Aldehydes & Ketones

The carbonyl group (C=O) is the most versatile functional group in organic chemistry. The carbon is sp2-hybridized, planar, and strongly electrophilic (C^$\delta$+=O^$\delta$-) due to the high electronegativity of oxygen. The nucleophilic addition mechanism proceeds: nucleophile (Nu-) attacks the electrophilic carbon → tetrahedral alkoxide intermediate → protonation to give the addition product. Reactivity decreases in the order: HCHO (C=O) > other aldehydes > ketones, because increasing substitution provides both steric hindrance and electron donation (+I effect of alkyl groups reduces the electrophilicity of C).

Reactions with NH₃ Derivatives are condensation reactions involving loss of H₂O: hydroxylamine (NH₂OH) gives oximes (C=NOH); phenylhydrazine (C₆H₅NHNH₂) gives phenylhydrazones (C=NNHC₆H₅); 2,4-DNP (2,4-dinitrophenylhydrazine) gives an orange/yellow precipitate (used to DETECT the carbonyl group); semicarbazide (H₂NCONHNH₂) gives semicarbazones (C=NNHCONH₂).

Distinction Tests: Tollens' test (ammoniacal AgNO₃) gives a silver mirror with aldehydes: RCHO + 2[Ag(NH₃)₂]+ + 2OH- → RCOO- + 2Ag (silver mirror) + 4NH₃ + H₂O. Ketones do NOT react. Fehling's test (alkaline Cu₂+ tartrate) gives a red precipitate of Cu₂O with aldehydes; ketones do NOT react. These tests distinguish aldehydes from ketones.

Reduction Methods: NaBH₄ and LiAlH₄ reduce C=O to C-OH (alcohol). Clemmensen reduction (Zn-Hg/conc. HCl — acidic conditions) reduces C=O completely to CH₂. Wolff-Kishner reduction (NH₂NH₂/KOH, ethylene glycol, heat — basic conditions) also reduces C=O to CH₂. The critical distinction: NaBH₄/LiAlH₄ give alcohol (C-OH retained); Clemmensen and Wolff-Kishner give the hydrocarbon (oxygen completely removed).

Aldol Condensation: Two molecules of acetaldehyde (CC=O) react with dilute NaOH to form $\beta$-hydroxy aldehyde (3-hydroxybutanal, the "aldol"): CH₃CHO + CH₃CHO → CH₃CH(OH)CH₂CHO.
On heating, this dehydrates to $\alpha$,$\beta$-unsaturated aldehyde (2-butenal, CC=CC=O).
Requirement: the aldehyde or ketone MUST have an $\alpha$-hydrogen (H on the carbon adjacent to C=O).

Cannizzaro Reaction: Aldehydes WITHOUT $\alpha$-hydrogen undergo disproportionation with concentrated NaOH: 2HCHO → HCOO-Na+ + CH₃OH (one molecule is oxidized to the acid salt, the other is reduced to alcohol). Only HCHO, C₆H₅CHO (), and (CH₃)₃CCHO undergo Cannizzaro. If $\alpha$-H is present, aldol condensation occurs instead.

Haloform Reaction (Iodoform Test): Methyl ketones (CH₃CO-R) + 3I₂ + 3NaOH → RCOONa + CHI₃ (iodoform, yellow precipitate). The test is also positive for compounds oxidizable to methyl ketones: ethanol (CCO) is oxidized to acetaldehyde then to CH₃CO-, and isopropanol (CC(O)C) is oxidized to acetone then gives iodoform.

The key testable concept is that Cannizzaro reaction requires NO $\alpha$-hydrogen (if $\alpha$-H is present, aldol occurs instead), and that the iodoform test works for ethanol and isopropanol because they are oxidized in situ to methyl carbonyl compounds.