NCERT Plus Two (Class 12) Chemistry • Chapter 8

Aldehydes, Ketones and Carboxylic Acids

Q: Give simple chemical tests to distinguish between: (a) Propanal and Propanone, (b) Benzaldehyde and Acetophenone.

(a) Propanal vs Propanone: Warm both with Tollens' reagent (or Fehling's solution). Propanal gives a bright silver mirror (or reddish-brown Cu 2 O ppt with Fehling's), whereas Propanone gives no reaction. (Alternatively, Propanone gives yellow ppt with I 2 /NaOH in Iodoform test; Propanal does not.) (b) Benzaldehyde vs Acetophenone: Warm with Tollens' reagent : Benzaldehyde forms a silver mirror while Acetophenone fails. (Or Iodoform test: Acetophenone gives yellow CHI 3 ppt while Benzaldehyde fails.)

Q: Explain Clemmensen reduction and Wolff–Kishner reduction with one chemical equation each.

Both convert the carbonyl group ( >C=O ) directly into a methylene group ( –CH 2 – ): Clemmensen Reduction: Reduction with Zn–Hg and conc. HCl . CH 3 COCH 3 + 4[H] —(Zn–Hg / conc. HCl)→ CH 3 CH 2 CH 3 + H 2 O Wolff–Kishner Reduction: Treatment with hydrazine (NH 2 NH 2 ) followed by heating with KOH in ethylene glycol . CH 3 CHO + NH 2 NH 2 → CH 3 CH=NNH 2 —(KOH / ethylene glycol, Δ)→ CH 3 CH 3 + N 2

Q: Distinguish between Aldol condensation and Cannizzaro reaction with suitable examples. Why does methanal undergo Cannizzaro reaction while ethanal undergoes Aldol?

Aldol Condensation: Given by aldehydes/ketones having at least one α-hydrogen on treatment with dilute alkali to form β-hydroxy aldehyde/ketone, which dehydrates on heating to an α,β-unsaturated carbonyl. 2 CH 3 CHO —(dil. NaOH)→ CH 3 CH(OH)CH 2 CHO —(Δ)→ CH 3 CH=CHCHO + H 2 O Cannizzaro Reaction: Given by aldehydes having NO α-hydrogen on heating with concentrated alkali ; undergoes disproportionation to an alcohol and a carboxylic acid salt. 2 HCHO + conc. KOH —(Δ)→ CH 3 OH + HCOOK Reason: Ethanal has 3 α-hydrogens (acidic), whereas Methanal lacks an α-carbon and thus has zero α-hydrogens .

Q: Although phenoxide ion has more resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give reason.

1. The conjugate base of a carboxylic acid ( carboxylate ion, RCOO - ) is stabilised by two equivalent resonance structures in which the negative charge is delocalised equally over two electronegative oxygen atoms . 2. In contrast, the phenoxide ion (C 6 H 5 O - ) has non-equivalent resonance structures in which the negative charge resides on less electronegative carbon atoms of the benzene ring. Hence carboxylate ion is more stable, making carboxylic acids stronger acids than phenols.

Q: Write the chemical equations for: (i) Rosenmund reduction, (ii) Etard reaction, and (iii) Gattermann–Koch reaction.

(i) Rosenmund Reduction: C 6 H 5 COCl + H 2 —(Pd / BaSO 4 )→ C 6 H 5 CHO + HCl (ii) Etard Reaction: C 6 H 5 CH 3 + CrO 2 Cl 2 (in CS 2 ) → Chromium complex —(H 3 O + )→ C 6 H 5 CHO (iii) Gattermann–Koch Reaction: C 6 H 6 + CO + HCl —(anhyd. AlCl 3 / CuCl)→ C 6 H 5 CHO

Q: Explain why aldehydes are generally more reactive than ketones towards nucleophilic addition reactions.

1. Steric Factor: Ketones have two relatively bulky alkyl/aryl groups attached to the carbonyl carbon, hindering the approach of an incoming nucleophile compared to aldehydes (which have only one substituent and a small H atom). 2. Electronic Factor: Two electron-releasing (+I) alkyl groups in ketones reduce the positive charge (electrophilicity) of the carbonyl carbon more strongly than the single alkyl group in aldehydes.

Q: Which test can distinguish Benzaldehyde from Ethanal even though both give a silver mirror with Tollens' reagent?

Fehling's test (or Iodoform test): Ethanal gives a reddish-brown precipitate of Cu 2 O on heating with Fehling's solution (and yellow CHI 3 ppt with I 2 /NaOH), whereas Benzaldehyde (an aromatic aldehyde) does NOT respond to Fehling's test.

Q: Arrange the following in increasing order of acid strength and justify: CH 3 COOH, HCOOH, FCH 2 COOH, ClCH 2 COOH.

Increasing Order: CH 3 COOH < HCOOH < ClCH 2 COOH < FCH 2 COOH Reason: The +I (electron-donating) methyl group destabilises CH 3 COO - , making acetic acid weaker than formic acid. Both –Cl and –F exert strong –I (electron-withdrawing) effects that stabilise the carboxylate anion; since Fluorine is more electronegative than Chlorine (stronger –I), FCH 2 COOH is the strongest acid among the four.

