Question

A student is synthesizing 1-bromobutane using the same procedure that you will in the laboratory. If the student started with 0.248 g of 1-butanol and recovers 0.379 g of 1-bromobutane, what is the student's percent yield? (provide your answer to the nearest tenth of a percent)

Answer 1

   3CH3-CH2-CH2-CH2-OH + PBr3 -----------> 3CH3-CH2-CH2-CH2-Br + H3PO3

3 moles of Butanol react with PBr3 to gives 3 moles of 1-bromobutane

3*74g of butanol react with PBr3 to gives 3*137g of 1-bromobutane

0.248g of butanol react with PBr3 to gives = 3*137*0.248/3*74   = 0.459g of 1-bromobutane

tehoritical yield = 0.459g

percentage yield = actual yield*100/theoritical yield

                              = 0.379*100/0.459   = 82.57%