C_6{H}_6+CO+HCl\stackrel{AlC{l}_3/CuCl}{\to }{C}_6{H}_{5}CHO+HCl$. (1)

Answer: (a) Rosenmund reduction (b) Etard reaction (c) Gatterman–Koch reaction

Nucleophilic addition: mechanism and reactivity

NCERT section 8.4 Chemical Reactions (1) Nucleophilic addition reactions (a) Mechanism (b) Reactivity

Mechanism: a nucleophile attacks the δ+ carbonyl carbon from a direction roughly perpendicular to the plane of the sp² orbitals. The carbon changes from sp² (trigonal planar) to sp³ (tetrahedral), giving a tetrahedral alkoxide intermediate. This intermediate then takes a proton from the medium to give the neutral addition product.

Why aldehydes are more reactive than ketones:

• Steric: a ketone carbon carries two alkyl groups, which crowd the approaching nucleophile; an aldehyde has one group and a small H.
• Electronic: two alkyl groups push electrons (+I) into the carbonyl carbon more than one does, making it less electrophilic.

So the order is HCHO > CH₃CHO > CH₃COCH₃. Aromatic aldehydes such as benzaldehyde are less reactive than aliphatic ones, because the benzene ring donates electrons by resonance, reducing the positive charge on the carbonyl carbon.

Worked example

Arrange in increasing order of reactivity towards nucleophilic addition and give reasons: ethanal, propanone, methanal, benzaldehyde. (3 marks)

Try it first, then show the solutionSolution

1. Methanal has two H atoms: least crowded and no electron-donating group, so it is the most reactive. (1)
2. Ethanal has one methyl group; benzaldehyde's ring reduces the electrophilicity of the carbonyl carbon by resonance, so it is less reactive than ethanal; propanone has two alkyl groups (steric and +I). (1)
3. Order: propanone < benzaldehyde < ethanal < methanal. (1)

Answer: Propanone < benzaldehyde < ethanal < methanal

Addition reactions: HCN, NaHSO₃, Grignard, alcohols, ammonia derivatives

NCERT section 8.4 Chemical Reactions (c) Some important examples of nucleophilic addition and nucleophilic addition-elimination reactions

• HCN: aldehydes and ketones add HCN to give cyanohydrins. The reaction is slow with pure HCN and is catalysed by a base, which makes CN⁻, a stronger nucleophile: $C{H}_{3}CHO+HCN\to C{H}_{3}CH(OH)CN$. Cyanohydrins are useful synthetic intermediates.
• Sodium hydrogensulphite: gives crystalline addition products. The equilibrium lies far to the right for most aldehydes and to the left for most ketones (steric reasons). The products are water-soluble and are decomposed by dilute acid or alkali to regenerate the carbonyl compound, so this is used to separate and purify aldehydes.
• Grignard reagents: give alcohols after hydrolysis (Chapter 7).
• Alcohols: with one equivalent of an alcohol and dry HCl, aldehydes give hemiacetals; with a second equivalent, acetals (gem-dialkoxy compounds). Ketones react with ethylene glycol under similar conditions to give cyclic ketals. Dry HCl protonates the carbonyl oxygen and increases the electrophilicity of the carbon. Acetals and ketals are hydrolysed back by aqueous mineral acids.
• Ammonia derivatives (Z–NH₂): addition followed by loss of water gives products with C=N–Z. The reaction is reversible and is catalysed by acid (a weakly acidic medium works best).
• Hydroxylamine (NH₂OH) → oxime
• Hydrazine (NH₂NH₂) → hydrazone
• Phenylhydrazine → phenylhydrazone
• 2,4-Dinitrophenylhydrazine (2,4-DNP) → 2,4-DNP derivative (yellow, orange or red solids used to characterise aldehydes and ketones)
• Semicarbazide → semicarbazone

Worked example

Write the products of (a) propanone with HCN, (b) ethanal with hydroxylamine, (c) ethanal with excess ethanol in dry HCl. (3 marks)

Try it first, then show the solutionSolution

1. (a) $C{H}_{3}COC{H}_3+HCN\to (C{H}_3{)}_{2}C(OH)CN$, acetone cyanohydrin. (1)
2. (b) $C{H}_{3}CHO+N{H}_{2}OH\to C{H}_{3}CH=NOH+H_{2}O$, ethanal oxime (acetaldoxime). (1)
3. (c) $C{H}_{3}CHO+2C_2{H}_{5}OH\stackrel{dryHCl}{\to }C{H}_{3}CH(O{C}_2{H}_5{)}_2+H_{2}O$, an acetal (1,1-diethoxyethane). (1)

