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.