Alcohol undergoes oxidation to methyl ketone for iodoform test

Steps:

• Decolourisation of warm acidified KMnO4 indicates primary or secondary alcohol (oxidisable).
• Yellow precipitate with alkaline I2 confirms iodoform test, positive for secondary alcohols oxidising to methyl ketones (R-CH(OH)-CH3).
• Molecular formula C5H12O fits all options; eliminate primaries and non-methyl ketone secondaries.
• Identify structure yielding CH3-C(O)-R upon oxidation.

Why B is correct:

• 3-Methylbutan-2-ol (CH3CH(OH)CH(CH3)2) oxidises to 3-methylbutan-2-one (CH3C(O)CH(CH3)2), a methyl ketone giving iodoform per the test's requirement for CH3-C(O)- group.

Why the others are wrong:

• A: Primary alcohol; oxidises to aldehyde (3-methylbutanal), not methyl ketone.
• C: Primary alcohol; oxidises to pentanal, not methyl ketone.
• D: Secondary alcohol; oxidises to pentan-3-one (CH3CH2C(O)CH2CH3), not methyl ketone.

Final answer: B