Aldehydes, Ketones & Carboxylic Acids

1. The Carbonyl Group

The C=O carbon is $sp^2$ hybridised and trigonal planar. Oxygen is more electronegative, so the carbon carries a partial positive charge and is attacked by nucleophiles. Aldehydes are more reactive than ketones for two reasons: a ketone has two electron-donating alkyl groups that reduce the positive charge on carbon (electronic), and two bulky groups that crowd the approaching nucleophile (steric).

2. Preparation

• Oxidation of alcohols: 1° alcohols give aldehydes with PCC; 2° alcohols give ketones.
• Rosenmund reduction: $RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO$. The catalyst is poisoned so the aldehyde is not reduced further.
• Stephen reaction: $RCN \xrightarrow{SnCl_2/HCl} RCH{=}NH \xrightarrow{H_3O^+} RCHO$.
• Etard reaction: toluene with chromyl chloride ($CrO_2Cl_2$) gives benzaldehyde.
• Friedel–Crafts acylation: benzene + $RCOCl$ with anhydrous $AlCl_3$ gives an aryl ketone.
• Ozonolysis of alkenes.

3. Nucleophilic Addition

• HCN: forms cyanohydrins.
• $NaHSO_3$: forms crystalline bisulphite adducts; used to separate aldehydes and methyl ketones.
• Grignard reagents: HCHO gives 1° alcohols, other aldehydes give 2° alcohols, ketones give 3° alcohols.
• Ammonia derivatives ($Z{-}NH_2$): addition–elimination gives oximes (with $NH_2OH$), hydrazones, 2,4-DNP derivatives and semicarbazones.

4. Reduction and Oxidation

• Clemmensen reduction ($Zn{-}Hg$/conc. HCl) and Wolff–Kishner reduction ($NH_2NH_2$, then KOH/ethylene glycol) convert C=O to $CH_2$. Choose Wolff–Kishner for acid-sensitive substrates.
• Tollens' reagent (silver mirror) and Fehling's solution (red $Cu_2O$) are reduced by aldehydes but not ketones. Benzaldehyde does not reduce Fehling's solution.
• Iodoform test: methyl ketones ($CH_3CO{-}$) and alcohols with $CH_3CH(OH){-}$ give yellow $CHI_3$ with $I_2$/NaOH.

5. Reactions Involving α-Hydrogen

Hydrogens on the carbon next to C=O are acidic because the resulting carbanion (enolate) is resonance-stabilised.

• Aldol condensation: two molecules with α-H, in dilute NaOH, give a β-hydroxy aldehyde or ketone, which dehydrates on heating to an α,β-unsaturated carbonyl compound.
• Cannizzaro reaction: aldehydes without α-H (HCHO, $C_6H_5CHO$) in concentrated NaOH disproportionate: one molecule is oxidised to the carboxylate, the other reduced to the alcohol. In a crossed Cannizzaro with HCHO, formaldehyde is always the one oxidised.

6. Carboxylic Acids

Acidity comes from resonance stabilisation of the carboxylate ion, where the negative charge is shared equally by two oxygens.

• Electron-withdrawing groups increase acidity: $CF_3COOH > CCl_3COOH > CHCl_2COOH > ClCH_2COOH > CH_3COOH$.
• Effect falls with distance: 2-chlorobutanoic acid is stronger than 4-chlorobutanoic acid.
• HCOOH is stronger than $CH_3COOH$ (no electron-donating alkyl group).
• HVZ reaction: $X_2$/red P introduces a halogen at the α-carbon.
• Decarboxylation: sodium salt + soda lime (NaOH + CaO) gives an alkane with one fewer carbon.
• Reduction with $LiAlH_4$ gives 1° alcohols; $NaBH_4$ does not reduce acids.

Practice questions

1. Which of these undergoes the Cannizzaro reaction?

   1. CH₃CHO
   2. C₆H₅CHO
   3. CH₃COCH₃
   4. CH₃CH₂CHO
   Answer

   B. Benzaldehyde has no α-hydrogen, so in concentrated alkali it disproportionates instead of undergoing aldol condensation.

2. Which reagent converts a ketone into a hydrocarbon under strongly basic conditions?

   1. Zn-Hg/HCl
   2. NH₂NH₂, KOH/ethylene glycol
   3. LiAlH₄
   4. NaBH₄
   Answer

   B. This is the Wolff–Kishner reduction. Clemmensen does the same under acidic conditions; LiAlH₄ and NaBH₄ stop at the alcohol.

3. Which compound gives a positive iodoform test?

   1. Pentan-3-one
   2. Propan-2-ol
   3. Methanol
   4. Benzaldehyde
   Answer

   B. Propan-2-ol, CH₃CH(OH)CH₃, is oxidised to acetone, a methyl ketone, which gives CHI₃.

4. The strongest acid among the following is:

   1. CH₃COOH
   2. ClCH₂COOH
   3. Cl₂CHCOOH
   4. Cl₃CCOOH
   Answer

   D. Three electron-withdrawing Cl atoms stabilise the carboxylate most.

5. Rosenmund reduction converts:

   1. Acid chlorides to aldehydes
   2. Nitriles to amines
   3. Ketones to alcohols
   4. Esters to acids
   Answer

   A. RCOCl + H₂ over Pd/BaSO₄ (poisoned) gives RCHO without further reduction.

6. Aldehydes are generally more reactive than ketones towards nucleophilic addition because:

   1. Aldehydes are more polar
   2. Ketones have more α-hydrogens
   3. Ketones have greater steric hindrance and +I effect
   4. Aldehydes have a lower molar mass
   Answer

   C. Two alkyl groups in ketones donate electrons (reducing the δ+ on carbon) and crowd the nucleophile.