Iodoform test identifies alcohols oxidizable to methyl ketones

Steps:

• Recognize alkaline I2(aq) as iodoform reagent, producing pale yellow CHI3 precipitate with methyl ketones or alcohols convertible to them.
• Recall iodoform-positive alcohols have CH3CH(OH)R structure (R = H, alkyl, aryl), oxidizing to CH3COR.
• Examine each option's structure for CH3CH(OH)- group.
• Identify pentan-2-ol as matching, while others do not.

Why B is correct:

• Pentan-2-ol (CH3CH(OH)CH2CH2CH3) oxidizes to pentan-2-one (CH3COCH2CH2CH3), a methyl ketone that reacts with I2/OH- to form CHI3 per iodoform mechanism.

Why the others are wrong:

• A: Pentan-1-ol (CH3CH2CH2CH2CH2OH) oxidizes to pentanal/acid, neither methyl carbonyl.
• C: Pentan-3-ol (CH3CH2CH(OH)CH2CH3) oxidizes to pentan-3-one (CH3CH2COCH2CH3), a symmetrical ketone without CH3CO-.
• D: 2-Methylpentan-2-ol ((CH3)2C(OH)CH2CH2CH3) is tertiary, cannot oxidize to ketone.

Final answer: B