Question

4.2 g of 1,4-di-t-butyl-2,5-dimethoxybenzene (250.37 g/mol) were synthesized by reacting 10.4 mL of t-butyl alcohol (MW 74.12 g/mol, D 0.79 g/mL), 25 mL of concentrated sulfuric acid (MW 98.08 g/mol, D 1.84 g/mL), and 5.6 g of 1,4-dimethoxybenzene (MW 138.17 g/mol) together. Calculate the percent yield of this reaction.

Answer 1

Answer:

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene is 41.40%.

Explanation:

Here, in the reaction sulfuric acid is playing the role of catalyst by donating its proton in initial stage of the reaction and in the end of the reaction the proton is returned back to sulfuric acid.

Mass = Density × Volume

Mass of t-butyl alcohol = $0.79 g/mL\times 10.4 mL=8.219 g$

Moles of t-butyl alcohol  =$\frac{8.219 g}{74.12 g/mol}=0.11084 mol$

Moles of 1,4-dimethoxybenzene = $\frac{5.6 g}{138.17 g/mol}=0.04052 mol$

According to reaction 2 mol of  t-butyl alcohol reacts  with 1 mol of 1,4-dimethoxybenzene.

Then 0.11084 moles of t-butyl alcohol will react with :

$\frac{1}{2}\times 0.11084 mole=0.05542 mol$ of 1,4-dimethoxybenzene.

This means that moles of 1,4-dimethoxybenzene are limited and moles of t-butyl alcohol are in excess.So, the moles of product will depend upon the moles of 1,4-dimethoxybenzene.

According top reaction 1 mol of 1,4-dimethoxybenzene gives 1 mol of  1,4-di-t-butyl-2,5-dimethoxybenzene.

Then 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene will give:

$\frac{1}{1}\times 0.04052 mol= 0.04052 mol$ of 1,4-di-t-butyl-2,5-dimethoxybenzene.

Mass of 0.04052 moles of 1,4-di-t-butyl-2,5-dimethoxybenzene:

0.04052 mol × 250.37 g/mol = 10.144 g

Percentage yield:

$\frac{Experimental}{Theoretical}\times 100$

Percentage yield of 1,4-di-t-butyl-2,5-dimethoxybenzene:

Experimental yield = 4.2 g

Theoretical yield = 10.144 g

$\frac{4.2 g}{10.144 g}\times 100=41.40\%$