CBSE Class 12 Chemistry Aldehydes Ketones And Carboxylic Acids Notes Set 07

This second set of Class 12 CBSE Chemistry notes on Aldehydes, Ketones and Carboxylic Acids deals with reduction, oxidation, the tests that separate aldehydes from ketones, aldol condensation, the Cannizzaro and haloform reactions, and then the nomenclature, preparation, acidity and reactions of carboxylic acids. Solved examples practise the named reactions.

Important Notes and Formulas

Reduction of Aldehydes and Ketones

1. To alcohols: aldehydes give primary alcohols and ketones give secondary alcohols with H₂ over Ni, Pt or Pd, or with NaBH₄ or LiAlH₄: RCHO —(NaBH₄)→ RCH₂OH; R₂C=O —(LiAlH₄)→ R₂CHOH.
2. Clemmensen reduction converts C=O to CH₂ using zinc amalgam and conc. HCl: C=O —(Zn–Hg, conc. HCl)→ CH₂.
3. Wolff–Kishner reduction converts C=O to CH₂ by forming the hydrazone with NH₂NH₂ and then heating it with KOH in ethylene glycol: C=O —(NH₂NH₂)→ C=N–NH₂ —(KOH, glycol, heat)→ CH₂ + N₂.
4. Use Clemmensen reduction for substrates that are stable to acid and Wolff–Kishner for substrates that are stable to base. For example, cyclohexanone gives cyclohexane by either method.

Oxidation of Aldehydes and Ketones

1. Aldehydes are easily oxidised to carboxylic acids with the same number of carbon atoms by mild oxidants as well as by KMnO₄ and K₂Cr₂O₇: RCHO —([O])→ RCOOH.
2. Ketones are oxidised only by strong oxidants under vigorous conditions (hot KMnO₄ or conc. HNO₃). The C–C bond next to C=O breaks, and a mixture of carboxylic acids with fewer carbon atoms forms.
3. Tollens' test: Tollens' reagent is ammoniacal silver nitrate, [Ag(NH₃)₂]⁺. Aldehydes reduce it to metallic silver, a silver mirror: RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 4NH₃ + 2H₂O.
4. Fehling's test: Fehling's solution is alkaline copper(II) sulphate with sodium potassium tartrate. Aliphatic aldehydes reduce the blue Cu²⁺ complex to a red-brown precipitate of Cu₂O: RCHO + 2Cu²⁺ + 5OH⁻ → RCOO⁻ + Cu₂O + 3H₂O. Aromatic aldehydes and ketones do not give the test.
5. Methanoic acid also gives Tollens' test because it contains an H–C(=O)– unit.

Reactions Due to α-Hydrogen: Aldol Condensation

1. The α-hydrogens of aldehydes and ketones are acidic because the carbanion formed on losing H⁺ is stabilised by resonance with the carbonyl group.
2. Aldehydes and ketones with at least one α-hydrogen, on treatment with dilute alkali, give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols). On heating, the aldol loses water to give an α,β-unsaturated carbonyl compound.
3. 2CH₃CHO —(dil. NaOH)→ CH₃CH(OH)CH₂CHO (3-hydroxybutanal) —(heat, –H₂O)→ CH₃CH=CHCHO (but-2-enal).
4. 2CH₃COCH₃ —(dil. base)→ (CH₃)₂C(OH)CH₂COCH₃ (4-hydroxy-4-methylpentan-2-one) —(heat)→ (CH₃)₂C=CHCOCH₃ (4-methylpent-3-en-2-one).
5. Mechanism: base removes an α-hydrogen to give an enolate ion; the enolate adds to the carbonyl carbon of a second molecule; the alkoxide is protonated by water.
6. A cross aldol condensation between two different carbonyl compounds that both have α-hydrogens gives a mixture of four products. If only one partner has α-hydrogen, e.g. benzaldehyde + ethanal, the product is cleaner (cinnamaldehyde, C₆H₅CH=CHCHO, after heating).

Cannizzaro Reaction

1. Aldehydes with no α-hydrogen undergo self oxidation and reduction (disproportionation) on heating with concentrated alkali. One molecule is oxidised to the carboxylate salt and another is reduced to the alcohol.
2. 2HCHO + conc. NaOH → CH₃OH + HCOONa.
3. 2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa.
4. Aldehydes with α-hydrogen, such as ethanal, give aldol condensation instead. Ketones do not give the Cannizzaro reaction.

