Answer:
0.2889 g brominated product
64.6 %
Step-by-step explanation:
This is a bromination chemical reaction of an alkene and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.
The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.
trans-cinnamic acid + pyridinium tribromide ⇒ 2,3-dibromo-3-
phenylpropanoic acid
Molar weight trans-cinnamic acid = 148.16 g/mol
mass trans-cinnamic acid = 139.0 mg x 1g/1000 mg = 0.139 g
# mol trans-cinnamic acid = 0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol
Since our reaction is 1 mol trans-cinnamic acid produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:
1 mol 2,3-dibromo-3-phenylpropanoic acid / trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid
= 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid
In grams the the theoretical yield is:
molar mass 2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol
The theoretical mass 2,3-dibromo-3-phenylpropanoic acid:
= 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol
= 0.2889 g
% yield = mass experimental/mass theoretical
= 0.1866 g / 0.2889 g x 100 = 64.6 %
