Haloform reaction on secondary alcohol

Steps:

• Identify CH3CH(OH)CH3 as propan-2-ol (likely butan-2-ol intended for product match).
• Oxidize secondary alcohol to methyl ketone (e.g., CH3COCH2CH3).
• Apply haloform reaction: treat methyl ketone with excess halogen (I2/Br2) and NaOH.
• Cleave to haloform and carboxylic acid salt: RCOCH3 → RCOONa + CHI3.

Why C is correct:

• Excess NaOH with halogen (e.g., I2) enables haloform oxidation of methyl ketone to RCOONa, per the reaction CH3COR + 3X2 + 4NaOH → RCOONa + CHX3 + 3NaX + 3H2O.

Why the others are wrong:

• A: Na2CO3 is a weak base, insufficient for oxidation or haloform cleavage.
• B: NaOH alone deprotonates but does not oxidize alcohol to ketone or perform haloform.
• D: Not enough information (option incomplete).

Final answer: C