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%

Answer 2

The theoretical yield and percent yield of 2‑naphthyl ether (C₁₃H₁₂O) obtained from the reaction between 0.55 g of 2‑naphthol (C₁₀H₈O) and excess allyl bromide (C₃H₅Br) are:

1. The theoretical yield of C₁₃H₁₂O is 0.703 g.

2. The percentage yield of C₁₃H₁₂O is 83.9%

We'll begin by writing the balanced equation for the reaction. This is given below:

C₁₀H₈O + C₃H₅Br → C₁₃H₁₂O + HBr

Molar mass of C₁₀H₈O = (12×10) + (1×8) + 16

= 144 g/mol

Mass of C₁₀H₈O from the balanced equation = 1 × 144 = 144 g

Molar mass of C₁₃H₁₂O = (12×13) + (1×12) + 16

= 184 g/mol

Mass of C₁₃H₁₂O from the balanced = 1 × 184 = 184 g

SUMMARY:

From the balanced equation above,

144 g of C₁₀H₈O reacted to produce 184 g of C₁₃H₁₂O

1. Determination of the theoretical yield of C₁₃H₁₂O.

From the balanced equation above,

144 g of C₁₀H₈O reacted to produce 184 g of C₁₃H₁₂O

Therefore,

0.55 g of C₁₀H₈O will react to produce = $\frac{0.55 * 184}{144}$ = 0.703 g of C₁₃H₁₂O

Thus, the theoretical yield of C₁₃H₁₂O is 0.703 g

2. Determination of the percentage yield of C₁₃H₁₂O.

Actual yield of C₁₃H₁₂O = 0.59 g

Theoretical yield C₁₃H₁₂O = 0.703 g

Percentage yield =?

$Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{0.59}{0.703} * 100$

= 83.9%

Thus, the percentage yield of C₁₃H₁₂O is 83.9%

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