Consider the nucleophilic addition reaction of 2-butanone with excess propyl magnesium bromide, prepared in situ by reacting 1-bromopropane with metallic magnesium, to synthesize 3-methyl-3-hexanol. - 2-Butanone: density = 0.81 g/mL - 1-Bromopropane: density = 1.35 g/mL - 3-Methyl-3-hexanol: density = 0.82 g/mL In a performed reaction, 0.40 mL of 2-butanone was reacted with an excess of propyl magnesium bromide to produce 0.38 g of 3-methyl-3-hexanol. To calculate the theoretical yield and percent yield for this reaction, proceed with the following steps: 1. Calculate the moles of 2-butanone: Density of 2-butanone = 0.81 g/mL Volume of 2-butanone = 0.40 mL Mass = Volume × Density = 0.40 mL × 0.81 g/mL = 0.324 g Molar mass of 2-butanone (C4H8O) = (4×12.01) + (8×1.008) + 16 = 72.11 g/mol Moles of 2-butanone = Mass ÷ Molar mass = 0.324 g ÷ 72.11 g/mol = 0.00449 mol 2. Determine the theoretical yield: Assuming the reaction proceeds with a 1:1 mole ratio, the theoretical yield of 3-methyl-3-hexanol will be the same as the moles of 2-butanone: Theoretical yield (in moles) = Moles of 2-butanone = 0.00449 mol To find the theoretical yield in grams: Molar mass of 3-methyl-3-hexanol (C7H16O) = (7×12.01) + (16×1.008) + 16 = 116.21 g/mol Theoretical yield (in grams) = Moles × Molar mass = 0.00449 mol × 116.21 g/mol = 0.522 g 3. Calculate the percent yield: Percent yield = (Actual yield ÷ Theoretical yield) × 100% Percent yield = (0.38 g ÷ 0.522 g) × 100% ≈ 72.80% Therefore, the theoretical yield is 0.522 g and the percent yield is approximately 72.80%.