# Reaction of cinnamic acid with Br2/Na2CO3

The first reaction produces benzaldehyde, and the next one (perkin's condensation)produces Cinnamic acid.(X)

Now the treatment of X with $$\ce{Br2/Na2CO3}$$ is whats troubling me. $$\ce{Na2CO3}$$ being a base, abstracts the hydrogen from the $$\ce{COOH}$$ group. $$\ce{Br2}$$ reacts with the alkene portion to yield a cyclic intermediate. What next?

The solution claims that somehow The $$\ce{CO2-}$$ group leaves and a bromoalkene forms. The addition Of moist KOH in the next step results in $$\ce{E2}$$ elimination to yield option (c).

I cant quite digest the $$\ce{CO2-}$$ group leaving. I thought of a mechanism, which is similar to the syn-elimination in esters: but that doesnt work out well..

Any hints/ insights will be appreciated. The correct answer is (c).

The sodium carbonate is just there to mop up any $$\ce{HBr}$$. The $$\ce{Br2}$$ adds across the double bond to give cinnamic acid dibromide.
The treatment of cinnamic acid dibromide with $$\ce{KOH}$$ at elevated temperature eliminates 2 eq. of $$\ce{HBr}$$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.