Chapter 17: Aldehydes, Ketones and Carboxylic Acids

Introduction to Carbonyl Compounds

Overview

• Aldehydes and ketones contain a carbonyl group (>C=O) with general formula CnH2nO.

• Aldehydes: Carbonyl attached to at least one hydrogen atom (–CHO group).

• Ketones: Carbonyl attached to two carbon atoms (alkyl/aryl groups).

• Carboxylic acids contain –COOH group, general formula CnH2nO2.

Solved Examples

• Ex 1: Identify the functional group in propanal.
  Solution: Propanal is an aldehyde with –CHO group.

• Ex 2: Write the general formula of ketones.
  Solution: CnH2nO.

Practice Set

• Q1: Which functional group is present in ketones?
  A. –OH
  B. –CHO
  C. >C=O
  D. –COOH

• Q2: The general formula of aldehydes is:
  A. CnH2nO
  B. CnH2n+2O
  C. CnH2nO2
  D. CnH2n-2O

• Q3: Which compound contains the –COOH group?
  A. Aldehyde
  B. Ketone
  C. Carboxylic acid
  D. Alcohol

Answer Key

• Q1: C – Ketones have >C=O group bonded to two carbons.

• Q2: A – Aldehydes and ketones have general formula CnH2nO.

• Q3: C – Carboxylic acids contain –COOH group.

Memory Aids

• Carbonyl group = C=O.

• Aldehyde = Carbonyl + at least one H.

• Ketone = Carbonyl between two carbons.

• Carboxylic acid = Carbonyl + –OH (–COOH).

Nomenclature and Structure of Carbonyl Group

Nomenclature

• Aldehydes: Common names from corresponding acid by replacing ‘-ic acid’ with ‘aldehyde’.

• Ketones: Named by listing alkyl/aryl groups attached to carbonyl.

• Substituent positions indicated by Greek letters α, β, γ, δ starting from carbon next to carbonyl.

• IUPAC names: Replace alkane ‘-e’ ending with ‘-al’ for aldehydes and ‘-one’ for ketones.

Structure

• Carbonyl carbon is sp2 hybridized.

• Forms three sigma (σ) bonds and one pi (π) bond with oxygen.

• Atoms attached to carbonyl carbon lie in one plane with bond angle ~120° (trigonal planar).

• π-electron cloud lies above and below the plane.

• Carbonyl group exhibits resonance: C=O ↔ C+–O–.

Solved Examples

• Ex 1: Name CH3CHO using IUPAC.
  Solution: Ethanal (acetaldehyde).

• Ex 2: Describe hybridization of carbonyl carbon.
  Solution: sp2 hybridized with trigonal planar geometry.

Practice Set

• Q1: The IUPAC suffix for aldehydes is:
  A. -one
  B. -al
  C. -ol
  D. -oic acid

• Q2: Which hybridization is present in carbonyl carbon?
  A. sp
  B. sp2
  C. sp3
  D. dsp2

• Q3: The resonance structure of carbonyl group shows:
  A. C=O double bond only
  B. C–O single bond only
  C. Partial positive charge on carbon and negative on oxygen
  D. No charge separation

Answer Key

• Q1: B – Aldehydes end with ‘-al’ in IUPAC nomenclature.

• Q2: B – Carbonyl carbon is sp2 hybridized.

• Q3: C – Resonance shows C+ and O– charges.

Memory Aids

• ‘-al’ suffix for aldehydes, ‘-one’ for ketones.

• Carbonyl carbon = sp2, trigonal planar.

• Resonance: C=O ↔ C+–O–.

• Greek letters α, β indicate substituent positions next to carbonyl.

Preparation of Aldehydes and Ketones

Methods of Preparation

• Oxidation of Alcohols: Primary alcohols oxidized by K2Cr2O7 form aldehydes; secondary alcohols form ketones.

• Catalytic Dehydrogenation: Alcohols heated with Cu lose H2 to give aldehydes or ketones.

• Reductive Ozonolysis: Alkenes cleaved by ozone yield aldehydes.

• Rosenmund Reduction: Acyl chlorides reduced with H2/Pd-BaSO4 produce aldehydes.

• Reduction of Nitriles: Controlled reduction converts nitriles to aldehydes.

• Esters Reduction: DIBAL-H reduces esters to aldehydes after hydrolysis.

• Acyl Chlorides + Organometallics: Reaction with RMgX or R2Cd forms ketones.

• Gatterman-Koch Reaction: Benzene + CO + HCl with catalysts forms benzaldehyde.

• Friedel-Crafts Acylation: Benzene + acyl chloride + AlCl3 yields aromatic ketones.

• From Alkynes: Hydration or hydroboration-oxidation forms aldehydes or ketones.

• Oxidation of 1,2-Glycols: Lead tetraacetate cleaves vicinal diols to aldehydes/ketones.

• Etard Reaction: Methylbenzenes oxidized by chromium reagents to benzaldehydes.

• Side Chain Chlorination: Methyl groups chlorinated then hydrolyzed to benzaldehydes.

Solved Examples

• Ex 1: Convert ethanol to ethanal using K2Cr2O7 (oxidation of primary alcohol).

• Ex 2: Prepare acetophenone by Friedel-Crafts acylation of benzene with acetyl chloride and AlCl3.

