Add 0.45 g of trans-cinnamic acid to a 25 mL round-bottom flask. Add 3.5 mL of m

Question

Add 0.45 g of trans-cinnamic acid 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

The reaction is an addition reaction .

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

1 moles of cinnamic acid (148g/mol)give one mole of dibromocinnamic acid(308g/mol)

0.45 g of cinnamic acid gives = 0.45g x308g /148g

= 0.9365 g

theoretical yield of dibromocinnamic acid = 0.9365g

% yield = experimental yield x 100/ theoretical yield

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