Iodoform test detects alcohols oxidizable to methyl ketones or acetaldehyde Steps:

• List C4H10O alcohol isomers: butan-1-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol (achiral primaries/tertiary); butan-2-ol (chiral secondary with 2 enantiomers).
• Recall iodoform positive for secondary alcohols CH3CH(OH)R (oxidizes to CH3COR) or primary ethanol (to CH3CHO).
• Check each: butan-1-ol → butanal (no); 2-methylpropan-1-ol → isobutyraldehyde (no); 2-methylpropan-2-ol (tertiary, no oxidation).
• Butan-2-ol (both enantiomers) → butan-2-one (CH3COCH2CH3, methyl ketone, yes). Why A is correct:
• Butan-2-ol's two stereoisomers both yield iodoform via oxidation to methyl ketone (haloform reaction on CH3C=O group). Why the others are wrong:
• B: Overcounts by ignoring stereoisomerism; only one structural but two total.
• C: Includes non-reactive isomers like primaries/tertiary that don't form methyl carbonyls.
• D: Counts all 5 isomers total, but only 2 react positively.

Final answer: A