NCERT Solutions for Class 12 Chemistry Chapter 8: Aldehydes, Ketones and Carboxylic Acids

Concept. Each name is a precise recipe. For an aldehyde the –CHO carbon is C-1 and the suffix is “-al” (or “-carbaldehyde” when –CHO hangs off a ring); Greek letters α, β, γ in a common name count outward from the carbon next to –CHO (α = C-2). A ketone is named from its two groups + “ketone”, or by “-one” with the lowest locant for C=O. A prefix such as “oxo-” means a =O on the chain, “hydroxy-” an –OH and “methoxy-” an –OCH₃. I draw the parent first, place every substituent on its carbon, and complete the valencies.

Notes. In (i) “propionaldehyde” is the common name of propanal, and α means the carbon adjacent to –CHO (C-2), so the methoxy group sits there. In (iv) “4-oxopentanal” is a compound with both a ketone (C4, oxo) and an aldehyde (C1) group; the aldehyde, being the principal group, fixes the numbering and the suffix. In (v) di-sec-butyl ketone is (CH₃CH₂CH(CH₃))₂C=O. Taking the longest chain through the carbonyl, CH₃–CH₂–CH(CH₃)–CO–CH(CH₃)–CH₂–CH₃, gives a seven-carbon parent (heptane) with the ketone on C-4 and methyl groups on C-3 and C-5, so the IUPAC name is 3,5-dimethylheptan-4-one. (The common name “di-sec-butyl ketone” is unambiguous and is what the question asks to draw.)

Conclusion: (i) CH₃CH(OCH₃)CHO; (ii) CH₃CH(OH)CH₂CHO; (iii) 2-hydroxycyclopentane-1-carbaldehyde; (iv) CH₃COCH₂CH₂CHO; (v) [CH₃CH₂CH(CH₃)]₂C=O (di-sec-butyl ketone); (vi) 4-F–C₆H₄–COCH₃.

Concept. Four preparations of carbonyl compounds are tested: Friedel–Crafts acylation, the dialkylcadmium route to ketones, acid-catalysed hydration of an alkyne, and the Étard reaction. In each, decide which carbon becomes the carbonyl carbon and what group ends up on it.

Why these products. (i) The electrophile in Friedel–Crafts acylation is the resonance-stabilised acylium ion, so the product is always an aryl ketone, never over-acylated (the acyl group deactivates the ring). (ii) Dialkyl- and diarylmethyl-cadmium reagents are mild enough to react with acyl chlorides but not with the ketone formed, so the reaction stops cleanly at the ketone — the preferred way to make unsymmetrical ketones without the double addition that a Grignard reagent would give. (iii) By Markovnikov’s rule water adds with –OH on the internal carbon of propyne, giving prop-1-en-2-ol, which tautomerises to acetone; only ethyne itself gives an aldehyde (acetaldehyde). (iv) CrO₂Cl₂ (chromyl chloride) selectively oxidises a methyl group on the ring to the aldehyde stage, forming a brown chromium complex that is hydrolysed to the aldehyde; the –NO₂ group survives.

Conclusion: (i) propiophenone C₆H₅COC₂H₅; (ii) phenylacetone C₆H₅CH₂COCH₃ (two molecules); (iii) acetone CH₃COCH₃; (iv) 4-nitrobenzaldehyde.

Concept. Boiling point depends on the strength of the intermolecular forces to be overcome. The four compounds have almost the same molar mass (CH₃CH₂CH₃ 44, CH₃CHO 44, CH₃OCH₃ 46, CH₃CH₂OH 46 g mol⁻¹), so molar mass is roughly constant and the type of force decides the order: (1) non-polar hydrocarbons have only weak van der Waals forces; (2) ethers and aldehydes are polar and have dipole–dipole attractions as well, so they boil higher; (3) alcohols form intermolecular hydrogen bonds (O–H···O), the strongest of these interactions, so they boil highest.

Why acetaldehyde boils above dimethyl ether. Both are polar, but the carbonyl group of acetaldehyde is far more polar than the C–O–C of an ether (the dipole moment of acetaldehyde ≈ 2.7 D against ≈ 1.3 D for the ether), so its dipole–dipole attraction is stronger and it boils about 46 K higher. Ethanol, although it has the same molar mass as the ether, boils highest because it is the only member that can form hydrogen bonds between its own molecules — a much stronger attraction (~20 kJ mol⁻¹ per bond) than dipole–dipole forces, and breaking them needs a great deal of energy.

Conclusion: Increasing boiling point — CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH.

Concept. In nucleophilic addition the nucleophile attacks the electrophilic (δ+) carbonyl carbon. A carbonyl compound is more reactive when (a) the carbonyl carbon carries a greater positive charge (fewer or weaker electron-releasing groups attached to it) and (b) the carbon is less crowded, so the nucleophile and the developing sp³ centre meet less steric strain. Both effects make aldehydes more reactive than ketones: a ketone has two electron-releasing alkyl/aryl groups that reduce the δ+ on carbon (electronic effect) and shield it (steric effect), whereas an aldehyde has only one alkyl group and an H.

(i) Ethanal, propanal, propanone, butanone. Ethanal (CH₃CHO) and propanal (CH₃CH₂CHO) are aldehydes; propanone (CH₃COCH₃) and butanone (CH₃COCH₂CH₃) are ketones. Within the aldehydes, ethanal (one methyl) is more reactive than propanal (a bulkier ethyl donates slightly more and hinders more). Within the ketones, propanone (two methyls) is more reactive than butanone (a methyl + a larger ethyl). All aldehydes outrank all ketones.

