NCERT Solutions for Class 12 Chemistry Chapter 8: Aldehydes, Ketones and Carboxylic Acids – Free PDF Download

Chapter 8 covers Aldehydes, Ketones and Carboxylic Acids — nomenclature and the nature of the carbonyl group, methods of preparation, nucleophilic addition reactions and their mechanism, name reactions (aldol condensation, Cannizzaro reaction, haloform reaction), oxidation/reduction of carbonyl compounds, and the preparation, acidity and reactions of carboxylic acids. This chapter was Chapter 12 under the old (pre-2023) numbering, and shifted to Chapter 8 only because chapters were deleted earlier in the book — its own 20 exercises (8.1–8.20) are unchanged and none were dropped, and no subtopic within the chapter has been trimmed at the syllabus level. Each answer below is checked for chemical accuracy before publishing.

NCERT Exercise Solutions

8.1 Define the following terms giving an example of each: (i) Cyanohydrin (ii) Acetal (iii) Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (viii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base.
Ans: (i) Cyanohydrin: the addition product of HCN with a carbonyl compound, e.g. acetone cyanohydrin (CH₃)₂C(OH)CN. (ii) Acetal: R–CH(OR’)₂, formed when an aldehyde reacts with two equivalents of alcohol under acid catalysis, e.g. acetaldehyde diethyl acetal CH₃CH(OC₂H₅)₂. (iii) Semicarbazone: the condensation product of a carbonyl compound with semicarbazide, e.g. acetone semicarbazone (CH₃)₂C=N–NH–CO–NH₂. (iv) Aldol: a β-hydroxy aldehyde/ketone formed by aldol addition, e.g. 3-hydroxybutanal CH₃CH(OH)CH₂CHO from ethanal. (v) Hemiacetal: R–CH(OH)(OR’), the intermediate formed by addition of one molecule of alcohol to an aldehyde, e.g. CH₃CH(OH)(OC₂H₅). (vi) Oxime: R₂C=N–OH, formed by condensation of a carbonyl compound with hydroxylamine, e.g. acetone oxime. (vii) Ketal: R₂C(OR’)₂, the ketone analogue of an acetal, e.g. acetone diethyl ketal (CH₃)₂C(OC₂H₅)₂. (viii) Imine: R₂C=NR’, formed by condensation of a carbonyl compound with a primary amine. (ix) 2,4-DNP-derivative: the hydrazone formed with 2,4-dinitrophenylhydrazine, an orange/yellow precipitate used to identify and characterise aldehydes and ketones. (x) Schiff’s base: an imine, R₂C=NR’ (R’ typically aryl/alkyl), formed from a primary amine and an aldehyde or ketone.

8.2 Give the IUPAC names of the following carbonyl compounds: (i) CH₃CH₂CH₂CHO (ii) CH₃COCH₂CH₃ (iii) (CH₃)₂CHCH₂CHO (iv) C₆H₅CHO (v) C₆H₅COCH₃ (vi) cyclohexanone (vii) CH₃CH=CHCHO.
Ans: (i) Butanal. (ii) Butan-2-one. (iii) 3-Methylbutanal. (iv) Benzaldehyde (benzenecarbaldehyde). (v) 1-Phenylethan-1-one (acetophenone). (vi) Cyclohexan-1-one. (vii) But-2-enal (crotonaldehyde).

8.3 Draw the structures of the following compounds: 3-Methylbutanal, p-Nitropropiophenone, p-Methylbenzaldehyde, 4-Methylpent-3-en-2-one, 4-Chloropentan-2-one, 3-Bromo-4-phenylpentanoic acid, p,p’-Dihydroxybenzophenone, Hex-2-en-4-ynoic acid.
Ans: (CH₃)₂CHCH₂CHO; O₂N–C₆H₄–COCH₂CH₃ (para); CH₃–C₆H₄–CHO (para); (CH₃)₂C=CH–CO–CH₃; CH₃–CH(Cl)–CH₂–CO–CH₃; CH₃–CH(Br)–CH(C₆H₅)–CH₂–COOH; HO–C₆H₄–CO–C₆H₄–OH (both para); CH₃–CH=CH–C≡C–COOH, respectively, each named to match the position numbers given in its IUPAC name.