Answer: (a) (CH₃)₂C(OH)CN (b) CH₃CH=NOH (c) CH₃CH(OC₂H₅)₂

Reduction, oxidation and tests for aldehydes

NCERT section 8.4 Chemical Reactions (2) Reduction (3) Oxidation

Reduction:

• To alcohols: NaBH₄, LiAlH₄ or catalytic hydrogenation; aldehydes give 1° and ketones give 2° alcohols.
• To hydrocarbons (C=O becomes CH₂):
• Clemmensen reduction: zinc amalgam and concentrated HCl.
• Wolff–Kishner reduction: hydrazine forms the hydrazone, which is heated with KOH (or potassium tert-butoxide) in a high-boiling solvent such as ethylene glycol.

Oxidation:

• Aldehydes are oxidised easily to carboxylic acids by HNO₃, KMnO₄ and K₂Cr₂O₇, and even by mild oxidising agents, because the H on the carbonyl carbon is easily removed. Ketones are oxidised only under vigorous conditions, with C–C bond cleavage, giving a mixture of acids with fewer carbons.
• Tollens' test: warming an aldehyde with freshly prepared ammoniacal silver nitrate gives a bright silver mirror: $RCHO+2[Ag(N{H}_3{)}_2{]}^{+}+3O{H}^{-}\to RCO{O}^{-}+2Ag+2H_{2}O+4N{H}_3$. Aromatic aldehydes also give this test.
• Fehling's test: Fehling A (aqueous CuSO₄) and Fehling B (alkaline sodium potassium tartrate, Rochelle salt) are mixed in equal amounts; on heating with an aldehyde a reddish brown precipitate of Cu₂O forms: $RCHO+2C{u}^{2+}+5O{H}^{-}\to RCO{O}^{-}+C{u}_{2}O+3H_{2}O$. Aromatic aldehydes do not respond to Fehling's test.
• Haloform reaction: aldehydes and ketones with at least one methyl group on the carbonyl carbon (methyl ketones, and ethanal) are oxidised by sodium hypohalite: the methyl group becomes a haloform. With I₂ and NaOH, a yellow precipitate of iodoform (CHI₃) forms: the iodoform test. The C=C, if present, is not affected.

Worked example

Give a chemical test to distinguish (a) propanal and propanone, (b) benzaldehyde and ethanal, (c) pentan-2-one and pentan-3-one. (3 marks)

Try it first, then show the solutionSolution

1. (a) Tollens' test: propanal gives a silver mirror; propanone does not (or Fehling's: propanal gives a red-brown precipitate). (1)
2. (b) Fehling's test: ethanal gives a reddish brown precipitate; benzaldehyde, an aromatic aldehyde, does not. (Iodoform test also works: ethanal gives yellow CHI₃.) (1)
3. (c) Iodoform test: pentan-2-one (a methyl ketone, CH₃CO–) gives a yellow precipitate of iodoform; pentan-3-one does not. (1)

Answer: (a) Tollens' or Fehling's (b) Fehling's or iodoform (c) iodoform test

α-Hydrogen reactions, Cannizzaro reaction and ring substitution

NCERT section 8.4 Chemical Reactions (4) Reactions due to α-hydrogen (5) Other reactions

Acidity of α-hydrogens: hydrogens on the carbon next to C=O are acidic, because the carbonyl group withdraws electrons and the resulting carbanion (enolate) is stabilised by resonance.

Aldol condensation: aldehydes and ketones with at least one α-hydrogen, with dilute alkali, give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols), which lose water easily on heating to give α,β-unsaturated carbonyl compounds:
$2C{H}_{3}CHO\stackrel{dil.NaOH}{\to }C{H}_{3}CH(OH)C{H}_{2}CHO$ (3-hydroxybutanal)
$C{H}_{3}CH(OH)C{H}_{2}CHO\stackrel{\Delta }{\to }C{H}_{3}CH=CHCHO+H_{2}O$ (but-2-enal)

Cross aldol condensation: between two different aldehydes and/or ketones. If both have α-hydrogens, a mixture of four products forms (for example ethanal + propanal). Ketones can also be one partner, for example benzaldehyde with acetophenone.

Cannizzaro reaction: aldehydes with no α-hydrogen (HCHO, C₆H₅CHO), heated with concentrated alkali, undergo self oxidation and reduction (disproportionation): one molecule is reduced to an alcohol and another is oxidised to the salt of a carboxylic acid.
$2HCHO+KOH\to C{H}_{3}OH+HCOOK$
$2C_6{H}_{5}CHO+NaOH\to C_6{H}_{5}C{H}_{2}OH+C_6{H}_{5}COONa$

Electrophilic substitution: in aromatic aldehydes and ketones, the carbonyl group is deactivating and meta directing: benzaldehyde with nitrating mixture gives m-nitrobenzaldehyde.

