Butan-2-ol, 14.8 g, is warmed with excess $I X_{2}$ in aqueous $NaOH$ and converted to iodoform in 80% overall yield. Take molar masses: butan-2-ol 74 g/mol, I 127, C 12, H 1. Calculate the mass of $CHI X_{3}$ obtained in grams, correct to two decimal places.

Numerical answer type, answer in g, tolerance 0.05.

Answer: 63.04 g

Butan-2-ol is first oxidised to butan-2-one and the methyl ketone then gives one

CHI X_{3}

per molecule. Moles of alcohol

= 14.8/74 = 0.200

M (CHI X_{3}) = 12 + 1 + 3 (127) = 394

. Mass

= 0.200 \times 394 \times 0.80 = 63.04