Question

After distilling your crude methyl benzoate, you set aside 4.93 grams of the purified ester. You then prepare the grignard reagent ( phenylmagnesium bromide ) by reacting 2.1 grams of magnesium with 14.12775 ml of bromobenzene. You add the 4.93 grams of methyl benzoate to the freshly prepared grignard reagent to form an addition product. Finally, after hydrolyzing the grignard addition product, you obtain 6.9 grams of the final product, triphenyl carbinol. What is the percent yield of triphenyl carbinol ? ( The density of bromobenzene is 1.495 g/ml )

Answer 1

Sol .

Firstly , formation of Grignard reagent takes place .

Reaction :

Magnesium + Bromobenzene ----> Phenylmagnesium bromide

As Mass of Magnesium = 2.1 g

Molar Mass of Magnesium = 24.305 g/mol

So , Moles of Magnesium = 2.1 / 24.305 = 0.0864  mol

Now , Volume of bromobenzene = 14.12775 ml

Density of bromobenzene = 1.495 g/ml

So , Mass of bromobenzene = 1.495 × 14.12775

= 21.1209 g

Also , Molar Mass of bromobenzene = 157.02 g/mol

So , Moles of bromobenzene = 21.1209 / 157.02 = 0.1345 mol

As from reaction , 1 mole of Magnesium combines with 1 mole of bromobenzene .

So , 0.0864 moles of Magnesium combines with 0.0864 moles of bromobenzene . But we have 0.1345 moles of bromobenzene .

So , Magnesium is the limiting reactant .

Now , 1 mole of Magnesium gives 1 mole of phenyl magnesium bromide .

Therefore , 0.0864 moles of Magnesium gives 0.0864 moles of phenylmagnesium bromide .

Now , Reaction of grignard reagent with ester :

2 Phenylmagnesium bromide + Methyl benzoate ----> Triphenyl carbinol

As Mass of Methyl benzoate = 4.93 g

Molar mass of Methyl benzoate = 136.15 g/mol

So , Moles of Methyl benzoate = 4.93 / 136.15

= 0.0362 mol

Now , from reaction , 2 moles of phenylmagnesium bromide combines with 1 mole of methyl benzoate .

So , 0.0864 moles of phenylmagnesium bromide combines with = 0.0864 / 2 = 0.0432 moles of methyl benzoate .

But we have 0.0362 moles of methyl benzoate .

So , methyl benzoate is the limiting reactant .

Now , 1 mole of methyl benzoate gives 1 mole of triphenyl carbinol

So , 0.0362 moles of methyl benzoate gives 0.0362 moles of triphenyl carbinol .

As Molar Mass of triphenyl carbinol = 260.33 g/mol

So , Mass of triphenyl carbinol = 0.0362 × 260.33

= 9.4239 g

Therefore , Theoretical yield = 9.4239 g

Now , Experimental yield = 6.9 g

So , Percentage yield = ( Experimental yield / Theoretical yield ) × 100

= ( 6.9 / 9.4239 ) × 100

=  73.22 %