Identify the product $C$ .

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Solution

1st ${C H B r}_{3}$ reacts with base $t_{B n} O K$ to $: {C B r}_{2}$ carbene. Now this : ${C B r}_{2}$ 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 ${B r}^{(-)}$ gives $C$ . Subsequently $C$ gives an aromatic compound $D$ which is bromo benzene.

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