Chapters

Everything NCERT Class 12 Unit 8 (Aldehydes, Ketones and Carboxylic Acids) says about aldehydes and ketones (its sections 8.1–8.5; the carboxylic-acid half is in Carboxylic Acids & its Derivatives), plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Organic chemistry has few formulas. These are the rules and tests the questions turn on, on one screen. Reagents are in the maps and the named-reaction table below.

Rule or test	What it says	Watch
Reactivity to nucleophiles	HCHO > RCHO > ArCHO > R₂CO > ArCOR	EWG on the ring raises it, EDG lowers it
Aldol	needs an α-H; dilute alkali	Gives β-hydroxy C=O, then α,β-unsaturated on heating
Cannizzaro	needs no α-H; concentrated alkali	Half → alcohol, half → carboxylate
Crossed Cannizzaro	with HCHO, HCHO is oxidised	HCHO → HCOO⁻; the other → alcohol
Haloform	CH₃CO– or CH₃CH(OH)–	One C less in the acid; CHI₃ yellow
Tollens'	all aldehydes (and HCOOH)	2 Ag per CHO, 2 e⁻
Fehling's	aliphatic aldehydes only	Red-brown Cu₂O; benzaldehyde negative
2,4-DNP	every aldehyde and ketone	Yellow-orange-red solid
Bisulphite adduct	aldehydes and small ketones	Acid or alkali gives the carbonyl back
Clemmensen	Zn-Hg / conc. HCl	Not for acid-sensitive molecules
Wolff–Kishner	N₂H₄, then KOH / glycol, heat	Not for base-sensitive molecules
Claisen–Schmidt	2 ArCHO + 1 acetone → dibenzalacetone	Mole ratio 2 : 1

0 · What JEE Main asks from this chapter

157 questions in JEE Main 2019–2026 (138 MCQ, 19 numerical), about 1 per paper. Grouped by type (counts are approximate; about 20 show the structures only as pictures):

Type	Share	Typical question
Multi-step: identify the compound or the product	~25%	C₉H₁₀O, iodoform positive, KMnO₄ → phthalic acid: find it
Preparation and its named reactions	~16%	Match Rosenmund, Stephen, Etard, Gattermann–Koch to reagents
Reduction, oxidation, Tollens' and Fehling's	~16%	How many of 7 compounds give Fehling's test
Aldol, cross aldol, Claisen–Schmidt	~14%	π bonds in the product of 2 PhCHO + acetone
Nucleophilic addition: order, acetals, C=N derivatives	~14%	Order of reactivity to HCN of four benzaldehydes
Haloform / iodoform	~9%	How many of 10 compounds do not give iodoform
Cannizzaro	~6%	PhCHO + HCHO with 50% NaOH: the products

The numericals count things (π electrons, N atoms, compounds that pass a test) or ask a yield. Draw the product before you count.

1 · Nomenclature and structure

Aldehyde: C=O bonded to a carbon and a hydrogen (HCHO has two H). Ketone: C=O bonded to two carbons.

Common names. Aldehydes take the common name of the acid they oxidise to, with -ic acid replaced by aldehyde (formic → formaldehyde). Substituent positions are Greek letters: α is the carbon next to the CHO, then β, γ, δ. Ketones name the two groups on the C=O (methyl n-propyl ketone); their positions are α, α′, β, β′ from the carbons next to the C=O. Alkyl phenyl ketones add the acyl name to -phenone: acetophenone is methyl phenyl ketone; benzophenone is diphenyl ketone.

IUPAC names.

• Replace the alkane's -e with -al or -one.
• Aldehyde: number from the CHO carbon. Ketone: number from the end nearer the C=O.
• Cyclic ketone: the C=O carbon is C-1.
• CHO on a ring: cycloalkane + carbaldehyde; number from the ring carbon carrying the CHO.
• C₆H₅CHO is benzenecarbaldehyde; IUPAC accepts benzaldehyde, and the others are named as substituted benzaldehydes.