Worked example

Which of the following undergo aldol condensation and which the Cannizzaro reaction? Methanal, ethanal, benzaldehyde, propanone, 2,2-dimethylpropanal. Give a reason and write the product for one Cannizzaro example. (3 marks)

Try it first, then show the solutionSolution

1. Aldol needs at least one α-hydrogen: ethanal and propanone. (1)
2. Cannizzaro needs no α-hydrogen: methanal, benzaldehyde and 2,2-dimethylpropanal ((CH₃)₃CCHO, whose α-carbon has no H). (1)
3. Example: $2C_6{H}_{5}CHO+NaOH\to C_6{H}_{5}C{H}_{2}OH+C_6{H}_{5}COONa$ (concentrated NaOH). (1)

Answer: Aldol: ethanal, propanone. Cannizzaro: methanal, benzaldehyde, 2,2-dimethylpropanal (2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa).

Carboxylic acids: naming, preparation and physical properties

NCERT section 8.6 Nomenclature and Structure of Carboxyl Group; 8.7 Methods of Preparation of Carboxylic Acids; 8.8 Physical Properties

Names: methanoic acid (formic acid, first got from red ants), ethanoic acid (acetic acid, in vinegar), benzoic acid; dicarboxylic acids such as ethanedioic (oxalic) acid. The carboxyl carbon is sp²; the –COOH group is planar, and the C=O is less electrophilic than in aldehydes because of resonance with the –OH oxygen.

Preparation:
1. Oxidation of 1° alcohols and aldehydes with KMnO₄ (neutral, acidic or alkaline), K₂Cr₂O₇ or CrO₃ in acid.
2. Oxidation of alkylbenzenes with chromic acid or acidic or alkaline KMnO₄: the entire side chain (whatever its length) becomes –COOH, giving an aromatic acid, provided the side chain has a benzylic H. Toluene gives benzoic acid.
3. Hydrolysis of nitriles and amides with H⁺ or OH⁻: nitrile → amide → acid. Mild conditions stop at the amide.
4. Grignard reagents react with CO₂ (dry ice) to give salts of carboxylic acids, which are acidified: $C{H}_{3}MgBr+C{O}_2\to C{H}_{3}COOMgBr$, then H₃O⁺. This adds one carbon to the chain, as does the nitrile route.
5. Hydrolysis of acyl halides and anhydrides (with water, or with alkali to give salts) and of esters (acidic hydrolysis gives acids; basic hydrolysis gives salts).

Physical properties: carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of similar mass, because of more extensive hydrogen bonding; most exist as hydrogen-bonded dimers in the vapour phase and in aprotic solvents. Simple acids up to four carbons are miscible with water; solubility falls as the hydrocarbon part grows. Benzoic acid is nearly insoluble in cold water.

Worked example

How will you obtain (a) benzoic acid from toluene, (b) ethanoic acid from methylmagnesium bromide, (c) propanoic acid from ethyl cyanide? (3 marks)

Try it first, then show the solutionSolution

1. (a) Heat toluene with alkaline KMnO₄, then acidify: C₆H₅CH₃ → C₆H₅COOK → C₆H₅COOH. (1)
2. (b) $C{H}_{3}MgBr+C{O}_2\to C{H}_{3}COOMgBr$; hydrolysis with H₃O⁺ gives CH₃COOH. (1)
3. (c) Hydrolysis with dilute acid: CH₃CH₂CN → CH₃CH₂CONH₂ → CH₃CH₂COOH. (1)

Answer: (a) Oxidation with alkaline KMnO₄ then H⁺ (b) CO₂ then H₃O⁺ (c) acid hydrolysis of the nitrile.

Acidity of carboxylic acids

NCERT section 8.9 Chemical Reactions (8.9.1 Reactions Involving Cleavage of O–H Bond: Acidity)

Carboxylic acids ionise in water: $RCOOH+H_{2}O\rightleftharpoons RCO{O}^{-}+H_3{O}^{+}$. Acid strength is given by $p{K}_a=-\log K_a$; a smaller pKa means a stronger acid. NCERT values: trifluoroacetic acid 0.23 (the strongest carboxylic acid), benzoic acid 4.19, acetic acid 4.76; ethanol about 16 and phenol 10.

Why stronger than phenols: the carboxylate ion has two equivalent resonance structures with the negative charge on two electronegative oxygen atoms. Phenoxide's resonance structures are non-equivalent and put the charge on less electronegative carbon atoms. So carboxylate is more stabilised.

