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 %

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the answer is option a.