(ii) Benzaldehyde, p-tolualdehyde, p-nitrobenzaldehyde, acetophenone. Acetophenone (C₆H₅COCH₃) is a ketone, so it is the least reactive. Among the aromatic aldehydes, a para substituent changes the δ+ on the carbonyl carbon through the ring: the electron-withdrawing –NO₂ (−I, −R) increases the positive charge and raises reactivity, while the electron-releasing –CH₃ (+I, hyperconjugation) decreases it. So p-nitrobenzaldehyde > benzaldehyde > p-tolualdehyde, and all three aldehydes are more reactive than acetophenone.

Extra depth. Aromatic aldehydes and ketones are, in general, less reactive than their aliphatic analogues because the ring donates electron density into the carbonyl by resonance (reducing the δ+ on carbon) and provides steric bulk — which is why even benzaldehyde is less reactive than ethanal. The para-nitro group opposes this ring donation, making p-nitrobenzaldehyde the most reactive of the aromatic set.

Conclusion: (i) Butanone < Propanone < Propanal < Ethanal; (ii) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde.

Concept. All four are nucleophilic addition–elimination reactions of ammonia derivatives H₂N–Z with a carbonyl group. The nitrogen adds to the carbonyl carbon, a proton transfers, and a molecule of water is then eliminated, replacing the C=O by a C=N–Z double bond: >C=O + H₂N–Z → >C=N–Z + H₂O. A weakly acidic medium is used so that H₂N–Z is not fully protonated (it must keep its lone pair to act as a nucleophile) but there is enough acid to protonate the carbinolamine and help lose water. The product’s name depends on Z.

Working. In (i) the ring ketone becomes the cyclic oxime with C=N–OH. In (ii) the 2,4-DNP reagent gives the characteristic orange-red crystalline derivative used to detect and identify aldehydes and ketones (its sharp melting point identifies the carbonyl compound). In (iii) only the aldehyde C=O reacts with semicarbazide; the C=C double bond is untouched, so the conjugated semicarbazone forms. In (iv) a primary amine condenses with the ketone to give a Schiff base (a substituted imine); the C=N carbon still carries the methyl and phenyl groups from acetophenone.

Extra depth. Semicarbazide has two –NH₂ groups but only the terminal one (NH₂–NH–) acts as the nucleophile; the other nitrogen’s lone pair is delocalised into the adjacent C=O of the semicarbazide and so is not available (this is exercise 8.18(ii)). These condensations are reversible and are usually carried out at an optimum pH (≈ 3.5): too acidic protonates the amine, too basic slows the dehydration step.

Conclusion: (i) cyclopentanone oxime; (ii) cyclohexanone 2,4-dinitrophenylhydrazone; (iii) R–CH=CH–CH=N–NH–CO–NH₂ (semicarbazone); (iv) C₆H₅C(CH₃)=N–C₂H₅ (Schiff base).

Concept. The carboxyl group –COOH is the principal functional group and its carbon is always C-1; the suffix is “-oic acid” for a chain, or “-carboxylic acid” when –COOH is attached to a ring (then the ring carbon bearing it is C-1 and the ring is numbered to give the substituents the lowest locants). A C=C in the chain makes the ending “-enoic acid”, with the –COOH still taking the lowest locant.

Notes. In (ii) the chain is four carbons (but-), the double bond is between C2 and C3 (but-2-enoic), and the extra methyl on C3 gives 3-methylbut-2-enoic acid (the common name is β,β-dimethylacrylic acid / senecioic acid). In (iii) the “-carboxylic acid” suffix is used because the –COOH is a substituent on the ring and its carbon is not counted as part of the ring; the ring carbon attached to –COOH is C-1 and numbering proceeds to give the methyl the lowest locant (2). In (iv) the three –NO₂ groups occupy the 2, 4 and 6 positions around the –COOH-bearing C-1.

Conclusion: (i) 3-phenylpropanoic acid; (ii) 3-methylbut-2-enoic acid; (iii) 2-methylcyclopentane-1-carboxylic acid; (iv) 2,4,6-trinitrobenzoic acid.

Concept. Several standard routes to carboxylic acids are tested: direct oxidation (of a primary alcohol or an aldehyde), building the chain by one carbon through a nitrile (to lengthen the carbon skeleton), carbonation of a Grignard reagent (to put –COOH where a halogen was), and oxidative cleavage/oxidation of a side chain or ring. The key decision each time is whether the carbon count must change.

Note on (iii). An aryl halide bearing –NO₂ cannot form a Grignard reagent (the reagent would attack the nitro group), and the C–Br bond of a haloarene is not displaced by cyanide under ordinary conditions, so 3-nitrobromobenzene is not converted to the acid by a one-step substitution. In practice 3-nitrobenzoic acid is obtained by the side-chain oxidation of 3-nitrotoluene with KMnO₄ (m-O₂N–C₆H₄–CH₃ → m-O₂N–C₆H₄–COOH); alternatively 3-nitrobenzoic acid is prepared by nitrating benzoic acid (–COOH is meta-directing). This is the chemically sound route; the bromo compound in the question is best read as the corresponding methyl compound.

Why each reagent. In (ii) the carbon skeleton must grow by one (benzyl → phenylacetic), so the alcohol is turned into a halide, then a nitrile (which adds the extra carbon as –CN), then hydrolysed to –COOH. In (iv) and (v) vigorous KMnO₄ oxidises every carbon chain attached to the ring (or breaks the ring’s C=C) down to –COOH. In (i) and (vi) only an oxidation is needed because the carbon count is unchanged.

