Mechanism for the conversion of dimethyl cyclohexa-1,4-diene-1,2-dicarboxylate to dimethyl benzene-1,3-dicarboxylate using bromine
Firstly, I didn't get which oxygen you are talking about to break the ring. If you are considering the oxygens that are part of esters, that is not possible. Secondly, $\ce{Br^-}$ is the conjugate base of s very strong acid $\ce{HBr}$. So, $\ce{Br^-}$ is a very weak base. Therefore, it is very very difficult for that $\ce{Br^-}$ to abstract the hydrogen atom and act as a base. 

Instead what you can simply do, you can first perform bromination i.e.  treating the substrate with $\ce{Br_2/ CHCl_3}$. You will first get a bromonium intermidiate, like this:



The intermidiate will form on the farther double bond as the $\pi$ -electron cloud of the other one is pulled by the electron withdrawing ester groups, and also to form that ring, there will be huge steric crowding. The other $\ce{Br^-}$ will act as nucleophile to break this ring, (normal mechanism) and you will get this product:



Now, the two $\ce{Br}$ are in anti position. So, if you now treat the compound with excess $\ce{EtO^-}$ in $\ce{EtOH}$ and heat it, anti-periplanar E2 elimination will occur, and finally you will get your desired product as the one mentioned above.



The elimination will occur from two different carbons as the bromines are in anti position. So, to follow anti elimination of $\ce{Br}$ with $\ce{H}$, the elimination has to be from two different non adjacent carbons.