Question

MiltIn Point: x 237 Theonital yield Percant yeld Stllbene: .c069 PYPdanminiromtde2.067 ZncHe acd 20ml Pndviei preductT erp): 2.746 Prodvct 2. 5719 Date Witness/TA Note: Insert Divider Under Copy Sheet Before Writing

Answer 1

The reaction is

4 Br H tri bromide + Bra Pyridinium + Brz - 1,2-dibrome 1,2-diphery lettore Shil bene

mass of Stilbene = 1.006 g (given)

molar mass of stilbene = 180.24 g

number of moles of stilbene = 5.58 x 10^-3 moles

Mass of Pyridinium tribromide = 2.067 g (given)

Molar mass of Pyridinium tribromide = 319.82

number of moles of Pyridinium tribromide = 6.46 x 10^-3 moles

This indicates stilbene is the limiting reagent (since one mole of pyridinium tribromide gives one mole of bromine which reacts with 1 mole of stilbene to give 1 mole of 1,2-dibromo-1,2-diphenylethane)

If the yield is 100%, 5.58 x 10^-3 moles of stilbene will give 5.58 x 10^-3 moles of 1,2-dibromo-1,2-diphenylethane

Molar mass of 1,2-dibromo-1,2-diphenylethane = 340.05 g

=> mass of 5.58 x 10^-3 moles of 1,2-dibromo-1,2-diphenylethane = 1.90 g

Theoretical Yield = 1.90 g

Since the mass of the product is 2.571 g

The yield is more than 100% (indicates presence of impurities)

Yield = 2.571*100/1.90

% Yield = 135.31%