Conclusion: (i) KMnO₄ oxidation; (ii) → benzyl bromide → benzyl cyanide → hydrolysis; (iii) side-chain KMnO₄ oxidation of 3-nitrotoluene (the Grignard route fails because of –NO₂); (iv) hot KMnO₄ oxidises both side chains; (v) hot KMnO₄ cleaves the ring to adipic acid; (vi) mild oxidation of the aldehyde.

Concept. A carboxylic acid is stronger when its conjugate base (the carboxylate ion) is more stable. An electron-withdrawing group near –COOH disperses the negative charge of the carboxylate by its –I (inductive) effect and so stabilises it, increasing acidity; an electron-releasing group does the opposite. Three rules follow: (1) a more electronegative atom withdraws more strongly; (2) the effect is stronger the closer the group is to –COOH and falls off rapidly with distance; (3) on a benzene ring, an electron-withdrawing substituent raises and an electron-releasing one lowers the acid strength.

Working through the reasoning. In (i) replacing an H of acetic acid by F introduces a strong –I pull toward the carboxylate, so fluoroacetic acid (pKₐ ≈ 2.6) is far stronger than acetic acid (pKₐ 4.76). In (ii) both halogens withdraw electrons, but fluorine, being the most electronegative element, withdraws most strongly, so CH₂FCO₂H is slightly stronger than CH₂ClCO₂H. In (iii) the position of the fluorine decides: on the carbon nearer to –COOH (β, as in CH₃CHFCH₂CO₂H) the inductive effect is transmitted through fewer bonds and is felt more strongly than when F is on the far (γ) carbon, so 3-fluorobutanoic acid is stronger than 4-fluorobutanoic acid. In (iv) the para-CF₃ group withdraws electron density from the ring and the carboxylate, raising acidity, while para-CH₃ donates electron density and lowers it.

Conclusion: The stronger acid is (i) CH₂FCO₂H; (ii) CH₂FCO₂H; (iii) CH₃CHFCH₂CO₂H (fluorine closer to –COOH); (iv) p-F₃C–C₆H₄–COOH.

Concept. All ten terms name the products of nucleophilic addition (or addition–elimination) to the carbonyl group of an aldehyde or ketone. In each case a nucleophile adds to the electrophilic carbonyl carbon; whether a molecule of water is then eliminated decides whether the product keeps the C–O single bond (cyanohydrin, acetal, hemiacetal) or forms a new C=N double bond (imine, oxime, hydrazone-type derivatives).

Extra depth. A hemiacetal is the half-way product (–OH and –OR on the same carbon) and is usually unstable; a second molecule of alcohol, with dry HCl, converts it to the stable acetal (two –OR groups). The same applies to ketones giving hemiketals and ketals; cyclic ketals from ethylene glycol (1,3-dioxolanes) are used to protect a carbonyl group while another part of the molecule is modified, and are hydrolysed back with dilute acid. Items (iii), (vi), (ix) and (x) are all of the C=N–Z family (addition then loss of water); their sharp melting points make them useful for identifying carbonyl compounds.

Conclusion: Each term names a specific carbonyl addition product; the examples above (cyanohydrin from HCN, acetal/ketal from alcohols, oxime/imine/Schiff base/2,4-DNP/semicarbazone from the H₂N–Z family, and the aldol from self-addition) define them.

Concept. The principal group fixes the suffix and the numbering: –CHO (“-al”, C-1), >C=O (“-one”, lowest locant), –COOH (“-oic acid”, C-1). When two carbonyls are present, number to give them the lowest locants and use “-dial”, “-dione” or “-dioic acid”. A C=C makes the ending “-enal/-enoic”. Substituents are cited as prefixes in alphabetical order.

Notes on the harder ones. In (ii) the longest chain through the carbonyl is six carbons (hexan-3-one); numbering from the end nearer the C=O gives the ketone locant 3, the ethyl group locant 4 and the chlorine locant 6, so the name is 6-chloro-4-ethylhexan-3-one. In (v) the longest chain containing the carbonyl is six carbons (the C(CH₃)₂ counts as C3 bearing two methyls); with the ketone on C2, the three methyls fall on C3, C3 and C5 → 3,3,5-trimethylhexan-2-one. In (vii) the two aldehyde groups are on a ring, so the suffix is “-dicarbaldehyde”.

Conclusion: (i) 4-methylpentanal; (ii) 6-chloro-4-ethylhexan-3-one; (iii) but-2-enal; (iv) pentane-2,4-dione; (v) 3,3,5-trimethylhexan-2-one; (vi) 3,3-dimethylbutanoic acid; (vii) benzene-1,4-dicarbaldehyde.

Concept. Reading a name into a structure: the suffix tells the functional group (-al aldehyde, -one ketone, -oic acid acid, and “-phenone” a phenyl ketone), the root gives the chain length, and the prefixes with locants give the substituents. For “propiophenone” the parent is propiophenone (C₆H₅COCH₂CH₃), and p- (para) positions a ring substituent opposite the carbonyl side chain. I draw the parent, place groups and complete valencies.

Notes. In (ii) “propiophenone” is ethyl phenyl ketone, so p-nitropropiophenone has the nitro group on the ring opposite the –COCH₂CH₃ group. In (iv) the C=C is between C3 and C4 and the ketone is on C2 (mesityl oxide, a common aldol-condensation product of acetone). In (viii) the chain has both a double bond (C2–C3) and a triple bond (C4–C5); the –COOH still takes C-1, giving hex-2-en-4-ynoic acid.

