NCERT Solutions for Class 12 Chemistry Chapter 8: Aldehydes, Ketones and Carboxylic Acids (NCERT 2026–27)
These Class 12 Chemistry Chapter 8 solutions cover Aldehydes, Ketones and Carboxylic Acids from the latest NCERT textbook (session 2026–27). Every Intext Question and every numbered Exercise question is reproduced verbatim and solved in full — with structures, IUPAC names, named reactions and step-by-step reasoning written in exam-ready CBSE style.
Chapter 8 deals with three closely related families of organic compounds that all contain the carbonyl group (>C=O). In aldehydes the carbonyl carbon is bonded to at least one hydrogen (R–CHO); in ketones it is bonded to two carbon atoms (R–CO–R′); and in carboxylic acids the carbonyl is attached to a hydroxyl group, giving the –COOH (carboxyl) group. The chapter covers their nomenclature (common and IUPAC), structure, methods of preparation, and the rich chemistry of these functional groups — nucleophilic addition, oxidation, reduction, α-hydrogen reactions (aldol and Cannizzaro), and the acidity of carboxylic acids. Mastering the many named reactions (Rosenmund, Etard, Gatterman–Koch, Stephen, Clemmensen, Wolff–Kishner, Hell–Volhard–Zelinsky and more) is the key to scoring well.
Key Concepts & Definitions
Carbonyl group: the >C=O group. The carbon is sp2-hybridised, trigonal planar with bond angles ≈ 120°. The C=O bond is polar (C is δ+, electrophilic; O is δ−, nucleophilic).
Nucleophilic addition: the characteristic reaction of aldehydes/ketones; a nucleophile attacks the electrophilic carbonyl carbon and the C changes from sp2 to sp3. Aldehydes are more reactive than ketones (less steric hindrance, less +I electron donation).
α-Hydrogen: hydrogen on the carbon adjacent to C=O; it is acidic because the resulting carbanion is resonance-stabilised. This drives the aldol condensation.
Carboxylic acid acidity: RCOOH is a stronger acid than alcohols/phenols because the carboxylate ion is stabilised by two equivalent resonance structures spreading the negative charge over two electronegative O atoms. Electron-withdrawing groups increase acidity; electron-donating groups decrease it.
Distinguishing tests: Tollens’ reagent (silver mirror) and Fehling’s reagent (red-brown ppt) oxidise aldehydes but not ketones; the iodoform test detects CH3CO– (methyl ketones) and CH3CH(OH)– groups.
Acidity (increasing): Ph < I < Br < Cl < F < CN < NO2 < CF3 (effect of substituent on the acid).
NCERT Intext Questions — Solutions
8.1 Write the structures of the following compounds.
(i) α-Methoxypropionaldehyde (ii) 3-Hydroxybutanal (iii) 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec. butyl ketone (vi) 4-Fluoroacetophenone
ANSWER(i) α-Methoxypropionaldehyde (2-methoxypropanal): CH3–CH(OCH3)–CHO. The α-carbon (next to CHO) carries the –OCH3 group.(ii) 3-Hydroxybutanal: CH3–CH(OH)–CH2–CHO. Number from CHO (C-1); OH is on C-3.(iii) 2-Hydroxycyclopentanecarbaldehyde: a cyclopentane ring with –CHO on C-1 and –OH on the adjacent C-2.(iv) 4-Oxopentanal: CH3–CO–CH2–CH2–CHO. A five-carbon chain with CHO at C-1 and a keto (=O) at C-4.(v) Di-sec-butyl ketone (2,4-dimethylpentan-3-one analogue): CH3CH2CH(CH3)–CO–CH(CH3)CH2CH3. Two sec-butyl groups on the carbonyl.(vi) 4-Fluoroacetophenone: a benzene ring bearing –COCH3 at C-1 and –F at the para (C-4) position: p-F–C6H4–COCH3.
