Identify the product $C$.

1st ${CHBr}_3$ reacts with base $t_{Bn}OK$ to $:{CBr}_2$ carbene. Now this : ${CBr}_2$ inserts into one of the double bond of cyclopentadiene to form $B$ which contains a three membered rings. Now breaking of three membered rings and simultaneous leaving of ${Br}^{\left(-\right)}$ gives $C$. Subsequently $C$ gives an aromatic compound $D$ which is bromo benzene.