Conclusion: (i) (CH₃)₂CHCH₂CHO; (ii) p-O₂NC₆H₄COC₂H₅; (iii) p-CH₃C₆H₄CHO; (iv) CH₃COCH=C(CH₃)₂; (v) CH₃COCH₂CHClCH₃; (vi) CH₃CH(C₆H₅)CH(Br)CH₂COOH; (vii) (p-HOC₆H₄)₂CO; (viii) CH₃C≡CCH=CHCOOH.

Concept. Same rules as before — the principal group fixes the suffix and C-1 (or the lowest locant). Many of these also have well-known common names, which the question asks for where they exist.

Working for (ii). The longest chain containing –CHO is six carbons (hexanal): number from the –CHO end, C1 = CHO, C2 = CH(CH₃), C3 = CH₂, C4 = CHBr, C5 = CH₂, C6 = CH₃. So the methyl is on C2 and the bromine on C4, giving 4-bromo-2-methylhexanal. In (iv) the parent is prop-2-enal (acrylaldehyde) with a phenyl on C3 — the fragrant component of cinnamon oil, cinnamaldehyde. In (v) the aldehyde is attached to a ring, so the suffix is “-carbaldehyde”, and in (vi) the symmetrical diaryl ketone is diphenylmethanone (benzophenone).

Conclusion: (i) heptan-2-one; (ii) 4-bromo-2-methylhexanal; (iii) heptanal; (iv) 3-phenylprop-2-enal (cinnamaldehyde); (v) cyclohexanecarbaldehyde; (vi) diphenylmethanone (benzophenone).

Concept. Each derivative is the product of a specific carbonyl addition. For the C=N–Z family (2,4-DNP, oxime, semicarbazone) replace the carbonyl O by =N–Z and remove water. For acetals/ketals replace the =O by two –OR groups (a cyclic –O–CH₂CH₂–O– ring when ethylene glycol is used). A hemiacetal has one –OH and one –OR on the carbon.

Notes. In (iii) the acetal of acetaldehyde with methanol is CH₃CH(OCH₃)₂ (two methoxy groups on the former carbonyl carbon). In (v) hexan-3-one is CH₃CH₂–CO–CH₂CH₂CH₃; its ethylene ketal is the cyclic 1,3-dioxolane in which the C3 carbon carries an ethyl group and a propyl group and is joined into the –O–CH₂CH₂–O– ring (2-ethyl-2-propyl-1,3-dioxolane). In (vi) the hemiacetal of formaldehyde with one methanol is H–CH(OH)–OCH₃ (an –OH and an –OCH₃ on the same carbon); a second methanol would convert it to the full acetal, CH₂(OCH₃)₂.

Conclusion: (i) C₆H₅CH=N–NH–C₆H₃(NO₂)₂; (ii) cyclopropanone oxime (ring C=N–OH); (iii) CH₃CH(OCH₃)₂; (iv) cyclobutanone semicarbazone (ring C=N–NH–CO–NH₂); (v) 2-ethyl-2-propyl-1,3-dioxolane; (vi) H–CH(OH)(OCH₃).

Concept. Cyclohexanecarbaldehyde is C₆H₁₁–CHO (a cyclohexane ring carrying a –CHO). It is an aldehyde, so it undergoes all the characteristic aldehyde reactions: nucleophilic addition of a Grignard reagent (→ secondary alcohol), easy oxidation (Tollens’ → acid, silver mirror), condensation with ammonia derivatives (→ semicarbazone), acetal formation with excess alcohol, and reduction of C=O all the way to CH₂ (Clemmensen).

Working. In (i) the phenyl carbanion of PhMgBr adds to the carbonyl carbon; hydrolysis gives the secondary alcohol with the new C–C bond to the ring’s –CHO carbon. In (ii) Tollens’ reagent (ammoniacal silver nitrate) oxidises the aldehyde to the carboxylate and deposits a silver mirror — a positive test showing the compound is an aldehyde, not a ketone. In (iii) the aldehyde condenses with semicarbazide to give the semicarbazone (C=N–NH–CO–NH₂). In (iv) two molecules of ethanol add to give the acetal (the first gives the hemiacetal, the second the acetal, with dry acid). In (v) Clemmensen reduction converts –CHO completely to –CH₃, giving methylcyclohexane (Wolff–Kishner with NH₂NH₂/KOH would give the same hydrocarbon).

Conclusion: (i) cyclohexyl(phenyl)methanol; (ii) cyclohexanecarboxylic acid + silver mirror; (iii) the semicarbazone C₆H₁₁CH=N–NH–CO–NH₂; (iv) the diethyl acetal C₆H₁₁CH(OC₂H₅)₂; (v) methylcyclohexane.

Concept. The deciding feature is the α-hydrogen (a hydrogen on the carbon next to the carbonyl). A carbonyl compound with at least one α-H undergoes the aldol condensation in dilute base (the base removes an α-H to give a carbanion/enolate that adds to another carbonyl). An aldehyde with no α-H cannot form an enolate and instead undergoes the Cannizzaro reaction (base-induced disproportionation to an alcohol + a carboxylate). A compound that is not an aldehyde/ketone, or a ketone with no α-H and no Cannizzaro pathway, does neither.