• Ex 3: Obtain benzaldehyde from toluene via side chain chlorination followed by hydrolysis.

Practice Set

• Q1: Which reagent is used in Rosenmund reduction to prepare aldehydes?

  • A. Pd on BaSO4

  • B. LiAlH4

  • C. K2Cr2O7

  • D. AlCl3

• Q2: Which method converts esters to aldehydes?

  • A. DIBAL-H reduction

  • B. Ozonolysis

  • C. Friedel-Crafts acylation

  • D. Catalytic dehydrogenation

• Q3: Which reaction sequence correctly forms benzaldehyde from benzene?

  • A. Gatterman-Koch reaction (CO + HCl + catalyst)

  • B. Friedel-Crafts acylation with acetyl chloride

  • C. Oxidation of methylbenzene with K2Cr2O7

  • D. Both A and C

Answer Key

• Q1: A. Pd on BaSO4 is the catalyst in Rosenmund reduction for acyl chlorides to aldehydes.

• Q2: A. DIBAL-H reduces esters to aldehydes under controlled conditions.

• Q3: D. Both Gatterman-Koch reaction and oxidation of methylbenzene yield benzaldehyde.

Memory Aids

• Rosenmund reduction uses Pd on BaSO4 to stop reduction at aldehyde stage.

• DIBAL-H is a mild reducing agent for esters to aldehydes.

• Friedel-Crafts acylation always gives ketones on aromatic rings.

• Gatterman-Koch reaction introduces formyl group on benzene.

• Oxidation of primary alcohols gives aldehydes; secondary gives ketones.

Physical Properties

Physical State

• Lower aldehydes and ketones (up to C10) are colourless, volatile liquids.

• Formaldehyde is a gas at room temperature.

• Higher aldehydes and ketones are solids with fruity odour.

• Lower aldehydes have unpleasant odour; ketones have pleasant smell.

Boiling Points

• Boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of similar molecular mass due to dipole-dipole interactions.

• Boiling points are lower than alcohols of similar molecular mass because they lack intermolecular hydrogen bonding.

• Among isomers, ketones have slightly higher boiling points than aldehydes due to two electron-releasing alkyl groups increasing polarity.

Solubility

• Lower aldehydes and ketones (up to C4) are soluble in water due to hydrogen bonding between carbonyl group and water.

• Solubility decreases with increasing molecular weight.

Hydrogen Bonding in Carboxyl Groups

• Hydrogen bond forms between –OH hydrogen (δ⁺) of one molecule and carbonyl oxygen (δ⁻) of another.

• Hydrogen bonds are weaker than covalent bonds but crucial for molecular structure and stability.

Solved Examples

• Ex 1: Compare boiling points of propanal and propanone.
  Step 1: Both have similar molecular mass.
  Step 2: Propanone (ketone) has two alkyl groups increasing polarity.
  Step 3: Propanone has higher boiling point due to stronger dipole interactions.

Practice Set

• Q1: Which of the following is a gas at room temperature?
  A. Acetaldehyde
  B. Formaldehyde
  C. Acetone
  D. Butanal

• Q2: Why do aldehydes and ketones have higher boiling points than ethers?
  A. Hydrogen bonding
  B. Dipole-dipole interactions
  C. Ionic bonding
  D. Van der Waals forces

• Q3: Which factor decreases solubility of aldehydes and ketones in water?
  A. Presence of carbonyl group
  B. Increase in molecular weight
  C. Hydrogen bonding
  D. Presence of alkyl groups

Answer Key

• Q1: B. Formaldehyde is a gas at room temperature; others are liquids.

• Q2: B. Dipole-dipole interactions increase boiling points compared to ethers.

• Q3: B. Increasing molecular weight reduces solubility due to decreased hydrogen bonding.

Memory Aids

• "Formaldehyde is a gas, others liquid up to C10".

• "Ketones boil higher than aldehydes due to two alkyl groups".

• "Solubility decreases as chain length increases".

Polarity of Carbonyl Group

Key Points

• Carbonyl group (>C=O) is polar due to electronegative oxygen.

• Carbonyl carbon is electrophilic (Lewis acid).

• Carbonyl oxygen is nucleophilic (Lewis base).

• Resonance structures show partial positive charge on carbon and negative on oxygen.

• High dipole moment (2.3-2.8 D) due to resonance and polarity.

• Aldehydes and ketones undergo nucleophilic addition, unlike alkenes which undergo electrophilic addition.

Solved Examples

• Ex 1: Explain why carbonyl carbon is electrophilic.
  - Oxygen's electronegativity pulls electron density, making carbon partially positive.

Practice Set

• Q1: What is the nature of carbonyl carbon?
  A. Nucleophilic B. Electrophilic C. Neutral D. Radical

• Q2: Which resonance structure contributes to carbonyl polarity?
  A. Double bond only B. Single bond with charges C. Both A and B D. None

• Q3: Carbonyl group polarity leads to which reaction type?
  A. Electrophilic addition B. Nucleophilic addition C. Radical substitution D. Elimination

Answer Key

• Q1: B - Carbonyl carbon is electrophilic due to oxygen's electronegativity.

• Q2: C - Both resonance forms contribute to polarity.

• Q3: B - Polarity causes nucleophilic addition reactions.

Memory Aids

• Carbonyl carbon = Positive center (Electrophile).

• Oxygen = Negative center (Nucleophile).

