Aldehydes, Ketones and Carboxylic Acids Class 12 Notes - CBSE Chemistry Chapter 12

Chapter summary

Aldehydes, Ketones and Carboxylic Acids covers the naming and structure of the polar carbonyl group, methods of preparing these compounds, nucleophilic addition reactions of the carbonyl, reactions involving the alpha-hydrogen such as aldol and Cannizzaro, identification tests, the acidity of carboxylic acids, and conversion of acids into their derivatives. It is a high-yield NEET chapter where questions test reactivity order, named reactions, distinguishing tests like Tollens and iodoform, and the factors affecting acid strength.

Chapter notes
🃏 Flash Cards: Aldehydes, Ketones and Carboxylic Acids

Class 12 Chemistry – swipe through all 9 cards to understand the whole chapter.

🧪Start here1/9

The Carbonyl Group >C=O

One polar C=O bond is the heart of this whole chapter.

C^δ+ = O^δ⁻ · sp2 carbon · trigonal planar · ∠ ≈ 120°

Oxygen is more electronegative, so the carbon is electron-poor (δ+).

  • Aldehyde R–CHO (carbonyl at chain end); ketone R–CO–R′ (in the middle); acid R–COOH
  • δ+ carbon is electrophilic → open to nucleophiles; oxygen is mildly basic
  • This single tilt drives nearly every reaction ahead
🏷️Naming2/9

IUPAC & Common Names

Number the chain so the carbonyl carbon gets the lowest locant.

–al (aldehyde) · –one (ketone) · –oic acid (acid) · –CHO & –COOH are always C-1

Seniority for suffix: carboxylic acid > ester > amide > aldehyde > ketone > alcohol.

  • Trivial names: HCHO formaldehyde, CH3CHO acetaldehyde, CH3COCH3 acetone
  • HCOOH formic acid, CH3COOH acetic acid
  • Winner of seniority becomes the suffix; the rest become prefixes
⚗️Preparation3/9

Making Aldehydes, Ketones & Acids

Match each reagent to its product — the favourite one-word questions.

Rosenmund: RCOCl + H2/Pd-BaSO4 → RCHO · Stephen: RCN + SnCl2/HCl → RCHO

Rosenmund uses POISONED Pd/BaSO4; plain Pd over-reduces to the alcohol.

  • Etard: toluene + CrO2Cl2 → benzaldehyde; Gattermann–Koch: C6H6 + CO/HCl/AlCl3 → benzaldehyde
  • Ketones: Friedel–Crafts acylation (RCOCl/AlCl3); nitrile + Grignard then hydrolysis
  • Ethyne is the ONLY alkyne giving an aldehyde on hydration (→ acetaldehyde)
➡️Core reaction4/9

Nucleophilic Addition to C=O

Electron-rich nucleophiles attack the δ+ carbonyl carbon — the signature reaction.

+HCN → cyanohydrin · +NaHSO3 → bisulphite adduct · +R′OH/HCl → acetal · +RMgX → alcohol

NH2–G derivatives: hydroxylamine → oxime, hydrazine → hydrazone, 2,4-DNP → orange ppt.

  • 2,4-DNP confirms a carbonyl but CANNOT tell aldehyde from ketone
  • NaHSO3 purifies aldehydes & small/methyl ketones; bulky ketones fail
  • Flat C=O becomes a tetrahedral carbon carrying two new groups
📊Reactivity order5/9

Aldehydes > Ketones

Fewer alkyl groups means more δ+ and less crowding, so faster addition.

HCHO > CH3CHO > CH3COCH3

Benzaldehyde is sluggish: the ring donates electrons by resonance, cutting the δ+.

  • Two +I alkyl groups in ketones shrink the δ+ on carbon
  • Alkyl groups also cause steric hindrance, blocking the nucleophile
  • Electron-withdrawing groups raise reactivity; donating groups lower it
🔗Alpha-H reactions6/9

Aldol, Cannizzaro & Haloform

The acidic α-hydrogen next to C=O decides which named reaction can run.

α-H present → Aldol & Iodoform · NO α-H → Cannizzaro (disproportionation)

Aldol gives a β-hydroxy carbonyl first; on heating it dehydrates to an α,β-unsaturated carbonyl.

  • Cannizzaro needs NO α-H (HCHO, C6H5CHO): one oxidised to acid, one reduced to alcohol
  • Iodoform (CHI3, yellow): CH3CO– groups, plus CH3CH(OH)– → tests methyl ketones, acetaldehyde, ethanol
  • 2 CH3CHO aldol → 3-hydroxybutanal, then → but-2-enal (crotonaldehyde)
🪞Identification tests7/9

Tollens & Fehling

Aldehydes oxidise easily; ketones resist — that gap powers the tests.

