WAEC Chemistry 2019 Paper

Concept:Propan-1-ol is a primary alcohol, while propan-2-ol is a secondary alcohol.The iodoform test distinguishes alcohols having the $\mathrm{CH_3CH(OH)-}$ group from those that do not.Explanation:Propan-2-ol contains the methyl carbinol group $\mathrm{CH_3CH(OH)CH_3}$.When treated with iodine in the presence of alkali ($\mathrm{I_2 / OH^-}$), it is oxidized to propanone, which gives a yellow precipitate of iodoform ($\mathrm{CHI_3}$).Propan-1-ol does not have the required $\mathrm{CH_3CH(OH)-}$ group, so it gives no iodoform precipitate.Acidified potassium dichromate ($\mathrm{H^+ / K_2Cr_2O_7}$) oxidizes both alcohols, so it cannot distinguish them clearly.Acetic acid and sodium chloride do not produce a characteristic observable difference.Answer:The correct reagent is $\mathrm{I_2 / OH^-}$, i.e., Option B.