8.2 Write the structures of products of the following reactions.
(i) Cyclohexanecarbaldehyde + CH3MgBr, then H3O+ (ii) (C6H5CH2)2Cd + 2 CH3COCl (iii) CH3–C≡C–H + H2O (Hg2+, H2SO4) (iv) m-nitrotoluene: 1. CrO2Cl2; 2. H3O+
ANSWER(i) Grignard adds to the –CHO; after hydrolysis a secondary alcohol forms → 1-cyclohexylethanol, C6H11–CH(OH)–CH3.(ii) Dialkylcadmium converts acid chlorides to ketones → 1-phenylpropan-2-one, C6H5CH2–CO–CH3 (the benzyl group joins the acetyl group).(iii) Markovnikov hydration of a higher alkyne gives a ketone → propan-2-one (acetone), CH3–CO–CH3.(iv) Etard reaction oxidises the –CH3 of m-nitrotoluene to –CHO → 3-nitrobenzaldehyde, m-O2N–C6H4–CHO.
8.3 Arrange the following compounds in increasing order of their boiling points.
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
ANSWERCH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH.Propane has only weak van der Waals forces (lowest). Dimethyl ether is weakly polar (no H-bonding). Acetaldehyde is more polar (strong dipole–dipole). Ethanol forms intermolecular hydrogen bonds, so it boils highest.
8.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.
(i) Ethanal, Propanal, Propanone, Butanone. (ii) Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone. (Hint: consider steric and electronic effects.)
ANSWER(i) Butanone < Propanone < Propanal < Ethanal. Aldehydes > ketones (less steric crowding, more +R/+I from two alkyls in ketones lowers electrophilicity); more/bulkier alkyl groups lower reactivity.(ii) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde. The ketone (acetophenone) is least reactive; the –CH3 of p-tolualdehyde donates electrons (less reactive than benzaldehyde); the –NO2 withdraws electrons, raising electrophilicity (most reactive).
8.5 Predict the products of the following reactions.
(i) Cyclohexanone with HCN (ii) Acetophenone (C6H5COCH3) with NaCN/H+ (iii) Pentan-3-one with 2,4-DNP reagent (iv) Methanal with hydroxylamine
ANSWER(i) HCN adds across C=O → the cyanohydrin 1-hydroxycyclohexane-1-carbonitrile (cyclohexanone cyanohydrin): a cyclohexane ring carrying –OH and –CN on the same carbon.(ii) → acetophenone cyanohydrin, C6H5C(OH)(CN)CH3 (2-hydroxy-2-phenylpropanenitrile).(iii) 2,4-DNP (Brady’s reagent) gives the 2,4-dinitrophenylhydrazone of pentan-3-one: (C2H5)2C=N–NH–C6H3(NO2)2 (orange/yellow solid).(iv) Hydroxylamine gives the oxime of methanal: H2C=N–OH (formaldoxime).
8.6 Give the IUPAC names of the following compounds.
(i) Ph CH2CH2COOH (ii) (CH3)2C=CHCOOH (iii) 2-methylcyclopentane carrying –COOH (methyl on C adjacent to COOH) (iv) 2,4,6-trinitro-substituted benzoic acid
8.7 Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene (ii) Acetophenone (iii) Bromobenzene (iv) Phenylethene (Styrene)
ANSWER(i) Ethylbenzene: alkaline KMnO4/Δ oxidises the entire side chain to –COOH, then H3O+ → C6H5COOH. C6H5CH2CH3 ⟶[KMnO4/KOH, Δ; H3O+] C6H5COOH.(ii) Acetophenone: the methyl ketone is oxidised by NaOI (or hot alkaline KMnO4) → benzoate, then acidify. C6H5COCH3 ⟶[I2/NaOH; H3O+] C6H5COOH + CHI3.(iii) Bromobenzene: form the Grignard, treat with CO2 (dry ice), then acidify. C6H5Br ⟶[Mg/ether] C6H5MgBr ⟶[1. CO2; 2. H3O+] C6H5COOH.(iv) Styrene: vigorous oxidation with hot alkaline KMnO4 cleaves the side chain. C6H5CH=CH2 ⟶[KMnO4/KOH, Δ; H3O+] C6H5COOH + CO2 + H2O.