Carbonyl Polarity, Rosenmund & Stephen, Nucleophilic Addition, Tollens/Fehling/Iodoform Tests, Aldol & Cannizzaro, Carboxylic Acidity & HVZ

📄 Official NCERT Textbook PDF

Topic 8.1: Nomenclature & Structure of the Carbonyl Group (>C=O)

Walk into a bakery or spice market in Kerala and breathe deeply: the rich aroma of vanilla (vanillin), sweet cinnamon (cinnamaldehyde), and wild meadowsweet (salicylaldehyde) all come from one superstar functional group—the carbonyl group (>C=O)! Even acetone (nail-polish remover) and vinegar share this same carbon–oxygen double bond. Because oxygen greedily pulls π-electrons toward itself, the carbonyl carbon turns hungry for nucleophiles (δ+) while oxygen turns negative (δ-)—a built-in molecular magnet that drives almost every reaction in this chapter!

Organic compounds containing the carbon–oxygen double bond (>C=O) are called carbonyl compounds. Depending on what attaches to the carbonyl carbon, we get:

Aldehydes (R–CHO): Carbonyl carbon is bonded to at least one hydrogen and one carbon (or another H in methanal).
Ketones (R–CO–R'): Carbonyl carbon is bonded to two carbon atoms (alkyl or aryl).
Carboxylic Acids (R–COOH): Carbonyl carbon bonded to a hydroxyl (–OH) group; their derivatives include acyl halides (–COX), amides (–CONH₂), esters (–COOR'), and anhydrides (–CO–O–CO–).

1. Common & IUPAC Nomenclature (NCERT Table 8.1)

Common Names of Aldehydes: Derived from the corresponding carboxylic acid by replacing –ic acid with –aldehyde (e.g. Formic acid → Formaldehyde; Acetic acid → Acetaldehyde). Substituent positions use Greek letters α, β, γ, δ starting from the carbon next to –CHO.
Common Names of Ketones: Name the two alkyl/aryl groups + ketone (e.g. dimethyl ketone = acetone; alkyl phenyl ketones end in –phenone like acetophenone, propiophenone, benzophenone). Positions use α, α', β, β'.
IUPAC Rules: Replace final –e of alkane with –al (aldehyde, –CHO is always C-1) or –one (ketone, number from end nearer >C=O). When –CHO is attached directly to a ring, use suffix carbaldehyde (e.g. cyclohexanecarbaldehyde, benzenecarbaldehyde / benzaldehyde).

Formula / Structure	Common Name	IUPAC Name
HCHO	Formaldehyde	Methanal
CH₃CHO	Acetaldehyde	Ethanal
CH₂=CH–CHO	Acrolein	Prop-2-enal
CH₃CH(OCH₃)CHO	α-Methoxypropionaldehyde	2-Methoxypropanal
(CH₃)₂C=CHCOCH₃	Mesityl oxide	4-Methylpent-3-en-2-one
C₆H₄(CHO)₂ (1,2-)	Phthaldehyde	Benzene-1,2-dicarbaldehyde

2. Structure & Polarity of the Carbonyl Group (NCERT Fig. 8.1)

Hybridisation: Carbonyl C is sp²-hybridised → 3 σ bonds + 1 π bond → Trigonal planar (~120°)
Resonance Polarity: >C=O ↔ >C⁺–O^- → C is Electrophilic (Lewis acid); O is Nucleophilic (Lewis base)

The π-electron cloud lies above and below the σ-plane. Due to oxygen's higher electronegativity and the dipolar resonance contributor >C⁺–O^-, carbonyl compounds possess substantial dipole moments (~2.3–2.8 D) and are much more polar than ethers.

NCERT Intext 8.1 Worked Key:
• 3-Hydroxybutanal: CH₃CH(OH)CH₂CHO | • 4-Oxopentanal: CH₃COCH₂CH₂CHO
• Di-sec-butyl ketone: [CH₃CH₂CH(CH₃)]₂CO | • 4-Fluoroacetophenone: p-F–C₆H₄–COCH₃

In common naming, the α-carbon is the carbon next to –CHO, whereas in IUPAC numbering the –CHO carbon itself is C-1! Thus α-methoxypropionaldehyde becomes 2-methoxypropanal, and CH₃CH(OH)CH₂CHO is 3-hydroxybutanal (β-hydroxybutyraldehyde). Also, never count a cyclic ring carbon inside the aldehyde suffix—use carbaldehyde!

Topic 8.2: Preparation of Aldehydes & Ketones: Named Reactions Mastery

Preparing an aldehyde is like toasting bread over a campfire: pull it out at the exact golden moment, or it burns straight into charcoal (a carboxylic acid)! Ordinary oxidants overshoot from alcohol to acid—so chemists invented clever "braking systems" like PCC, Rosenmund's poisoned catalyst, DIBAL-H, and Etard's chromyl chloride complex that freeze the reaction right at the aldehyde stage!

1. Common Methods for Both Aldehydes & Ketones

Oxidation / Dehydrogenation of Alcohols: 1° alcohols + PCC → Aldehyde; 2° alcohols + CrO₃ → Ketone. Passing alcohol vapours over heated Cu or Ag (573 K) dehydrogenates 1° → Aldehyde and 2° → Ketone.
Ozonolysis of Alkenes: Alkene + O₃ followed by Zn + H₂O cleaves C=C into aldehydes and/or ketones.
Hydration of Alkynes (H₂SO₄ / HgSO₄, 333 K): Ethyne is the ONLY alkyne that gives an aldehyde (Ethanal); all higher alkynes follow Markovnikov addition + tautomerism to give Ketones (e.g. Propyne → Propanone).

