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

0.70 g

41 %

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 %