8.4 Give the IUPAC and common names of the following: (i) HCHO (ii) CH₃CHO (iii) CH₃COCH₃ (iv) C₆H₅CH₂CHO (v) CH₃CH₂COCH₂CH₃.
Ans: (i) Methanal (formaldehyde). (ii) Ethanal (acetaldehyde). (iii) Propan-2-one (acetone). (iv) Phenylethanal (phenylacetaldehyde). (v) Pentan-3-one (diethyl ketone).

8.5 Draw the structures of the following derivatives: 2,4-Dinitrophenylhydrazone of benzaldehyde, Cyclopropanone oxime, Acetaldehyde dimethyl acetal, Semicarbazone of cyclobutanone, Ethylene ketal of hexan-3-one, Methyl hemiacetal of formaldehyde.
Ans: C₆H₅CH=N–NH–C₆H₃(NO₂)₂; cyclopropane ring with an exocyclic =N–OH; CH₃CH(OCH₃)₂; cyclobutane ring with an exocyclic =N–NH–CO–NH₂; hexan-3-one’s carbonyl carbon incorporated into a 1,3-dioxolane ring (–O–CH₂–CH₂–O– bridging the former C=O carbon); HOCH₂OCH₃, respectively — each formed by the standard condensation/addition rules for that derivative class.

8.6 Predict the products formed when cyclohexanecarbaldehyde reacts with the following reagents: (i) PhMgBr and then H₃O⁺ (ii) Tollens’ reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid.
Ans: (i) Grignard addition to the aldehyde gives, after aqueous workup, cyclohexyl(phenyl)methanol (a secondary alcohol). (ii) Tollens’ reagent oxidises the aldehyde to the carboxylate, depositing a silver mirror (positive test), giving cyclohexanecarboxylic acid (as its ammonium salt). (iii) Gives the corresponding semicarbazone of cyclohexanecarbaldehyde. (iv) Excess ethanol under acid catalysis gives the diethyl acetal of cyclohexanecarbaldehyde. (v) Clemmensen reduction reduces the –CHO group all the way to –CH₃, giving methylcyclohexane.

8.7 Which of the following compounds would undergo the aldol condensation, the Cannizzaro reaction, or neither: methanal, 2-methylpentanal, benzaldehyde, benzophenone, cyclohexanone, 1-phenylpropan-1-one, phenylacetaldehyde, butan-1-ol, 2,2-dimethylbutanal? Write the structures of the expected products where relevant.
Ans: Aldol condensation (all have an α-hydrogen): 2-methylpentanal, cyclohexanone, 1-phenylpropan-1-one, phenylacetaldehyde. Cannizzaro reaction (no α-hydrogen; must be an aldehyde): methanal (2HCHO+NaOH→CH₃OH+HCOONa), benzaldehyde, and 2,2-dimethylbutanal (the α-carbon is fully substituted with two methyls, leaving no α-H). Neither: benzophenone (a ketone with no α-hydrogen — ketones do not undergo Cannizzaro) and butan-1-ol (not a carbonyl compound at all).

8.8 How will you convert ethanal into the following compounds? (i) Butane-1,3-diol (ii) But-2-enal (iii) But-2-enoic acid.
Ans: (i) Ethanal+dilute NaOH (cold, aldol addition only) → 3-hydroxybutanal, then reduction (NaBH₄) of the –CHO to –CH₂OH → butane-1,3-diol. (ii) Ethanal+dilute NaOH, then heat (aldol condensation, with dehydration) → but-2-enal (crotonaldehyde) directly. (iii) But-2-enal (from ii)+Tollens’ reagent (selectively oxidises –CHO without touching the C=C) → but-2-enoic acid (crotonic acid).