2. Exclusive Methods for Aldehydes (NCERT 8.2.2)

Named Reaction	Reactant & Reagent	Product & Key Point
Rosenmund Reduction	`R–COCl + H₂` over Pd / BaSO₄ (poisoned with S/quinoline)	`R–CHO` (Cannot make HCHO since HCOCl is unstable)
Stephen Reaction	`R–CN + SnCl₂ + HCl` → Imine·HCl → `H₃O⁺`	`R–CHO` (Reduction of nitriles)
DIBAL-H Reduction	`R–CN` or `R–COOR'` + 1. AlH(i-Bu)₂, 2. H₂O	`R–CHO` (Does NOT reduce C=C double bonds!)
Etard Reaction	Toluene + CrO₂Cl₂ (in CS₂) → Cr complex → `H₃O⁺`	Benzaldehyde (Chromyl chloride stops oxidation)
CrO₃ / Ac₂O Method	Toluene + CrO₃ + (CH₃CO)₂O (273–283 K) → `H₃O⁺`	Benzylidene diacetate → Benzaldehyde
Side-Chain Chlorination	Toluene + `Cl₂ / hν` → Benzal chloride (`C₆H₅CHCl₂`) → `H₂O (373 K)`	Benzaldehyde (Commercial manufacturing route)
Gattermann–Koch Reaction	Benzene + CO + HCl / anhyd. AlCl₃ (or CuCl)	Benzaldehyde (Direct formylation of arene)

3. Exclusive Methods for Ketones (NCERT 8.2.3)

From Acyl Chlorides + Dialkylcadmium (R₂Cd): Prepared by 2 RMgX + CdCl₂ → R₂Cd + 2 MgXCl. Then 2 R'COCl + R₂Cd → 2 R'COR + CdCl₂. Milder than Grignard, so it stops cleanly at the ketone!
From Nitriles + Grignard Reagent: R–CN + R'MgX (dry ether) → imine magnesium salt → H₃O⁺ → R–CO–R'.
Friedel–Crafts Acylation: Benzene (or activated arene) + RCOCl (or anhydride) + anhyd. AlCl₃ → Aromatic ketone (e.g. Benzene + CH₃COCl → Acetophenone; Benzene + C₆H₅COCl → Benzophenone).

NCERT Example 8.1 & Intext 8.2 Conversions:
• Hexan-1-ol → Hexanal: PCC | • Allyl alcohol → Propenal: PCC (protects C=C!)
• p-Fluorotoluene → p-Fluorobenzaldehyde: CrO₂Cl₂ / H₃O⁺ (or CrO₃/Ac₂O)
• But-2-ene → Ethanal: O₃ / Zn–H₂O | • CH₃–C≡CH + Hg²⁺/H₂SO₄: Propanone

Do NOT confuse Stephen reaction (SnCl₂ + HCl converts nitriles only to aldehydes) with DIBAL-H (reduces both nitriles and esters to aldehydes without touching C=C). Also remember: in hydration of alkynes (Hg²⁺/H₂SO₄), ethyne is the ONLY alkyne giving an aldehyde—every other alkyne gives a ketone!

Topic 8.3 & 8.5: Physical Properties & Uses: Dipole–Dipole Boiling Points & Water Solubility

Picture molecules at a crowded bus stop. Butane molecules barely wave to each other (weak van der Waals forces) so they fly apart easily as gas at 273 K. Propanal and Acetone hold onto each other by electrostatic dipole hugs (boiling at 322–329 K). But Propan-1-ol molecules lock arms tightly through intermolecular hydrogen bonds, refusing to boil until 370 K! Yet drop acetone into water, and water happily reaches out its own H-bonds to dissolve it completely.

1. Physical State & Odour

Methanal is a gas at room temperature; Ethanal is a volatile liquid; higher aldehydes and ketones are liquids or solids. Lower aldehydes have sharp pungent odours, whereas larger molecules turn pleasant and fragrant (used in perfumes & flavouring).

2. Boiling Point Comparison (Comparable Molecular Masses)

Aldehydes and ketones boil higher than hydrocarbons and ethers (due to stronger dipole–dipole interactions between polar >C=O groups), yet lower than alcohols of similar molar mass (because aldehydes and ketones lack an –OH proton and cannot form intermolecular hydrogen bonds among themselves).

Compound (Molar Mass ~58–60)	Dominant Intermolecular Force	Boiling Point (K)
n-Butane (CH₃CH₂CH₂CH₃, M = 58)	Weak London dispersion (van der Waals)	273 K
Methoxyethane (CH₃OCH₂CH₃, M = 60)	Weak dipole–dipole	281 K
Propanal (CH₃CH₂CHO, M = 58)	Strong carbonyl dipole–dipole	322 K
Acetone / Propanone (CH₃COCH₃, M = 58)	Strong carbonyl dipole–dipole (slightly higher μ)	329 K
Propan-1-ol (CH₃CH₂CH₂OH, M = 60)	Extensive intermolecular H-bonding	370 K

3. Solubility in Water

Lower members—methanal, ethanal, and propanone—are miscible with water in all proportions because their carbonyl oxygen accepts hydrogen bonds from water (>C=O···H–O–H). As the non-polar alkyl chain lengthens, solubility drops sharply. All are soluble in organic solvents like benzene, ether, and chloroform.

NCERT Intext 8.3 & Example 8.2:
Arrange in increasing order of boiling points:
• Intext 8.3 (M ≈ 44–46): CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH
• Example 8.2 (M ≈ 72–74): CH₃(CH₂)₂CH₃ < C₂H₅OC₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH

4. Key Industrial Uses (NCERT 8.5)

Formalin (40% aqueous formaldehyde): Preserving biological specimens; making Bakelite (phenol–formaldehyde resin) and urea–formaldehyde glues.
Acetaldehyde: Starting material for acetic acid, ethyl acetate, polymers, and drugs.
Acetone & Ethyl Methyl Ketone (EMK): Widely used industrial solvents; Benzaldehyde in perfumery & dyes.