• Resonance stabilizes polarity.

Acidity of α-Hydrogen

Key Points

• α-Hydrogens in carbonyl compounds are acidic.

• Acidity due to electron-withdrawing carbonyl group.

• Conjugate base (enolate ion) stabilized by resonance.

• Base removes α-H, forming resonance-stabilized enolate.

Solved Examples

• Ex 1: Show resonance stabilization of enolate ion.
  - Negative charge delocalizes between oxygen and α-carbon.

Practice Set

• Q1: Why are α-hydrogens acidic?
  A. Inductive effect B. Resonance stabilization C. Both A and B D. None

• Q2: What species forms after α-H removal?
  A. Carbocation B. Radical C. Enolate ion D. Carbene

• Q3: Which stabilizes the enolate ion?
  A. Inductive effect B. Resonance C. Hyperconjugation D. None

Answer Key

• Q1: C - Both inductive effect and resonance stabilize the conjugate base.

• Q2: C - Enolate ion forms after α-H removal.

• Q3: B - Resonance delocalizes negative charge.

Memory Aids

• α-H acidic due to resonance-stabilized enolate.

• Base removes α-H to form enolate.

Nucleophilic Addition Reactions

Key Points

• Nucleophile attacks electrophilic carbonyl carbon.

• First step: slow nucleophilic attack forming tetrahedral intermediate.

• Second step: fast protonation forming alcohol or derivative.

• Common nucleophiles: Grignard reagents, hydride ions, cyanide, hydroxide, alkoxides.

• Aldehydes more reactive than ketones due to steric and electronic factors.

• Reactivity order: HCHO > RCHO > PhCHO > RCOR > RCOPh > PhCOPh.

• Reactions form alcohols, cyanohydrins, hydrates, hemiacetals, acetals, ketals.

Solved Examples

• Ex 1: Reaction of aldehyde with HCN.
  - Nucleophile CN⁻ attacks carbonyl carbon forming cyanohydrin after protonation.

• Ex 2: Formation of hemiacetal.
  - Alcohol attacks carbonyl carbon, followed by protonation.

Practice Set

• Q1: Which reagent forms cyanohydrin with aldehydes?
  A. HCN B. RMgX C. NaBH4 D. NH3

• Q2: Why are aldehydes more reactive than ketones?
  A. Less steric hindrance B. More electrophilic C. Both A and B D. None

• Q3: Product of aldehyde + RMgX followed by acid workup?
  A. Alcohol B. Ketone C. Alkane D. Ester

Answer Key

• Q1: A - HCN provides CN⁻ nucleophile forming cyanohydrin.

• Q2: C - Aldehydes have less steric hindrance and more electrophilic carbon.

• Q3: A - Grignard reagent forms alcohol after protonation.

Memory Aids

• Nucleophile attacks carbonyl carbon forming tetrahedral intermediate.

• Aldehydes > Ketones in reactivity.

• Common nucleophiles: CN⁻, RMgX, H⁻.

Nucleophilic Addition-Elimination Reactions

Key Points

• Carbonyl compounds react with nitrogen nucleophiles (NH3, amines) to form imines (Schiff bases).

• Reaction involves nucleophilic addition followed by elimination of water.

• Different nitrogen reagents form derivatives: imines, oximes, hydrazones, semicarbazones.

• Useful for identification and characterization of aldehydes and ketones.

Solved Examples

• Ex 1: Formation of imine from aldehyde and NH3.
  - Nucleophilic attack by NH3, elimination of water forms C=NH.

• Ex 2: Formation of oxime with hydroxylamine.
  - Similar mechanism forming C=NOH.

Practice Set

• Q1: Product of aldehyde + NH3 + heat?
  A. Oxime B. Imine C. Hydrazone D. Semicarbazone

• Q2: Which reagent forms hydrazone?
  A. NH2OH B. NH2NH2 C. NH3 D. RNH2

• Q3: Purpose of these derivatives?
  A. Identification B. Reduction C. Oxidation D. Polymerization

Answer Key

• Q1: B - Imine (Schiff base) forms with NH3.

• Q2: B - Hydrazine (NH2NH2) forms hydrazone.

• Q3: A - Used for identification and characterization.

Memory Aids

• NH3 → imines; NH2OH → oximes; NH2NH2 → hydrazones.

• Reaction: addition then elimination of water.

Oxidation and Reduction Reactions

Key Points

• Aldehydes oxidize to carboxylic acids; ketones generally resist oxidation.

• Oxidizing agents: K2Cr2O7, Tollens’ reagent, Fehling’s, Benedict’s solutions.

• Tollens’ test: silver mirror with aldehydes.

• Reduction of aldehydes/ketones to alcohols by H2 (Ni, Pt, Pd) or LiAlH4, NaBH4.

• Further reduction of aldehydes to hydrocarbons by Clemmensen or Wolff-Kishner methods.

Solved Examples

• Ex 1: Tollens’ test for aldehydes.
  - Aldehyde oxidized, silver mirror formed.

• Ex 2: Reduction of ketone with NaBH4.
  - Ketone converted to secondary alcohol.

