Identification of Amine Product R from Dialkylbenzene Derivatives

To determine the structure of compound R, let us trace the complete reaction sequence step-by-step:

1. Oxidation of Compound O to P: Compound O is a 1,2-dialkylbenzene (1,2-dipropylbenzene). Since each alkyl side chain possesses benzylic hydrogen atoms, treatment with vigorous oxidizing agents such as acidic potassium permanganate ($\text{KMnO}_4/\text{H}^+$) results in complete side-chain oxidation to yield phthalic acid (benzene-1,2-dicarboxylic acid) as compound P. $\text{Compound O} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{C}_6\text{H}_4(\text{COOH})_2 \quad (\text{Compound P: Phthalic acid})$

2. Formation of Compound Q: When phthalic acid (P) is heated with ammonia ($\text{NH}_3$), it forms phthalamide as compound Q: $\text{C}_6\text{H}_4(\text{COOH})_2 + 2\text{NH}_3 \xrightarrow{\Delta} \text{C}_6\text{H}_4(\text{CONH}_2)_2 + 2\text{H}_2\text{O} \quad (\text{Compound Q: Phthalamide})$

3. Hofmann Bromamide Degradation to form Compound R: Treatment of phthalamide (Q) with bromine in the presence of sodium hydroxide ($\text{Br}_2/\text{NaOH}$) undergoes the Hofmann bromamide degradation. In this reaction, both primary amide groups ($-\text{CONH}_2$) lose their carbonyl carbon atoms to form primary amino groups ($-\text{NH}_2$): $\text{C}_6\text{H}_4(\text{CONH}_2)_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{C}_6\text{H}_4(\text{NH}_2)_2 \quad (\text{Compound R: Benzene-1,2-diamine})$

Thus, compound R is 1,2-diaminobenzene (o-phenylenediamine), which corresponds to structure (A).

Correct Answer: A