Question

After distilling your crude methyl benzoate, you set aside 5.12 grams of the purified ester. You then prepare the grignard reagent ( phenylmagnesium bromide ) by reacting 2.3 grams of magnesium with 9.45 ml of bromobenzene. You add the 5.12 grams of methyl benzoate to the freshly prepared grignard reagent to form an addition product. Finally, after hydrolyzing the grignard addition product, you obtain 5.3 grams of the final product, triphenyl carbinol. What is the percent yield of triphenyl carbinol ? ( The density of bromobenzene is 1.495 g/ml, triphenyl carbinol = 260.33 g/mol, bromobenzene = 157.01 g/mol, Mg = 24.3 g/mol )

Answer 1

Percent yield is 54%

Step-by-step explanation:

The Grignard reaction is:

Mg + bromobenzene → phenylmagnesium bromide

Moles of Mg are:

2,3g×(1mol/24.3g) = 0.095 moles Mg.

Moles of bromobenzene are:

9,45mL×(1.495g/1mL)×(1mol/157.01g)= 0.0900 of bromobenzene.

The limiting reactant is bromobenzene. That means moles of phenylmagnesium bromide are 0.0900 moles.

To produce triphenyl carbinol you require:

2 phenylmagnesium bromide + methyl benzoate → triphenyl carbinol

Moles of methyl benzoate are:

5.12g×(1mol/136,15g) = 0.0376 moles of methyl benzoate.

The complete reaction of 0.0376 moles of methyl benzoate requires:

0.0376 moles of methyl benzoate×
`(2mol PhenylmagnesiumBromide)/(1molMethylBenzoate)` = 0.0752 moles of phenylmagnesium bromide. As you have 0.900 moles of phenylmagnesium bromide, limiting reagent is Methyl benzoate and moles of triphenyl carbinol are 0.0376. In grams:

0.0376 moles of triphenyl carbinol×(260.33g/mol) = 9,79g of triphenyl carbinol -This is the teorethical yield-

Percent yield is the ratio of actual yield to the theoretical yield. That is:

5,3g / 9,79g ×100 = 54%

I hope it helps!