Carboxylic acids react with Na (H₂), NaOH (salt + water) and even with the weak bases sodium carbonate and sodium hydrogencarbonate, releasing CO₂ with brisk effervescence. Phenols do not react with NaHCO₃, which is a useful test.

Substituent effects:

• Electron-withdrawing groups (–NO₂, –CN, halogens, –CF₃) stabilise the carboxylate and increase acidity. More such groups, and groups closer to –COOH, have a bigger effect: CF₃COOH > CCl₃COOH > CHCl₂COOH > ClCH₂COOH, and FCH₂COOH > ClCH₂COOH > BrCH₂COOH. Order of effect: Ph < I < Br < Cl < F < CN < NO₂ < CF₃.
• Electron-donating groups (alkyl) decrease acidity: HCOOH > CH₃COOH > CH₃CH₂COOH.
• A phenyl or vinyl group directly attached to –COOH increases acidity, because the sp² carbon is more electronegative: C₆H₅COOH (4.19) is stronger than CH₃COOH (4.76).

Worked example

Arrange in decreasing order of acidity and explain: CH₃COOH, ClCH₂COOH, Cl₂CHCOOH, FCH₂COOH. (2 marks)

Try it first, then show the solutionSolution

1. Halogens withdraw electrons by the −I effect and stabilise the carboxylate; two Cl atoms withdraw more than one, and F (more electronegative) more than Cl. (1)
2. Order: Cl₂CHCOOH > FCH₂COOH > ClCH₂COOH > CH₃COOH. (1)

Answer: Cl₂CHCOOH > FCH₂COOH > ClCH₂COOH > CH₃COOH

Reactions of carboxylic acids

NCERT section 8.9.2 Reactions Involving Cleavage of C–OH Bond; 8.9.3 Reactions Involving –COOH Group; 8.9.4 Substitution Reactions in the Hydrocarbon Part

Cleavage of C–OH:

• Anhydrides: heating with a mineral acid such as H₂SO₄, or with P₂O₅: $2C{H}_{3}COOH\to (C{H}_{3}CO{)}_{2}O+H_{2}O$.
• Esterification: with alcohols or phenols in the presence of concentrated H₂SO₄ or dry HCl. Mechanism: the carbonyl oxygen is protonated; the alcohol adds to the carbonyl carbon (nucleophilic addition); a proton transfer turns –OH into –OH₂⁺, which leaves as water; loss of H⁺ gives the ester. It is reversible.
• Acid chlorides: with PCl₅, PCl₃ or SOCl₂. SOCl₂ is preferred because SO₂ and HCl escape as gases: $RCOOH+SOC{l}_2\to RCOCl+S{O}_2+HCl$.
• Amides: with ammonia an ammonium salt forms, which on strong heating loses water to give an amide: $C{H}_{3}COON{H}_4\stackrel{\Delta }{\to }C{H}_{3}CON{H}_2+H_{2}O$. Phthalic acid with NH₃ gives phthalimide on strong heating.

Reactions of the –COOH group:

• Reduction to 1° alcohols by LiAlH₄, or better by diborane (which does not easily reduce esters, nitro or halo groups). NaBH₄ does not reduce –COOH.
• Decarboxylation: sodium salts heated with soda lime (NaOH and CaO in a 3 : 1 ratio) lose CO₂ to give an alkane with one carbon fewer: $C{H}_{3}COONa+NaOH\stackrel{CaO,\Delta }{\to }C{H}_4+N{a}_{2}C{O}_3$. Kolbe electrolysis of aqueous sodium or potassium salts gives an alkane with twice the number of carbons of the alkyl group.

Substitution in the hydrocarbon part:

• Hell–Volhard–Zelinsky (HVZ) reaction: acids with an α-hydrogen, treated with Cl₂ or Br₂ and a small amount of red phosphorus, give α-halocarboxylic acids: $C{H}_{3}COOH+C{l}_2\stackrel{redP}{\to }ClC{H}_{2}COOH+HCl$.
• Ring substitution: –COOH is deactivating and meta directing. Aromatic acids do not undergo Friedel–Crafts reactions, because the carboxyl group deactivates the ring and the catalyst AlCl₃ binds to it.

