Add 0.30 g of trans-stilbene to a 25 mL round-bottom flask. Add 3.5 mL of methyl

Question

Add 0.30 g of trans-stilbene to a 25 mL round-bottom flask. Add 3.5 mL of methylene chloride to the flask and dissolve the solid with swirling. Add a boiling chip and take the round-bottom flask to the hood. Add 2.0 mL of 10% Br_2/CH_2Cl_2 to the flask with swirling. Take the flask back to the lab bench and heat the mixture at reflux for 20 minutes. Cool the mixture to room temperature and collect the solid by vacuum filtration. Rinse the solid with a small amount of methylene chloride and allow to air-dry for several minutes. Weigh the solid and take the melting point. Determine the percent yield of the reaction and the identity (and therefore the addition mechanism) of the product.

Explanation / Answer

Ph -CH=CH-Ph + Br2 ----> PhCHBr-CHBr-Ph

180.25g/mol 340 g/mol

Thus one mole of stilbene (180g/mol) give one moles of dibromide(340g/mol)

0.30 g of stilbene gives = 0.30g x [340g/mol ] /180g/mol

= 0.567 g

The teoretical yield of stilbenedibromide = 0.567g

percentage yild = experimental yield x100/theoretical yield

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