8.8 Which acid of each pair shown here would you expect to be stronger?
(i) CH3CO2H or CH2FCO2H (ii) CH2FCO2H or CH2ClCO2H (iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv) benzoic acid or 4-nitrobenzoic acid
ANSWER(i) CH2FCO2H is stronger — the electron-withdrawing F (−I) stabilises the carboxylate.(ii) CH2FCO2H is stronger — F is more electronegative than Cl, so its −I effect is greater.(iii) CH3CHFCH2CO2H is stronger — the F is closer to –COOH (on the β-carbon vs the γ-carbon), so the −I effect on the acid group is stronger.(iv) 4-Nitrobenzoic acid is stronger — the –NO2 group withdraws electrons and stabilises the carboxylate ion.
NCERT Exercises — Solutions
8.1 What is meant by the following terms? Give an example of the reaction in each case.
(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
ANSWER(i) Cyanohydrin: an α-hydroxynitrile formed when HCN adds to a carbonyl. CH3CHO + HCN → CH3CH(OH)CN.(ii) Acetal: a gem-dialkoxy compound, RCH(OR′)2, from an aldehyde + 2 mol alcohol (dry HCl). CH3CHO + 2 C2H5OH → CH3CH(OC2H5)2 + H2O.(iii) Semicarbazone: >C=N–NHCONH2, from carbonyl + semicarbazide. CH3CHO + H2N–NHCONH2 → CH3CH=N–NHCONH2 + H2O.(iv) Aldol: a β-hydroxy aldehyde/ketone from self-addition of carbonyl with α-H (dil. base). 2 CH3CHO ⟶[dil. NaOH] CH3CH(OH)CH2CHO (3-hydroxybutanal).(v) Hemiacetal: RCH(OH)(OR′), an aldehyde + 1 mol alcohol. CH3CHO + C2H5OH → CH3CH(OH)(OC2H5).(vi) Oxime: >C=N–OH, from carbonyl + hydroxylamine. CH3CHO + NH2OH → CH3CH=N–OH + H2O.(vii) Ketal: a gem-dialkoxy compound from a ketone (R2C(OR′)2); ketones with ethylene glycol give cyclic ketals. (CH3)2CO + HOCH2CH2OH → cyclic ethylene ketal + H2O.(viii) Imine: >C=N–R (or >C=NH), from carbonyl + ammonia/primary amine. CH3CHO + NH3 → CH3CH=NH + H2O.(ix) 2,4-DNP-derivative: the 2,4-dinitrophenylhydrazone, >C=N–NH–C6H3(NO2)2; orange/red solid used to identify carbonyls. CH3CHO + 2,4-DNP → CH3CH=N–NH–C6H3(NO2)2.(x) Schiff’s base: a substituted imine >C=N–R from a carbonyl + a primary (aryl) amine. C6H5CHO + C6H5NH2 → C6H5CH=N–C6H5 + H2O.
8.2 Name the following compounds according to IUPAC system of nomenclature.
(i) CH3CH(CH3)CH2CH2CHO (ii) CH3CH2COCH(C2H5)CH2CH2Cl (iii) CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CHO-p
8.3 Draw the structures of the following compounds.
(i) 3-Methylbutanal (ii) p-Nitropropiophenone (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p′-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid
ANSWER(i) 3-Methylbutanal: (CH3)2CH–CH2–CHO.(ii) p-Nitropropiophenone:p-O2N–C6H4–CO–CH2CH3 (a benzene ring with –COCH2CH3 and a para –NO2).(iii) p-Methylbenzaldehyde:p-CH3–C6H4–CHO.(iv) 4-Methylpent-3-en-2-one: CH3–CO–CH=C(CH3)–CH3 i.e. (CH3)2C=CH–CO–CH3.(v) 4-Chloropentan-2-one: CH3–CO–CH2–CHCl–CH3.(vi) 3-Bromo-4-phenylpentanoic acid: CH3–CH(C6H5)–CHBr–CH2–COOH.(vii) p,p′-Dihydroxybenzophenone: (p-HO–C6H4)2C=O (two para-hydroxyphenyl groups on a central carbonyl).(viii) Hex-2-en-4-ynoic acid: CH3–C≡C–CH=CH–COOH (C-1 = COOH, double bond at C-2, triple bond at C-4).
