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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).

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