Wait—can aldehydes or ketones form hydrogen bonds? YES with water (which explains why propanone dissolves completely in water!), but NO with themselves (because they have no H atom attached to oxygen—hence lower boiling point than alcohols)!

Topic 8.4 (1): Nucleophilic Addition Reactions: Reactivity Order & Key Derivatives

Picture trying to land a helicopter on two helipads. Ethanal's helipad has one small tree (–CH₃) on one side and tiny grass (–H) on the other—wide open and inviting! Ketone's helipad (like propanone or di-tert-butyl ketone) has two giant towers blocking both approaches AND two electron-pumping groups soothing the carbonyl's positive hunger. Naturally, incoming nucleophiles land far faster on aldehydes than on ketones!

1. Mechanism of Nucleophilic Addition (NCERT Fig. 8.2)

Unlike alkenes (electrophilic addition), polar carbonyls undergo nucleophilic addition. The nucleophile (Nu^-) attacks the electrophilic sp² carbonyl carbon perpendicular to its planar σ-framework. Carbon rehybridises from sp² → sp³ forming a tetrahedral alkoxide intermediate, which captures a proton (H⁺) to yield the neutral addition product.

2. Why Aldehydes are More Reactive than Ketones

Steric Reason: Two alkyl/aryl substituents in ketones crowd the tetrahedral transition state and hinder the nucleophile's approach, whereas aldehydes have one tiny hydrogen atom.
Electronic Reason: Two alkyl groups exert stronger +I electron-releasing effect than one, reducing the positive charge (electrophilicity) on the carbonyl carbon.

NCERT Example 8.3 & Intext 8.4 Reactivity Rankings:
• Aliphatic series: Butanone < Propanone < Propanal < Ethanal
• Benzaldehyde vs Propanal: Benzaldehyde is LESS reactive than propanal because +R resonance from the benzene ring reduces electrophilicity of the –CHO carbon!
• Aromatic series: Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde (EWG –NO₂ boosts reactivity!).

3. Five Essential Nucleophilic Addition / Condensation Reactions

Reagent	Product Formed	Crucial Exam Note
HCN (base-catalysed, `OH^-`)	Cyanohydrin `>C(OH)(CN)`	Pure HCN reacts too slowly; base generates strong nucleophile `CN^-`.
NaHSO₃	Crystalline bisulphite adduct	Equilibrium favours product for aldehydes & methyl ketones; hydrolysed back by dil. acid/alkali → used for purification!
RMgX (dry ether, then H₃O⁺)	Alcohol (1°, 2°, or 3°)	HCHO → 1° alcohol; other RCHO → 2° alcohol; Ketones → 3° alcohol.
ROH / dry HCl	Hemiacetal → Acetal; Ketone + glycol → Cyclic ketal	Dry HCl protonates carbonyl O to activate C=O; acetals/ketals resist bases and hydrolyse back in aq. acid.
H₂N–Z (weak acid, pH ~3.5)	`>C=N–Z` (addition–elimination)	–OH → Oxime; –NH₂ → Hydrazone; 2,4-DNP → Yellow/orange/red solid; –NHCONH₂ → Semicarbazone.

NCERT Exercise 8.18(ii) Favorite: Semicarbazide (H₂N–NH–CO–NH₂) has TWO –NH₂ groups, yet ONLY the hydrazine-end –NH₂ attacks the carbonyl! Why? The other –NH₂ is directly attached to >C=O, so its lone pair is delocalised by resonance and unavailable for nucleophilic attack! Also, H₂N–Z condensation needs weak acid—too much acid protonates H₂N–Z into non-nucleophilic H₃N⁺–Z!

Topic 8.4 (2–3): Clemmensen vs Wolff–Kishner Reduction & Diagnostic Tests (Tollens', Fehling's, Iodoform)

Welcome to the Forensic Chemistry Test Bench! You hold three clear colourless bottles: Propanal, Benzaldehyde, and Acetophenone. Which is which? First add 2,4-DNP—all three flash orange-red crystals (confirming >C=O!). Next warm with Tollens' reagent: Propanal and Benzaldehyde coat the tube with a gorgeous gleaming Silver Mirror—Acetophenone fails! Finally heat with Fehling's solution: Propanal drops a brick-red Cu₂O precipitate while aromatic Benzaldehyde refuses to budge! One two-minute bench test identifies every bottle with 100% certainty.

1. Reduction of Aldehydes & Ketones (NCERT 8.4.2)

To Alcohols: NaBH₄, LiAlH₄, or H₂ / Ni (or Pd) reduces aldehydes → 1° alcohols and ketones → 2° alcohols.
To Hydrocarbons (>C=O → –CH₂–):
- Clemmensen Reduction: Zn–Hg / conc. HCl (strongly acidic—avoid when acid-sensitive groups like –OH or C=C are present).
- Wolff–Kishner Reduction: NH₂NH₂ followed by heating with KOH in ethylene glycol (strongly basic—use when acid-sensitive groups are present!).

2. Oxidation & The Three Classic Diagnostic Tests (NCERT 8.4.3)

Aldehydes are easily oxidised to carboxylic acids with the same number of carbons (even by mild oxidants). Ketones resist mild oxidants and require strong oxidising agents at elevated temperatures, undergoing C–C bond cleavage to give a mixture of shorter carboxylic acids.