Worked example

Write the equations for (a) the Hell–Volhard–Zelinsky reaction of ethanoic acid, (b) decarboxylation of sodium ethanoate, and (c) the reaction of ethanoic acid with thionyl chloride. (3 marks)

Try it first, then show the solutionSolution

1. (a) $C{H}_{3}COOH+C{l}_2\stackrel{redP}{\to }ClC{H}_{2}COOH+HCl$ (chloroethanoic acid). (1)
2. (b) $C{H}_{3}COONa+NaOH\stackrel{CaO,\Delta }{\to }C{H}_4+N{a}_{2}C{O}_3$ (soda lime). (1)
3. (c) $C{H}_{3}COOH+SOC{l}_2\to C{H}_{3}COCl+S{O}_2+HCl$. (1)

Answer: (a) ClCH₂COOH + HCl (b) CH₄ + Na₂CO₃ (c) CH₃COCl + SO₂ + HCl

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.

Topics in this chapter

• Nucleophilic Addition Reactions of Aldehydes and Ketones
• Oxidation, Reduction and Tests for Aldehydes and Ketones
• Aldol and Cannizzaro Reactions

Aldehydes, Ketones and Carboxylic Acids formula sheet

Rosenmund reduction

$C_6{H}_{5}COCl+H_2\stackrel{Pd-BaS{O}_4}{\to }{C}_6{H}_{5}CHO+HCl$

Poisoned catalyst stops at the aldehyde.

Stephen reaction

$RCN\stackrel{{\text{(i) SnCl}}_2{\text{, HCl (ii) H}}_3{\text{O}}^{+}}{\to }RCHO$

Via the imine RCH=NH. DIBAL-H does the same for nitriles and esters.

Etard reaction

$C_6{H}_{5}C{H}_3\stackrel{{\text{(i) CrO}}_2{\text{Cl}}_2{\text{, CS}}_2{\text{ (ii) H}}_3{\text{O}}^{+}}{\to }{C}_6{H}_{5}CHO$

Through a chromium complex.

Gatterman–Koch reaction

$C_6{H}_6+CO+HCl\stackrel{AlC{l}_3/CuCl}{\to }{C}_6{H}_{5}CHO+HCl$

Formylation of benzene.

Ketone from dialkylcadmium

$2C_6{H}_{5}COCl+(C{H}_3{)}_{2}Cd\to 2C_6{H}_{5}COC{H}_3+CdC{l}_2$

$(C{H}_3{)}_{2}Cd$ from $2C{H}_{3}MgBr+CdC{l}_2$.

Friedel–Crafts acylation

$C_6{H}_6+C{H}_{3}COCl\stackrel{anhyd.AlC{l}_3}{\to }{C}_6{H}_{5}COC{H}_3+HCl$

Gives aryl ketones (acetophenone).

Hydration of ethyne

$C_2{H}_2+H_{2}O\stackrel{H_{2}S{O}_4,HgS{O}_4}{\to }C{H}_{3}CHO$

Other alkynes give ketones.

Cyanohydrin formation

$C{H}_{3}CHO+HCN\to C{H}_{3}CH(OH)CN$

Base-catalysed; CN⁻ is the real nucleophile.

Acetal formation

$C{H}_{3}CHO+2C_2{H}_{5}OH\stackrel{dryHCl}{\to }C{H}_{3}CH(O{C}_2{H}_5{)}_2+H_{2}O$

Ketones with ethylene glycol give cyclic ketals.

Oxime formation

$C{H}_{3}CHO+N{H}_{2}OH\to C{H}_{3}CH=NOH+H_{2}O$

Hydrazine, phenylhydrazine, 2,4-DNP and semicarbazide react similarly (C=N–Z).

Clemmensen and Wolff–Kishner reductions

$C=O\stackrel{\text{Zn-Hg, conc. HCl}}{\to }C{H}_2;C=O\stackrel{{\text{NH}}_2{\text{NH}}_2\text{, then KOH, ethylene glycol}}{\to }C{H}_2$

Both convert the carbonyl group into CH₂.

Tollens' test

$RCHO+2[Ag(N{H}_3{)}_2{]}^{+}+3O{H}^{-}\to RCO{O}^{-}+2Ag+2H_{2}O+4N{H}_3$

Silver mirror; given by all aldehydes.

Fehling's test

$RCHO+2C{u}^{2+}+5O{H}^{-}\to RCO{O}^{-}+C{u}_{2}O+3H_{2}O$

Red-brown Cu₂O; aromatic aldehydes do not respond.

Iodoform reaction

$C{H}_{3}COC{H}_3+3I_2+4NaOH\to CH{I}_3+C{H}_{3}COONa+3NaI+3H_{2}O$

Yellow CHI₃; given by CH₃CO– compounds (and ethanal).

Aldol reaction and condensation

$2C{H}_{3}CHO\stackrel{dil.NaOH}{\to }C{H}_{3}CH(OH)C{H}_{2}CHO$

Heating gives $C{H}_{3}CH(OH)C{H}_{2}CHO\to C{H}_{3}CH=CHCHO+H_{2}O$.