Structure	Common name	IUPAC name
(CH₃)₂CHCHO	Isobutyraldehyde	2-Methylpropanal
CH₃CH(OCH₃)CHO	α-Methoxypropionaldehyde	2-Methoxypropanal
CH₃(CH₂)₃CHO	Valeraldehyde	Pentanal
CH₂=CHCHO	Acrolein	Prop-2-enal
o-C₆H₄(CHO)₂	Phthaldehyde	Benzene-1,2-dicarbaldehyde
CH₃COCH₂CH₂CH₃	Methyl n-propyl ketone	Pentan-2-one
(CH₃)₂CHCOCH(CH₃)₂	Diisopropyl ketone	2,4-Dimethylpentan-3-one
(CH₃)₂C=CHCOCH₃	Mesityl oxide	4-Methylpent-3-en-2-one

Example (2021) · "Mesityl oxide is the common name of" → 4-methylpent-3-en-2-one. Another 2021 question asked its C–C σ bonds: (CH₃)₂C=CH–CO–CH₃ has 5 (the C=C counts once).

Structure of the C=O group.

• The carbonyl carbon is sp²: three σ bonds in one plane at about 120°; its p orbital overlaps an O p orbital to make the π bond, with the π cloud above and below the plane.
• O carries two lone pairs.
• O is more electronegative, so the bond is polar: the carbon is electrophilic (a Lewis acid) and the oxygen nucleophilic (a Lewis base). Resonance C=O ⟷ C⁺–O⁻ explains the polarity.
• Carbonyl compounds have large dipole moments; they are more polar than ethers.

2 · Preparation

Both aldehydes and ketones:

• Oxidation of alcohols: 1° → aldehyde, 2° → ketone. To stop at the aldehyde use PCC (C₅H₅NH⁺CrO₃Cl⁻); acidified dichromate or Jones reagent goes on to the acid.
• Dehydrogenation: alcohol vapour over Ag or Cu (Cu at 573 K). For volatile alcohols; industrial.
• Ozonolysis of alkenes: O₃, then Zn dust and water. Aldehydes, ketones or both, depending on the alkene's substitution.
• Hydration of alkynes: H₂O with H₂SO₄ and HgSO₄. Ethyne gives ethanal; every other alkyne gives a ketone.

Aldehydes only:

• Rosenmund reduction: RCOCl + H₂ over Pd on BaSO₄ → RCHO. The poisoned catalyst stops at the aldehyde.
• Stephen reaction: RCN + SnCl₂/HCl → imine (RCH=NH) → H₃O⁺ → RCHO.
• DIBAL-H (diisobutylaluminium hydride), then water: nitrile → aldehyde, and ester → aldehyde. It leaves a C=C alone: CH₃CH=CH(CH₂)₂CN → CH₃CH=CH(CH₂)₂CHO.
• From toluene (strong oxidants go on to benzoic acid, so the CH₃ is trapped as something hard to oxidise):
  • Etard reaction: CrO₂Cl₂ (chromyl chloride) in CS₂ gives a chromium complex; H₃O⁺ gives benzaldehyde.
  • CrO₃ in acetic anhydride, 273–283 K → benzylidene diacetate C₆H₅CH(OCOCH₃)₂; aqueous acid → benzaldehyde.
  • Side-chain chlorination: Cl₂/hν → benzal chloride C₆H₅CHCl₂; H₂O at 373 K → benzaldehyde. The commercial method.
• Gattermann–Koch reaction: benzene + CO + HCl with anhydrous AlCl₃ or CuCl → benzaldehyde.

Ketones only:

• Acyl chloride + dialkylcadmium: 2R′MgX + CdCl₂ → R′₂Cd; then R′₂Cd + 2RCOCl → 2RCOR′ + CdCl₂. Cadmium is too weak to attack the ketone again.
• Nitrile + Grignard, then hydrolysis → ketone. CH₃CH₂CN + C₆H₅MgBr → propiophenone (C₆H₅COCH₂CH₃).
• Friedel–Crafts acylation: ArH + RCOCl (or an anhydride) with anhydrous AlCl₃ → ArCOR.

Example (NCERT Example 8.1) · Hexan-1-ol → hexanal: PCC. Cyclohexanol → cyclohexanone: anhydrous CrO₃. p-Fluorotoluene → p-fluorobenzaldehyde: CrO₂Cl₂ then H₂O, or CrO₃ in acetic anhydride. Ethanenitrile → ethanal: DIBAL-H. Allyl alcohol → propenal: PCC. But-2-ene → ethanal: O₃ then Zn/H₂O.

Example (2021) · Which carbonyl compound cannot come from an alkyne with HgSO₄/H₂SO₄? Propanal. Propyne gives acetone; only ethyne gives an aldehyde.

Example (2021) · CH₃CH₂CN → (H₃O⁺, Δ) propanoic acid → (SOCl₂) propanoyl chloride → (H₂, Pd/BaSO₄) propanal.

