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State Board · Class 12 · Chemistry

Lesson 12: Aldehydes, Ketones and Carboxylic Acids

7 concepts · 56 practice questions with answers · Free to read

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Classification and nomenclature

From daily life

Formalin (biology lab), nail-polish remover (acetone) and vinegar (acetic acid) all contain a carbonyl group.

Simple definition

Aldehydes have –CHO (at the chain end); ketones have >C=O between two carbons; carboxylic acids have –COOH.

Carbonyl classesCarbonyl compounds: classes and namesHCHOmethanal (formaldehyde)CH₃CHOethanal (acetaldehyde)CH₃COCH₃propanone (acetone)CH₃COC₂H₅butanoneC₆H₅CHObenzaldehydeCH₃COOHethanoic acid (acetic)C₆H₅COOHbenzoic acidaldehyde –al (–CHO), ketone –one (>C=O), acid –oic acid (–COOH)

In the figure : Classes and names.

Rule 1: Naming aldehydes — Replace the final e of the alkane with 'al'; the CHO carbon is C-1.

  • Easy : HCHO = methanal.
  • Medium : CH₃CH₂CHO = propanal.
  • Exam-type : C₆H₅CHO = benzaldehyde.

Rule 2: Naming ketones — Replace e with 'one'; lowest locant to C=O.

  • Easy : CH₃COCH₃ = propanone.
  • Medium : CH₃COCH₂CH₃ = butanone.
  • Exam-type : CH₃COCH₂CH₂CH₃ = pentan-2-one.

Rule 3: Naming acids — Replace e with 'oic acid'; the COOH carbon is C-1.

  • Easy : HCOOH = methanoic acid.
  • Medium : CH₃COOH = ethanoic acid.
  • Exam-type : C₆H₅COOH = benzoic acid.

Reasoning: name CH₃CH(CH₃)CH₂CHO

  • Step 1: CHO carbon is C-1; four-carbon chain ⇒ butanal.
  • Step 2: The methyl branch is on C-3.
  • Step 3: Name: 3-methylbutanal.

CET level (MCQ)

  • CET 1) Name of CH₃COCH₃? 1) ethanal 2) propanone 3) propanal 4) butanone — Answer: 2).
  • CET 2) Name of CH₃CH₂CHO? 1) propanone 2) propanal 3) ethanal 4) butanal — Answer: 2).

Real-life use: Formalin, acetone, vinegar, flavours.

Remember

CHO at the end (al); C=O inside (one); COOH = oic acid.

Common mistake

Writing 'al' for a ketone name.

CET trap

The aldehyde carbon is always C-1, so no locant is written.

Shortcut

Memory clue: "al = end, one = middle".

Spot the difference

ClassGroupSuffix
Aldehyde–CHO-al
Ketone>C=O-one
Acid–COOH-oic acid

Practice questions

Easy
  1. 1Aldehyde vs ketone.
  2. 2Name HCHO.
  3. 3Name CH₃COC₂H₅.
Medium
  1. 4Formula of benzoic acid.
  2. 5Formula of 3-methylbutanal.
  3. 6What is acetone?
Challenging
  1. 7Locant of the COOH carbon?
  2. 8Formula of pentan-2-one.
Show answers (try them yourself first)
  1. 1–CHO at the end; >C=O inside.
  2. 2Methanal.
  3. 3Butanone.
  4. 4C₆H₅COOH.
  5. 5CH₃CH(CH₃)CH₂CHO.
  6. 6Propanone.
  7. 71.
  8. 8CH₃COCH₂CH₂CH₃.

Preparation of aldehydes and ketones

From daily life

Almond flavour (benzaldehyde) and vanilla (vanillin) are made in factories by such reactions.

Simple definition

Aldehydes: PCC oxidation of 1° alcohols, Rosenmund, Stephen, DIBAL-H. Ketones: 2° alcohols, Friedel–Crafts acylation, calcium salts.