Cannizzaro reaction

$2C_6{H}_{5}CHO+NaOH\to C_6{H}_{5}C{H}_{2}OH+C_6{H}_{5}COONa$

Concentrated alkali; aldehydes without α-H. $2HCHO+KOH\to C{H}_{3}OH+HCOOK$.

Acids and bicarbonate

$C{H}_{3}COOH+NaHC{O}_3\to C{H}_{3}COONa+H_{2}O+C{O}_2$

Brisk effervescence; phenols do not react.

Esterification

$C{H}_{3}COOH+C_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_2{H}_5+H_{2}O$

Conc. H₂SO₄ or dry HCl catalyst.

Acid chloride formation

$RCOOH+PC{l}_5\to RCOCl+POC{l}_3+HCl$

Also $3RCOOH+PC{l}_3\to 3RCOCl+H_{3}P{O}_3$ and with SOCl₂.

Decarboxylation

$C{H}_{3}COONa+NaOH\stackrel{CaO,\Delta }{\to }C{H}_4+N{a}_{2}C{O}_3$

Soda lime NaOH : CaO = 3 : 1.

Hell–Volhard–Zelinsky reaction

$C{H}_{3}COOH+C{l}_2\stackrel{redP}{\to }ClC{H}_{2}COOH+HCl$

α-Halogenation; needs an α-hydrogen.

Acidity

$p{K}_a=-\log K_a$

CF₃COOH 0.23 < C₆H₅COOH 4.19 < CH₃COOH 4.76 (smaller pKa, stronger acid).

Aldehydes, Ketones and Carboxylic Acids: diagrams to practise

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

How the board asks Aldehydes, Ketones and Carboxylic Acids

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

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.

• Rosenmund Reduces Chlorides; Stephen Starts with Nitriles

  R-R: Rosenmund Reduces acid chlorides (Pd-BaSO₄). S-N: Stephen works on Nitriles (SnCl₂/HCl).

  Use it when: Matching named reactions to starting materials.

• Etard = Toluene + chromyl chloride

  'E-T-C': Etard, Toluene, CrO₂Cl₂. Gatterman–Koch: G-K = Gas (CO) + HCl.

  Use it when: Routes to benzaldehyde.

• α-H decides aldol or Cannizzaro

  Has α-H → aldol with dilute base. No α-H → Cannizzaro with concentrated base.

  Use it when: Predicting which reaction occurs.

• Test ladder

  2,4-DNP: any C=O. Tollens': any aldehyde. Fehling's: aliphatic aldehydes. Iodoform: CH₃CO– (and ethanal, ethanol, propan-2-ol). NaHCO₃: acids.

  Use it when: Distinguish-pairs questions.

• Clemmensen in Acid, Wolff–Kishner in base

  C for Concentrated HCl (Zn-Hg); W-K for KOH with hydrazine. Both make C=O into CH₂.

  Use it when: Choosing the reduction.

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.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

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.

Where students lose marks in Aldehydes, Ketones and Carboxylic Acids

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

• ✗Mistake: Saying benzaldehyde gives Fehling's test

  ✓Do this: Aromatic aldehydes give Tollens' test but not Fehling's test.

  Usually costs: 1 mark in a distinguish question.

• ✗Mistake: Using concentrated alkali for aldol

  Wrong working

  2CH₃CHO + conc. NaOH → CH₃CH₂OH + CH₃COONa

  Right working

  2CH₃CHO + dil. NaOH → CH₃CH(OH)CH₂CHO (aldol); Cannizzaro needs an aldehyde without α-H

  ✓Do this: Check for α-hydrogen first, then pick dilute (aldol) or concentrated (Cannizzaro) alkali.

  Usually costs: The equation mark.

• ✗Mistake: Writing the iodoform test as positive for all ketones

  ✓Do this: Only compounds with CH₃CO– (methyl ketones and ethanal), or alcohols oxidised to them (ethanol, propan-2-ol), give iodoform.

  Usually costs: 1 mark.

• ✗Mistake: Wrong acidity order for halogenated acids

  Wrong working

  BrCH₂COOH > ClCH₂COOH > FCH₂COOH

  Right working

  FCH₂COOH > ClCH₂COOH > BrCH₂COOH (stronger −I effect of F)

  ✓Do this: More electronegative and closer substituents increase acidity more.

  Usually costs: 1 mark.

• ✗Mistake: Forgetting hydrolysis steps

  Students stop at the intermediate in Etard, Stephen, Grignard and nitrile reactions.

  ✓Do this: Write '(ii) H₃O⁺' on the arrow; the aldehyde, ketone or acid appears only after hydrolysis.