8.4 Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO (iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO (v) cyclopentane–CHO (cyclopentane carbaldehyde) (vi) PhCOPh
8.5 Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde (ii) Cyclopropanone oxime (iii) Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde
ANSWER(i) C6H5CH=N–NH–C6H3(NO2)2 (the 2,4-dinitrophenyl group bears NO2 at positions 2 and 4).(ii) a cyclopropane ring with the ring carbon double-bonded to N–OH: cyclopropylidene =N–OH.(iii) Acetaldehyde dimethyl acetal: CH3CH(OCH3)2.(iv) Cyclobutanone semicarbazone: the cyclobutane ring carbon =N–NHCONH2.(v) Ethylene ketal of hexan-3-one: a five-membered 1,3-dioxolane ring (–O–CH2–CH2–O–) with the C-3 carbon bearing two ethyl groups, (C2H5)2C(–O–CH2CH2–O–).(vi) Methyl hemiacetal of formaldehyde: H2C(OH)(OCH3).
8.6 Predict the products formed when cyclohexanecarbaldehyde reacts with 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
ANSWERLet C6H11– denote the cyclohexyl group; the substrate is C6H11CHO.(i) Grignard addition then hydrolysis gives a secondary alcohol: C6H11–CH(OH)–C6H5 (phenyl(cyclohexyl)methanol).(ii) Tollens’ reagent oxidises the aldehyde to the carboxylate; on acidification → cyclohexanecarboxylic acid, C6H11COOH, plus a silver mirror.(iii) Semicarbazide gives the semicarbazone: C6H11CH=N–NHCONH2.(iv) Excess ethanol + dry acid gives the diethyl acetal: C6H11CH(OC2H5)2.(v) Clemmensen reduction converts –CHO to –CH3 → methylcyclohexane, C6H11CH3.
8.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal (ii) 2-Methylpentanal (iii) Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal
ANSWERAldol condensation (have α-H): (ii) 2-methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, (vii) phenylacetaldehyde.Cannizzaro reaction (no α-H, an aldehyde): (i) methanal, (iii) benzaldehyde, (ix) 2,2-dimethylbutanal.Neither: (iv) benzophenone (a ketone with no α-H) and (viii) butan-1-ol (an alcohol, not a carbonyl).Sample aldol product — cyclohexanone self-aldol then −H2O gives 2-(cyclohexylidene)cyclohexan-1-one (an α,β-unsaturated ketone formed by joining two rings at the α-carbon).Sample aldol product — 2-methylpentanal → 2-methyl-2-(1-hydroxy…) aldol, which on dehydration gives the α,β-unsaturated aldehyde 2,4-dimethyl-2-heptenal.Cannizzaro products: benzaldehyde → C6H5CH2OH (benzyl alcohol) + C6H5COO−Na+ (sodium benzoate); methanal → CH3OH + HCOO−Na+.
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
8.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
ANSWERCross-aldol of propanal (CH3CH2CHO) and butanal (CH3CH2CH2CHO) gives four β-hydroxy aldehydes (shown before dehydration). The α-carbanion of one acts as nucleophile, the CHO of the other as electrophile.(1) Propanal (Nu) + propanal (E) → 3-hydroxy-2-methylpentanal, CH3CH2CH(OH)CH(CH3)CHO.(2) Butanal (Nu) + butanal (E) → 2-ethyl-3-hydroxyhexanal, CH3CH2CH2CH(OH)CH(C2H5)CHO.(3) Propanal (Nu) + butanal (E) → 3-hydroxy-2-methylhexanal, CH3CH2CH2CH(OH)CH(CH3)CHO.(4) Butanal (Nu) + propanal (E) → 2-ethyl-3-hydroxypentanal, CH3CH2CH(OH)CH(C2H5)CHO.(On heating, each loses water to give the corresponding α,β-unsaturated aldehyde.)