Ald/ketone prepPreparation of aldehydes and ketonesRCOCl + H₂, Pd–BaSO₄Rosenmund ⇒ RCHORCN + SnCl₂/HCl ⇒ H₃O⁺Stephen ⇒ RCHORCN / ester + DIBAL-HRCHO1° ROH + PCCRCHO (stops before acid)2° ROH + CrO₃ / K₂Cr₂O₇R₂COArH + RCOCl / AlCl₃Friedel–Crafts ⇒ ArCOR(RCOO)₂Ca, heatR₂CO (ketone)alkene + O₃ ⇒ Zn/H₂Oozonolysis ⇒ RCHO / R₂CO

In the figure : Preparation routes.

Rule 1: For aldehydes — RCOCl + H₂/Pd–BaSO₄ (Rosenmund). RCN + SnCl₂/HCl ⇒ H₃O⁺ (Stephen). RCN + DIBAL-H.

  • Easy : CH₃COCl ⇒ CH₃CHO.
  • Medium : C₆H₅CN ⇒ C₆H₅CHO.
  • Exam-type : Poisoned Pd (BaSO₄) stops at the aldehyde.

Rule 2: For ketones — 2° ROH + CrO₃; ArH + RCOCl/AlCl₃; heating (RCOO)₂Ca; nitrile + RMgX.

  • Easy : Propan-2-ol ⇒ acetone.
  • Medium : Benzene + CH₃COCl ⇒ acetophenone.
  • Exam-type : (CH₃COO)₂Ca ⇒ acetone.

Rule 3: Ozonolysis — Alkene + O₃ ⇒ Zn/H₂O ⇒ two carbonyl compounds (the C=C breaks).

  • Easy : CH₃CH=CH₂ ⇒ CH₃CHO + HCHO.
  • Medium : (CH₃)₂C=CH₂ ⇒ acetone + HCHO.
  • Exam-type : C=C ⇒ 2 C=O.

Reasoning: acetophenone from benzene

  • Step 1: CH₃COCl + anhydrous AlCl₃ ⇒ CH₃CO⁺ electrophile.
  • Step 2: It substitutes on the ring.
  • Step 3: C₆H₅COCH₃ + HCl.

CET level (MCQ)

  • CET 1) Catalyst in Rosenmund reduction? 1) Pd–BaSO₄ 2) Ni 3) Pt 4) Fe — Answer: 1).
  • CET 2) Oxidation of propan-2-ol gives? 1) propanal 2) propanone 3) propanoic acid 4) CO₂ — Answer: 2).

Real-life use: Flavours, vanilla, drug synthesis.

Remember

Aldehydes: Rosenmund/Stephen/PCC; ketones: 2° ROH, FC, Ca salts.

Common mistake

Using unpoisoned Pd in Rosenmund ⇒ over-reduction to the alcohol.

CET trap

Friedel–Crafts acylation gives a ketone with no rearrangement.

Shortcut

Memory clue: "Rosenmund = chloride + H₂".

Spot the difference

ReactionStartProduct
RosenmundRCOClRCHO
StephenRCNRCHO
Friedel–CraftsArH + RCOClArCOR

Practice questions

Easy
  1. 1Write Rosenmund reduction.
  2. 2Write Stephen reaction.
  3. 3Why poisoned Pd?
Medium
  1. 4What is ozonolysis?
  2. 5Acetophenone from benzene.
  3. 62° ROH + CrO₃?
Challenging
  1. 7Heating (CH₃COO)₂Ca?
  2. 8Role of PCC?
Show answers (try them yourself first)
  1. 1RCOCl + H₂ ⇒ RCHO + HCl.
  2. 2RCN ⇒ RCHO.
  3. 3To stop at the aldehyde.
  4. 4Cleaving C=C with O₃.
  5. 5CH₃COCl/AlCl₃.
  6. 6A ketone.
  7. 7Acetone.
  8. 81° ROH ⇒ RCHO.

Preparation of carboxylic acids

From daily life

The sting of ant bites (methanoic acid) and the acetic acid in pickles — carboxylic acids are everywhere.

Simple definition

Acids come from oxidation, nitrile hydrolysis, Grignard + CO₂ and side-chain oxidation.