  Usually costs: ½ mark per missing step.

NCERT exercise guide: Aldehydes, Ketones and Carboxylic Acids

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Aldehydes, Ketones and Carboxylic Acids: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.9 marks per paper; 30 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Aldehydes, Ketones and Carboxylic Acids

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Aldehydes, Ketones and Carboxylic Acids question, by year

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.

1. Q1 · 1 mark · MCQBoard level

   Which of the following does NOT undergo aldol condensation?
   (a) Ethanal (b) Propanone (c) Methanal (d) Propanal

   Hint

   Look for α-hydrogen.

   Answer

   (c) Methanal, which has no α-hydrogen (it undergoes the Cannizzaro reaction instead).

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Formaldehyde undergoes the Cannizzaro reaction.
   Reason (R): Formaldehyde has no α-hydrogen.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   What decides aldol versus Cannizzaro?

   Answer

   (a)

3. Q3 · 2 marks · Very short answerBoard level

   Write the equations for (a) the Etard reaction and (b) the Stephen reaction. (2 marks)

   Hint

   Chromyl chloride; SnCl₂/HCl.

   Answer

   (a) Toluene + CrO₂Cl₂ (in CS₂) gives a chromium complex; H₃O⁺ gives benzaldehyde (1). (b) RCN + SnCl₂ + HCl gives RCH=NH, which is hydrolysed by H₃O⁺ to RCHO, e.g. CH₃CN → CH₃CHO (1).

4. Q4 · 2 marks · Very short answerBoard level

   Why is the boiling point of ethanoic acid higher than that of propan-1-ol, although their molar masses are equal (60 g mol⁻¹)? (2 marks)

   Hint

   Think dimers.

   Answer

   Ethanoic acid forms more extensive hydrogen bonds (1); its molecules exist as hydrogen-bonded dimers even in the vapour, so more energy is needed to separate them (1).

5. Q5 · 3 marks · Short answerHarder board level

   Write the products when benzaldehyde reacts with (a) HCN, (b) hydroxylamine, (c) concentrated NaOH, (d) nitrating mixture. (3 marks)

   Hint

   No α-H; –CHO is meta directing.

   Answer

   (a) C₆H₅CH(OH)CN (benzaldehyde cyanohydrin) (½+½). (b) C₆H₅CH=NOH (benzaldoxime) (½). (c) C₆H₅CH₂OH + C₆H₅COONa (Cannizzaro) (1). (d) m-nitrobenzaldehyde (½).

6. Q6 · 3 marks · Short answerHarder board level

   Arrange in increasing order of acid strength: CH₃CH₂COOH, CH₃COOH, ClCH₂COOH, FCH₂COOH, C₆H₅COOH, HCOOH. (3 marks)

   Hint

   Alkyl groups weaken; halogens and sp² carbon strengthen.

   Answer

   CH₃CH₂COOH < CH₃COOH < C₆H₅COOH < HCOOH < ClCH₂COOH < FCH₂COOH (2); alkyl (+I) lowers acidity, the sp² phenyl carbon raises it, and halogens (−I, F most) raise it most (1).

7. Q7 · 3 marks · Short answerHarder board level

   Write the mechanism of the nucleophilic addition of HCN to ethanal and explain why the reaction is faster in the presence of a little base. (3 marks)

   Hint

   CN⁻ is the nucleophile.

   Answer

   Base removes H⁺ from HCN to give CN⁻, a stronger nucleophile (1). CN⁻ attacks the δ+ carbonyl carbon; the π electrons move to oxygen, giving a tetrahedral alkoxide ion (1). The alkoxide takes H⁺ from HCN or water to give CH₃CH(OH)CN, regenerating CN⁻ (1).

8. Q8 · 5 marks · Long answerHarder board level

   (a) How are the following prepared? Give equations: (i) acetophenone from benzene (ii) benzaldehyde from benzene by a one-step reaction. (2) (b) Give two tests to distinguish acetophenone from benzaldehyde. (2) (c) Why does acetophenone give the iodoform test? (1) (5 marks)

   Hint

   Friedel–Crafts acylation; Gatterman–Koch.

   Answer

   (a)(i) C₆H₆ + CH₃COCl (anhyd. AlCl₃) → C₆H₅COCH₃ + HCl (1). (ii) C₆H₆ + CO + HCl (AlCl₃/CuCl) → C₆H₅CHO + HCl (1). (b) Tollens': benzaldehyde gives a silver mirror, acetophenone does not (1); iodoform: acetophenone gives yellow CHI₃, benzaldehyde does not (1). (c) It contains the CH₃CO– group (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Suppose a compound A (C₅H₁₀O) forms a 2,4-DNP derivative, does not reduce Tollens' reagent, and gives a positive iodoform test. On Clemmensen reduction it gives pentane. Identify A and give its IUPAC name. Why can it not be pentan-3-one? (3 marks)

   Hint

   Straight chain (pentane), methyl ketone.