Test & Reagent	Visible Positive Result	Who Gives Positive?	Who Fails?
Tollens' Test Fresh ammoniacal AgNO₃ `[Ag(NH₃)₂]⁺ OH^-`, Δ	Bright Silver Mirror (Ag↓) `RCHO + 2[Ag(NH₃)₂]⁺ + 3OH^- → RCOO^- + 2Ag↓ + 2H₂O + 4NH₃`	ALL Aldehydes (both aliphatic AND aromatic like Benzaldehyde!)	All Ketones (Propanone, Acetophenone)
Fehling's Test Fehling A (aq. CuSO₄) + Fehling B (alk. Rochelle salt), Δ	Reddish-brown ppt (Cu₂O↓) `RCHO + 2Cu²⁺ + 5OH^- → RCOO^- + Cu₂O↓ + 3H₂O`	Aliphatic Aldehydes ONLY (HCHO, CH₃CHO, propanal)	Aromatic aldehydes (Benzaldehyde) & Ketones!
Iodoform Test `I₂ + NaOH` (`NaOI`), Δ	Yellow ppt of CHI₃↓ + `RCOONa` (1 C less)	CH₃CO– (ethanal, propanone, pentan-2-one, acetophenone) & CH₃CH(OH)– alcohols	Methanal, Propanal, Pentan-3-one, Benzophenone!

NCERT Example 8.4 (Classic Board Detective Problem):
Compound (A) C₈H₈O gives an orange-red 2,4-DNP ppt (⇒ carbonyl), yellow ppt with I₂/NaOH (⇒ CH₃CO– methyl ketone!), negative Tollens'/Fehling's (⇒ ketone, not aldehyde), doesn't decolourise Br₂ water (⇒ aromatic ring, no alkene!), and vigorous CrO₃ oxidation gives benzoic acid C₇H₆O₂.
Conclusion: (A) = Acetophenone (C₆H₅COCH₃); (B) = Benzoic acid!

Top Kerala Board Trap: Benzaldehyde gives a gleaming Silver Mirror with Tollens' reagent, yet gives NO precipitate with Fehling's solution! Never forget: Fehling's distinguishes aliphatic aldehydes from aromatic aldehydes! Also remember: Ethanal (CH₃CHO) is the ONLY aldehyde giving a positive Iodoform test—allowing instant separation from Propanal (CH₃CH₂CHO)!

Topic 8.4 (4–5): α-Hydrogen Acidity: Aldol, Cross-Aldol & Cannizzaro Reactions

Stand at a fork in the alkali road! Hand a carbonyl molecule to aqueous base and ask one single diagnostic question: "Do you have at least one α-hydrogen next to your C=O?" If YES (like ethanal or propanone), even dilute NaOH gently plucks that acidic α-H, turning one molecule into a nucleophile that attacks its twin—Aldol Condensation! If NO (like formaldehyde or benzaldehyde), concentrated NaOH forces two twins into a dramatic sacrifice where one oxidises into acid salt while the other reduces into alcohol—the Cannizzaro Reaction!

1. Why are α-Hydrogens Acidic?

Hydrogens attached to the carbon immediately adjacent to >C=O (the α-carbon) are weakly acidic due to: (i) strong electron-withdrawing effect of the carbonyl group, and (ii) resonance stabilisation of the conjugate base (enolate ion).

2. Aldol Reaction & Aldol Condensation (Needs ≥ 1 α-Hydrogen)

Dilute alkali (dil. NaOH / Ba(OH)₂): Two molecules condense to form a β-hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol).
On warming (Δ): Loses water easily to yield an α,β-unsaturated carbonyl compound.

• 2 CH₃CHO —(dil. NaOH)→ CH₃CH(OH)CH₂CHO (3-Hydroxybutanal) —(Δ, –H₂O)→ CH₃CH=CHCHO (But-2-enal)
• 2 CH₃COCH₃ —(Ba(OH)₂)→ (CH₃)₂C(OH)CH₂COCH₃ —(Δ, –H₂O)→ (CH₃)₂C=CHCOCH₃ (4-Methylpent-3-en-2-one)

3. Cross-Aldol Condensation

When two different carbonyls both possess α-hydrogens (e.g. Ethanal + Propanal), cross-aldol gives a mixture of 4 products: 2 self-aldol (But-2-enal, 2-Methylpent-2-enal) + 2 cross-aldol (2-Methylbut-2-enal, Pent-2-enal). Mixing an aromatic aldehyde with no α-H (Benzaldehyde) + Acetophenone cleanly gives a single major cross product (Benzalacetophenone).

4. Cannizzaro Reaction (Aldehydes with NO α-Hydrogen!)

Aldehydes lacking α-hydrogens undergo self-oxidation and reduction (disproportionation) on heating with concentrated alkali (50% NaOH/KOH): one molecule reduces to an alcohol while the other oxidises to a carboxylic acid salt.

• 2 HCHO + conc. KOH —(Δ)→ CH₃OH (Methanol) + HCOOK (Potassium formate)
• 2 C₆H₅CHO + conc. NaOH —(Δ)→ C₆H₅CH₂OH (Benzyl alcohol) + C₆H₅COONa (Sodium benzoate)

5. Electrophilic Substitution in Aromatic Aldehydes & Ketones

–CHO and –COR groups are strongly ring-deactivating and meta-directing (Benzaldehyde + conc. HNO₃ / H₂SO₄ at 273–283 K → m-Nitrobenzaldehyde).