Acid prepPreparation of carboxylic acids1° ROH / RCHO + KMnO₄ or K₂Cr₂O₇oxidation ⇒ RCOOHRCN + H₃O⁺ / OH⁻, heathydrolysis ⇒ RCOOH (same C)RMgX + CO₂ ⇒ H₃O⁺Grignard ⇒ RCOOH (+1 C)C₆H₅–CH₃ + KMnO₄ / H⁺side chain ⇒ C₆H₅COOHester / amide hydrolysisRCOOH + ROH / NH₃A benzylic H is needed; tert-butylbenzene is not oxidised

In the figure : Acid preparation.

Rule 1: Oxidation — 1° ROH or RCHO + KMnO₄ / K₂Cr₂O₇ ⇒ RCOOH (same carbon count).

  • Easy : C₂H₅OH ⇒ CH₃COOH.
  • Medium : CH₃CHO ⇒ CH₃COOH.
  • Exam-type : Toluene ⇒ benzoic acid.

Rule 2: Nitrile and Grignard — RCN + H₂O/H⁺ ⇒ RCOOH (same C). RMgX + CO₂ ⇒ H₃O⁺ ⇒ RCOOH (one extra C).

  • Easy : CH₃CN ⇒ CH₃COOH.
  • Medium : CH₃MgBr + CO₂ ⇒ CH₃COOH.
  • Exam-type : C₂H₅MgBr + CO₂ ⇒ propanoic acid.

Rule 3: Side chain and hydrolysis — ArCH₃ + KMnO₄ ⇒ ArCOOH (needs a benzylic H). Ester/amide hydrolysis ⇒ RCOOH.

  • Easy : Ethyl acetate ⇒ CH₃COOH + C₂H₅OH.
  • Medium : C₆H₅CH₂CH₃ ⇒ C₆H₅COOH.
  • Exam-type : tert-butylbenzene ⇒ no reaction.

Reasoning: propanoic acid from C₂H₅Br

  • Step 1: C₂H₅Br + Mg (dry ether) ⇒ C₂H₅MgBr.
  • Step 2: C₂H₅MgBr + CO₂ ⇒ C₂H₅COOMgBr.
  • Step 3: H₃O⁺ ⇒ C₂H₅COOH (one carbon added).

CET level (MCQ)

  • CET 1) Carbon count in Grignard + CO₂? 1) +1 2) same 3) −1 4) doubles — Answer: 1).
  • CET 2) Toluene + KMnO₄ ⇒ ? 1) benzoic acid 2) phenol 3) benzaldehyde 4) benzene — Answer: 1).

Real-life use: Food preservatives, drugs, polymers.

Remember

Nitrile: same C; Grignard + CO₂: +1 C; needs a benzylic H.

Common mistake

Assuming the carbon count stays the same with Grignard + CO₂.

CET trap

tert-Butylbenzene has no benzylic H so KMnO₄ leaves it unchanged.

Shortcut

Memory clue: "Grignard + CO₂ = one more carbon".

Spot the difference

RouteCarbons
Oxidationsame
Nitrilesame
Grignard + CO₂+1

Practice questions

Easy
  1. 1Three routes to acids.
  2. 2Write nitrile hydrolysis.
  3. 3Write Grignard + CO₂.
Medium
  1. 4Toluene ⇒ benzoic acid.
  2. 5Why not tert-butylbenzene?
  3. 6CH₃CHO ⇒ ?
Challenging
  1. 7Ester hydrolysis.
  2. 8C₂H₅MgBr + CO₂ ⇒ ?
Show answers (try them yourself first)
  1. 1Oxidation, nitrile, Grignard.
  2. 2RCN + 2H₂O ⇒ RCOOH + NH₃.
  3. 3RMgX + CO₂ ⇒ RCOOH.
  4. 4Side-chain oxidation.
  5. 5No benzylic H.
  6. 6CH₃COOH.
  7. 7RCOOH + R′OH.
  8. 8Propanoic acid.

Polarity of the carbonyl group and physical properties

From daily life

Acetone mixes with water yet evaporates fast — polar, but no H-bonding with itself.

Simple definition

C=O is polar: C δ+, O δ–. A nucleophile attacks the carbon. Reactivity: HCHO > RCHO > R₂CO.

