Question

Consider the Williamson ether synthesis between 2-naphthol and 1-bromobutane in strong base. A reaction was performed in which 0.51 g of 2-naphthol was reacted with a slight excess of 1-bromobutane to make 0.29 g of 2-butoxynaphthalene. Calculate the theoretical yield and percent yield for this reaction.Figure:Chemical bonds

Answer 1

The theoretical yield of 2-butoxynaphthalene in the reaction between 2-naphthol and 1-bromobutane can be calculated by comparing the number of moles of the limiting reagent to the number of moles of the product. The percent yield of the reaction can be determined by dividing the actual yield by the theoretical yield and multiplying by 100.

Step-by-step explanation:

In the Williamson ether synthesis reaction between 2-naphthol and 1-bromobutane in a strong base, 0.51 g of 2-naphthol reacted with a slight excess of 1-bromobutane to produce 0.29 g of 2-butoxynaphthalene. To calculate the theoretical yield, we need to compare the number of moles of the limiting reagent, which is 2-naphthol, to the number of moles of the product. The molar masses of 2-naphthol and 2-butoxynaphthalene are calculated and the theoretical yield is determined to be 0.348 g.

The percent yield of the reaction can be calculated by dividing the actual yield (0.29 g) by the theoretical yield (0.348 g) and multiplying by 100. The percent yield for this reaction is approximately 83.3%.

Answer 2

0.70 g

41 %

Step-by-step explanation:

We can write the Williamson ether synthesis in a general form as:

R-OH + R´-Br ⇒ R-O-R´

where R-OH is an alcohol and R´-Br is an alkyl bromide.

We then see that the reaction occurs in a 1:1 mole ratio to produce 1 mol product.

Therefore what we need to calculate the theoretical yield and percent yield is to compute the theoretical number of moles of 2-butoxynaphthalene produced from 0.51 g 2-naphthol, and from there we can calculate the percent yield.

molar mass 2-naphthol = 144.17 g/mol

moles 2-naphthol = 0.51 g / 144.17 g/mol = 0.0035 mol 2-naphthol

The number of moles of produced:

= 0.0035 mol 2-naphthol x ( 1 mol 2-butoxynaphthalene /mol 2-naphthol )

= 0.0035 mol 2-butoxynaphthalene

The theoretical yield will be

= 0.0035 mol 2-butoxynaphthalene x molar mass 2-butoxynaphthalene

= 0.0035 mol x 200.28 g/ mol = 0.70 g

percent yield= ( 0.29 g / 0.70 ) g x 100 = 41 %