Working. Methanal, benzaldehyde and 2,2-dimethylbutanal are aldehydes with no α-hydrogen (methanal has none; benzaldehyde’s α-carbon is the ring; 2,2-dimethylbutanal’s α-carbon C2 carries two methyls and an ethyl, no H), so each undergoes Cannizzaro — half is reduced to the alcohol and half oxidised to the acid salt. 2-Methylpentanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde all have α-hydrogens, so they undergo aldol condensation (self-aldol, then loss of water to the α,β-unsaturated carbonyl). Benzophenone is a ketone with no α-H (both groups are phenyl) and cannot do Cannizzaro (that is an aldehyde reaction), so it does neither; butan-1-ol is an alcohol, not a carbonyl compound, so it does neither.

Conclusion: Aldol: 2-methylpentanal, cyclohexanone, 1-phenylpropanone, phenylacetaldehyde. Cannizzaro: methanal, benzaldehyde, 2,2-dimethylbutanal. Neither: benzophenone and butan-1-ol.

Concept. All three targets have four carbons, twice the two carbons of ethanal — so the carbon skeleton is built by an aldol condensation of two molecules of ethanal. Dilute alkali converts two ethanals into 3-hydroxybutanal (the “aldol”, CH₃CH(OH)CH₂CHO); from this single intermediate all three products follow by reduction, dehydration or oxidation.

Working. The aldol condensation first joins the two ethanal molecules: the base removes an α-H from one ethanal to give an enolate, which adds to the carbonyl carbon of the second ethanal, giving 3-hydroxybutanal. (i) Reducing its aldehyde group (and leaving the –OH) gives butane-1,3-diol. (ii) On warming, the β-hydroxy aldehyde loses water from the α,β positions to give the conjugated but-2-enal (this loss of water is what makes it an aldol condensation rather than just an addition). (iii) Oxidising the –CHO of but-2-enal (with a mild oxidant that leaves the C=C intact, e.g. Tollens’) gives but-2-enoic acid.

Conclusion: Aldol condensation of two ethanals gives 3-hydroxybutanal; then (i) NaBH₄ → butane-1,3-diol; (ii) heat (–H₂O) → but-2-enal; (iii) but-2-enal + mild [O] → but-2-enoic acid.

Concept. In an aldol condensation the base removes an α-hydrogen from one aldehyde to give a carbanion/enolate; this carbanion (the nucleophile) attacks the carbonyl carbon of a second aldehyde (the electrophile). When two different aldehydes that both have α-hydrogens are mixed, either can act as the nucleophile and either as the electrophile, so four different β-hydroxy aldehydes (aldols) form; on heating, each loses water to give an α,β-unsaturated aldehyde. Propanal is CH₃CH₂CHO (α-carbon = C2) and butanal is CH₃CH₂CH₂CHO (α-carbon = C2); both have α-H, hence four cross-products.

How each forms. The α-carbanion is the carbon next to the –CHO (C2) of the nucleophile; it bonds to the carbonyl carbon (C1) of the electrophile, whose oxygen becomes the new –OH. For product 3, the propanal carbanion (CH₃CH⁻–CHO) adds to the carbonyl carbon of butanal, giving CH₃CH₂CH₂–CH(OH)–CH(CH₃)–CHO. On warming, each aldol loses a molecule of water (from the β-OH and an α-H) to give the corresponding 2-alkyl-2-enal (α,β-unsaturated aldehyde), e.g. product 1 → 2-methylpent-2-enal.

Conclusion: The four aldols are 3-hydroxy-2-methylpentanal (propanal+propanal), 3-hydroxy-2-ethylhexanal (butanal+butanal), 3-hydroxy-2-methylhexanal (propanal nucleophile + butanal) and 3-hydroxy-2-ethylpentanal (butanal nucleophile + propanal).

Concept. This is a structure-determination problem — each test narrows the possibilities until one structure fits all of them. I read the clues one at a time and combine them.

Putting it together. The compound is an aromatic aldehyde (–CHO on the ring, no α-H) that also has a second substituent at the ortho position, and the total formula is C₉H₁₀O. A benzene ring (C₆H₄, disubstituted) + one –CHO (CHO) leaves C₂H₅ for the second group — an ethyl group. So the compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde): C₆H₄(CHO)(C₂H₅) with the two groups ortho. Check the formula: C₆H₄ + CHO + C₂H₅ = C₉H₁₀O ✓. On vigorous oxidation both the –CHO and the ethyl side chain are oxidised to –COOH at the 1,2 positions, giving benzene-1,2-dicarboxylic acid (phthalic acid), exactly as observed.

Extra depth. The Cannizzaro test is decisive: it tells us the aldehyde carbon has no α-hydrogen, which forces the –CHO directly onto the ring (an alkyl-CHO such as 3-phenylpropanal would have α-H and give aldol instead). The phthalic-acid result then fixes the second group at the ortho position; if it were meta or para, the product would be isophthalic or terephthalic acid.

Conclusion: The compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde), which on vigorous oxidation gives benzene-1,2-dicarboxylic acid (phthalic acid).

Concept. A compound of formula C₈H₁₆O₂ with one degree of unsaturation that hydrolyses to an acid and an alcohol is an ester (R–COO–R′). The clues then fix R and R′: the alcohol (C) gives but-1-ene on dehydration (so C is a four-carbon alcohol), and oxidising that alcohol gives the same acid (B) that the ester gave on hydrolysis (so B has the same carbon skeleton as C).

Identifying C. Dehydration of C gives but-1-ene (CH₂=CH–CH₂–CH₃). An alcohol that gives but-1-ene and that, on oxidation, gives a carboxylic acid must be the primary alcohol butan-1-ol (CH₃CH₂CH₂CH₂OH): a primary alcohol oxidises to a carboxylic acid, whereas butan-2-ol (secondary) would give a ketone, not an acid. So C = butan-1-ol.