8.10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound.
ANSWER2,4-DNP positive → a carbonyl; reduces Tollens’ → an aldehyde; undergoes Cannizzaro → no α-hydrogen, so the CHO must be attached directly to the ring. Oxidation to phthalic acid (benzene-1,2-dicarboxylic acid) means there are two side chains in the ortho positions, one being –CHO and the other an ethyl group (C9H10O = ring + CHO + C2H5).Therefore the compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde), o-(C2H5)C6H4CHO.
8.11 An organic compound (A) (molecular formula 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.
ANSWER(C) dehydrates to but-1-ene → (C) is butan-1-ol, CH3CH2CH2CH2OH. Oxidation of (C) gives (B), so (B) = butanoic acid, CH3CH2CH2COOH. The ester of (B) and (C) (C8H16O2) is (A) = butyl butanoate, CH3CH2CH2COOCH2CH2CH2CH3.Hydrolysis: CH3CH2CH2COOC4H9 + H2O ⟶[dil. H2SO4] CH3CH2CH2COOH (B) + C4H9OH (C).Oxidation: CH3CH2CH2CH2OH ⟶[H2CrO4] CH3CH2CH2COOH (B).Dehydration: CH3CH2CH2CH2OH ⟶[conc. H2SO4, Δ] CH3CH2CH=CH2 (but-1-ene) + H2O.
8.12 Arrange the following compounds in increasing order of their property as 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)
ANSWER(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde. (More/bulkier groups → greater steric hindrance and +I, so lower reactivity to HCN.)(ii) (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH. (−I of Br increases acidity, and is strongest when Br is on the α-carbon; the branched acid with +I is weakest.)(iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid. (–OCH3 donates electrons (weaker acid); –NO2 withdraws; two –NO2 groups give the strongest.)
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
ANSWER(i) Tollens’/Fehling’s test: propanal (aldehyde) gives a silver mirror / red-brown ppt; propanone does not.(ii) Iodoform test: acetophenone (a methyl ketone) gives a yellow ppt of CHI3; benzophenone does not.(iii) NaHCO3 test: benzoic acid gives brisk effervescence (CO2); phenol does not. (Phenol gives a violet colour with neutral FeCl3.)(iv) NaHCO3 test: benzoic acid liberates CO2; ethyl benzoate (an ester) does not.(v) Iodoform test: pentan-2-one (a methyl ketone, CH3CO–) gives yellow CHI3; pentan-3-one does not.(vi) Tollens’ test: benzaldehyde (aldehyde) gives a silver mirror; acetophenone does not. (Acetophenone gives a positive iodoform test, benzaldehyde does not.)(vii) Iodoform test: ethanal (CH3CHO) gives a yellow ppt of iodoform; propanal does not.
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.
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
8.16 Describe the following.
(i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation (iv) Decarboxylation
ANSWER(i) Acetylation: introduction of an acetyl group (CH3CO–) onto an –OH or –NH2 group, usually with acetic anhydride or acetyl chloride (in pyridine). e.g. C6H5OH + (CH3CO)2O → C6H5OCOCH3 + CH3COOH.(ii) Cannizzaro reaction: aldehydes with no α-H, on heating with concentrated alkali, undergo self oxidation–reduction (disproportionation) — one molecule is reduced to an alcohol and another oxidised to a carboxylate. 2 HCHO ⟶[conc. NaOH] CH3OH + HCOONa.(iii) Cross aldol condensation: aldol condensation between two different aldehydes/ketones. If both have α-H, a mixture of four products results; it is synthetically useful when one component (e.g. benzaldehyde) has no α-H. e.g. C6H5CHO + CH3CHO ⟶[dil. NaOH] C6H5CH=CHCHO + H2O.(iv) Decarboxylation: loss of CO2 from the sodium salt of a carboxylic acid on heating with sodalime (NaOH + CaO, 3:1), giving a hydrocarbon. CH3COONa + NaOH ⟶[CaO, Δ] CH4 + Na2CO3.
