The product (C) in the following sequence of | vedclass

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(D) $1$. The starting material is propylbenzene. Oxidation with $KMnO_4-KOH$ followed by acidification $(H_3O^+)$ converts the alkyl side chain into a

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$4$

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(D) $1$. The starting material is propylbenzene. Oxidation with $KMnO_4-KOH$ followed by acidification $(H_3O^+)$ converts the alkyl side chain into a carboxylic acid group,yielding benzoic acid $(B)$.$2$. Benzoic acid is then subjected to electrophilic aromatic substitution with $Br_2$ in the presence of $FeBr_3$. Since the $-COOH$ group is meta-directing,the product $(C)$ is $m$-bromobenzoic acid.$3$. The structure of $m$-bromobenzoic acid contains a benzene ring (which has $3$ $\pi$ bonds) and a carbonyl group ($C=O$,which has $1$ $\pi$ bond).$4$. Total $\pi$ bonds $= 3 + 1 = 4$.

The product $(C)$ in the following sequence of reactions has . . . . . . . . $\pi$ bonds.

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