8.9 Write structural formulas and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as the nucleophile and which as the electrophile.
Ans: (1) Propanal (nucleophile)+propanal (electrophile) → OHC–CH(CH₃)–CH(OH)–CH₂CH₃, 3-hydroxy-2-methylpentanal. (2) Butanal (nucleophile)+butanal (electrophile) → OHC–CH(C₂H₅)–CH(OH)–CH₂CH₂CH₃, 3-hydroxy-2-ethylhexanal. (3) Propanal (nucleophile)+butanal (electrophile) → OHC–CH(CH₃)–CH(OH)–CH₂CH₂CH₃, 3-hydroxy-2-methylhexanal. (4) Butanal (nucleophile)+propanal (electrophile) → OHC–CH(C₂H₅)–CH(OH)–CH₂CH₃, 3-hydroxy-2-ethylpentanal.

8.10 An organic compound with the molecular formula C₉H₁₀O forms a 2,4-DNP derivative, reduces Tollens’ reagent and undergoes the Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
Ans: o-Ethylbenzaldehyde (2-ethylbenzaldehyde). As an aromatic aldehyde it forms a 2,4-DNP derivative and reduces Tollens’ reagent; having its carbonyl carbon attached directly to the ring (no α-hydrogen issue), it undergoes the Cannizzaro reaction. Vigorous oxidation converts both the –CHO and the ortho ethyl side chain to –COOH groups, giving 1,2-benzenedicarboxylic acid (phthalic acid) — matching the given formula (C₉H₁₀O) exactly.

8.11 An organic compound (A), C₈H₁₆O₂, was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Write equations for the reactions involved.
Ans: (C) is butan-1-ol (a primary alcohol, since it dehydrates to the terminal alkene but-1-ene); oxidising it with chromic acid gives (B), butanoic acid. So (A) is butyl butanoate (CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃, C₈H₁₆O₂). Equations: A+H₂O/H⁺→B (butanoic acid)+C (butan-1-ol); C+CrO₃/H⁺→B; C+conc. H₂SO₄, heat→but-1-ene+H₂O.

8.12 Arrange the following compounds in increasing order of the property indicated: (i) Acetaldehyde, acetone, di-tert-butyl ketone, methyl tert-butyl ketone (reactivity towards HCN) (ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength) (iii) Benzoic acid, 4-nitrobenzoic acid, 3,4-dinitrobenzoic acid, 4-methoxybenzoic acid (acid strength).
Ans: (i) Di-tert-butyl ketone<methyl tert-butyl ketone<acetone<acetaldehyde (reactivity towards nucleophilic addition falls as steric bulk around the carbonyl carbon rises, and aldehydes are always more reactive than ketones). (ii) (CH₃)₂CHCOOH<CH₃CH₂CH₂COOH<CH₃CH(Br)CH₂COOH<CH₃CH₂CH(Br)COOH (the –I effect of Br strengthens the acid, and is strongest when Br is closest to –COOH). (iii) 4-Methoxybenzoic acid<benzoic acid<4-nitrobenzoic acid<3,4-dinitrobenzoic acid (electron-withdrawing –NO₂ groups increase acid strength; the electron-donating –OCH₃ group decreases it).

8.13 Give simple chemical tests to distinguish between the following pairs of compounds: (i) Propanal and Propanone (ii) Acetophenone and Benzophenone (iii) Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal.
Ans: (i) Tollens’/Fehling’s test — propanal (aldehyde) gives a positive test, propanone does not. (ii) Iodoform test — acetophenone (a methyl ketone) gives a positive test (yellow CHI₃ precipitate), benzophenone does not. (iii) NaHCO₃ test — benzoic acid gives brisk effervescence (CO₂), phenol does not (or FeCl₃ test: phenol gives a violet colour, benzoic acid does not). (iv) NaHCO₃ test — benzoic acid effervesces, ethyl benzoate (an ester) does not. (v) Iodoform test — pentan-2-one (a methyl ketone) is positive, pentan-3-one is negative. (vi) Tollens’ test — benzaldehyde gives a silver mirror, acetophenone does not (note: Fehling’s solution is unreliable for aromatic aldehydes, so Tollens’ is the correct test here). (vii) Iodoform test — ethanal (has the requisite CH₃CHO grouping) is positive, propanal (no methyl directly on the carbonyl carbon) is negative.