PolarityPolar C=O and boiling points (K)C=Oδ+δ–Nu⁻ attacks C δ+Reactivity: HCHO > RCHO > R₂CO273butane329propanone348butanal353butanone370propan-1-ol391butan-1-ol391aceticCarbonyls: dipoles ⇒ above alkanes, below alcohols (no H-bond)

In the figure : Polarity and b.p.

Rule 1: Boiling points — Alkane < carbonyl < alcohol ≈ acid. Acids form dimers through two H-bonds.

  • Easy : Butane 273 K, butanal 348 K.
  • Medium : Butanone 353 K, butan-1-ol 391 K.
  • Exam-type : Acetic acid 391 K.

Rule 2: Solubility — Small (≤ 4 C) aldehydes/ketones mix with water (H-bond to O); longer chains are less soluble.

  • Easy : HCHO and acetone mix.
  • Medium : Benzaldehyde is poorly soluble.
  • Exam-type : Acids up to 4 C mix.

Rule 3: Aldehyde vs ketone reactivity — Ketones have two +I groups and steric hindrance ⇒ less reactive; ArCHO is less reactive due to resonance.

  • Easy : HCHO is the most reactive.
  • Medium : Acetone < acetaldehyde.
  • Exam-type : Benzaldehyde < ethanal.

Reasoning: why is butanal's b.p. below butan-1-ol's?

  • Step 1: Similar molar masses (72 and 74).
  • Step 2: The aldehyde has no O–H, so no H-bonds between molecules.
  • Step 3: Only dipole attractions ⇒ 348 K < 391 K.

CET level (MCQ)

  • CET 1) Most reactive? 1) HCHO 2) CH₃CHO 3) CH₃COCH₃ 4) C₆H₅CHO — Answer: 1).
  • CET 2) Highest b.p.? 1) butanal 2) butan-1-ol 3) butanone 4) butane — Answer: 2).

Real-life use: Solvents (acetone), disinfectant (HCHO).

Remember

b.p.: alkane < carbonyl < alcohol; HCHO most reactive.

Common mistake

Assuming intermolecular H-bonding in aldehydes.

CET trap

An acid's b.p. exceeds that of an alcohol of similar M (dimer).

Shortcut

Memory clue: "HCHO first, ketones last".

Spot the difference

Compoundb.p. (K)
Butanal348
Butan-1-ol391
Acetic acid391

Practice questions

Easy
  1. 1Why is C=O polar?
  2. 2Why aldehydes > ketones?
  3. 3Why aldehyde b.p. < alcohol?
Medium
  1. 4Why do acids dimerise?
  2. 5Why is benzaldehyde less reactive?
  3. 6Is acetone miscible with water?
Challenging
  1. 7Reactivity order.
  2. 8What are dipole attractions?
Show answers (try them yourself first)
  1. 1C δ+, O δ–.
  2. 2Fewer +I groups and less hindrance.
  3. 3No O–H.
  4. 4Two H-bonds.
  5. 5Resonance.
  6. 6Yes.
  7. 7HCHO > RCHO > R₂CO.
  8. 8δ+ end attracts δ– end.

Tests and nucleophilic addition

From daily life

The same aldehyde test is used to put a silver layer on a mirror.

Simple definition

Nucleophilic addition: Nu⁻ attacks C δ+. Tests: 2,4-DNP, Tollens, Fehling, iodoform.

TestsAldehydes and ketones: tests and nucleophilic addition2,4-DNPboth ⇒ orange-yellow pptTollens (Ag⁺/NH₃)only RCHO ⇒ silver mirrorFehling (Cu²⁺)aliphatic RCHO ⇒ red Cu₂OI₂ / NaOHCH₃CO– / CH₃CH(OH)– ⇒ yellow CHI₃NaHSO₃RCHO, small ketones ⇒ crystalsHCN⇒ cyanohydrinNH₂OH⇒ oxime (C=N–OH)Benzaldehyde and ketones: no Fehling result

In the figure : Tests.

Rule 1: Aldehyde vs ketone — Tollens: RCHO ⇒ silver mirror. Fehling: aliphatic RCHO ⇒ red Cu₂O. Ketones give nothing.