Haloform Reaction and Iodoform Test

1. Methyl ketones (CH₃CO–R) and ethanal, and alcohols that can be oxidised to them (ethanol and CH₃CH(OH)–R), react with a halogen and alkali (for iodine, I₂ and NaOH, i.e. NaOI) to give a haloform.
2. CH₃COCH₃ + 3NaOI → CH₃COONa + CHI₃ + 2NaOH. The yellow precipitate of iodoform (CHI₃) with a characteristic smell is the iodoform test.
3. Ketones such as pentan-3-one and benzophenone, and alcohols like propan-1-ol, do not respond.

Electrophilic Substitution in Aromatic Aldehydes and Ketones

The carbonyl group is deactivating and meta directing, so aromatic aldehydes and ketones undergo nitration, sulphonation and halogenation mainly at the meta position. Benzaldehyde with conc. HNO₃ and H₂SO₄ gives 3-nitrobenzaldehyde.

Carboxylic Acids: Nomenclature and Structure

1. IUPAC names replace the final –e of the alkane by –oic acid. The COOH carbon is C-1. Dicarboxylic acids end in –dioic acid, and a ring attached to COOH is named carboxylic acid.
2. The carboxyl carbon is sp² hybridised, planar, with bond angles about 120°. The acid is a resonance hybrid of the two carboxylate forms after loss of H⁺.

Preparation of Carboxylic Acids

1. Oxidation of primary alcohols and aldehydes with KMnO₄ (neutral, acidic or alkaline) or K₂Cr₂O₇ in acid: RCH₂OH —([O])→ RCHO —([O])→ RCOOH.
2. Oxidation of alkylbenzenes with hot alkaline or acidic KMnO₄ oxidises the entire side chain to –COOH whatever its length, provided the benzylic carbon has a hydrogen: toluene → benzoic acid; ethylbenzene → benzoic acid.
3. Hydrolysis of nitriles with dilute acid or alkali: RCN + 2H₂O —(H+ or OH⁻)→ RCOOH + NH₃ (the amide forms as an intermediate).
4. Grignard reagent + dry ice: RMgX + CO₂ → RCOOMgX —(H₃O⁺)→ RCOOH. This adds one carbon atom.
5. Hydrolysis of acyl chlorides, anhydrides, esters and amides gives the acid: RCOCl + H₂O → RCOOH + HCl; (RCO)₂O + H₂O → 2RCOOH; RCOOR' + H₂O ⇌ RCOOH + R'OH; RCONH₂ + H₂O → RCOOH + NH₃.

Physical Properties and Acidity

1. Aliphatic acids up to C₉ are colourless liquids with unpleasant smells; higher acids are waxy solids. Benzoic acid is a solid that is almost insoluble in cold water.
2. Carboxylic acids boil higher than aldehydes, ketones and even alcohols of comparable molar mass because they form extensive intermolecular hydrogen bonds. In the vapour phase and in aprotic solvents they exist as hydrogen-bonded dimers.
3. Acids with up to four carbon atoms are miscible with water by hydrogen bonding. Solubility falls as the hydrocarbon chain lengthens.
4. Acidity: RCOOH + H₂O ⇌ RCOO⁻ + H₃O⁺. The carboxylate ion is stabilised by resonance (equal C–O bonds, charge shared by two oxygens), whereas the alkoxide from an alcohol is not. Carboxylic acids (pKa about 4–5) are therefore much stronger acids than alcohols and phenols.
5. They liberate CO₂ from sodium hydrogencarbonate, which alcohols and phenols do not, so this is the test for a –COOH group: RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂.
6. Substituent effect: electron-withdrawing groups (–Cl, –F, –NO₂) stabilise the anion and increase acid strength; electron-donating groups (–CH₃) decrease it. The effect grows with the number of groups and falls with distance from –COOH.

Lower pKa means a stronger acid, so the acidity order is CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > HCOOH > C₆H₅COOH > CH₃COOH.

