Correct answer: I2/NaOH
The fragment CH3-CH(OH)- is a methyl carbinol that gives a positive iodoform (haloform) reaction. I2/NaOH cleaves it to a carboxylate (losing CHI3), converting CH3-CH(OH)- into -COOH while leaving the C=C double bond intact. Alkaline KMnO4 would instead cleave the double bond.