8.17 Complete each synthesis by giving missing starting material, reagent or products.
[The NCERT prints these as nine lettered structure schemes, (i)–(ix). Worked logically below.]
ANSWERThis question gives partial reaction schemes in the printed book (structures only). The reasoning to complete them is:(i) Aldehyde/ketone + the shown ylide/reagent → supply the carbonyl product or the alcohol obtained on reduction.(ii) An α,β-unsaturated carbonyl formed by aldol condensation → trace it back to the two carbonyls bearing α-H.(iii) Conversion of –COOH to –COCl with SOCl2, then Rosenmund (H2, Pd/BaSO4) to the aldehyde.(iv) Grignard + carbonyl → alcohol; identify R–MgX and the carbonyl.(v) Clemmensen / Wolff–Kishner reduction of >C=O to >CH2.(vi) HVZ α-halogenation: RCH2COOH + Br2/red P → RCHBrCOOH.(vii) Esterification: RCOOH + R′OH/H+ → ester.(viii) Nitrile hydrolysis or Tollens’ oxidation to give the carboxylic acid.(ix) Decarboxylation of the sodium salt with sodalime to the hydrocarbon.Refer to the structures printed alongside each scheme in your textbook; the named reactions above complete every blank.
8.18 Give plausible explanation for each of the following.
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
ANSWER(i) Cyanohydrin formation is a nucleophilic addition needing access to the carbonyl carbon. In 2,2,6-trimethylcyclohexanone the bulky methyl groups flanking C=O cause severe steric hindrance, blocking the approach of CN−, so little cyanohydrin forms.(ii) In semicarbazide H2N–NH–CO–NH2, one –NH2 is attached to the carbonyl (C=O of urea) and its lone pair is delocalised into that carbonyl by resonance, so it is a poor nucleophile. The other (terminal) –NH2 is a free, good nucleophile and is the one that condenses with the carbonyl to give the semicarbazone.(iii) Esterification is reversible; by Le Chatelier’s principle, continuously removing the water (or distilling off the ester) drives the equilibrium to the right and increases the yield of ester.
8.19 An organic compound contains 69.77% carbon, 11.63% hydrogen and 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 positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
ANSWERFind the empirical formula. C: 69.77/12 = 5.81; H: 11.63/1 = 11.63; O: (100−69.77−11.63 = 18.60)/16 = 1.16. Ratio = 5.81:11.63:1.16 = 5:10:1 → empirical formula C5H10O (mass 86). Since molecular mass = 86, molecular formula = C5H10O.Interpret the tests. NaHSO3 addition + does not reduce Tollens’ → a ketone (not an aldehyde). Positive iodoform → it contains a CH3CO– group (a methyl ketone). Oxidation gives ethanoic + propanoic acids → the C–C bonds on either side of C=O cleave to give a 2-carbon and a 3-carbon acid.Therefore the compound is pentan-2-one, CH3COCH2CH2CH3.
8.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
ANSWERAcid strength depends on the stability of the conjugate base, which depends on how effectively the negative charge is delocalised, not merely on the number of resonance structures.In the carboxylate ion the two resonance structures are equivalent, and the negative charge is delocalised over two electronegative oxygen atoms — very effective stabilisation.In the phenoxide ion the major contributing structures are non-equivalent; in three of them the negative charge sits on the less electronegative carbon atoms of the ring, which is far less stabilising. So although phenoxide has more resonance structures, they stabilise the charge less effectively.Hence the carboxylate ion is more stabilised than the phenoxide ion, and carboxylic acids are stronger acids than phenols.
