Question

You are preparing triphenyl carbinol by a grginard synthesis. First you prepare methyl benzoate, and set aside 4.40 grams of the purified ester for later use. You then prepare the grignard reagent ( phenylmagnesium bromide ) by reacting 1.78 grams of magnesium with 12.01 ml of bromobenzene. You add the 4.40 grams of methyl benzoate to the freshly prepared grignard reagent to form an addition product. Finally, after hydrolyzing the grignard addition product, you obtain 5.27 grams of the final product, triphenyl carbinol. What is the percent yield of triphenyl carbinol ? ( The density of bromobenzene is 1.495 g/ml ) (molecular weights: bromobenzene = 157, methyl benzoate=136, magnesium =24.3, triphenyl carbinol =260)

Answer 1

Solution :-

Given data

Mass of Mg =1.78 g

Volume of bromobenzene = 12.01 ml (d=1.495 g / ml)

Mass of bromobenzene = 12.01 ml * 1.495 g per ml = 17.95 g

Mass of the methyl benzoate = 4.40 g

Balanced reaction equation for the reaction is shown in the following image

Lets first calculate the moles of the each reactant

Moles of Mg= 1.78 g / 24.3 g per mol = 0.0732 mol Mg

Moles of Bromobenzene = 17.95 g / 157 g per mol = 0.1143 mol

Since mole ratio of the Bromobenzene and Mg is 1 :1 therefore Mg is the limiting reactant because its moles are less than moles of the bromobenzene.

Therefore moles of the phenyl magnesium bromide that can be produced are same as moles of Mg that is moles of the phenyl magnesium bromide = 0.0732 mol

Now lets calculate the moles of the methyl benzoate

Moles of methyl benzoate =4.40 g / 136 g per mol = 0.03235 mol methyl benzoate

Now lets calculate the moles of the phenyl magnesium bromide needed to react with the 4.40 g methyl benzoate.

Mole ratio of the methyl benzoate to phenyl magnesium bromide is 1 :2

(0.03235 mol methyl benzoate * 2 mol phenyl magnesium bromide / 1 mol methyl benzoate) = 0.0647 mol phenyl magnesium bromide

Moles of phenyl magnesium bromide are excess than required for the reaction with the methyl benzoate

Therefore the limiting reactant is the methyl benzoate

Now lets calculate the moles of the triphenyl carbinol using the mole ratio of the methyl benzoate and triphenyl carbinol

(0.03235 mol methyl benzoate * 1 mole triphenyl carbinol / 1 mol methyl benzoate) = 0.03235 mol triphenyl carbinol

Now lets convert moles of triphenyl carbinol to its mass

Mass = moles * molar mass

Mass of triphenyl carbinol = 0.03235 mol * 260 g per mol = 8.411 g

Therefore the theoretical yield of the product is 8.411 g

Now lets calculate the percent yield

Formula to calculate the percent yield is as follows

% yield = (actual yield / theoretical yield )*100 %

Lets put the values in the formula

% yield = (5.27 g / 8.411 g)*100%

             = 62.66 %

Therefore the percent yield of the product triphenyl carbinol = 62.66 %