Practice Set

• Q1: Which reagent oxidizes aldehydes to acids?
  A. NaBH4 B. K2Cr2O7 C. LiAlH4 D. RMgX

• Q2: Tollens’ reagent detects?
  A. Ketones B. Aldehydes C. Alcohols D. Alkenes

• Q3: Clemmensen reduction converts aldehydes to?
  A. Alcohols B. Hydrocarbons C. Acids D. Esters

Answer Key

• Q1: B - K2Cr2O7 oxidizes aldehydes to acids.

• Q2: B - Tollens’ reagent detects aldehydes.

• Q3: B - Clemmensen reduces aldehydes to hydrocarbons.

Memory Aids

• Aldehydes oxidize; ketones resist oxidation.

• Tollens’ test = silver mirror for aldehydes.

• NaBH4 and LiAlH4 reduce carbonyl to alcohol.

Haloform Reaction

Key Points

• Occurs with compounds having CH3CO– or CH3CH(OH)– groups.

• Reaction with halogen and base produces haloform (CHI3, CHBr3, CHCl3).

• Used as a test for methyl ketones and related compounds.

• Produces carboxylate salt and haloform precipitate.

Solved Examples

• Ex 1: Iodoform test for acetyl group.
  - Methyl ketone reacts with NaOI to give yellow CHI3 precipitate.

Practice Set

• Q1: Haloform reaction requires?
  A. Aldehyde B. Methyl ketone C. Alcohol D. Carboxylic acid

• Q2: Product of haloform reaction?
  A. Alcohol B. Haloform C. Alkene D. Ester

• Q3: Iodoform test positive compound?
  A. CH3COCH3 B. Benzaldehyde C. Ethanol D. Acetic acid

Answer Key

• Q1: B - Methyl ketones give haloform reaction.

• Q2: B - Haloform (CHI3, etc.) is formed.

• Q3: A - Acetone (CH3COCH3) gives positive iodoform test.

Memory Aids

• Haloform test detects methyl ketones.

• Positive test: yellow precipitate of CHI3.

Aldol Condensation

Key Points

• Aldehydes/ketones with at least one α-H undergo aldol condensation in dilute alkali.

• Forms β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol).

• Upon heating, aldol loses water to form α,β-unsaturated carbonyl compounds.

• Intramolecular aldol condensation forms cyclic products from diketones.

• Cross aldol condensation between two different carbonyls with α-H gives mixture of products.

Solved Examples

• Ex 1: Aldol condensation of acetaldehyde.
  - Two molecules form β-hydroxy aldehyde, then dehydrate to crotonaldehyde.

Practice Set

• Q1: Aldol condensation requires?
  A. No α-H B. At least one α-H C. Acidic medium D. No base

• Q2: Product after dehydration is?
  A. Alcohol B. α,β-unsaturated carbonyl C. Ketone D. Alkane

• Q3: Intramolecular aldol condensation forms?
  A. Linear product B. Cyclic product C. Alcohol D. Ester

Answer Key

• Q1: B - At least one α-H is needed.

• Q2: B - α,β-unsaturated carbonyl compound forms.

• Q3: B - Cyclic products form from diketones.

Memory Aids

• Aldol = Aldehyde + Alcohol intermediate.

• Heating causes dehydration to α,β-unsaturated product.

• Intramolecular = cyclic products.

Cannizzaro Reaction

Key Points

• Occurs with aldehydes lacking α-H.

• Self-oxidation and reduction in concentrated alkali.

• One aldehyde molecule oxidized to carboxylate salt; another reduced to alcohol.

• Cross and intramolecular Cannizzaro reactions also occur.

• Some aromatic aldehydes undergo electrophilic substitution instead.

Solved Examples

• Ex 1: Cannizzaro reaction of benzaldehyde.
  - Benzaldehyde forms benzyl alcohol and benzoate salt.

Practice Set

• Q1: Cannizzaro reaction requires aldehydes with?
  A. α-H B. No α-H C. Ketones D. Alcohols

• Q2: Products of Cannizzaro reaction?
  A. Alcohol + acid salt B. Alcohol + ketone C. Acid + ketone D. Alcohol + ester

• Q3: Which aldehyde does NOT undergo Cannizzaro?
  A. Benzaldehyde B. Formaldehyde C. Acetaldehyde D. p-Nitrobenzaldehyde

Answer Key

• Q1: B - Aldehydes without α-H undergo Cannizzaro.

• Q2: A - Alcohol and carboxylate salt form.

• Q3: C - Acetaldehyde has α-H, does not undergo Cannizzaro.

Memory Aids

• No α-H → Cannizzaro reaction.

• One molecule oxidized, one reduced.

• Aromatic aldehydes may undergo substitution instead.

Distinction Between Aldehydes and Ketones

Key Points

• Schiff’s reagent: Aldehydes turn pink; ketones no change.

• Fehling’s solution: Aldehydes give red precipitate; ketones do not.

• Tollens’ reagent: Aldehydes form silver mirror; ketones do not.

• 2,4-Dinitrophenylhydrazine: Both form colored crystals; melting points identify compounds.

• α-Hydroxy ketones reduce Tollens’, Fehling’s, and Benedict’s solutions.

• Haloform reaction oxidizes α,β-unsaturated methyl ketones to acids without affecting double bond.

Solved Examples

• Ex 1: Test to distinguish aldehyde and ketone.
  - Use Tollens’ reagent; aldehyde gives silver mirror.

