Positive iodoform test $\frac{yellow CHI3 precipitate with I2}{NaOH}$:

1. All methyl ketones: RCOCH3 (acetone, acetophenone, etc.)
2. Acetaldehyde: CH3CHO
3. Ethanol: CH3CH2OH (oxidized to CH3CHO → then iodoform)
4. Secondary alcohols with methyl: RCH(OH)CH3 (oxidized to RCOCH3)

Negative iodoform:

• Diethyl ketone (CH3CH2COCH2CH3) — no methyl on C=O
• Benzaldehyde (PhCHO) — no methyl on C=O
• 3-Pentanone — no methyl on C=O
• Methanol — oxidized to HCHO, not to a methyl ketone

Mechanism: Base abstracts alpha-H → halogenation of all 3 alpha-H's → CX3 group cleaved by OH- → CHX3 + RCOO-