Why a Grignard stops at the ketone · Beyond NCERT

RMgX adds to a nitrile once, giving a C=N–MgX salt. The ketone appears only at the H₃O⁺ work-up, when the Grignard is already destroyed. With an ester two RMgX add and you get a 3° alcohol: C₆H₅COOC₂H₅ + 2CH₃MgBr gives 2-phenylpropan-2-ol, not acetophenone (2022). Every acidic H (–OH, –NH, –COOH, ≡C–H) uses up one RMgX first.

3 · Physical properties

• Methanal is a gas; ethanal a volatile liquid; the rest liquids or solids.
• Boiling point: higher than hydrocarbons and ethers of similar mass (dipole–dipole attraction), lower than alcohols (no H-bond between carbonyl molecules).

Compound (mass 58–60)	b.p. (K)
n-Butane	273
Methoxyethane	281
Propanal	322
Acetone	329
Propan-1-ol	370

• Solubility: methanal, ethanal and propanone mix with water in all proportions (they H-bond to water). Solubility falls fast as the chain grows. All dissolve in benzene, ether, methanol, chloroform.
• Odour: the lower aldehydes are sharp and pungent; larger ones are more fragrant (perfumes, flavours).

Example (NCERT Example 8.2) · Masses 72–74: n-pentane < ethoxyethane < butanal < butan-1-ol. Intext 8.3: CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH.

4 · Nucleophilic addition

Alkenes undergo electrophilic addition; aldehydes and ketones undergo nucleophilic addition.

Mechanism:

1. The nucleophile attacks the carbonyl carbon roughly perpendicular to the plane of its sp² orbitals.
2. Carbon goes sp² → sp³; a tetrahedral alkoxide intermediate forms.
3. The alkoxide takes a proton from the medium. Net result: Nu⁻ and H⁺ add across C=O.

Reactivity: aldehydes > ketones, for two reasons.

• Steric: two groups on a ketone crowd the approach more than one.
• Electronic: two alkyl groups (+I) cut the carbon's δ+ more than one.
• Aromatic carbonyls are slower: the ring's resonance reduces the C=O polarity, so benzaldehyde is less reactive than propanal (NCERT Example 8.3).
• On the ring, an EWG (–NO₂) raises reactivity; an EDG (–CH₃, –OCH₃) lowers it.

Example (NCERT Intext 8.4, asked in 2020 and 2025) · Butanone < propanone < propanal < ethanal. Acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde.

Example (2020) · Towards HCN: 2-methoxybenzaldehyde < 4-methoxybenzaldehyde < benzaldehyde < 4-nitrobenzaldehyde. Ortho-OCH₃ adds crowding to its +R effect, so it is slowest.

(a) HCN → cyanohydrin. Slow with pure HCN, so a base is added: CN⁻ is the stronger nucleophile. Cyanohydrins are useful intermediates: hydrolysis gives α-hydroxy acids.

Example (NCERT exercise 8.18) · Cyclohexanone forms its cyanohydrin in good yield; 2,2,6-trimethylcyclohexanone does not. The three methyls block the attack.

(b) NaHSO₃ → bisulphite adduct. The equilibrium lies to the right for most aldehydes and to the left for most ketones (steric). The adduct can be turned back into the carbonyl compound with dilute mineral acid or alkali, so it is used to separate and purify aldehydes.

Example (2025) · Toluene → (CrO₂Cl₂, CS₂; H₃O⁺) benzaldehyde → excess NaHSO₃ → the solid filtered off is C₆H₅CH(OH)SO₃Na; dilute HCl gives back benzaldehyde.

(c) Grignard reagents → alcohols (from Unit 7): HCHO gives a 1° alcohol, other aldehydes 2°, ketones 3°.

Example (2025) · X (C₃H₆O) is not readily oxidised, so it is acetone. Reduction gives propan-2-ol (Y); HBr gives 2-bromopropane (Z). Its Grignard + acetone → 2,3-dimethylbutan-2-ol.

(d) Alcohols → hemiacetal → acetal. An aldehyde with one equivalent of monohydric alcohol and dry HCl gives a hemiacetal (alkoxy alcohol); a second alcohol gives the acetal, a gem-dialkoxy compound. Ketones with ethylene glycol give cyclic ethylene glycol ketals. Dry HCl protonates the carbonyl O and makes the carbon more electrophilic. Aqueous mineral acid hydrolyses acetals and ketals back.

