Iodoform test distinguishes alcohols oxidizable to acetaldehyde or methyl ketones

Steps:

• Alkaline aqueous iodine acts as hypoiodite oxidant for the iodoform reaction, yielding yellow CHI3 precipitate if the product is CH3CHO or CH3COR.
• Positive for ethanol (oxidizes to CH3CHO) and secondary alcohols with CH3CH(OH)R structure (oxidize to CH3COR).
• Negative for primary alcohols like propan-1-ol (oxidizes to CH3CH2CHO) and tertiary alcohols (no oxidation).
• Select the pair showing different test results (one positive, one negative) to allow distinction.

Why B is correct:

• Propan-1-ol oxidizes to propanal (no CH3C=O group, negative test); propan-2-ol oxidizes to acetone (CH3COCH3, positive test per haloform reaction on methyl ketones).

Why the others are wrong:

• A: Both ethane-1,2-diol (oxidizes to acetaldehyde equivalent) and ethanol give positive test.
• C: Both ethanol and butan-2-ol (oxidizes to CH3COCH2CH3) give positive test.
• D: Both propan-2-ol and methylpropan-2-ol (both oxidize to methyl ketones) give positive test.

Final answer: B