Watch out for NCERT Exercise 8.7! Does Benzophenone (Ph₂CO) undergo Cannizzaro because it has no α-H? NO! Cannizzaro works ONLY for aldehydes without α-H (like HCHO, C₆H₅CHO, (CH₃)₃C–CHO, 2,2-dimethylbutanal)—ketones without α-H like benzophenone undergo neither Aldol nor Cannizzaro!

Topic 8.6 – 8.8: Carboxylic Acids: Nomenclature, Resonance, Preparation & Dimerisation

Did you know ancient chemists named carboxylic acids after the surprises where they first found them? Red ants gave formic acid (formica), sour vinegar gave acetic acid (acetum), and rancid butter gave butyric acid (butyrum)! Even stranger: boil acetic acid into vapour or dissolve it in benzene, and pairs of acid molecules lock face-to-face through two strong hydrogen bonds, masquerading as giant twin cyclic dimers with double the expected molar mass!

1. Nomenclature & Structure of the Carboxyl Group (`–COOH`)

IUPAC Naming: Replace –e of alkane with –oic acid (–COOH carbon is always C-1). For dicarboxylic acids keep –e: remember the classic chain series "O My Such Good Apple": HOOC–COOH (Oxalic / Ethanedioic), HOOCCH₂COOH (Malonic / Propanedioic), HOOC(CH₂)₂COOH (Succinic / Butanedioic), HOOC(CH₂)₃COOH (Glutaric / Pentanedioic), HOOC(CH₂)₄COOH (Adipic / Hexanedioic).
Resonance in –COOH: Carboxyl carbon is planar sp² (~120°) and is LESS electrophilic than aldehyde/ketone carbonyl carbon because the lone pair on –OH donates electron density via resonance (–C(=O)–OH ↔ –C(O^-)=O⁺H).

2. Six Important Methods of Preparation (NCERT 8.7)

From 1° Alcohols & Aldehydes: Strong oxidants like KMnO₄, K₂Cr₂O₇/H₂SO₄, or Jones reagent (CrO₃–H₂SO₄).
Side-Chain Oxidation of Alkylbenzenes: Heating with alkaline KMnO₄ followed by H₃O⁺ oxidises the entire 1° or 2° alkyl chain (regardless of chain length—methyl, ethyl, propyl, isopropyl—AND acyl/vinyl side chains!) straight to –COOH! Exception: Tertiary alkyl groups (e.g. tert-butylbenzene) lack benzylic H and are NOT oxidised!
Hydrolysis of Nitriles & Amides: R–CN + H₂O (H⁺ or OH^-, Δ) → RCONH₂ → RCOOH + NH₃/NH₄⁺ (increases carbon chain by +1 C!).
Grignard Reagent + Dry Ice (CO₂): RMgX + CO₂ (dry ether) → RCOOMgX —(H₃O⁺)→ RCOOH (also ascends chain by +1 C; works even for aryl halides!).
Hydrolysis of Acyl Halides & Anhydrides: Rapid hydrolysis with water or aq. base + acidification.
Hydrolysis of Esters: Acidic hydrolysis directly yields RCOOH; basic saponification yields RCOO^- which acidifies to RCOOH.

3. Physical Properties & Cyclic Dimerisation (NCERT 8.8)

Boiling Point Order: Carboxylic Acids > Alcohols > Ketones / Aldehydes > Ethers > Alkanes
Why? Stronger, more extensive intermolecular H-bonding—most carboxylic acids exist as 8-membered cyclic dimers even in the vapour phase or aprotic solvents (like benzene)!

Aliphatic acids up to C₄ are water-miscible via H-bonding; solubility drops as the hydrophobic chain grows. Benzoic acid is nearly insoluble in cold water.

In side-chain oxidation of alkylbenzenes using hot alkaline KMnO₄, it doesn't matter whether the chain is –CH₃, –CH₂CH₃, or –CH(CH₃)₂—ALL yield Benzoic acid (with extra carbons lost as CO₂)! Only a 3° alkyl group like –C(CH₃)₃ refuses oxidation! And remember: cyclic dimerisation in benzene makes acetic acid show an apparent molar mass of ~120 g/mol (van't Hoff factor i ≈ 0.5)!

Topic 8.9 – 8.10: Acidity of Carboxylic Acids (±I Effects), HVZ, Decarboxylation & Uses

Drop acetic acid onto baking soda (NaHCO₃) and—FIZZ!—carbon dioxide bubbles race to the top. Drop phenol (an acid too!) into the same baking soda—complete silence! Why does carboxylic acid win that proton tug-of-war hands down? Because when carboxylic acid gives up H⁺, the leftover negative charge relaxes equally across two identical electronegative oxygen atoms, whereas poor phenoxide forces its negative charge onto ring carbons!

1. Acidity & Why Carboxylic Acids > Phenols > Alcohols (NCERT 8.9.1)

Bicarbonate Diagnostic Test: Unlike phenols, carboxylic acids react with NaHCO₃ / Na₂CO₃ evolving brisk effervescence of CO₂↑!
Carboxylate vs Phenoxide (NCERT Ex. 8.20): Even though phenoxide has more resonance structures, carboxylate ion (RCOO^-) has two equivalent resonance structures with negative charge delocalised over two electronegative oxygen atoms. Phenoxide has non-equivalent structures putting negative charge on less electronegative ring carbons!

2. Substituent Effects on Acid Strength (Smaller pK_a = Stronger Acid!)

Electron-Withdrawing Groups (EWG, –I / –R) disperse negative charge on RCOO^- and strengthen the acid; Electron-Donating Groups (EDG, +I) intensify negative charge and weaken the acid.

