Answer:
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 %
