Question

The product Ⓒ in the following sequence of reactions has _______ π bonds.

Answer 1

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.