EWG Strength Order: Ph < I < Br < Cl < F < CN < NO₂ < CF₃
NCERT pK_a Series (Decreasing Acidity):
CF₃COOH (0.23) > CCl₃COOH > CHCl₂COOH > NO₂CH₂COOH > NCCH₂COOH > FCH₂COOH > ClCH₂COOH > BrCH₂COOH > HCOOH > ClCH₂CH₂COOH > C₆H₅COOH (4.19) > C₆H₅CH₂COOH > CH₃COOH (4.76) > CH₃CH₂COOH

Distance & Count: Inductive pull weakens rapidly with distance (α-haloakanoic > β- > γ-) and grows with number of EWGs (CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH).
Aromatic Acids: 4-Nitrobenzoic acid (pK_a 3.41) > Benzoic acid (4.19) > 4-Methoxybenzoic acid (4.46). Direct attachment of sp² phenyl/vinyl carbon makes benzoic acid stronger than acetic acid!

3. Reactions of C–OH, –COOH & Hydrocarbon Part (NCERT 8.9.2–8.9.4)

Reaction Type	Reagent & Conditions	Product & Board Highlight
Anhydride / Ester / Amide	`H₂SO₄` or `P₂O₅` (Δ); `R'OH / H⁺`; `NH₃ + Δ`	Acid anhydride; Ester (remove H₂O to shift forward); Amide (Phthalic acid → Phthalimide)
Acid Chloride	`PCl₅`, `PCl₃`, or SOCl₂	`RCOCl` — SOCl₂ preferred because by-products `SO₂↑` & `HCl↑` escape as gases!
Reduction to 1° Alcohol	`LiAlH₄/ether` or B₂H₆, then `H₃O⁺`	`RCH₂OH` (NaBH₄ does NOT reduce –COOH! B₂H₆ spares esters/nitro/halo)
Sodalime Decarboxylation	`RCOONa + NaOH + CaO (3:1), Δ`	Alkane (R–H) with 1 carbon LESS + Na₂CO₃! (Kolbe electrolysis → R–R, doubles alkyl C)
Hell–Volhard–Zelinsky (HVZ)	`RCH₂COOH + X₂ (Cl₂/Br₂) / Red P`, then H₂O	α-Halocarboxylic acid `RCH(X)COOH` (requires at least one α-H!)
Ring Substitution	`HNO₃/H₂SO₄` or `Br₂/FeBr₃`	`–COOH` is meta-directing & deactivating; NO Friedel–Crafts reaction!

Uses (NCERT 8.10): HCOOH (rubber/textile/leather/electroplating), CH₃COOH (vinegar/solvent), Hexanedioic/Adipic acid (Nylon-6,6), Sodium benzoate (food preservative), Fatty acids (soaps/detergents).

Three Classic Exam Traps: (1) Benzoic acid does NOT undergo Friedel–Crafts reaction because –COOH is strongly deactivating AND complexes with the Lewis acid catalyst AlCl₃! (2) Formic acid (HCOOH) and Benzoic acid have NO α-hydrogen—so they fail the HVZ reaction! (3) NaBH₄ reduces aldehydes/ketones smoothly, yet cannot reduce carboxylic acids!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 2 Marks): Give simple chemical tests to distinguish between: (a) Propanal and Propanone, (b) Benzaldehyde and Acetophenone.

(a) Propanal vs Propanone: Warm both with Tollens' reagent (or Fehling's solution). Propanal gives a bright silver mirror (or reddish-brown Cu₂O ppt with Fehling's), whereas Propanone gives no reaction. (Alternatively, Propanone gives yellow ppt with I₂/NaOH in Iodoform test; Propanal does not.)

(b) Benzaldehyde vs Acetophenone: Warm with Tollens' reagent: Benzaldehyde forms a silver mirror while Acetophenone fails. (Or Iodoform test: Acetophenone gives yellow CHI₃ ppt while Benzaldehyde fails.)

Q2 (Kerala HSE SAY / NCERT • 2 Marks): Explain Clemmensen reduction and Wolff–Kishner reduction with one chemical equation each.

Both convert the carbonyl group (>C=O) directly into a methylene group (–CH₂–):

Clemmensen Reduction: Reduction with Zn–Hg and conc. HCl.
CH₃COCH₃ + 4[H] —(Zn–Hg / conc. HCl)→ CH₃CH₂CH₃ + H₂O

Wolff–Kishner Reduction: Treatment with hydrazine (NH₂NH₂) followed by heating with KOH in ethylene glycol.
CH₃CHO + NH₂NH₂ → CH₃CH=NNH₂ —(KOH / ethylene glycol, Δ)→ CH₃CH₃ + N₂

Q3 (Kerala HSE March • 3 Marks): Distinguish between Aldol condensation and Cannizzaro reaction with suitable examples. Why does methanal undergo Cannizzaro reaction while ethanal undergoes Aldol?

Aldol Condensation: Given by aldehydes/ketones having at least one α-hydrogen on treatment with dilute alkali to form β-hydroxy aldehyde/ketone, which dehydrates on heating to an α,β-unsaturated carbonyl.
2 CH₃CHO —(dil. NaOH)→ CH₃CH(OH)CH₂CHO —(Δ)→ CH₃CH=CHCHO + H₂O

Cannizzaro Reaction: Given by aldehydes having NO α-hydrogen on heating with concentrated alkali; undergoes disproportionation to an alcohol and a carboxylic acid salt.
2 HCHO + conc. KOH —(Δ)→ CH₃OH + HCOOK

Reason: Ethanal has 3 α-hydrogens (acidic), whereas Methanal lacks an α-carbon and thus has zero α-hydrogens.