Practice Set

• Q1: Which reagent gives silver mirror with aldehydes?
  A. Fehling’s B. Tollens’ C. 2,4-DNP D. Schiff’s

• Q2: 2,4-DNP test detects?
  A. Only aldehydes B. Only ketones C. Both aldehydes and ketones D. Neither

• Q3: Which ketone reduces Tollens’ reagent?
  A. Methyl ketone B. α-Hydroxy ketone C. Aromatic ketone D. No ketone

Answer Key

• Q1: B - Tollens’ reagent forms silver mirror with aldehydes.

• Q2: C - 2,4-DNP reacts with both aldehydes and ketones.

• Q3: B - α-Hydroxy ketones reduce Tollens’ reagent.

Memory Aids

• Tollens’ and Fehling’s positive for aldehydes.

• 2,4-DNP positive for both.

• α-Hydroxy ketones are exceptions, reduce Tollens’.

Uses of Aldehydes and Ketones

Applications in Industry and Synthesis

• Aldehydes and ketones serve as solvents in the chemical industry due to their polarity and ability to dissolve various compounds.

• They act as key starting materials and reagents for synthesizing numerous organic compounds.

• Formalin (40% formaldehyde solution) is widely used to preserve biological specimens by cross-linking proteins.

• Formaldehyde is essential in producing bakelite (phenol-formaldehyde resin), urea-formaldehyde adhesives, and other polymers.

• Acetaldehyde (CH3CHO) is a precursor for acetic acid, ethyl acetate, vinyl acetate, polymers, and pharmaceuticals.

• Benzaldehyde is utilized in perfumery and dye manufacturing due to its aromatic properties.

• Compounds like butyraldehyde, vanillin, acetophenone, and camphor are employed as flavoring agents.

Solved Examples

• Ex 1: Identify the product formed when formaldehyde reacts with phenol.
  Step 1: Formaldehyde + Phenol → Phenol-formaldehyde resin (Bakelite)

• Ex 2: Name the compound used as a starting material for vinyl acetate.
  Answer: Acetaldehyde (CH3CHO)

Practice Set

• Q1: Which aldehyde is used to preserve biological specimens?
  A. Acetaldehyde
  B. Formaldehyde
  C. Benzaldehyde
  D. Butyraldehyde

• Q2: Bakelite is synthesized from phenol and which compound?
  A. Acetaldehyde
  B. Formaldehyde
  C. Benzaldehyde
  D. Vanillin

• Q3: Acetaldehyde is NOT used in the manufacture of:
  A. Acetic acid
  B. Ethyl acetate
  C. Nylon
  D. Vinyl acetate

Answer Key

• Q1: B. Formaldehyde
  Used as formalin (40% solution) for preservation.

• Q2: B. Formaldehyde
  Phenol reacts with formaldehyde to form bakelite.

• Q3: C. Nylon
  Acetaldehyde is not involved in nylon synthesis.

Memory Aids

• Formalin = Formaldehyde solution for preservation.

• Bakelite = Phenol + Formaldehyde polymer.

• Acetaldehyde → Acetic acid, esters, vinyl acetate.

• Benzaldehyde → Perfumes and dyes.

• Flavor agents: Butyraldehyde, vanillin, camphor.

Nomenclature and Structure of Carboxyl Group

Nomenclature

• Common names end with suffix -ic acid, derived from Latin/Greek natural sources.

• IUPAC names: replace alkane ending -e with -oic acid.

• Numbering starts from carboxyl carbon as carbon 1.

• For multiple carboxyl groups, use prefixes: dicarboxylic acid, tricarboxylic acid, etc.

• Positions of –COOH groups indicated by Arabic numerals before prefix.

• Examples include formic acid (HCOOH) = methanoic acid, acetic acid = ethanoic acid.

Structure

• Carboxyl carbon bonds lie in one plane, separated by ~120°.

• Carboxylic carbon less electrophilic than carbonyl carbon due to resonance.

• Resonance stabilizes the –COOH group by delocalizing electrons.

Solved Examples

• Ex 1: Name CH3–COOH using IUPAC.
  Solution: Alkane is methane; replace -e with -oic acid → methanoic acid.

• Ex 2: Name HOOC–CH2–COOH.
  Solution: Two –COOH groups, prefix dicarboxylic acid, numbering from one end → ethane-1,2-dicarboxylic acid.

Practice Set

• Q1: What is the IUPAC name of CH3–CH2–COOH?
  A. Propanoic acid
  B. Propionic acid
  C. Acetic acid
  D. Methanoic acid

• Q2: How many carboxyl groups are present in benzene-1,2,3-tricarboxylic acid?
  A. One
  B. Two
  C. Three
  D. Four

• Q3: Which statement is correct about carboxyl carbon?
  A. It is more electrophilic than carbonyl carbon
  B. Bonds lie in one plane separated by 90°
  C. Resonance decreases its electrophilicity
  D. It has no resonance structures

Answer Key

• Q1: A. Propanoic acid – IUPAC name replaces alkane ending with -oic acid.

• Q2: C. Three – indicated by prefix tri in tricarboxylic acid.

• Q3: C. Resonance delocalizes electrons, reducing electrophilicity of carboxyl carbon.

Memory Aids

• Common acid names end with -ic acid, IUPAC with -oic acid.

• Number carboxyl carbon as 1 always.

• Multiple –COOH groups use prefixes: di-, tri-, etc.