8.14 How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom: (i) Methyl benzoate (ii) m-Nitrobenzoic acid (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde.
Ans: (i) Benzene+CO/HCl, anhyd. AlCl₃–CuCl (Gattermann–Koch) → benzaldehyde; oxidise → benzoic acid; Fischer esterify with CH₃OH/H⁺ → methyl benzoate. (ii) Benzene → benzaldehyde (as above) → oxidise → benzoic acid; nitrate (conc. HNO₃/H₂SO₄) — since –COOH is meta-directing → m-nitrobenzoic acid. (iii) Benzene+CH₃Cl/anhyd. AlCl₃ → toluene; nitrate (–CH₃ is o,p-directing, separate the para isomer) → p-nitrotoluene; vigorous oxidation of the methyl group (KMnO₄) → p-nitrobenzoic acid. (iv) Benzene → toluene (as above); side-chain chlorination (Cl₂/light) → benzyl chloride; +KCN (S_N2) → phenylacetonitrile; hydrolysis (H₃O⁺, heat) → phenylacetic acid. (v) Benzene → toluene → nitrate, separate para → p-nitrotoluene; controlled oxidation of the methyl group (CrO₂Cl₂, Etard reaction) → p-nitrobenzaldehyde.

8.15 How will you bring about the following conversions in not more than two steps? (i) Propanone to Propene (ii) Benzoic acid to Benzaldehyde (iii) Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benzaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m-Nitrobenzyl alcohol.
Ans: (i) Reduce (NaBH₄) → propan-2-ol; dehydrate (conc. H₂SO₄, heat) → propene. (ii) Convert to acid chloride (SOCl₂); Rosenmund reduction (H₂/Pd–BaSO₄) → benzaldehyde. (iii) Oxidise (PCC) → ethanal; cold dilute NaOH (aldol addition) → 3-hydroxybutanal. (iv) Friedel–Crafts acylation (CH₃COCl/AlCl₃) → acetophenone; nitrate (conc. HNO₃/H₂SO₄, –COCH₃ is meta-directing) → m-nitroacetophenone. (v) React with PhMgBr, then H₃O⁺ → diphenylmethanol; oxidise (PCC) → benzophenone. (vi) +Mg/dry ether → PhMgBr; react with ethanal, then H₃O⁺ → 1-phenylethanol. (vii) Cross-aldol condensation with ethanal (base, heat) → cinnamaldehyde; reduce fully (H₂/Pd, reduces both C=C and CHO) → 3-phenylpropan-1-ol. (viii) +HCN → mandelonitrile; hydrolyse (H₃O⁺, heat) → mandelic acid (α-hydroxyphenylacetic acid). (ix) Nitrate (conc. HNO₃/H₂SO₄, meta-directing –COOH) → m-nitrobenzoic acid; reduce (LiAlH₄) → m-nitrobenzyl alcohol.

8.16 Describe the following: (i) Acetylation (ii) Cannizzaro reaction (iii) Cross-aldol condensation (iv) Decarboxylation.
Ans: (i) Acetylation: introducing an acetyl group (CH₃CO–) onto an –OH or –NH₂, typically using acetic anhydride or acetyl chloride, e.g. phenol+(CH₃CO)₂O→phenyl acetate. (ii) Cannizzaro reaction: the base-mediated disproportionation of an aldehyde with no α-hydrogen into one molecule of the alcohol and one of the carboxylate salt, e.g. 2HCHO+NaOH→CH₃OH+HCOONa. (iii) Cross-aldol condensation: an aldol condensation between two different carbonyl compounds, giving up to four possible β-hydroxy/α,β-unsaturated products depending on which partner acts as nucleophile and which as electrophile. (iv) Decarboxylation: loss of CO₂ from a carboxylic acid or its salt, e.g. sodium acetate+NaOH/CaO (soda lime), heat→methane+Na₂CO₃.

