Correct Answer: Option B

Explanation

The reagent I\(_2\) / OH\(^-\) (iodine in alkaline solution) is used for the iodoform test, which identifies compounds containing a methyl ketone group or a methyl hydroxy group (specifically a secondary alcohol with the structure CH\(_3\)CH(OH)R). 

Propan-2-ol is a secondary alcohol (CH\(_3\)CH(OH)CH\(_3\)) that possesses the required methyl group adjacent to the hydroxyl-bearing carbon. It reacts with I\(_2\) / OH\(^-\) to produce a characteristic yellow precipitate of iodoform (CHI\(_3\).

Propan-1-ol is a primary alcohol (CH\(_3\)CH\(_2\)CH\(_2\)OH) and does not have this specific structure. Consequently, it gives a negative result (no yellow precipitate)

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