  • Easy : 0.1 mol RCHO ⇒ 21.6 g Ag.
  • Medium : Benzaldehyde with Fehling ⇒ none.
  • Exam-type : Acetone with Tollens ⇒ none.

Rule 2: Addition reactions — HCN ⇒ cyanohydrin; NaHSO₃ ⇒ crystals; NH₂OH ⇒ oxime; 2,4-DNP ⇒ hydrazone.

  • Easy : CH₃CHO + HCN ⇒ CH₃CH(OH)CN.
  • Medium : RCHO + NH₂OH ⇒ RCH=NOH.
  • Exam-type : 2,4-DNP ⇒ orange ppt.

Rule 3: Iodoform test — CH₃CO– (or CH₃CH(OH)–) + I₂/NaOH ⇒ yellow CHI₃.

  • Easy : Acetone ⇒ CHI₃.
  • Medium : Ethanol ⇒ CHI₃.
  • Exam-type : 5.8 g acetone ⇒ 39.4 g CHI₃.

Calculation: CHI₃ from 5.8 g acetone

  • Step 1: n = 5.8 ÷ 58 = 0.1 mol.
  • Step 2: 1 : 1 ⇒ 0.1 mol CHI₃.
  • Step 3: M = 393.7 ⇒ 39.4 g.

CET level (MCQ)

  • CET 1) Test only for aldehydes? 1) Tollens 2) 2,4-DNP 3) iodoform 4) NaHSO₃ — Answer: 1).
  • CET 2) Which gives iodoform? 1) CH₃COCH₃ 2) C₆H₅CHO 3) HCHO 4) CH₃CH₂CHO — Answer: 1).

Real-life use: Mirrors, identification tests.

Remember

Tollens = Ag mirror; 2,4-DNP = both; iodoform = CH₃CO–.

Common mistake

Thinking 2,4-DNP is only for aldehydes; ketones give it too.

CET trap

Benzaldehyde gives Tollens but not Fehling.

Shortcut

Memory clue: "Tollens = the aldehyde mirror".

Spot the difference

ReagentRCHOR₂CO
Tollensmirrornone
Fehlingrednone
2,4-DNPpptppt

Practice questions

Easy
  1. 1Describe Tollens test.
  2. 2Fehling test.
  3. 3Iodoform test is for?
Medium
  1. 4Write the HCN addition.
  2. 5What does 2,4-DNP give?
  3. 6Ag from 0.1 mol RCHO?
Challenging
  1. 7CHI₃ from 5.8 g acetone?
  2. 8Benzaldehyde + Fehling.
Show answers (try them yourself first)
  1. 1Silver mirror.
  2. 2Red Cu₂O.
  3. 3CH₃CO– / CH₃CH(OH)–.
  4. 4Cyanohydrin.
  5. 5Orange hydrazone.
  6. 621.6 g.
  7. 739.4 g.
  8. 8No reaction.

Aldol, Cannizzaro, haloform and reduction

From daily life

Drug makers use the aldol reaction to build new C–C bonds.

Simple definition

With α-H ⇒ aldol; without α-H ⇒ Cannizzaro. Reductions: Clemmensen, Wolff–Kishner (C=O ⇒ CH₂), NaBH₄ (⇒ ROH).

Named reactionsNamed reactionsAldolneeds α-H; dil. NaOH ⇒ β-hydroxyCannizzarono α-H; conc. NaOH ⇒ ROH + RCOO⁻HaloformCH₃CO– + X₂/NaOH ⇒ CHX₃ + RCOO⁻ClemmensenZn–Hg / conc. HCl ⇒ C=O → CH₂Wolff–KishnerNH₂NH₂, KOH / glycol ⇒ C=O → CH₂NaBH₄ / LiAlH₄RCHO ⇒ 1°; R₂CO ⇒ 2° alcoholHVZRCH₂COOH + X₂/P ⇒ α-halo acidAldol needs α-H; Cannizzaro needs none

In the figure : Named reactions.

Rule 1: Aldol condensation — 2 CH₃CHO + dil. NaOH ⇒ CH₃CH(OH)CH₂CHO (3-hydroxybutanal); heating loses H₂O ⇒ crotonaldehyde.

