Question

A reaction was performed in which 0.55 g of 2‑naphthol was reacted with a slight excess of allyl bromide to make 0.59 g of allyl 2‑naphthyl ether. Calculate the theoretical yield and percent yield for this reaction. Allyl bromide has a density of 1.40 g/mL .

Answer 1

Answer:

• Theoretical yield = 0.7028 g

• Percent yield = 83.95%

Explanation:

The reaction that takes places is:

• C₁₀H₈O + C₃H₅Br → C₁₃H₁₂O + HBr

The theoretical yield is calculated when assuming that all of the limiting reactant transformed into the product:

• 0.55 g naphthol ÷ 144gNaphthol/mol * $\frac{1molAllyl-Naphthyl Ether}{1molNaphthol}$ * 184 gEther/mol = 0.7028 g

The percent yield is calculated using the theoretical yield and the experimental yield:

• 0.59g / 0.7028 g * 100% = 83.95%