Extra Practice Questions
Short Answer Type (Q1–Q5)
Q1. Why are aldehydes more reactive than ketones towards nucleophilic addition?
ANSWERKetones have two alkyl/aryl groups on the carbonyl carbon. These cause greater steric hindrance to the attacking nucleophile and donate electron density (+I/+R), lowering the electrophilicity of the carbonyl carbon. Aldehydes have only one such group, so they are less hindered and more electrophilic, hence more reactive.
Q2. Why does benzaldehyde not give the Fehling’s test though it gives the Tollens’ test?
ANSWERAromatic aldehydes such as benzaldehyde do not respond to Fehling’s test (the milder, alkaline Cu2+ reagent is unable to oxidise them), but the stronger Tollens’ reagent (ammoniacal AgNO3) does oxidise benzaldehyde, giving a silver mirror.
Q3. What is the Hell–Volhard–Zelinsky (HVZ) reaction? Give an example.
ANSWERA carboxylic acid with an α-H, on treatment with Cl2 or Br2 in the presence of a small amount of red phosphorus, is halogenated at the α-position to give an α-halo acid. e.g. CH3COOH + Cl2 ⟶[red P] ClCH2COOH + HCl.
ANSWERCH3COOH < HCOOH < ClCH2COOH < Cl3CCOOH. The +I of CH3 makes acetic acid weakest; the −I of Cl increases acidity, and three Cl atoms (trichloroacetic acid) give the strongest acid (pKa lowest).
Q5. Why is the boiling point of a carboxylic acid higher than that of an alcohol of comparable molecular mass?
ANSWERCarboxylic acid molecules associate through stronger and more extensive intermolecular hydrogen bonding, even forming stable cyclic dimers (held by two H-bonds). These dimers are not fully broken even in the vapour phase, so more energy is needed to vaporise them than for alcohols, giving carboxylic acids the higher boiling point.
Long Answer Type (Q6–Q8)
Q6. Explain the mechanism of nucleophilic addition of HCN to a carbonyl compound, and state why a trace of base accelerates it.
ANSWERThe carbonyl carbon is electrophilic (δ+). A nucleophile (CN−) attacks it from a direction roughly perpendicular to the plane of the sp2 carbon. The carbon rehybridises from sp2 to sp3, and a tetrahedral alkoxide intermediate is formed (negative charge now on oxygen). This intermediate then captures a proton (from HCN or the medium) to give the neutral product — the cyanohydrin, RR′C(OH)CN. The net change is addition of CN− and H+ across C=O. Pure HCN reacts very slowly because it is only weakly ionised; a trace of base generates a higher concentration of the much stronger nucleophile CN−, which speeds up the rate-determining nucleophilic attack.
Q7. Describe four important methods for preparing carboxylic acids, with one equation each.
ANSWER(a) Oxidation of primary alcohols/aldehydes: CH3CH2OH ⟶[KMnO4/H+] CH3COOH (or RCHO ⟶[mild oxidant] RCOOH).(b) From alkylbenzenes (side-chain oxidation): C6H5CH2CH3 ⟶[KMnO4/KOH, Δ; H+] C6H5COOH.(c) From nitriles (hydrolysis): RCN ⟶[H2O/H+ or OH−] RCOOH (via the amide).(d) From Grignard reagents + CO2: RMgX ⟶[1. CO2(dry ice); 2. H3O+] RCOOH. (Methods c and d add one carbon, ascending the series.)
Q8. Discuss the effect of substituents on the acidity of carboxylic acids with suitable examples.