Tollens [Ag(NH3)2]⁺ → silver mirror (aldehydes) · Fehling → red Cu2O (aliphatic aldehydes)

NEET trap: benzaldehyde gives Tollens but does NOT give Fehling’s.

  • Ketones give neither Tollens nor Fehling
  • Aromatic aldehydes do not respond to Fehling’s
  • Aldehydes are easily oxidised to acids; ketones cleave only under strong oxidation
🧲Acidity8/9

Why Carboxylic Acids Are Acidic

The carboxylate ion shares its negative charge equally over two oxygens.

R–C(=O)–O⁻ ↔ R–C(–O⁻)=O · resonance-stabilised RCOO⁻ · lower pKa = stronger acid

Stronger than phenol: phenoxide spreads charge onto less electronegative carbons.

  • EWG (–Cl, –F, –NO2) stabilise the anion → stronger: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH
  • Effect fades with distance: 2-chloro > 3-chloro > 4-chloro acid
  • HCOOH > CH3COOH (formic has no electron-donating +I alkyl group)
🔧Acid reactions9/9

Reactions of –COOH

Swap the –OH and the single –COOH unfolds into many derivatives.

+SOCl2 → acyl chloride · +R′OH/H2SO4 → ester · +NH3,Δ → amide · +P2O5 → anhydride

SOCl2 is preferred — its byproducts SO2 and HCl are gases that escape, leaving a clean product.

