$word_{count}$: 240

The alpha-H (adjacent to C=O, pKa ~20 for ketones) is acidic due to enolate anion resonance stabilization. The presence or absence of alpha-H determines which reaction pathway a carbonyl follows. Decision tree: Does the aldehyde have alpha-H? YES → Aldol condensation $\frac{dilute NaOH}{base}$; NO → Cannizzaro reaction (conc. NaOH). Aldol: enolate of one molecule attacks C=O of another → beta-hydroxy carbonyl → dehydration → alpha,beta-unsaturated carbonyl. Cannizzaro: hydride transfer between two molecules → one oxidized to carboxylate, one reduced to alcohol. For crossed versions: crossed aldol works best when one partner has no alpha-H (acts as electrophile only). Crossed Cannizzaro with HCHO: formaldehyde always gets oxidized (best hydride donor). The haloform reaction is a special alpha-H reaction: methyl ketones (RCOCH3) + X2/NaOH → trihalomethane + carboxylate. Iodoform test (I2/NaOH → yellow CHI3) detects: all RCOCH3, CH3CHO, ethanol (oxidized to CH3CHO), and RCHOHCH3 (oxidized to RCOCH3). Other alpha-H reactions: Claisen condensation (ester self-condensation → beta-keto ester), Michael addition (conjugate 1,4-addition), malonic/acetoacetic ester syntheses $\frac{for building acids}{ketones}$, and alpha-halogenation.