Example (2019) · Fastest acetal: HCHO with MeOH (least crowded carbonyl and alcohol). Another 2019 paper: propanal with excess methanol beats acetone, and beats a stoichiometric amount.

Acetals as protecting groups · Beyond NCERT · Acetals and ketals are stable in base and break only in aqueous acid. So a ketone is protected as its glycol ketal while a Grignard or LiAlH₄ acts elsewhere, then freed with H₃O⁺. (2023: the assertion "stable in base" is true; the reason "alkoxide is a good leaving group" is false.) The more reactive C=O is protected first: an aldehyde before a ketone, a ketone before an ester.

(e) Ammonia and its derivatives, H₂N–Z → >C=N–Z. Reversible and acid-catalysed. The equilibrium goes to the product because the intermediate loses water fast (addition–elimination).

Reagent H₂N–Z	Product
Ammonia (Z = H)	Imine
Amine (Z = R)	Substituted imine (Schiff's base)
Hydroxylamine (Z = OH)	Oxime
Hydrazine (Z = NH₂)	Hydrazone
Phenylhydrazine (Z = NHC₆H₅)	Phenylhydrazone
2,4-Dinitrophenylhydrazine	2,4-DNP derivative: yellow, orange or red solid
Semicarbazide (Z = NHCONH₂)	Semicarbazone

• 2,4-DNP characterises aldehydes and ketones. Esters, acids and phenols don't react (2021: acetophenone gives the orange precipitate; ethyl benzoate and salicylic acid don't).
• Semicarbazide H₂N–NH–CO–NH₂ has two NH₂, but only the one on the NH reacts. The other is amide-like: its lone pair is delocalised into the C=O (NCERT exercise 8.18).

Example (2021, 2024) · Semicarbazone of acetone or ethanal, R₂C=N–NH–CO–NH₂: 3 N atoms.

Weak acid, about pH 4.5 · Beyond NCERT · Too much acid protonates the H₂N–Z and kills the nucleophile; too little fails to activate the C=O.

Enamines and hydrates · Beyond NCERT

• A 2° amine gives an enamine (C=C–NR₂) and needs an α-H. Di-tert-butyl ketone has none, so it gives no enamine (2022).
• Water adds reversibly to give a gem-diol (hydrate). The equilibrium is large for HCHO (K ≈ 2280) and for Cl₃CCHO (chloral, −I of Cl; K ≈ 2000), both stated true in 2025. Ninhydrin's hydrate is its stable form: indane-1,2,3-trione + H₂O → 2,2-dihydroxyindane-1,3-dione (2023).

5 · Reduction

• To alcohols: NaBH₄, LiAlH₄ or H₂/catalyst. Aldehyde → 1° alcohol; ketone → 2° alcohol.
• To hydrocarbons (C=O → CH₂):
  • Clemmensen: zinc amalgam + conc. HCl.
  • Wolff–Kishner: hydrazine (the hydrazone forms), then KOH or NaOH heated in a high-boiling solvent such as ethylene glycol.

Which reagent, which group · Beyond NCERT

Reagent	Reduces	Leaves alone
NaBH₄	aldehydes, ketones	esters, acids, isolated C=C
LiAlH₄	also esters, acids, amides	isolated C=C
DIBAL-H (1 equiv, cold)	ester or nitrile → aldehyde	—
H₂/Pd	C=C first	—
Clemmensen (acid)	C=O → CH₂; keeps esters, amines	breaks acid-sensitive groups
Wolff–Kishner (base)	C=O → CH₂; keeps acid-sensitive groups	breaks base-sensitive groups (C–Cl, esters)

Example (2019) · CH₃CH=CHCO₂CH₃ + LiAlH₄ → CH₃CH=CHCH₂OH; the C=C stays. NaBH₄ in MeOH on a keto-ester reduces only the ketone.

Example (2023) · 4-Hydroxy-4-methylpentan-2-one with Zn-Hg/HCl does not give 2-methylpentan-2-ol: the 3° alcohol dehydrates in hot acid. Assertion false, reason true.

Example (2024) · Styrene → (H₂O, H⁺) 1-phenylethanol → (CrO₃) acetophenone → (N₂H₄, KOH, heat) ethylbenzene.

