Question

If you start with 75 mg of anisole and excess bromine, and the bromination occurs 3 times, then what is your theoretical yield (in mg) of tribromoanisole?

Answer 1

Answer:

239.6 mg is the  theoretical yield of tribromoanisole.

Explanation:

$C_6H_5-OCH_3+3Br_2\rightarrow C_6H_2Br_3-OCH_3+3HBr$

Mass of anisole $C_6H_5-OCH_3$ = 75 mg = 0.075 g

1 mg = 0.001 g

Moles of anisole = $\frac{0.075 g}{108 g/mol}=0.0006944 mol$

According to reaction, 1 mole anisole gives 1 mole of tribromoanisole.

Then 0.0006944 mole of anisole will give:

$\frac{1}{1}\times 0.0006944 mol=0.0006944 mol$ of tribromoanisole

Mass of 0.0006944 mole of tribromoanisole:

0.0006944 mol × 345 g/mol = 0.2396 g = 239.6 mg

239.6 mg is the  theoretical yield of tribromoanisole.