ANSWERSubstituents alter acidity by changing the stability of the carboxylate conjugate base. Electron-withdrawing groups (EWG) (e.g. –NO2, –CN, halogens) disperse and stabilise the negative charge by −I/−R effects, so they increase acidity — e.g. ClCH2COOH (pKa 2.86) is stronger than CH3COOH (4.76), and the order of substituent effect is Ph < I < Br < Cl < F < CN < NO2 < CF3. Electron-donating groups (EDG) (e.g. –CH3, –OCH3) intensify the charge and destabilise the carboxylate, so they decrease acidity — 4-methoxybenzoic acid is weaker than benzoic acid. The effect is stronger the closer the group is to –COOH (α > β > γ). On an aromatic ring, –NO2 at ortho/para raises acidity most (resonance + induction).
MCQs & Assertion–Reason
1. The carbonyl carbon in aldehydes and ketones is:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: Aldehydes are more reactive than ketones in nucleophilic addition.
Reason: Ketones have two alkyl groups that cause steric hindrance and reduce the electrophilicity of the carbonyl carbon.
A-R 2. Assertion: Carboxylic acids are stronger acids than phenols.
Reason: The carboxylate ion is stabilised by two equivalent resonance structures with charge on two oxygen atoms.
A-R 3. Assertion: Benzaldehyde does not undergo aldol condensation.
Reason: Benzaldehyde has no α-hydrogen atom.
A-R 4. Assertion: Formaldehyde gives the iodoform test.
Reason: The iodoform test is given by compounds containing the CH3CO– or CH3CH(OH)– group.
A-R 5. Assertion: 4-Nitrobenzoic acid is a stronger acid than benzoic acid.
Reason: The –NO2 group is electron-withdrawing and stabilises the carboxylate ion.
Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A). A-R 4: Assertion is false — formaldehyde (HCHO) has no CH3CO– group, so it does NOT give the iodoform test; the Reason is a correct general statement.
Common Mistakes & Exam Tips
Avoid these errors
Numbering an aldehyde chain from the wrong end — the –CHO carbon is always C-1.
Saying aromatic aldehydes give Fehling’s test — they give Tollens’ but not Fehling’s.
Claiming HCHO gives the iodoform test — it has no CH3CO– group, so it does not.
Mixing up Clemmensen (Zn–Hg/HCl, acidic) with Wolff–Kishner (NH2NH2/KOH, basic) — both reduce C=O to CH2, but use opposite media.
Ranking acidity by “number of resonance structures” — rank by how effectively the charge is delocalised.
Forgetting that –COOH is meta-directing and deactivating, so it does not undergo Friedel–Crafts reactions.
How to score full marks
Memorise the reagent for each named reaction (Rosenmund, Etard, Gatterman–Koch, Stephen, Clemmensen, Wolff–Kishner, HVZ, Cannizzaro, aldol) — one-mark and conversion questions test these directly. For “distinguish between” questions, quote a single decisive test with the observation (silver mirror, red-brown ppt, yellow CHI3, CO2 with NaHCO3). For acidity ordering, justify with −I/+I and resonance. Always show the reagent over the arrow and balance simple by-products (H2O, HCl).
Frequently Asked Questions
What is Class 12 Chemistry Chapter 8 about?
Chapter 8, Aldehydes, Ketones and Carboxylic Acids, studies three families of carbonyl-containing compounds — their nomenclature, structure, preparation, physical properties, and reactions including nucleophilic addition, oxidation, reduction, aldol and Cannizzaro reactions, and the acidity of carboxylic acids.
How do you distinguish an aldehyde from a ketone?
Use Tollens’ reagent (ammoniacal AgNO3) or Fehling’s reagent: an aldehyde gives a silver mirror / red-brown precipitate because it is oxidised, while a ketone gives no reaction.
Why are carboxylic acids more acidic than phenols?
Because the carboxylate conjugate base is stabilised by two equivalent resonance structures spreading the negative charge over two electronegative oxygen atoms, whereas the phenoxide ion places much of its charge on less electronegative ring carbons. Effective charge delocalisation, not the number of resonance structures, decides acid strength.
Are these Class 12 Chemistry Chapter 8 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27, with every Intext Question and Exercise solved.
