A mixture of 5.8 g propanone, 3.6 g butan-2-one, 3.0 g propan-2-ol and 7.2 g butanal is treated with excess I2/NaOH. Every compound that can give the…

The iodoform test is positive for compounds containing the $\mathrm{CH_3CO\!-\!}$ group and for alcohols having $\mathrm{CH_3CH(OH)\!-\!}$, which are first oxidised to the methyl ketone by hypoiodite. Propanone ($\mathrm{CH_3COCH_3}$), butan-2-one ($\mathrm{CH_3COC_2H_5}$) and propan-2-ol ($\mathrm{CH_3CH(OH)CH_3}$) qualify. Butanal, $\mathrm{CH_3CH_2CH_2CHO}$, has no $\mathrm{CH_3CO}$ unit and no $\mathrm{CH_3CH(OH)}$ unit, so it does not. Moles: propanone $5.8/58=0.100$; butan-2-one $3.6/72=0.050$; propan-2-ol $3.0/60=0.050$. Each gives one mole of $\mathrm{CHI_3}$, so the total is $0.200\ \mathrm{mol}$. Mass at full conversion $=0.200\times394=78.8\ \mathrm{g}$; at $80\%$ yield $=0.8\times78.8=63.04\ \mathrm{g}$. The mass of iodoform therefore comes only from three of the four compounds; including butanal wrongly would raise the total to $0.272\ \mathrm{mol}$ of product and give a wrong answer.