The reaction sequence proceeds as follows: Step 1: Oxidation of \(\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\) with \(\text{KMnO}_4\) and \(\text{KOH}\) The side chain ethyl group of ethylbenzene is oxidized to a carboxylate salt: \[\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \rightarrow [\text{KMnO}_4, \text{KOH}, \Delta] \text{C}_6\text{H}_5\text{COOK} (A).\] Here, \(A\) is \(\text{C}_6\text{H}_5\text{COOK}\), potassium benzoate. Step 2: Acidification with \(\text{H}_3\text{O}^+\) The carboxylate salt is acidified to form benzoic acid: \[\text{C}_6\text{H}_5\text{COOK} + \text{H}_3\text{O}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH} (B).\] Here, \(B\) is \(\text{C}_6\text{H}_5\text{COOH}\), benzoic acid. Step 3: Bromination with \(\text{Br}_2/\text{FeBr}_3\) Bromination occurs at the para-position of the benzene ring relative to the carboxylic acid group, yielding: \[\text{C}_6\text{H}_5\text{COOH} \rightarrow [\text{Br}_2, \text{FeBr}_3] p-\text{BrC}_6\text{H}_4\text{COOH} (C).\] Here, \(C\) is para-bromobenzoic acid (\(p-\text{BrC}_6\text{H}_4\text{COOH}\)). Step 4: Count the \(\pi\)-bonds in \(C\) - The benzene ring contributes 3 \(\pi\)-bonds. - The \(\text{C} = \text{O}\) group in the carboxylic acid contributes 1 \(\pi\)-bond. Thus, the total number of \(\pi\)-bonds in \(C\) is: \[\pi\text{-bonds} = 3 + 1 = 4.\] Final Answer: 4.
