A Levels Chemistry (9701)•9701/11/M/J/20
Explanation
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
Topic: Hydroxy compounds
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