The question is :
Now, this is what I have done :
So I dont know how to proceed to the final product. In some book they tell we need to break the cyclopropane to get the desirable answer but how ?
Please help.
This paper (JCS Perkin 1 (1975) 1638-40) reports that dichlorocarbene reacts with cyclopentadiene at 180C to give chlorobenzene. The strained cyclopropyl ring opens to give insertion with loss of chloride anion. I see no reason why dibromocarbene should give a different outcome, so the answer is C.
See this SE answer How does chloroform react with cyclopentadiene in base?
Here OTs being a good leaving group will be eliminated by alc KOH making cyclopent-1,3-diene, t-BuO- being a solvent will make :CBr2 which will act as a Base and abstract proton from C5. Hence option D will be the answer.