Chemical Reactions of Carboxylic Acids

1. Reaction with metals and alkalis: 2RCOOH + 2Na → 2RCOONa + H₂; RCOOH + NaOH → RCOONa + H₂O.
2. Esterification: RCOOH + R'OH ⇌ RCOOR' + H₂O in the presence of conc. H₂SO₄ (or dry HCl) as catalyst. The reaction is reversible; removing water or using excess alcohol pushes it forward. Example: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (ethyl ethanoate).
3. Acid chlorides: RCOOH + PCl₅ → RCOCl + POCl₃ + HCl; RCOOH + SOCl₂ → RCOCl + SO₂ + HCl. Thionyl chloride is preferred because the by-products are gases.
4. Anhydride formation: two molecules of acid lose water on heating with P₂O₅ (or conc. H₂SO₄): 2CH₃COOH → (CH₃CO)₂O + H₂O.
5. Amide formation: RCOOH + NH₃ → RCOONH₄ —(heat)→ RCONH₂ + H₂O.
6. Reduction: LiAlH₄ or diborane reduces RCOOH to the primary alcohol RCH₂OH. NaBH₄ does not reduce carboxylic acids.
7. Decarboxylation: heating the sodium salt with soda lime gives an alkane with one carbon fewer: CH₃COONa + NaOH —(CaO, heat)→ CH₄ + Na₂CO₃.
8. Kolbe electrolysis: electrolysis of an aqueous sodium or potassium salt gives an alkane with twice the carbon atoms at the anode: 2CH₃COO⁻ → CH₃CH₃ + 2CO₂ + 2e⁻.
9. Hell–Volhard–Zelinsky (HVZ) reaction: carboxylic acids having an α-hydrogen react with Cl₂ or Br₂ in the presence of red phosphorus, followed by hydrolysis, to give α-halo acids: RCH₂COOH --(i) X₂/red P, (ii) H₂O-→ RCHXCOOH. Example: CH₃CH₂COOH → CH₃CHBrCOOH (2-bromopropanoic acid).
10. Ring substitution: in benzoic acid the –COOH group is deactivating and meta directing. Nitration gives 3-nitrobenzoic acid. Benzoic acid does not undergo Friedel–Crafts reactions.

Uses of Carboxylic Acids

1. Methanoic acid is used in rubber, textile, dyeing, leather and electroplating industries.
2. Ethanoic acid is used as a solvent and as vinegar in food; it is also used to make dyes, perfumes and plastics.
3. Hexanedioic acid is used to make nylon-6,6, and esters of long-chain fatty acids occur in fats and oils.
4. Benzoic acid and its salts are used as food preservatives.

Solved Examples

Example 1. Write the products of the Cannizzaro reaction of benzaldehyde with concentrated NaOH and explain why ethanal does not give this reaction.
Answer:
2C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH (benzyl alcohol) + C₆H₅COONa (sodium benzoate).
Ethanal has α-hydrogen atoms, which concentrated alkali removes to start aldol condensation. Benzaldehyde has no α-hydrogen, so it can only disproportionate.

Example 2. Which of CH₃CHO, CH₃COCH₂CH₃, C₆H₅CHO and CH₃CH₂CH₂OH gives the iodoform test? Give the reason.
Answer:
The test needs a CH₃CO– group or an alcohol that is oxidised to one, such as CH₃CH(OH)–.
CH₃CHO and CH₃COCH₂CH₃ contain CH₃CO– and give yellow CHI₃.
C₆H₅CHO has no CH₃CO– group, and propan-1-ol is oxidised to propanal, not to a methyl carbonyl, so neither responds.

Example 3. Arrange HCOOH, CH₃COOH, ClCH₂COOH and Cl₂CHCOOH in the increasing order of acid strength.
Answer:
Chlorine is electron withdrawing (–I), so more chlorine atoms stabilise the carboxylate ion more.
CH₃ releases electrons and weakens the acid; HCOOH has no alkyl group.
CH₃COOH < HCOOH < ClCH₂COOH < Cl₂CHCOOH (pKa 4.76, 3.75, 2.86, 1.29).

Example 4. Write the sequence of reactions to convert ethanoic acid into 2-chloroethanoic acid and then show how HVZ reaction is used.
Answer:
CH₃COOH + Cl₂ —(red P)→ ClCH₂COOH + HCl. This is the Hell–Volhard–Zelinsky reaction: red phosphorus forms PCl₃, which makes the acid chloride whose α-hydrogen is replaced by chlorine; water then gives the α-chloro acid.
Product: chloroethanoic acid (ClCH₂COOH, pKa 2.86), which is a stronger acid than ethanoic acid.