Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic acid. After the reaction and workup, the student ended up with 0.1866 g of brominated product. Calculate the student\'s theoretical and percent yields.

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 %

mx+89b(-98)nb

step-by-step explanation: