Aldehydes, Ketones and Carboxylic Acids

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Exercise 8.1Carbonyl Compounds — Preparation and Reactions

Write the structures of the following compounds: (i) α-methoxypropionaldehyde (ii) 3-hydroxybutanal (iii) 2-hydroxycyclopentane-1-carbaldehyde (iv) 4-oxopentanal (v) Di-sec-butyl ketone (vi) 4-chloroacetophenone

Solution

(i) α-methoxypropionaldehyde: CH₃CH(OCH₃)CHO (methoxy group on α-carbon relative to the aldehyde) (ii) 3-hydroxybutanal: CH₃CH(OH)CH₂CHO (iii) 2-hydroxycyclopentane-1-carbaldehyde: Cyclopentane ring with CHO at position 1 and OH at position 2 (iv) 4-oxopentanal: OHC–CH₂–CH₂–CO–CH₃ (CH₃COCH₂CH₂CHO) (v) Di-sec-butyl ketone: CH₃CH₂CH(CH₃)–CO–CH(CH₃)CH₂CH₃ (vi) 4-chloroacetophenone: ClC₆H₄COCH₃ with Cl at para position

Write the IUPAC names of the following ketones and aldehydes: (i) PhCOPh (ii) CH₃CO(CH₂)₄CH₃ (iii) CH₃COCH₂COCH₃ (iv) PhCHO (v) Ph(CH₂)₃CHO

Solution

(i) PhCOPh (benzophenone): Diphenylmethanone (ii) CH₃CO(CH₂)₄CH₃: The C=O is on C2 of a 7-carbon chain Heptan-2-one (iii) CH₃COCH₂COCH₃: Pentane-2,4-dione (acetylacetone) (iv) PhCHO (benzaldehyde): Benzaldehyde (IUPAC: benzenecarbaldehyde) (v) Ph(CH₂)₃CHO: 4-phenylbutanal

Predict the products of the following reactions: (i) CH₃CHO + H₂ → (Ni, Δ) (ii) PhCHO + CH₃CHO → (dilute NaOH)

Solution

(i) CH₃CHO + H₂ → (Ni catalyst, heat): Reduction of acetaldehyde: CH₃CHO + H₂ → CH₃CH₂OH (Ethanol — primary alcohol) (ii) PhCHO + CH₃CHO + dilute NaOH: This is an aldol-type reaction. However, PhCHO has no α-hydrogen, while CH₃CHO has 3 α-hydrogens. CH₃CHO forms enolate under NaOH: CH₃CHO ⇌ ⁻CH₂CHO (enolate) Enolate attacks PhCHO (Claisen-Schmidt condensation): PhCHO + CH₃CHO → Ph–CH(OH)–CH₂CHO (aldol product) → dehydration → Ph–CH=CH–CHO (cinnamaldehyde) Product: Cinnamaldehyde (3-phenylprop-2-enal) via Claisen-Schmidt condensation

Which of the following compounds undergoes aldol condensation? (i) Ethanal (ii) Propanal (iii) Benzaldehyde (iv) Pentan-3-one

Solution

Aldol condensation requires α-hydrogen (H atoms on the carbon adjacent to C=O): (i) Ethanal (CH₃CHO): has 3 α-hydrogens → undergoes aldol condensation ✓ (ii) Propanal (CH₃CH₂CHO): has 2 α-hydrogens → undergoes aldol condensation ✓ (iii) Benzaldehyde (PhCHO): no α-hydrogen (CHO directly attached to ring) → does NOT undergo aldol condensation ✗ (iv) Pentan-3-one (CH₃COCH₂CH₂CH₃): Wait — pentan-3-one is CH₃CH₂COCH₂CH₃, which has α-hydrogens → undergoes aldol condensation ✓ Compounds (i), (ii), and (iv) undergo aldol condensation. Benzaldehyde does NOT.

Distinguish between the following with suitable chemical tests: (i) Acetaldehyde and benzaldehyde (ii) Propan-2-ol and propanal

Solution

(i) Acetaldehyde vs Benzaldehyde: Test 1 — Tollens' reagent (ammoniacal AgNO₃): • Both give silver mirror (both are aldehydes) Test 2 — Fehling's solution: • Acetaldehyde: gives brick-red precipitate (positive) • Benzaldehyde: NO brick-red precipitate (negative) Aromatic aldehydes do not reduce Fehling's solution Test 3 — Iodoform test: • Acetaldehyde: gives yellow iodoform precipitate (positive — has CH₃CO– group) • Benzaldehyde: negative (ii) Propan-2-ol vs Propanal: Test 1 — Fehling's solution: • Propanal (aldehyde): gives brick-red precipitate ✓ • Propan-2-ol (secondary alcohol): NO reaction ✗ Test 2 — Oxidation with K₂Cr₂O₇/H₂SO₄: • Propan-2-ol: changes orange to green (chromium reduced) • Propanal: also gives positive Most distinguishing: Fehling's test or Tollens' reagent: • Propanal gives positive (silver mirror or red ppt) • Propan-2-ol gives negative

How will you prepare the following from benzene: (i) benzoic acid (ii) m-nitroacetophenone

Solution

(i) Benzene → Benzoic acid: Benzene → Friedel-Crafts alkylation → Toluene (C₆H₅CH₃) Toluene → (KMnO₄/H⁺, or K₂Cr₂O₇/H₂SO₄) → Benzoic acid (C₆H₅COOH) Or: Benzene → Grignard → PhMgBr → CO₂ → PhCOO⁻Mg→ PhCOOH (requires PhBr first) (ii) m-Nitroacetophenone from benzene: Benzene → (Friedel-Crafts acylation: CH₃COCl, AlCl₃) → Acetophenone (C₆H₅COCH₃) Acetophenone → (conc. HNO₃ + conc. H₂SO₄) → m-Nitroacetophenone Note: COCH₃ is a meta-director (electron-withdrawing by induction), so nitration gives the meta product.

Arrange the following compounds in increasing order of their acidic strength: formic acid, acetic acid, propanoic acid, benzoic acid.

Solution

Acidity of carboxylic acids depends on stability of carboxylate ion (conjugate base): Factors affecting acidity: • Electron-withdrawing groups on α-carbon increase acidity • More electron density on carboxylate ion → less stable → less acidic Formic acid (HCOOH): no alkyl group, H is electron-withdrawing compared to alkyl pKa ≈ 3.74 Acetic acid (CH₃COOH): one methyl group (electron-donating via induction) pKa ≈ 4.76 Propanoic acid (CH₃CH₂COOH): ethyl group (slightly more electron-donating) pKa ≈ 4.87 Benzoic acid (C₆H₅COOH): phenyl group partially withdraws electrons by resonance pKa ≈ 4.20 Increasing order of acidity (weakest to strongest): Propanoic acid < Acetic acid < Benzoic acid < Formic acid Note: Formic acid is strongest because H (vs alkyl) does not donate electrons to carboxylate.

Explain why carboxylic acids are stronger acids than alcohols, although both contain –OH groups.

Solution

Both carboxylic acids and alcohols contain the –OH group, but their acidity differs dramatically: pKa: CH₃COOH ≈ 4.76 vs CH₃CH₂OH ≈ 16 Reason — Resonance stabilisation of carboxylate ion: When a carboxylic acid loses H⁺: RCOOH → RCOO⁻ + H⁺ The carboxylate ion (RCOO⁻) is stabilised by resonance: R–C(=O)–O⁻ ↔ R–C(–O⁻)=O The negative charge is delocalised over both oxygen atoms equally (equivalent resonance structures). This resonance stabilisation greatly lowers the energy of the conjugate base. When an alcohol loses H⁺: ROH → RO⁻ + H⁺ The alkoxide ion (RO⁻) has NO resonance stabilisation. The negative charge remains localised on one oxygen. Greater stability of carboxylate ion (vs alkoxide) → equilibrium shifts more toward ionisation → higher Ka → lower pKa → stronger acid. Conclusion: The resonance stabilisation of –COO⁻ makes carboxylic acids much stronger acids than alcohols.

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