Question

Calculate the theoretical yield for the bromination of bothstilbenes and cinnamic acid, assuming the presence of excesspyridinium tribromide.

cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg

Answer 1

Molar mass of pyridinium trichloride=319.82 g/mol

Molar mass of cis and trans –stilbene=180.25 g/mol

Molar mass of cinnamic acid=148.16 g/mol

Now , as dibromide is formed (cis-stilbene dibromide,trans-stilbene dibromide,cinnamic acid dibromide)

1) So 1 mole of cis-stilbene reacts with 1 mole of pyridinium trichloride to give 1 mole of cis-stilbene dibromide(as per the reaction)

Given mass of cis-stilbene=density*volume=1.01 g/ml*100*10^-6 L=1.01 g/ml*100*10^- 3 ml=0.101 g

moles of cis-stilbene=mass/molar mass=0.101 g/180.25 g/mol=5.60*10^-4 moles=moles of cis-stilbene dibromide formed

molar mass of (cis-stilbene dibromide=180.25+2*80=340.25 g/mol

so mass of product with 0.101 g(5.60*10^-4 moles) of reactant cis-stilbene=5.60*10^-4 moles*340.25g/mol=0.19 g

theoretical yield=0.19 g

2) moles of trans-stilbene reacted=moles of trans-stilbene dibromide formed=100*10^-3 g/180.25 g/mol=5.56*10^-4 moles

Mass of product =moles of trans-stilbene dibromide*molar mass=5.56*10^-4 moles *340.25 g/mol=0.19 g(theoretical yield)

3) moles of cinnamic acid reacted=moles of cinnamic acid dibromide formed=150*10^-3 g/148.16 g/mol=0.00101 moles

Mass of cinnamic acid dibromide formed=0.00101 moles*molar mass of cinnamic acid dibromide formed=0.00101 moles*308.16 g/mol=0.31 g(theoretical yield)