• Carboxyl carbon resonance lowers electrophilicity.

Methods of Preparation of Carboxylic Acids

Overview of Reactions

• Primary alcohols (RCH2OH) oxidized to carboxylic acids (RCOOH) using strong oxidizers like KMnO4 or K2Cr2O7.

• Aldehydes (RCHO) oxidized to carboxylic acids by similar oxidizing agents.

• Alkyl cyanides (RCN) hydrolyzed under acidic or basic conditions to yield carboxylic acids.

• Grignard reagents react with CO2 followed by acid workup to form carboxylic acids.

• Amides hydrolyzed under acidic or basic conditions to carboxylic acids.

• Alkenes and alkynes oxidized to carboxylic acids using strong oxidizers.

• Acyl halides, acid anhydrides, and esters hydrolyzed to carboxylic acids.

• Alkyl benzenes oxidized by KMnO4 to benzoic acid if side chain has benzylic hydrogens.

Solved Examples

• Ex 1: Oxidation of ethanol to acetic acid using KMnO4.

  • CH3CH2OH + [O] → CH3COOH

• Ex 2: Hydrolysis of ethyl cyanide to acetic acid.

  • CH3CN + 2H2O + H+ → CH3COOH + NH4+

• Ex 3: Reaction of phenylmagnesium bromide with CO2.

  • C6H5MgBr + CO2 → C6H5COOMgBr → C6H5COOH (after acid workup)

Practice Set

• Q1: Which reagent is used to oxidize primary alcohols to carboxylic acids?

  • A. LiAlH4

  • B. KMnO4

  • C. NaBH4

  • D. H2SO4

• Q2: Hydrolysis of which compound yields a carboxylic acid?

  • A. Alkyl halide

  • B. Alkyl cyanide

  • C. Alkene

  • D. Ether

• Q3: What is the product when benzyl chloride is oxidized with KMnO4?

  • A. Benzaldehyde

  • B. Benzoic acid

  • C. Benzene

  • D. Phenol

Answer Key

• Q1: B. KMnO4 - Strong oxidizer converting primary alcohols to carboxylic acids.

• Q2: B. Alkyl cyanide - Hydrolyzed under acidic/basic conditions to carboxylic acids.

• Q3: B. Benzoic acid - Side chain oxidation of benzyl chloride yields benzoic acid.

Memory Aids

• Primary alcohols and aldehydes → oxidize with KMnO4 or K2Cr2O7 to carboxylic acids.

• Alkyl cyanides → hydrolysis → carboxylic acids.

• Grignard + CO2 + acid → carboxylic acid.

• Alkyl benzene side chains with benzylic H → oxidized to benzoic acid.

Physical Properties

Physical State

• Lower fatty acids (up to C9) are colourless liquids.

• Higher fatty acids are colourless waxy solids.

Odour

• First three members have a sharp pungent odour.

• Members from C4 to C9 smell like rancid butter.

• Higher members are odourless.

Boiling Points

• Carboxylic acids have higher boiling points than aldehydes, ketones, and alcohols of similar molecular mass.

• Due to strong intermolecular hydrogen bonding forming dimers.

• Hydrogen bonds persist even in the vapour phase.

This image shows a dimer formed by two carboxylic acid molecules connected through hydrogen bonds:

• Each molecule has a carboxyl group (R-COOH).

• Hydrogen bonds form between the -OH hydrogen of one molecule and the C=O oxygen of the other.

• This creates a stable ring-like dimer structure.

• Common in vapor state or aprotic solvents.

• Dotted lines represent hydrogen bonds.

Solubility

• Simple aliphatic carboxylic acids (up to 4 carbons) are miscible with water due to hydrogen bonding.

• Solubility decreases as carbon chain length increases.

• Higher acids are practically insoluble due to hydrophobic hydrocarbon chains.

Solved Examples

Ex 1: Explain why acetic acid has a higher boiling point than ethanol.

• Acetic acid forms dimers via hydrogen bonding, increasing intermolecular forces.

• Ethanol forms hydrogen bonds but does not dimerize as strongly.

• Result: Acetic acid has a higher boiling point.

Ex 2: Predict solubility of hexanoic acid in water.

• Hexanoic acid has 6 carbons, longer hydrocarbon chain.

• Less soluble due to increased hydrophobic interactions.

• Expected to be sparingly soluble or insoluble.

Practice Set

Q1: Which of the following carboxylic acids is a colourless liquid at room temperature?

• A. Nonanoic acid (C9H18O2)

• B. Decanoic acid (C10H20O2)

• C. Butanoic acid (C4H8O2)

• D. Stearic acid (C18H36O2)

Q2: Why do carboxylic acids have higher boiling points than alcohols of similar molecular mass?

• A. Due to ionic bonding

• B. Due to dimer formation via hydrogen bonding

• C. Due to van der Waals forces only

• D. Due to lower molecular weight

Q3: Which statement about solubility of carboxylic acids is correct?

• A. Solubility increases with increasing carbon chain length

• B. All carboxylic acids are insoluble in water

• C. Carboxylic acids up to 4 carbons are miscible with water

• D. Higher carboxylic acids are more soluble than lower ones

Answer Key

Q1: C. Butanoic acid is a colourless liquid; acids up to C9 are liquids.

Q2: B. Carboxylic acids form dimers via hydrogen bonding, increasing boiling points.