6 · Oxidation and the tests

• Aldehydes oxidise easily to carboxylic acids: HNO₃, KMnO₄, K₂Cr₂O₇, and even the mild Tollens' and Fehling's reagents.
• Ketones need strong oxidants and heat; C–C bonds break and give a mixture of acids with fewer carbons than the ketone.

Popoff's rule · Beyond NCERT · In an unsymmetrical ketone, the C=O stays mostly with the smaller alkyl group as the chain breaks.

Tollens' test. Warm the aldehyde with freshly prepared ammoniacal silver nitrate: a bright silver mirror. Alkaline medium; the aldehyde becomes the carboxylate.

Example (2021) · Electrons passed to Tollens' reagent per CHO: 2 (two Ag⁺ → two Ag).

Fehling's test. Fehling A is aqueous CuSO₄; Fehling B is alkaline sodium potassium tartrate (Rochelle salt). Mix equal volumes just before use. Heating with an aldehyde gives a reddish-brown precipitate of Cu₂O. Aromatic aldehydes do not respond.

Example (2024) · Fehling's positive among benzaldehyde, acetaldehyde, acetone, acetophenone, methanal, 4-nitrobenzaldehyde and cyclohexanecarbaldehyde: 3 (acetaldehyde, methanal, cyclohexanecarbaldehyde).

Other positives · Beyond NCERT · Formic acid gives Tollens' (it has an H–C=O; 2024). α-Hydroxy ketones such as fructose give both tests (2025: fructose, ethanal and phenylacetaldehyde positive with Fehling's; benzaldehyde and acetophenone negative).

Haloform reaction. A methyl ketone (or ethanal), i.e. any CH₃CO– group, with sodium hypohalite → the sodium salt of the acid with one carbon less, plus haloform CHX₃. A C=C in the molecule is not touched.

• Iodoform test (NaOI, or I₂/NaOH) gives yellow CHI₃. It detects CH₃CO– and also CH₃CH(OH)–, which is oxidised to CH₃CO– first.
• Ethanal is the only aldehyde and ethanol the only 1° alcohol that pass.
• Acetic acid, its esters and its amides fail, although they hold CH₃CO.

Example (2023) · Iodoform positive among 1-phenylbutan-2-one, 2-methylbutan-2-ol, 3-methylbutan-2-ol, 1-phenylethanol, 3,3-dimethylbutan-2-one and 1-phenylpropan-2-ol: 4: 3-methylbutan-2-ol, 1-phenylethanol, 3,3-dimethylbutan-2-one and 1-phenylpropan-2-ol.

Example (2022) · Hex-4-en-2-ol → (PCC) hex-4-en-2-one → (NaOI) CH₃CH=CHCH₂COO⁻ → (soda lime) but-2-ene. One C lost at each of the last two steps.

KOI from bleach · Beyond NCERT · NaOCl + KI → KOI, an oxidising agent. An α,β-unsaturated methyl ketone still gives iodoform (2026).

Example (NCERT Example 8.4) · A (C₈H₈O) gives an orange-red 2,4-DNP derivative and iodoform. It reduces neither Tollens' nor Fehling's and decolourises neither Br₂ water nor Baeyer's reagent. Chromic acid gives B (C₇H₆O₂). So A is a methyl ketone with an aromatic ring: acetophenone; B is benzoic acid.

7 · Reactions of the α-hydrogen

The α-H is acidic because the C=O withdraws electrons and the conjugate base (the enolate) is resonance-stabilised, with the charge shared onto O.

Aldol reaction. An aldehyde or ketone with at least one α-H, in dilute alkali, gives a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol). These lose water easily to give the α,β-unsaturated carbonyl: aldol condensation. The name covers ketones too.

3-Hydroxybutanal, then but-2-enal. Acetone with Ba(OH)₂: 4-hydroxy-4-methylpentan-2-one, then mesityl oxide (4-methylpent-3-en-2-one).

NCERT's chapter summary says aldol gives "α-hydroxyaldehydes". It is β: the OH sits two carbons from the C=O.

Cross aldol. Two different carbonyls. If both have α-H you get four products. Ethanal + propanal → but-2-enal and 2-methylpent-2-enal (self), pent-2-enal and 2-methylbut-2-enal (cross). A ketone can be one partner: benzaldehyde + acetophenone → 1,3-diphenylprop-2-en-1-one (benzalacetophenone).

Example (2026) · Propanal (Stephen, from propanenitrile) + ethanal (ozonolysis of but-2-ene), alkali, heat. Which is not formed? 3-Methylbut-2-enal. Write each enolate attacking each carbonyl and you get the four above.

