Identify compounds A and E in the following reaction sequence.

The starting compound is 1-ethyl-4-nitrobenzene.

Reaction with $\text{Br}_2/\text{AlBr}_3$ is an electrophilic aromatic substitution. The $-\text{C}_2\text{H}_5$ group is activating and ortho/para-directing, while the $-\text{NO}_2$ group is deactivating and meta-directing. Both groups direct the incoming electrophile ($\text{Br}^+$) to the position ortho to the $-\text{C}_2\text{H}_5$ group. Thus, compound A is 2-bromo-1-ethyl-4-nitrobenzene.

Reaction with $\text{Sn}/\text{HCl}$ reduces the $-\text{NO}_2$ group to an $-\text{NH}_2$ group, forming compound B (4-amino-2-bromo-1-ethylbenzene).

Reaction with $\text{NaNO}_2/\text{HCl}$ at $273-278\text{ K}$ converts the $-\text{NH}_2$ group into a diazonium salt, forming compound C (3-bromo-4-ethylbenzenediazonium chloride).

Reaction with ethanol ($\text{C}_2\text{H}_5\text{OH}$) reduces the diazonium group, replacing it with a hydrogen atom. This forms compound D (1-bromo-2-ethylbenzene).

Reaction with alkaline $\text{KMnO}_4$ followed by acidic hydrolysis oxidizes the ethyl side chain ($-\text{C}_2\text{H}_5$) to a carboxylic acid group ($-\text{COOH}$). The bromine atom remains unaffected. This forms compound E (2-bromobenzoic acid).

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