Iodoform test for secondary alcohols with CH3CH(OH)R structure

Steps:

• Recognize yellow precipitate with I2/NaOH as iodoform test, positive for alcohols oxidizing to CH3COR.
• Recall secondary alcohols CH3CH(OH)R oxidize to methyl ketones, giving iodoform; primaries and tertiaries do not unless specific cases.
• Analyze structures: A is primary (no methyl carbinol), B is tertiary (no oxidation), C is secondary CH3CH(OH)CH(CH3)2, D is secondary CH3CH2CH(OH)CH2CH3 (no methyl carbinol).
• Confirm C oxidizes to CH3C(O)CH(CH3)2, a methyl ketone.

Why C is correct:

• 3-Methylbutan-2-ol is CH3CH(OH)CH(CH3)2, a secondary alcohol with CH3CH(OH)- group, oxidizes to methyl ketone per iodoform criteria.

Why the others are wrong:

• A: Primary alcohol, does not oxidize to methyl ketone.
• B: Tertiary alcohol, resists oxidation entirely.
• D: Secondary but symmetric (Et2CHOH), oxidizes to non-methyl ketone (pentan-3-one).

Final answer: C