Q3: C. Carboxylic acids with up to 4 carbons are miscible due to hydrogen bonding with water.

Memory Aids

• "Up to C4, mix with H2O" – carboxylic acids soluble up to 4 carbons.

• "Dimer doubles boiling" – dimer formation raises boiling points.

• "Short chain liquid, long chain solid" – physical state trend.

Reactions Involving Cleavage of O — H Bond

Reactions with metals and alkalies

• Carboxylic acid (R-COOH) reacts with sodium (Na) to form sodium salt (RCOONa) and hydrogen gas (H2).

• Sodium salt (RCOONa) reacts with sodium carbonate (Na2CO3) or sodium bicarbonate (NaHCO3) to produce CO2, H2O, and regenerate sodium salt.

• Sodium salt reacts with water and sodium hydroxide (NaOH) to revert back to carboxylic acid.

• These reactions demonstrate acid-base behavior and salt formation of carboxylic acids.

Acidic strength of carboxylic acids

• Carboxylic acids dissociate in water: RCOOH + H2O ⇌ RCOO− + H3O+

• Acidity arises from proton liberation due to positive charge on oxygen; carboxylate ion is resonance stabilized.

• Equilibrium constant (Keq) and acid dissociation constant (Ka) define acid strength; pKa = -log Ka.

• Acid strength ∝ 1/pKa.

• Carboxylic acids are weaker than mineral acids but stronger than phenols and alcohols.

• Carboxylate ion has two equivalent resonance structures with negative charge on electronegative oxygen, stabilizing it more than phenoxide ion.

Effect of substituents on acidity

• Electron withdrawing groups (EWG) increase acidity by stabilizing conjugate base via inductive/resonance effects.

• Acidity order: Ph < I < Br < Cl < F < CN < NO2 < CF3.

• Examples: Chloroacetic acid > Acetic acid; Benzoic acid > Acetic acid.

• Acidity increases with number and electronegativity of EWGs.

• Acidity increases as halogen atom gets closer to carboxyl group.

• Electron donating groups (EDG) decrease acidity by destabilizing conjugate base; e.g., formic acid > acetic acid.

• In aromatic acids, EWGs on phenyl ring increase acidity; EDGs decrease acidity.

Solved Examples

• Ex 1: Compare acidity of trifluoroacetic acid and trichloroacetic acid.
  Answer: Trifluoroacetic acid is stronger due to higher electronegativity of F than Cl.

• Ex 2: Predict product when chloroacetic acid reacts with NaOH.
  Answer: Sodium chloroacetate salt forms.

Practice Set

• Q1: Which is the strongest acid?
  A. Acetic acid
  B. Chloroacetic acid
  C. Benzoic acid
  D. Formic acid

• Q2: What is the product when R-COOH reacts with NaHCO3?
  A. R-COONa + CO2 + H2O
  B. R-COOH + NaOH
  C. R-COONa + H2
  D. R-COONa + H2O

• Q3: Which substituent increases acidity the most?
  A. -NO2
  B. -CH3
  C. -OH
  D. -OCH3

Answer Key

• Q1: B. Chloroacetic acid; EWG Cl increases acidity over acetic and benzoic acids.

• Q2: A. R-COONa + CO2 + H2O; acid reacts with bicarbonate releasing CO2.

• Q3: A. -NO2; strong EWG stabilizes conjugate base, increasing acidity.

Memory Aids

• EWG increase acidity; EDG decrease acidity.

• Closer and more electronegative substituents increase acid strength.

• Carboxylate ion resonance stabilizes conjugate base, enhancing acidity.

Reactions Involving Cleavage of C—OH Bond

Derivatives formation from carboxylic acids

• Carboxylic acid (R-COOH) reacts to form:

  • Acid anhydrides by heating with acid catalyst (removal of water).

  • Esters by reaction with alcohol (R'OH) under acid catalyst and heat (removal of water).

  • Acid chlorides by reaction with PCl3 or SOCl2 in pyridine.

  • Amides by reaction with NH3 and heat (removal of water).

Mechanism of esterification

• Nucleophilic acyl substitution.

• (a) Protonation of carbonyl oxygen increases electrophilicity.

• (b) Nucleophilic attack by alcohol forms tetrahedral intermediate.

• (c) Elimination of water forms protonated ester.

• (d) Deprotonation yields ester.

Solved Examples

• Ex 1: Write the product of esterification of acetic acid with methanol.
  Answer: Methyl acetate (CH3COOCH3) + H2O.

• Ex 2: Identify reagent to convert acetic acid to acetyl chloride.
  Answer: SOCl2 in pyridine.

Practice Set

• Q1: Which reagent converts R-COOH to R-COCl?
  A. PCl3
  B. NaOH
  C. NH3
  D. H2SO4

• Q2: Esterification requires:
  A. Base catalyst
  B. Acid catalyst and heat
  C. No catalyst
  D. Reducing agent

• Q3: In esterification, the leaving group is:
  A. OH−
  B. H2O
  C. H+
  D. R'

Answer Key

• Q1: A. PCl3; converts acid to acid chloride.

• Q2: B. Acid catalyst and heat; required for ester formation.

• Q3: B. H2O; water is eliminated during esterification.

Memory Aids

• Esterification: acid + alcohol + acid catalyst + heat → ester + water.