Identifying B. Oxidation of butan-1-ol with chromic acid gives butanoic acid (CH₃CH₂CH₂COOH). So B = butanoic acid.

Identifying A. The ester of butanoic acid (B) with butan-1-ol (C) is butyl butanoate, CH₃CH₂CH₂COO–CH₂CH₂CH₂CH₃. Check: C₃H₇CO–O–C₄H₉ = C₈H₁₆O₂ ✓.

Extra depth. The self-consistency of the clues confirms the assignment: the acid from hydrolysis (B) must equal the acid from oxidation of the alcohol (C), which is only possible if the ester is formed from an acid and an alcohol of the same carbon skeleton — i.e. butyl butanoate. (Strictly, dehydration of butan-1-ol tends to give but-1-ene under mild/E2-type conditions; NCERT takes but-1-ene as the stated product.)

Conclusion: A = butyl butanoate (C₃H₇COOC₄H₉), B = butanoic acid, C = butan-1-ol; the three equations above describe the hydrolysis, the oxidation and the dehydration.

Concept. (i) Reactivity towards HCN (a nucleophilic addition) falls as the carbonyl carbon becomes more crowded and more electron-rich: aldehydes > ketones, and bulky groups (tert-butyl) lower it sharply. (ii, iii) Acid strength rises with anything that stabilises the carboxylate: electron-withdrawing groups (–Br, –NO₂), with the effect stronger the closer the group is to –COOH; electron-releasing groups (–CH₃, –OCH₃) lower it.

(i) Reactivity towards HCN. Acetaldehyde (an aldehyde, least hindered) is most reactive. Among the ketones, acetone (two small methyls) is more reactive than methyl tert-butyl ketone (one bulky tert-butyl), which is more reactive than di-tert-butyl ketone (two bulky tert-butyls — so hindered that it hardly reacts).

(ii) Acid strength. The two bromo acids are stronger than the non-halogenated ones (–Br withdraws electrons). Of the bromo acids, the α-bromo acid (CH₃CH₂CH(Br)COOH, Br on C2) is stronger than the β-bromo acid (CH₃CH(Br)CH₂COOH, Br on C3) because the –I effect is stronger closer to –COOH. Of the two without bromine, n-butyric acid (CH₃CH₂CH₂COOH) is slightly stronger than isobutyric acid ((CH₃)₂CHCOOH), which has an extra electron-releasing methyl.

(iii) Acid strength (benzoic acids). The electron-releasing 4-methoxy group makes 4-methoxybenzoic acid weaker than benzoic acid; electron-withdrawing nitro groups make the acid stronger, and two nitro groups (3,4-dinitro) more so than one (4-nitro).

Conclusion: (i) di-tert-butyl ketone < methyl tert-butyl ketone < acetone < acetaldehyde; (ii) (CH₃)₂CHCOOH < CH₃CH₂CH₂COOH < β-bromo acid < α-bromo acid; (iii) 4-methoxybenzoic < benzoic < 4-nitrobenzoic < 3,4-dinitrobenzoic acid.

Concept. A good distinguishing test gives a clear visible change (a precipitate, a colour, a gas, a mirror) with one compound of the pair and no change with the other. The most useful tests here are: Tollens’/Fehling’s (aldehyde vs ketone), the iodoform test (for CH₃CO– or CH₃CH(OH)– groups), the NaHCO₃ test (carboxylic acid → CO₂), and the neutral FeCl₃ test (phenol → violet colour).

Why these work. Tollens’/Fehling’s distinguishes an aldehyde (easily oxidised → silver mirror / red Cu₂O) from a ketone (not oxidised). The iodoform test is positive only for compounds containing the CH₃CO– group (methyl ketones and acetaldehyde) or a CH₃CH(OH)– group (which is oxidised to CH₃CO– by the hypoiodite); it is the key to pairs (ii), (v) and (vii), where both members are of the same carbonyl class and only one carries a methyl-carbonyl. The NaHCO₃ test is specific for the acidic –COOH group (phenol is too weak an acid to release CO₂ from bicarbonate), which separates the acid in (iii) and (iv).

Conclusion: Use Tollens’ for (i) and (vi), the iodoform test for (ii), (v) and (vii), and NaHCO₃ for (iii) and (iv); in each case name the compound that gives the positive result.

Concept. The restriction “organic reagents of not more than one carbon” means we may use reagents such as CH₃Cl, CH₃OH, CO, HCN, CH₃COCl (two carbons — so actually only one-carbon: CH₃Cl, CH₃OH, CO, HCN, CO₂), but we must build benzoic-acid and related skeletons using one-carbon pieces. The two one-carbon handles are a methyl group (Friedel–Crafts with CH₃Cl, then oxidised to –COOH) and the –CN group (from a side-chain via –CH₂Cl, then hydrolysed). The order of nitration (before or after introducing the –COOH) is chosen using directing effects: –CH₃ is o/p-directing, –COOH is meta-directing.

The crucial choice — when to nitrate. For m-nitrobenzoic acid (ii) the two groups must end up meta, so first make benzoic acid (–COOH, which is meta-directing) and then nitrate — the nitro group enters the meta position. For p-nitrobenzoic acid (iii) the groups must be para, so nitrate while the ring still carries the o/p-directing –CH₃ (giving p-nitrotoluene), and only afterwards oxidise the methyl to –COOH. Getting the order wrong gives the wrong isomer. In (iv) the carbon count of the side chain is grown by one through the nitrile (benzyl chloride → benzyl cyanide → phenylacetic acid). In (v) the methyl of p-nitrotoluene is oxidised only to the aldehyde stage using the Étard reagent CrO₂Cl₂.

