Identify Compounds A and E in Reaction Sequence

To determine the correct structures of compounds A and E, we analyze the given reaction sequence step-by-step:

Step 1: Electrophilic Aromatic Bromination ($\text{Br}_2 / \text{AlBr}_3$)

The starting material is 1-ethyl-4-nitrobenzene:

• The ethyl group ($-\text{C}_2\text{H}_5$) is an activating and $ortho/para$-directing group via hyperconjugation and inductive effects ($+I$).
• The nitro group ($-\text{NO}_2$) is a strongly deactivating and $meta$-directing group via resonance ($-R$) and inductive effects ($-I$).

Since the orientation of electrophilic substitution is governed by the stronger activating group ($-\text{C}_2\text{H}_5$), and the $para$-position relative to $-\text{C}_2\text{H}_5$ is blocked by $-\text{NO}_2$, the electrophile ($\text{Br}^+$) attacks at the position $ortho$ to $-\text{C}_2\text{H}_5$ (which is also $meta$ to $-\text{NO}_2$).

$\text{1-ethyl-4-nitrobenzene} \xrightarrow{\text{Br}_2 / \text{AlBr}_3} \text{A: 2-bromo-1-ethyl-4-nitrobenzene}$

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Step 2: Reduction of Nitro Group ($\text{Sn} / \text{HCl}$)

Tin and hydrochloric acid selectively reduce the nitro group ($-\text{NO}_2$) to an amino group ($-\text{NH}_2$):

$\text{A} \xrightarrow{\text{Sn} / \text{HCl}} \text{B: 3-bromo-4-ethylaniline}$

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Step 3: Diazotization ($\text{NaNO}_2 / \text{HCl}, 273\text{--}278\text{ K}$)

Treatment of the primary aromatic amine with nitrous acid at low temperature converts the $-\text{NH}_2$ group into a diazonium salt:

$\text{B} \xrightarrow{\text{NaNO}_2 / \text{HCl}, 273\text{--}278\text{ K}} \text{C: 3-bromo-4-ethylbenzenediazonium chloride}$

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Step 4: Deamination ($\text{C}_2\text{H}_5\text{OH}$)

Ethanol acts as a mild reducing agent, replacing the diazonium group ($-\text{N}_2^+\text{Cl}^-$) with a hydrogen atom ($-\text{H}$):

$\text{C} \xrightarrow{\text{C}_2\text{H}_5\text{OH}} \text{D: 1-bromo-2-ethylbenzene}$

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Step 5: Side-Chain Oxidation ((i) $\text{KMnO}_4 / \text{KOH}$, (ii) $\text{H}_3\text{O}^+$)

Alkaline potassium permanganate oxidizes the alkyl side chain ($-\text{C}_2\text{H}_5$) containing benzylic hydrogens into a carboxylate salt, which upon acidification yields the carboxylic acid group ($-\text{COOH}$):

$\text{D} \xrightarrow{\text{(i) }\text{KMnO}_4/\text{KOH} \quad \text{(ii) }\text{H}_3\text{O}^+} \text{E: 2-bromobenzoic acid}$

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Conclusion:

• Compound A: 2-bromo-1-ethyl-4-nitrobenzene (with $\text{Br}$ $ortho$ to $\text{C}_2\text{H}_5$)
• Compound E: 2-bromobenzoic acid (with $\text{Br}$ $ortho$ to $\text{COOH}$)

This corresponds to Option B.