• Acid chlorides formed using PCl3, SOCl2.

• Amides formed by reaction with ammonia and heat.

Reactions Involving COOH Group

Reduction, decarboxylation, and electrolysis

• Reduction of R-COOH to R-CH2OH using LiAlH4 or B2H6 followed by acid workup.

• Decarboxylation: Heating R-COOH with NaOH and CaO produces alkane (RH) and Na2CO3.

• Kolbe electrolysis: Electrolysis of RCOONa with NaOH produces coupled hydrocarbon (R-R), CO2, and NaOH.

• LiAlH4 is stronger reducing agent than NaBH4, reducing more functional groups.

• In α,β-unsaturated aldehydes/ketones, LiAlH4 reduces only C=O; NaBH4 reduces both C=O and C=C but C=O reduction predominates.

Substitution reactions in hydrocarbon part

• Hell-Volhard-Zelinsky reaction: RCH2COOH + X2/Red P → α-halocarboxylic acid.

• Aromatic carboxylic acids undergo electrophilic substitution; COOH is deactivating and meta-directing.

Distinction test between phenol and carboxylic acid

• NaHCO3 test: Phenol no reaction; carboxylic acid gives effervescence (CO2).

• FeCl3 test: Phenol gives violet color; carboxylic acid gives buff precipitate.

Abnormal behaviour of formic acid

• Due to aldehydic group presence.

• Oxidizes Tollens’ reagent and Fehling solution like aldehydes.

• Reactions:

  • HCOOH + Ag2O → CO2 + H2O + 2Ag (silver mirror)

  • HCOOH + Cu2+ + OH− → CO2 + H2O + Cu2O (red ppt.)

  • With mercuric chloride:

    • 2HCOOH + 2HgCl2 → CO2 + Hg2Cl2 (white ppt.) + 2HCl

    • HCOOH + Hg2Cl2 → CO2 + 2Hg (grey ppt.) + 2HCl

Solved Examples

• Ex 1: Write the product of Kolbe electrolysis of sodium acetate.
  Answer: Ethane (C2H6), CO2, and NaOH.

• Ex 2: How to distinguish phenol and carboxylic acid using NaHCO3?
  Answer: Carboxylic acid effervesces; phenol shows no reaction.

Practice Set

• Q1: Which reagent reduces carboxylic acid to alcohol?
  A. NaBH4
  B. LiAlH4
  C. H2SO4
  D. NaOH

• Q2: Hell-Volhard-Zelinsky reaction introduces halogen at:
  A. Carboxyl carbon
  B. α-Carbon
  C. β-Carbon
  D. Aromatic ring

• Q3: Formic acid reduces Tollens’ reagent to:
  A. Silver mirror
  B. Red ppt.
  C. White ppt.
  D. No reaction

Answer Key

• Q1: B. LiAlH4; strong reducing agent for carboxylic acids.

• Q2: B. α-Carbon; halogen introduced next to COOH group.

• Q3: A. Silver mirror; formic acid reduces Tollens’ reagent.

Memory Aids

• LiAlH4 reduces carboxylic acids; NaBH4 does not.

• Hell-Volhard-Zelinsky: α-halogenation of carboxylic acids.

• Formic acid behaves like aldehyde in oxidation tests.

Uses of Carboxylic Acids

Industrial and Food Applications

• Methanoic acid: Used in rubber, textile, dyeing, leather, and electroplating industries.

• Ethanoic acid: Acts as a solvent and is the main component of vinegar in the food industry.

• Hexanedioic acid: Used in manufacturing nylon-6,6 polymer.

• Esters of benzoic acid: Employed in perfumery for fragrance production.

• Sodium benzoate: Serves as a food preservative to inhibit microbial growth.

• Higher fatty acids: Utilized in making soaps and detergents.

Solved Examples

• Ex 1: Identify the carboxylic acid used in nylon-6,6 production.
  Solution: Hexanedioic acid (adipic acid) is used in nylon-6,6 synthesis.

• Ex 2: Name the carboxylic acid component of vinegar.
  Solution: Ethanoic acid (acetic acid) is the main acid in vinegar.

Practice Set

• Q1: Which carboxylic acid is commonly used in the leather industry?
  A. Ethanoic acid
  B. Methanoic acid
  C. Hexanedioic acid
  D. Benzoic acid

• Q2: Sodium benzoate is primarily used as:
  A. Solvent
  B. Food preservative
  C. Polymer precursor
  D. Detergent

• Q3: Which acid is a key raw material for nylon-6,6 synthesis?
  A. Methanoic acid
  B. Ethanoic acid
  C. Hexanedioic acid
  D. Benzoic acid

Answer Key

• Q1: B. Methanoic acid - Used in leather and related industries for its chemical properties.

• Q2: B. Food preservative - Sodium benzoate inhibits microbial growth in foods.

• Q3: C. Hexanedioic acid - Also known as adipic acid, essential for nylon-6,6 manufacture.

Memory Aids

• Methanoic acid = "Meth" for "Manufacturing" industries (rubber, leather).

• Ethanoic acid = "Eat" for food (vinegar solvent).

• Hexanedioic acid = "Hexa" = 6 carbons, nylon-6,6 precursor.

• Sodium benzoate = "Preserves" food by preventing spoilage.

• Higher fatty acids = Soap and detergent production.