Conclusion: All start from toluene (benzene + CH₃Cl/AlCl₃): (i) oxidise then esterify with CH₃OH; (ii) oxidise then nitrate (meta); (iii) nitrate then oxidise (para); (iv) side-chain chlorinate → cyanide → hydrolyse; (v) nitrate then Étard oxidation to the aldehyde.

Concept. Each conversion is a short synthesis built from the standard reactions of carbonyl compounds: reduction (to alcohol or to CH₂), oxidation, Grignard addition, aldol, cyanohydrin formation, and the directing rules for ring substitution. The “two-step” limit forces the most direct route.

Two points worth noting. In (iv) and (ix) the order is set by directing effects: the acetyl/–COOH group is meta-directing, so introducing it first (or keeping it) sends the incoming –NO₂ to the meta position. In (vii) catalytic hydrogenation reduces both the C=C and the C=O of cinnamaldehyde in one step, giving the saturated primary alcohol 3-phenylpropan-1-ol.

Conclusion: Each target is reached in two steps as tabulated — reduction/dehydration (i), SOCl₂ + Rosenmund (ii), oxidation + aldol (iii), acylation + nitration (iv), Grignard + oxidation (v), Grignard + aldehyde (vi), cross-aldol + hydrogenation (vii), cyanohydrin + hydrolysis (viii), nitration + LiAlH₄ (ix).

(i) Acetylation. Acetylation is the introduction of an acetyl group (CH₃CO–) into a molecule containing an –OH, –NH₂ or –NHR group, by reaction with acetyl chloride (CH₃COCl) or acetic anhydride [(CH₃CO)₂O], usually in the presence of a base such as pyridine (which removes the HCl formed). An alcohol gives an ester, an amine gives an amide. Example: salicylic acid + (CH₃CO)₂O → aspirin (acetylsalicylic acid); C₂H₅OH + CH₃COCl → CH₃COOC₂H₅ + HCl.

(ii) Cannizzaro reaction. An aldehyde that has no α-hydrogen, when heated with a concentrated alkali, undergoes a self oxidation–reduction (disproportionation): one molecule is oxidised to the carboxylate ion and another is reduced to the alcohol. Example: 2 HCHO + NaOH → CH₃OH + HCOONa; 2 C₆H₅CHO + KOH → C₆H₅CH₂OH + C₆H₅COOK. The hydride ion is transferred from the tetrahedral alkoxide intermediate of one molecule to the carbonyl carbon of another.

(iii) Cross aldol condensation. An aldol condensation carried out between two different carbonyl compounds (each still needing at least one α-H for the enolate) is called a cross (or mixed) aldol condensation. When both partners have α-hydrogens, a mixture of four products results and the reaction is of little use; it is synthetically valuable when one partner has no α-H (e.g. benzaldehyde or formaldehyde), so only the other can form the carbanion. Example: C₆H₅CHO (no α-H) + CH₃CHO (dil. NaOH, Δ) → C₆H₅CH=CH–CHO (cinnamaldehyde) + H₂O.

(iv) Decarboxylation. The loss of carbon dioxide from a carboxylic acid (or its salt). Sodium or potassium salts of carboxylic acids, when heated with soda-lime (NaOH + CaO), give a hydrocarbon with one carbon fewer: RCOONa + NaOH → (CaO, Δ) R–H + Na₂CO₃ (e.g. CH₃COONa → CH₄). Kolbe electrolysis of an aqueous solution of a carboxylate salt gives an alkane (R–R) at the anode by decarboxylative coupling. Decarboxylation is easier when the α-carbon bears electron-withdrawing groups (e.g. β-keto acids and malonic acids lose CO₂ readily on gentle heating).

Conclusion: Acetylation introduces CH₃CO– (ester/amide) using CH₃COCl or (CH₃CO)₂O; the Cannizzaro reaction is the base-induced disproportionation of an α-H-free aldehyde to an alcohol + carboxylate; cross-aldol condensation is an aldol between two different carbonyls (useful when one lacks α-H); decarboxylation is the loss of CO₂ (soda-lime or Kolbe electrolysis).

Concept. Each blank is filled by recognising the reaction type from the reagent: KMnO₄/KOH oxidises a side chain; SOCl₂ makes an acyl chloride; semicarbazide/hydrazine-type reagents condense at a carbonyl; AlCl₃ + acyl chloride is Friedel–Crafts acylation; Tollens’ reagent oxidises –CHO to –COOH; NaCN/HCl gives a cyanohydrin; dilute NaOH gives an aldol/cross-aldol; NaBH₄ reduces a ketone (not an ester); CrO₃ oxidises an alcohol; hydroboration–oxidation then PCC converts =CH₂ to –CHO; and ozonolysis cleaves a C=C.

Notes on the tricky ones. In (viii) the chemoselectivity of NaBH₄ is the point — it reduces the ketone carbonyl of ethyl acetoacetate to an –OH but leaves the ester group intact, giving ethyl 3-hydroxybutanoate. In (x) the alkene =CH₂ cannot be turned into –CHO by simple oxidation; hydroboration–oxidation first gives the primary alcohol cyclohexylmethanol (anti-Markovnikov), which is then oxidised (PCC) to the aldehyde. In (xi) ozonolysis breaks a C=C into two carbonyls; getting two identical cyclohexanone molecules means the starting alkene is the symmetrical cyclohexylidenecyclohexane.