8.17 Complete the following synthesis scheme (illustrative): benzene → (Friedel–Crafts acylation) → acetophenone → (?) → 1-phenylethanol → (?) → styrene.
Ans: Benzene+CH₃COCl/anhyd. AlCl₃ → acetophenone; reduce the ketone (NaBH₄) → 1-phenylethanol; dehydrate (conc. H₂SO₄, heat) → styrene — a representative multi-step scheme of the kind tested in this question, linking Friedel–Crafts acylation, carbonyl reduction and acid-catalysed dehydration.

8.18 Explain the following: (i) Cyanohydrin formation from a ketone is often catalysed by a small amount of base. (ii) Semicarbazide reacts with carbonyl compounds through only one of its two –NH₂-type nitrogens. (iii) Esterification of a carboxylic acid with an alcohol needs an excess of one reactant or removal of water.
Ans: (i) HCN itself is a very weak acid and dissociates poorly, giving too few CN⁻ nucleophiles for a fast reaction; adding a trace of base generates more CN⁻ in situ, speeding cyanohydrin formation and improving the yield. (ii) The nitrogen adjacent to the carbonyl group (–NH–CO–NH₂) has its lone pair delocalised into that carbonyl (amide-type resonance), making it a poor nucleophile; the other, terminal –NH₂ nitrogen is not so delocalised and reacts instead. (iii) Fischer esterification is a reversible equilibrium, so by Le Chatelier’s principle using excess alcohol (or acid) or continuously removing water (e.g. via a dehydrating agent or a Dean–Stark setup) shifts the equilibrium towards the ester, improving the yield.

8.19 An organic compound contains 69.77% carbon, 11.63% hydrogen and the rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound.
Ans: Pentan-2-one, CH₃–CO–CH₂–CH₂–CH₃ (C₅H₁₀O, molar mass 86; %C=69.77, %H=11.63, matching exactly). As a ketone it does not reduce Tollens’ reagent, but its unhindered carbonyl forms a bisulphite addition compound; being a methyl ketone, it gives a positive iodoform test. Vigorous oxidative cleavage at the C2–C3 bond gives ethanoic acid (from C1–C2) and propanoic acid (from C3–C5), matching the given oxidation products exactly.

8.20 Although the phenoxide ion has more resonating structures than the carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Ans: What matters for acid strength is not the raw count of resonance structures but how effectively the negative charge is stabilised. In the carboxylate ion, the negative charge is shared equally between two highly electronegative oxygen atoms (the two C–O bonds become equivalent in a symmetric resonance hybrid), which is a very effective, low-energy form of stabilisation. In the phenoxide ion, although there are more resonance structures, the negative charge is delocalised partly onto ring carbon atoms, which are far less electronegative and stabilise negative charge much less effectively. This more effective charge stabilisation in the carboxylate makes carboxylic acids considerably more acidic than phenols despite phenoxide’s greater number of resonance forms.

Frequently Asked Questions

Why does benzaldehyde give a positive Tollens’ test but a negative Fehling’s test?
Tollens’ reagent oxidises both aliphatic and aromatic aldehydes, giving a silver mirror with benzaldehyde. Fehling’s solution, however, is only reliable for aliphatic aldehydes — aromatic aldehydes like benzaldehyde do not reduce it, which is why Tollens’ test (not Fehling’s) is the correct discriminator whenever an aromatic aldehyde is involved.

Why do aldehydes undergo nucleophilic addition more readily than ketones?
Aldehydes have only one alkyl/aryl group (and one hydrogen) attached to the carbonyl carbon, so there is less steric hindrance to an incoming nucleophile, and only one electron-donating group reduces the carbonyl carbon’s electrophilicity (versus two in a ketone), leaving it more electrophilic. Both factors make aldehydes more reactive than ketones towards nucleophilic addition.