Question

When heated with phosphoric acid, 2-butanol undergoes dehydration to give a
mixture of the three isomeric alkenes. The alkenes are shown below along with
the area of the peak obtained when the mixture is injected into a gas
chromatograph.
a. If 2.50 mL of 2-butanol is used, calculate the theoretical yield of the
alkene mixture. (answer: 1.53 g)
b. Using the area listed above, if 1.31 g of the alkene mixture is obtained,
calculate the mass of the major isomer obtained and its percent yield.
(answer: 0.848 g, 55.4%)

please shpw the work for how to get the answers

Answer 1

Dehydration of 2-butanol with phosphoric acid is an E1 type of reaction. It gives three isomeric alkene with major product being trans-2-butene. The other two products formed are cis-2-butene and 1-butene is minor product.

a. moles of 2-butanol = 2.50 ml x 0.808 g/ml/74.12 g/mol = 0.0273 mol

Theoretical yield of alkene mixture = 0.0273 mol x 56.106 = 1.53 g

b. Total mass of alkene formed = 1.31 g

area selectivity for 2-butene (major product) = area of 2-butene/total area of all three alkenes

Mass of major isomer product = selectivity fraction as found above x 1.31 g

Percent yield = mass of major product x 100/1.53

Feeding the area values from chromatogram, we can calculated the remaining mas and % yield for major product.