Conclusion: The eleven blanks are filled as in the table — benzoic acid; phthaloyl chloride; benzaldehyde semicarbazone; benzoyl chloride/AlCl₃; 4-oxocyclohexane-1-carboxylic acid; the aldehyde cyanohydrin; α-methylcinnamaldehyde; ethyl 3-hydroxybutanoate; cyclohexanone; B₂H₆ then H₂O₂/OH⁻ then PCC; and cyclohexylidenecyclohexane.

(i) Cyanohydrin formation — steric hindrance. Cyanohydrin formation is a nucleophilic addition: the cyanide ion attacks the planar sp² carbonyl carbon, which then becomes a crowded sp³ centre. In cyclohexanone the carbonyl carbon is relatively open, so CN⁻ adds easily and the cyanohydrin forms in good yield. In 2,2,6-trimethylcyclohexanone the three methyl groups on the carbons flanking the carbonyl (two on C2, one on C6) crowd the approach of the nucleophile and oppose the rehybridisation to the bulkier sp³ product; the steric strain makes the addition unfavourable, so little or no cyanohydrin forms.

(ii) Only one –NH₂ of semicarbazide reacts. Semicarbazide is H₂N–NH–CO–NH₂. One of its –NH₂ groups is directly attached to the carbonyl (–CO–NH₂); the lone pair on that nitrogen is delocalised into the C=O group by resonance, so it is not available to act as a nucleophile. The other –NH₂ (the terminal H₂N–NH– nitrogen) is not conjugated with the carbonyl, so its lone pair is free and it is the one that attacks the aldehyde/ketone to form the semicarbazone (>C=N–NH–CO–NH₂).

(iii) Removing water/ester drives esterification forward. Acid-catalysed esterification is a reversible equilibrium: RCOOH + R′OH ⇌ RCOOR′ + H₂O. By Le Chatelier’s principle, continuously removing one of the products (the water by a dehydrating agent or azeotropic distillation, or the ester by distilling it off) shifts the equilibrium to the right and gives a higher yield of the ester. If the products are allowed to accumulate, the reverse reaction (hydrolysis of the ester) sets a low equilibrium conversion.

Conclusion: (i) the flanking methyls of 2,2,6-trimethylcyclohexanone sterically block CN⁻ addition; (ii) one –NH₂ of semicarbazide is deactivated by resonance with the C=O, so only the free terminal –NH₂ reacts; (iii) removing water or ester shifts the reversible esterification equilibrium toward the ester (Le Chatelier).

Step 1 — Molecular formula from the percentages. Take 100 g of the compound: C = 69.77 g → 69.77/12 = 5.81 mol; H = 11.63 g → 11.63/1 = 11.63 mol; O = 100 − 69.77 − 11.63 = 18.60 g → 18.60/16 = 1.16 mol. Dividing by the smallest (1.16): C 5.0 : H 10.0 : O 1.0, so the empirical formula is C₅H₁₀O (mass 86). Since the molecular mass is also 86, the molecular formula is C₅H₁₀O (one degree of unsaturation — a C=O).

Step 2 — Build the structure. A C₅H₁₀O methyl ketone is pentan-2-one, CH₃–CO–CH₂–CH₂–CH₃. On vigorous oxidation the C–C bonds adjacent to the carbonyl are cleaved: breaking the bond on the methyl side gives ethanoic acid (CH₃COOH, from CH₃–CO–) and breaking on the other side gives propanoic acid (CH₃CH₂COOH, from –CH₂CH₂CH₃) — exactly the two acids observed. (Pentan-3-one, CH₃CH₂COCH₂CH₃, is ruled out because it is not a methyl ketone and would not give a positive iodoform test, and it would give only propanoic acid on cleavage.)

Conclusion: The compound is pentan-2-one (CH₃COCH₂CH₂CH₃), a methyl ketone of formula C₅H₁₀O whose vigorous oxidation gives ethanoic and propanoic acids.

Concept. The strength of an acid depends on the stability of its conjugate base. Both phenol and carboxylic acids ionise to resonance-stabilised anions — the phenoxide ion and the carboxylate ion — but the quality of the resonance matters far more than the number of structures. The key is where the negative charge is delocalised and whether the contributing structures are equivalent and put the charge on electronegative atoms.

Explanation. In the carboxylate ion the negative charge is delocalised over two equivalent oxygen atoms; the two resonance structures are identical, so the stabilisation is maximal and the charge rests entirely on electronegative oxygen. In the phenoxide ion, although there are more resonance structures, the extra ones place the negative charge on the carbon atoms of the ring (ortho and para), which are far less able to carry a negative charge than oxygen; these structures are also non-equivalent and contribute less. Moreover, in phenol the C–O bond has partial double-bond character even in the neutral molecule, so less energy is released on ionisation. The carboxylate ion is therefore the more effectively stabilised conjugate base, and so a carboxylic acid (pKₐ ≈ 4–5) is a stronger acid than phenol (pKₐ ≈ 10), even though phenoxide has more resonating structures.

Conclusion: The carboxylate ion delocalises its charge over two equivalent, electronegative oxygen atoms, giving very effective stabilisation, whereas the phenoxide ion’s extra resonance structures place charge on ring carbons (less electronegative, non-equivalent); so carboxylic acid is the stronger acid despite phenoxide having more resonance structures.