Example (2024) · "Benzaldehyde and ethanal will not give a cross-aldol product" is false. Benzaldehyde has no α-H, but it is a fine electrophile; ethanal's enolate attacks it (cinnamaldehyde).

Claisen–Schmidt condensation · Beyond NCERT (the name) · An aromatic aldehyde (no α-H) + a ketone with α-H. 2 PhCHO + acetone → dibenzalacetone PhCH=CH–CO–CH=CHPh (C₁₇H₁₄O, 234 g mol⁻¹). Only the ketone can form an enolate and the aldehyde is the better electrophile, so one main product forms.

Example (2024) · 351 g dibenzalacetone = 1.5 mol, from 87 g acetone (1.5 mol). Benzaldehyde = 2 × 1.5 = 3 mol × 106 = 318 g.

Example (2025) · 5.3 g benzaldehyde = 0.05 mol → theoretical 0.025 mol × 234 = 5.85 g. Actual 3.51 g → 60%.

Example (2024, 2026) · π bonds in dibenzalacetone: 2 rings × 3 + 2 C=C + 1 C=O = 9. Benzaldehyde (Gattermann–Koch) + acetophenone (Friedel–Crafts) in alkali → chalcone PhCH=CH–CO–Ph: π electrons = 12 + 2 + 2 = 16.

Intramolecular aldol · Beyond NCERT · A dicarbonyl closes into the ring that is 5- or 6-membered. Count the ring atoms from the enolate carbon to the carbonyl carbon it attacks.

Example (2025) · Target 1-(cyclopent-1-en-1-yl)ethan-1-one. Ozonolysis of 1-methylcyclohexene gives 6-oxoheptanal, CH₃CO(CH₂)₄CHO. The CH₂ next to the ketone attacks the CHO, closing a 5-ring.

Which α-C reacts, and when it stops · Beyond NCERT

• Base: an unsymmetrical methyl ketone reacts at its CH₃ (the less substituted α-C). Acid: it reacts at the more substituted one.
• No dehydration without a second α-H. Isobutyraldehyde gives (CH₃)₂CH–CH(OH)–C(CH₃)₂–CHO and stops: the α-C has no H left (2022).
• Excess HCHO adds once for every α-H, then a crossed Cannizzaro reduces the CHO. Propanal → CH₃C(CH₂OH)₃ (2025); acetaldehyde gives pentaerythritol the same way.

8 · Other reactions

Cannizzaro reaction. An aldehyde with no α-H, heated with concentrated alkali, disproportionates: one molecule is reduced to the alcohol, another is oxidised to the carboxylate.

Crossed and intramolecular Cannizzaro · Beyond NCERT

• With HCHO as one partner, HCHO is oxidised (to HCOO⁻) and the other aldehyde becomes the alcohol. PhCHO + HCHO, 50% NaOH → PhCH₂OH + HCOOH after acid (2019).
• The key step is a hydride (H⁻) transfer from one aldehyde carbon to the other. In D₂O/NaOD the alcohol's C–H stay H and only the O picks up D: PhCH₂OD (2022).
• A dialdehyde does it inside one molecule. Glyoxal (C₂H₂O₂, M = 58) → HOCH₂COOH (2026). PhCOCHO → PhCH(OH)COO⁻K⁺ (2025).
• Other aldehydes without an α-H also do it: m-chlorobenzaldehyde, vanillin, anisaldehyde, 2,2-dimethylpropanal.

Example (2026) · A (C₈H₈O₂) gives a single cross-aldol product with acetophenone and a substituted benzyl alcohol with conc. NaOH: no α-H, an aldehyde → 4-methoxybenzaldehyde.

Electrophilic substitution. On aromatic aldehydes and ketones the C=O group deactivates the ring and directs meta. Benzaldehyde + HNO₃/H₂SO₄ (273–283 K) → m-nitrobenzaldehyde.

Example (NCERT exercise 8.7) · Aldol: 2-methylpentanal, cyclohexanone, 1-phenylpropanone, phenylacetaldehyde. Cannizzaro: methanal, benzaldehyde, 2,2-dimethylbutanal. Neither: benzophenone (a ketone with no α-H), butan-1-ol (not a carbonyl).