  • Easy : Ethanal and propanone undergo it.
  • Medium : HCHO / C₆H₅CHO do not.
  • Exam-type : An α-hydrogen is essential.

Rule 2: Cannizzaro — 2 RCHO (no α-H) + conc. NaOH ⇒ RCH₂OH + RCOO⁻Na⁺ (one oxidised, one reduced).

  • Easy : 10.6 g benzaldehyde ⇒ 5.4 g benzyl alcohol.
  • Medium : + 7.2 g sodium benzoate.
  • Exam-type : 2 HCHO ⇒ CH₃OH + HCOONa.

Rule 3: Haloform and reduction — CH₃COR + 3 X₂/NaOH ⇒ CHX₃ + RCOONa. Zn–Hg/HCl or NH₂NH₂/KOH ⇒ C=O ⇒ CH₂.

  • Easy : Acetone ⇒ CHI₃ + CH₃COONa.
  • Medium : C₆H₅COCH₃ ⇒ C₆H₅CH₂CH₃.
  • Exam-type : Acidic: Clemmensen; basic: Wolff–Kishner.

Calculation: 10.6 g benzaldehyde + conc. NaOH

  • Step 1: n = 10.6 ÷ 106 = 0.1 mol.
  • Step 2: 2 : 1 : 1 ⇒ 0.05 mol alcohol and 0.05 mol salt.
  • Step 3: 0.05 × 108 = 5.4 g; 0.05 × 144 = 7.2 g.

CET level (MCQ)

  • CET 1) Cannizzaro is given by? 1) HCHO 2) CH₃CHO 3) CH₃COCH₃ 4) C₂H₅CHO — Answer: 1).
  • CET 2) C=O ⇒ CH₂ (acidic)? 1) Clemmensen 2) Wolff–Kishner 3) NaBH₄ 4) LiAlH₄ — Answer: 1).

Real-life use: Drugs, polymers, fragrances.

Remember

α-H ⇒ aldol; no α-H ⇒ Cannizzaro; Zn–Hg = acidic; NH₂NH₂ = basic.

Common mistake

Thinking benzaldehyde gives aldol; it has no α-H.

CET trap

Aldol uses dilute alkali; Cannizzaro uses concentrated.

Shortcut

Memory clue: "Aldol needs α-H; Cannizzaro needs none".

Spot the difference

Reactionα-HBase
Aldolneededdilute
Cannizzaroabsentconc.
HaloformCH₃CO–X₂/NaOH

Practice questions

Easy
  1. 1Write the aldol reaction.
  2. 2Write the Cannizzaro reaction.
  3. 3What is α-H?
Medium
  1. 4Write Clemmensen.
  2. 5Write Wolff–Kishner.
  3. 610.6 g benzaldehyde ⇒ ?
Challenging
  1. 7Haloform is for?
  2. 8Role of NaBH₄?
Show answers (try them yourself first)
  1. 12 CH₃CHO ⇒ 3-hydroxybutanal.
  2. 22 RCHO ⇒ RCH₂OH + RCOONa.
  3. 3H on the carbon next to C=O.
  4. 4Zn–Hg / conc. HCl.
  5. 5NH₂NH₂, KOH, heat.
  6. 65.4 g and 7.2 g.
  7. 7Compounds with CH₃CO–.
  8. 8RCHO ⇒ 1°; R₂CO ⇒ 2° ROH.

Acidity and reactions of carboxylic acids

From daily life

The sour taste of lemon and tamarind, and vinegar dissolving limescale in a kettle — acidity.

Simple definition

RCOOH ⇌ RCOO⁻ + H⁺; the carboxylate is resonance-stabilised. –I groups (Cl) raise acidity; +I groups lower it.

AcidsAcidity (pKa) and reactionsCl₃CCOOH0.65Cl₂CHCOOH1.29ClCH₂COOH2.86HCOOH3.75C₆H₅COOH4.20CH₃COOH4.76lower pKa ⇒stronger acidRCOOH + R′OH, H⁺⇌ ester + H₂ORCOOH + SOCl₂⇒ RCOCl + SO₂ + HClRCOOH + NaHCO₃⇒ CO₂ (test)RCOOH + LiAlH₄⇒ RCH₂OHRCOONa + NaOH/CaO, heat⇒ RH (decarboxylation)

In the figure : pKa and reactions

Rule 1: Acidity order — Cl₃CCOOH (0.65) > Cl₂CHCOOH (1.29) > ClCH₂COOH (2.86) > HCOOH (3.75) > C₆H₅COOH (4.20) > CH₃COOH (4.76).