Q4 (Kerala HSE Model / March • 2 Marks): Although phenoxide ion has more resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Give reason.

1. The conjugate base of a carboxylic acid (carboxylate ion, RCOO^-) is stabilised by two equivalent resonance structures in which the negative charge is delocalised equally over two electronegative oxygen atoms.

2. In contrast, the phenoxide ion (C₆H₅O^-) has non-equivalent resonance structures in which the negative charge resides on less electronegative carbon atoms of the benzene ring. Hence carboxylate ion is more stable, making carboxylic acids stronger acids than phenols.

Q5 (Kerala HSE SAY • 2 Marks): Write the chemical equations for: (i) Rosenmund reduction, (ii) Etard reaction, and (iii) Gattermann–Koch reaction.

(i) Rosenmund Reduction: C₆H₅COCl + H₂ —(Pd / BaSO₄)→ C₆H₅CHO + HCl

(ii) Etard Reaction: C₆H₅CH₃ + CrO₂Cl₂ (in CS₂) → Chromium complex —(H₃O⁺)→ C₆H₅CHO

(iii) Gattermann–Koch Reaction: C₆H₆ + CO + HCl —(anhyd. AlCl₃ / CuCl)→ C₆H₅CHO

Q6 (Kerala HSE March (High Yield) • 2 Marks): Explain why aldehydes are generally more reactive than ketones towards nucleophilic addition reactions.

1. Steric Factor: Ketones have two relatively bulky alkyl/aryl groups attached to the carbonyl carbon, hindering the approach of an incoming nucleophile compared to aldehydes (which have only one substituent and a small H atom).

2. Electronic Factor: Two electron-releasing (+I) alkyl groups in ketones reduce the positive charge (electrophilicity) of the carbonyl carbon more strongly than the single alkyl group in aldehydes.

Q7 (Kerala HSE March • 1 Marks): Which test can distinguish Benzaldehyde from Ethanal even though both give a silver mirror with Tollens' reagent?

Fehling's test (or Iodoform test): Ethanal gives a reddish-brown precipitate of Cu₂O on heating with Fehling's solution (and yellow CHI₃ ppt with I₂/NaOH), whereas Benzaldehyde (an aromatic aldehyde) does NOT respond to Fehling's test.

Q8 (Kerala HSE SAY • 2 Marks): Arrange the following in increasing order of acid strength and justify: CH₃COOH, HCOOH, FCH₂COOH, ClCH₂COOH.

Increasing Order: CH₃COOH < HCOOH < ClCH₂COOH < FCH₂COOH

Reason: The +I (electron-donating) methyl group destabilises CH₃COO^-, making acetic acid weaker than formic acid. Both –Cl and –F exert strong –I (electron-withdrawing) effects that stabilise the carboxylate anion; since Fluorine is more electronegative than Chlorine (stronger –I), FCH₂COOH is the strongest acid among the four.

Q9 (Kerala HSE March (Structure Elucidation) • 3 Marks): An organic compound (A) with molecular formula C₉H₁₀O forms a 2,4-DNP derivative, reduces Tollens' reagent, and undergoes Cannizzaro reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid. Identify (A) and explain.

1. Forms 2,4-DNP derivative and reduces Tollens' reagent ⇒ (A) is an aldehyde (–CHO).
2. Undergoes Cannizzaro reaction ⇒ No α-hydrogen (–CHO attached directly to aromatic ring).
3. Vigorous oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid) ⇒ benzene ring has two substituents at 1,2- (ortho) positions.
4. Formula C₉H₁₀O = C₆H₄(CHO)(C₂H₅) ⇒ (A) is 2-Ethylbenzaldehyde!

Q10 (Kerala HSE Model • 2 Marks): Write equations for: (a) Hell–Volhard–Zelinsky (HVZ) reaction of ethanoic acid, and (b) Sodalime decarboxylation of sodium propanoate.

(a) HVZ Reaction: CH₃COOH —(1. Cl₂ / Red P, 2. H₂O)→ ClCH₂COOH (2-Chloroethanoic acid)

(b) Sodalime Decarboxylation: CH₃CH₂COONa + NaOH —(CaO, Δ)→ CH₃–CH₃ (Ethane) + Na₂CO₃

Q11 (Kerala HSE Improvement • 1 Marks): Why do carboxylic acids show higher boiling points than alcohols of comparable molecular masses?

Carboxylic acids form stronger and more extensive intermolecular hydrogen bonds than alcohols, associating as stable cyclic dimers that persist even in the vapour phase.

Q12 (Kerala HSE / NCERT Exemplar • 3 Marks): Give reasons for the following: (i) Cyclohexanone forms cyanohydrin in good yield, whereas 2,2,6-trimethylcyclohexanone does not. (ii) Semicarbazide has two –NH₂ groups, yet only one takes part in semicarbazone formation. (iii) Benzoic acid does not undergo Friedel–Crafts reaction.

(i) Three methyl groups at C-2 and C-6 create severe steric hindrance blocking nucleophilic attack by CN^- at C=O.

(ii) One –NH₂ group is directly attached to the carbonyl (>C=O), so its lone pair is delocalised by resonance and loses nucleophilicity; only the distant hydrazine –NH₂ reacts.

(iii) The –COOH group strongly deactivates the benzene ring AND binds to the Lewis acid catalyst AlCl₃.