9 · Reaction map and named reactions

Name	Reagents	Change
Rosenmund reduction	H₂, Pd–BaSO₄	RCOCl → RCHO
Stephen reaction	SnCl₂, HCl; H₃O⁺	RCN → RCHO
Etard reaction	CrO₂Cl₂ in CS₂; H₃O⁺	C₆H₅CH₃ → C₆H₅CHO
Gattermann–Koch	CO, HCl, anhyd. AlCl₃ / CuCl	C₆H₆ → C₆H₅CHO
Friedel–Crafts acylation	RCOCl, anhyd. AlCl₃	ArH → ArCOR
Clemmensen reduction	Zn-Hg, conc. HCl	C=O → CH₂
Wolff–Kishner reduction	N₂H₄; KOH, ethylene glycol, Δ	C=O → CH₂
Tollens' test	[Ag(NH₃)₂]⁺, OH⁻	RCHO → RCOO⁻ + Ag mirror
Fehling's test	Cu²⁺, tartrate, OH⁻	RCHO → RCOO⁻ + Cu₂O
Haloform reaction	NaOX (X₂ + NaOH)	RCOCH₃ → RCOO⁻ + CHX₃
Aldol condensation	dil. NaOH; Δ	2 RCH₂CHO → α,β-unsaturated aldehyde
Claisen–Schmidt (Beyond NCERT)	ArCHO + ketone, NaOH	→ ArCH=CH–CO–R
Cannizzaro reaction	conc. NaOH / KOH	2 RCHO (no α-H) → RCH₂OH + RCOO⁻

Match-the-list questions (2022, 2025, 2026) pair exactly these names and reagents.

10 · Traps

Trap	What goes wrong	Do this
Fehling's and benzaldehyde	Counting aromatic aldehydes as positive (2024, 2025)	Fehling's: aliphatic aldehydes and α-hydroxy ketones only
Iodoform	Counting propanal, butanal, pentan-3-one or CH₃COOH (2025)	Needs CH₃CO– or CH₃CH(OH)– bonded to C or H, not to O or N
Benzaldehyde in cross aldol	"No α-H, so no aldol" (2024)	It can't be the enolate, but it is the electrophile
Clemmensen on a 3° alcohol	Expecting the OH to survive (2023)	Hot acid dehydrates it; pick Wolff–Kishner
Wolff–Kishner on C–Cl	Expecting the Cl to stay (2023)	Hot base substitutes or eliminates it
DIBAL-H on a nitrile	Writing the amine (2019, 2021)	DIBAL-H then water → aldehyde; LiAlH₄ → amine
Ester + 2 RMgX	Stopping at the ketone (2022)	Two equivalents add: 3° alcohol
Jones reagent on a 1° alcohol	Writing the aldehyde (2021)	CrO₃–H₂SO₄ goes on to the acid; PCC stops
Crossed Cannizzaro	Oxidising the other aldehyde	HCHO is always the one oxidised
Cannizzaro in D₂O	Putting D on carbon (2022)	H moves C → C as hydride; D lands only on O
Aldol vs Cannizzaro	Using conc. alkali for aldol	α-H + dilute → aldol; no α-H + conc. → Cannizzaro
KOI	Calling it a reducing agent (2026)	Hypoiodite oxidises

11 · Numbers worth knowing

Compound	g mol⁻¹
HCHO	30
CH₃CHO	44
acetone	58
benzaldehyde	106
cyclohexanecarbaldehyde	112
acetophenone	120
benzophenone	182
chalcone C₁₅H₁₂O	208
dibenzalacetone C₁₇H₁₄O	234
semicarbazide	75
CHI₃	394

• Formalin is a 40% solution of HCHO.
• The C=O bond angle is about 120°.
• Tollens': 2 Ag per CHO. Claisen–Schmidt for dibenzalacetone: 2 mol PhCHO per mol acetone.

12 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Natural sources: vanillin (vanilla beans), salicylaldehyde (meadow sweet), cinnamaldehyde (cinnamon).
• Formaldehyde: formalin (40%) preserves biological specimens; makes bakelite (phenol–formaldehyde resin), urea–formaldehyde glues and other polymers.
• Acetaldehyde: starting material for acetic acid, ethyl acetate, vinyl acetate, polymers and drugs.
• Benzaldehyde: perfumery and dyes.
• Acetone and ethyl methyl ketone: common industrial solvents.
• Known for odour and flavour: butyraldehyde, vanillin, acetophenone, camphor.
• Tollens was Bernhard Tollens (1841–1918), professor at Göttingen.