  • LiAlH4 (or B2H6) reduces –COOH → 1° alcohol; NaBH4 does NOT reduce acids
  • HVZ: acid with α-H + X2/red P → α-halo acid; HCOOH & benzoic acid (no α-H) fail
  • Decarboxylation: RCOONa + soda lime → alkane (CH3COONa → CH4, loses one carbon)
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📝 Practice Aldehydes, Ketones and Carboxylic Acids — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Q1NEET 2021
Acetone reacts with C₂H₅MgBr in dry ether and the adduct is hydrolysed with H₂O/H⁺. The IUPAC name of the organic product is:
Correct answer: D. Ethylmagnesium bromide adds across the C=O of acetone. (CH₃)₂C=O + C₂H₅MgBr gives (CH₃)₂C(OMgBr)(C₂H₅), which on acidic hydrolysis gives (CH₃)₂C(OH)C₂H₅. Counting the longest chain through the –OH carbon (C₂H₅ + that carbon = 4 carbons, with a methyl branch) gives 2-methylbutan-2-ol.
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Q2NEET 2021
Match each reaction with the product it forms: (A) Benzene + CO + HCl with anhydrous AlCl₃/CuCl, (B) R–CO–CH₃ + X₂/NaOH, (C) R–COOH + R’OH with conc. H₂SO₄, (D) R–CH₂–COOH + X₂/Red P then H₂O. The reactions are, respectively:
Correct answer: B. (A) Benzene + CO + HCl/AlCl₃·CuCl is the Gattermann–Koch synthesis of benzaldehyde. (B) A methyl ketone + halogen/base is the Haloform reaction. (C) Acid + alcohol/conc. H₂SO₄ is Fischer Esterification. (D) An acid with α-H + X₂/red P is the Hell–Volhard–Zelinsky reaction giving an α-halo acid.
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Q3NEET 2020
The reaction between benzaldehyde and acetophenone in the presence of dilute NaOH is known as:
Correct answer: C. Benzaldehyde has no α-hydrogen but acetophenone (C₆H₅COCH₃) does. Dilute NaOH generates the enolate of acetophenone, which adds to benzaldehyde’s carbonyl; dehydration gives an α,β-unsaturated ketone (chalcone). Because two different carbonyls are involved, it is a cross (mixed) aldol condensation, also called the Claisen–Schmidt reaction.
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Q4NEET 2020
An α,β-dicarboxylic acid that can form a cyclic anhydride on heating and a cyclic imide on strong heating with ammonia is:
Correct answer: A. Only when the two –COOH groups are close (ortho / same-side, syn) can they cyclise. Phthalic acid (the 1,2-isomer) loses water on heating to give phthalic anhydride, and on strong heating with ammonia gives the cyclic imide phthalimide. The 1,3- and 1,4-isomers are too far apart to cyclise.
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Q5NEET 2018
Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. This is due to:
Correct answer: D. Carboxylic acids form strong intermolecular hydrogen bonds, even existing as cyclic dimers (two –COOH groups H-bonded together). This extensive association raises the boiling point above that of aldehydes/ketones (no H-bonding) and alcohols (weaker, single H-bond) of similar mass.
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Q6NEET 2015
Reaction of a carbonyl compound with which reagent involves nucleophilic addition followed by elimination of water?
Correct answer: B. Ammonia derivatives such as hydrazine (H₂N–NH₂) add to the carbonyl and then eliminate water to form a C=N product (a hydrazone): R₂C=O + H₂N–NH₂ → R₂C=N–NH₂ + H₂O. A feebly acidic medium is optimal. Grignard, HCN and NaHSO₃ all give simple addition products with no loss of water.
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Q7NEET 2015
Which one of the following esters gets hydrolysed most easily under alkaline conditions?
Correct answer: A. Alkaline hydrolysis is faster when the leaving aryloxide is more stable (more electron-poor ring). The strongly electron-withdrawing p-NO₂ group best stabilises the developing negative charge on the phenoxide leaving group (and the tetrahedral intermediate), so 4-nitrophenyl acetate hydrolyses fastest. Electron-donating –OCH₃ slows it the most.
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Q8NEET 2015
Treatment of cyclopentanone with methyllithium (CH₃Li) gives which species (the intermediate before work-up)?
Correct answer: A. Methyllithium is a strong carbon nucleophile. CH₃⁻ adds to the carbonyl carbon of cyclopentanone, and the oxygen becomes an alkoxide (–O⁻Li⁺) — a lithium cyclopentanonyl (alkoxide) anion. Aqueous work-up then protonates it to 1-methylcyclopentanol; the intermediate is the alkoxide anion.
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Q9NEET 2015
An organic compound X (molecular formula C₅H₁₀O) gives a phenylhydrazone, gives a negative response to the iodoform and Tollens’ tests, and produces n-pentane on reduction. X is:
Correct answer: C. Phenylhydrazone formation shows a carbonyl group. Negative Tollens’ rules out an aldehyde, so X is a ketone (eliminates pentanal and the alcohol). Negative iodoform rules out the CH₃–CO– pattern, so it is not 2-pentanone. Pentan-3-one (CH₃CH₂–CO–CH₂CH₃, C₅H₁₀O) fits all clues and reduces to n-pentane.
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Q10NEET 2009
Propionic acid with Br₂ and red P (Hell–Volhard–Zelinsky) yields a dibromo product. Its structure is:
Correct answer: D. HVZ halogenates only the α-carbon (the one next to –COOH), leaving the –OH of the acid intact. Propionic acid is CH₃–CH₂–COOH; its α-carbon is the CH₂. With excess Br₂/red P both α-hydrogens are replaced, giving CH₃–CBr₂–COOH (2,2-dibromopropanoic acid).
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Frequently Asked Questions

What is the carbonyl group and why is it reactive?

The carbonyl group is the carbon-oxygen double bond (C=O) found in aldehydes, ketones and carboxylic acids. It is polar because oxygen is more electronegative, giving the carbon a partial positive charge. This makes the carbonyl carbon electrophilic, so it is readily attacked by nucleophiles.

What is the reactivity order of carbonyl compounds towards nucleophilic addition?

The order is formaldehyde (HCHO) greater than other aldehydes greater than ketones. Reactivity decreases as more or bulkier alkyl groups are attached to the carbonyl carbon, because they cause steric hindrance and reduce the partial positive charge through their electron-donating effect.

What is the Cannizzaro reaction and which compounds undergo it?

The Cannizzaro reaction is a base-induced disproportionation of aldehydes that have no alpha-hydrogen, such as formaldehyde and benzaldehyde. One molecule is oxidised to a carboxylic acid salt and another is reduced to an alcohol.

Why are carboxylic acids more acidic than alcohols and phenols?

Carboxylic acids are more acidic because the carboxylate ion formed after losing a proton is stabilised by resonance, with the negative charge delocalised equally over two oxygen atoms. In phenoxide the charge is spread onto less electronegative carbon atoms, so it is less stabilised and phenol is a weaker acid.

How can you distinguish an aldehyde from a ketone?

Aldehydes are easily oxidised and give positive tests with Tollens’ reagent (forming a silver mirror) and Fehling’s solution (forming a red-brown precipitate), whereas ketones do not. The iodoform test is positive for methyl ketones and acetaldehyde but is not a general aldehyde-versus-ketone test.

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