Aldehydes Ketones

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your JEE Main attempt.

Aldehydes (R–CHO) and ketones (R–CO–R′) both contain the carbonyl >C=O group, but an aldehyde has at least one H on the carbonyl carbon while a ketone has two alkyl/aryl groups. The carbonyl carbon is sp² hybridised, trigonal planar, and electrophilic — so nucleophilic addition dominates their chemistry.

Must-know reagents for JEE Main:

Tollens’ (ammoniacal Ag⁺) and Fehling’s (Cu²⁺/tartrate) — positive only for aldehydes (and the methyl group test 2,4-DNP is universal for both).
Iodoform test (I₂/NaOH → yellow CHI₃) — positive only for the CH₃–CO– group.
NaBH₄ reduces only aldehydes/ketones; LiAlH₄ also reduces esters, acids, amides.

Quick mnemonic: “Tollen = aldehyde silver mirror; Iodoform = methyl ketone yellow.”

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few weeks to months of preparation.

Nomenclature, Structure, and Reactivity

The carbonyl carbon carries a partial positive charge (δ⁺) because oxygen pulls π-electron density. Aldehydes are more reactive than ketones toward nucleophiles — less steric crowding and stronger δ⁺ (only one alkyl group donates electrons by hyperconjugation, versus two in ketones).

Preparation Routes

Oxidation: primary alcohols → aldehydes (PCC, CrO₃/pyridine); secondary alcohols → ketones (K₂Cr₂O₇).
Rosenmund reduction: RCOCl + H₂ (Pd/BaSO₄) → RCHO.
Stephen reaction: RC≡N + SnCl₂/HCl → RCH=NH → RCHO.
Friedel–Crafts acylation: RCOCl + ArH + AlCl₃ → aryl ketone.
Ozonolysis of alkenes: gives aldehydes and/or ketones depending on substitution.

Nucleophilic Addition Products

Reagent	Product	Use
HCN (or NaCN + dil. H₂SO₄)	Cyanohydrin R₂C(OH)(CN)	Extends carbon chain
NaHSO₃	Bisulfite adduct (white crystalline)	Separation/purification
2,4-DNP	Orange-yellow hydrazone	Identification (mp)
NH₂OH	Oxime	Identification

α-Hydrogen Chemistry

The enol/enolate tautomer is reachable because α-hydrogens (pKa ≈ 19–20) are acidic. This drives the aldol condensation: 2 CH₃CHO + dil. NaOH → CH₃CH(OH)CH₂CHO → (Δ) → CH₃CH=CHCHO (crotonaldehyde, conjugated α,β-unsaturated aldehyde).

Qualitative Tests at a Glance

Compound	Tollens’	Fehling’s	Iodoform	2,4-DNP
HCHO	✓	✗	✗	✓
CH₃CHO	✓	✓	✓	✓
CH₃COCH₃	✗	✗	✓	✓
PhCHO	✓	✗	✗	✓
PhCOCH₃	✗	✗	✓	✓

JEE trap: benzaldehyde gives Tollens’ but not Fehling’s; acetophenone gives iodoform but not Tollens’.

Reduction Toolbox

Clemmensen (Zn–Hg / conc. HCl) — acid medium; >C=O → CH₂.
Wolff–Kishner (H₂N–NH₂ / KOH / ethylene glycol, ~470 K) — base medium; >C=O → CH₂.
Choose Clemmensen for base-sensitive substrates and Wolff–Kishner for acid-sensitive ones.
NaBH₄ (mild, works in EtOH/H₂O): aldehydes + ketones only.
LiAlH₄ (strong, dry ether mandatory): aldehydes, ketones, esters, acids, amides.

Special Aldehyde Reactions

Cannizzaro (no α-H): 2 PhCHO + conc. NaOH → PhCH₂OH + PhCOO⁻Na⁺ (disproportionation).
Tischenko: 2 PhCHO + Al(OR)₃ → PhCOOCH₂Ph (ester).
Schiff base: RCHO + R′NH₂ → RCH=NR′ + H₂O (imine, pH ≈ 4–5).

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline aiming for top rank.

Mechanism Nuances That Appear in JEE

Cyanohydrin formation is simultaneously acid- and base-catalysed: free CN⁻ is too basic and would deprotonate an α-H (pKa ≈ 20) faster than it attacks the carbonyl. Generating HCN in situ with NaCN + dilute H₂SO₄ (pH ≈ 4–5) keeps CN⁻ concentration low enough for nucleophilic attack while H⁺ protonates the alkoxide oxygen. Drawing CN⁻ attacking first is the common mechanistic error.

Aldol condensation has two kinetic stages: (1) NaOH removes the α-H forming the enolate, which attacks a second carbonyl to give the β-hydroxy carbonyl (aldol); (2) on heating, E1cb dehydration yields the conjugated α,β-unsaturated carbonyl — the thermodynamic product because conjugation stabilises the C=C.

Haloform Stoichiometry and Selectivity

For every CH₃–CO– group, exactly 3 moles of I₂ and 4 moles of NaOH are consumed, giving 1 mole of CHI₃, the sodium salt of the acid, NaI, and water. Ethanol and isopropanol (CH₃CH(OH)–) also give a positive iodoform because NaOH oxidises them first to the corresponding carbonyl. Methanol, formaldehyde, and benzaldehyde are negative.

Worked Example — Iodoform on Acetophenone

PhCOCH₃ + 3 I₂ + 4 NaOH → PhCOO⁻Na⁺ + CHI₃↓ (yellow) + 3 NaI + 3 H₂O. Moles of I₂ used per mole of ketone = 3.

Common Mistakes

Writing “AgNO₃” for Tollens’ — it must be ammoniacal AgNO₃; neutral AgNO₃ gives AgCl/AgBr/AgI with halides but not the silver mirror with aldehydes.
Assuming all aldehydes work with Fehling’s — aromatic aldehydes and formaldehyde do not, because they are either too weakly reducing or react by Cannizzaro.
Picking Clemmensen on a phenol-containing substrate — conc. HCl would protonate/decompose it; switch to Wolff–Kishner.
Forgetting that NaBH₄ tolerates protic solvents while LiAlH₄ must be in anhydrous ether (water ignites it).
Drawing aldol as a single-molecule product — it always forms between two carbonyl molecules (self-aldol or cross/Claisen–Schmidt with a different aldehyde).

Practice Prompts

Predict the products of (a) PhCHO + HCN, (b) CH₃COCH₃ + 2,4-DNP, (c) PhCHO + conc. NaOH. Identify which is a cyanohydrin, which is a hydrazone, and which is Cannizzaro.
An unknown carbonyl compound (C₃H₆O) gives a positive iodoform test but a negative Tollens’ test. Name it and justify each result using the structural argument for δ⁺ at the carbonyl carbon and the requirement of a CH₃–CO– motif.

Exam Weightage and Strategy

JEE Main draws 2–3 Chemistry questions (≈3% weightage) from this unit, mostly single-correct MCQs on reagent selection, identification tests, and named reactions, plus occasional assertion-reason items comparing NaBH₄ vs LiAlH₄, Clemmensen vs Wolff–Kishner, and Tollens’ vs Fehling’s. Budget ≈90 seconds per question; memorise the qualitative-test table above because it resolves 80% of these items.

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Aldehydes Ketones