   Answer

   Carbonyl compound, not an aldehyde, so a ketone (1); Clemmensen gives n-pentane, so the chain is unbranched; iodoform positive, so CH₃CO–: A is CH₃COCH₂CH₂CH₃, pentan-2-one (1). Pentan-3-one, CH₃CH₂COCH₂CH₃, has no CH₃CO– group and would not give iodoform (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    Why do aromatic carboxylic acids not undergo Friedel–Crafts reaction? (2 marks)

    Hint

    Two reasons: the ring and the catalyst.

    Answer

    The –COOH group is deactivating, making the ring too electron-poor (1), and the Lewis acid catalyst AlCl₃ binds to the carboxyl group, so it is not available (1).

11. Q11 · 2 marks · Very short answerHarder board level

    Convert (a) propanone to propene, (b) benzoic acid to benzaldehyde. (2 marks)

    Hint

    Reduce, then dehydrate; make the acid chloride, then Rosenmund.

    Answer

    (a) CH₃COCH₃ + H₂ (Ni) or NaBH₄ → CH₃CH(OH)CH₃; conc. H₂SO₄, heat → CH₃CH=CH₂ (1). (b) C₆H₅COOH + SOCl₂ → C₆H₅COCl; H₂/Pd–BaSO₄ → C₆H₅CHO (1).

12. Q12 · 3 marks · Short answerHOTS / competency

    Suppose you mix ethanal and benzaldehyde with dilute NaOH. (a) Which compound can form the carbanion (enolate)? (b) Write the main cross aldol condensation product. (c) Why is this mixture more selective than ethanal with propanal? (3 marks)

    Hint

    Only one partner has α-hydrogens.

    Answer

    (a) Ethanal (it has α-H; benzaldehyde has none) (1). (b) The enolate of ethanal adds to benzaldehyde, and dehydration gives C₆H₅CH=CHCHO (3-phenylprop-2-enal, cinnamaldehyde) (1). (c) Benzaldehyde can only accept, so fewer combinations form; with ethanal and propanal both can donate and accept, giving four products (1).

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. 1.What is the hybridisation of the carbonyl carbon?Show answer

   sp² (trigonal planar, about 120°).

2. 2.What catalyst is used in Rosenmund reduction?Show answer

   Palladium on barium sulphate (Pd–BaSO₄).

3. 3.Name the reagent in the Etard reaction.Show answer

   Chromyl chloride, CrO₂Cl₂.

4. 4.Why are aldehydes more reactive than ketones towards nucleophilic addition?Show answer

   Less steric hindrance and fewer electron-donating alkyl groups on the carbonyl carbon.

5. 5.What does Clemmensen reduction use?Show answer

   Zinc amalgam and concentrated HCl.

6. 6.Which aldehydes undergo the Cannizzaro reaction?Show answer

   Those without an α-hydrogen, such as methanal and benzaldehyde, with concentrated alkali.

7. 7.What is the product of aldol condensation of ethanal after heating?Show answer

   But-2-enal, CH₃CH=CHCHO.

8. 8.Why are carboxylic acids more acidic than phenols?Show answer

   The carboxylate ion has two equivalent resonance structures with the charge on two oxygen atoms; phenoxide's charge goes onto less electronegative carbons.

9. 9.What is soda lime?Show answer

   A mixture of NaOH and CaO in the ratio 3 : 1.

10. 10.What is the HVZ reaction?Show answer

    α-Halogenation of a carboxylic acid with Cl₂ or Br₂ and a small amount of red phosphorus.

Quick revision checklist

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

• ☐Can you name CH₃CH(CH₃)CH₂CHO and C₆H₅COCH₃ in IUPAC?
• ☐Can you write the Rosenmund, Stephen, Etard and Gatterman–Koch reactions?
• ☐Can you draw the mechanism of nucleophilic addition of HCN?
• ☐Can you write the products of an aldehyde with hydroxylamine, hydrazine, 2,4-DNP and semicarbazide?
• ☐Can you distinguish propanal/propanone, ethanal/benzaldehyde and pentan-2-one/pentan-3-one?
• ☐Can you write aldol, cross aldol and Cannizzaro reactions?
• ☐Can you order substituted acetic acids by acidity and explain?
• ☐Can you write HVZ, decarboxylation and acid chloride formation equations?

Aldehydes, Ketones and Carboxylic Acids: frequently asked questions