In a set of reactions, ethyl benzene yielded a product D: would be:

1. Step 1: Alkaline ${\text{KMnO}}_4$ oxidizes any alkyl side chain on benzene to a carboxylic acid. Product $B$ is benzoic acid (${\text{C}}_6{\text{H}}_5\text{COOH}$).2. Step 2: The $-\text{COOH}$ group is meta-directing. Halogenation gives $m$ -bromobenzoic acid ($C$).3. Step 3: Esterification with ethanol gives ethyl $m$ -bromobenzoate ($D$).