  • Easy : More Cl ⇒ more acidic.
  • Medium : Formic > acetic (CH₃ is +I).
  • Exam-type : Carboxylic > phenol (10).

Rule 2: Reactions of acids — NaHCO₃ ⇒ CO₂ (test). SOCl₂ ⇒ RCOCl. LiAlH₄ ⇒ RCH₂OH. NH₃ ⇒ amide.

  • Easy : CH₃COOH + NaHCO₃ ⇒ CO₂ fizz.
  • Medium : CH₃COOH + SOCl₂ ⇒ CH₃COCl.
  • Exam-type : LiAlH₄ ⇒ ethanol.

Rule 3: Esterification and decarboxylation — RCOOH + R′OH ⇌ RCOOR′ + H₂O (conc. H₂SO₄). RCOONa + NaOH/CaO, heat ⇒ RH + Na₂CO₃.

  • Easy : 60 g + 46 g ⇒ 88 g ethyl acetate (100 %).
  • Medium : CH₃COONa ⇒ CH₄.
  • Exam-type : Halogenation: HVZ ⇒ α-halo acid.

Calculation: 60 g acetic acid + 46 g ethanol

  • Step 1: n = 60 ÷ 60 = 1 mol; 46 ÷ 46 = 1 mol.
  • Step 2: 1 : 1 ⇒ 1 mol ester (100 % yield).
  • Step 3: M = 88 ⇒ 88 g.

CET level (MCQ)

  • CET 1) Strongest acid? 1) CH₃COOH 2) ClCH₂COOH 3) Cl₃CCOOH 4) C₆H₅COOH — Answer: 3).
  • CET 2) Gives CO₂ with NaHCO₃? 1) RCOOH 2) phenol 3) alcohol 4) ether — Answer: 1).

Real-life use: Vinegar, drugs, ester flavours, aspirin.

Remember

–Cl ⇒ stronger; –CH₃ ⇒ weaker; NaHCO₃ ⇒ CO₂.

Common mistake

Thinking a larger pKa means a stronger acid; it is the reverse.

CET trap

Carboxylic acids beat phenols, so NaHCO₃ separates them.

Shortcut

Memory clue: "more chlorine = stronger acid".

Spot the difference

AcidpKa
Cl₃CCOOH0.65
HCOOH3.75
CH₃COOH4.76

Practice questions

Easy
  1. 1State acidity order.
  2. 2Why does Cl raise acidity?
  3. 3NaHCO₃ test.
Medium
  1. 4Write esterification.
  2. 560 g + 46 g ⇒ ?
  3. 6What is decarboxylation?
Challenging
  1. 7RCOOH + SOCl₂?
  2. 8RCOOH + LiAlH₄?
Show answers (try them yourself first)
  1. 1Cl₃CCOOH strongest.
  2. 2–I effect stabilises the anion.
  3. 3CO₂ fizz ⇒ an acid.
  4. 4RCOOH + R′OH ⇌ RCOOR′ + H₂O.
  5. 588 g.
  6. 6Loss of CO₂.
  7. 7RCOCl + SO₂ + HCl.
  8. 8RCH₂OH.

Quick revision of the whole lesson

ConceptKey formula / fact
Aldehyde–CHO, -al
Ketone>C=O, -one
RosenmundRCOCl + H₂ ⇒ RCHO
ReactivityHCHO > RCHO > R₂CO
Tollensonly RCHO ⇒ Ag mirror
Aldolneeds α-H
Cannizzarono α-H
AcidityCl₃CCOOH > … > CH₃COOH
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Nomenclature.
  • Ald/ketone preparation.
  • Acid preparation.
  • Polarity, b.p.
  • Tests, addition.
  • Aldol, Cannizzaro, reduction.
  • Acidity, reactions.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.