Hofmann Bromamide Reaction and Nitration of Aniline

To determine the correctness of Statements I and II, let us analyze them individually:

Analysis of Statement I: When benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$) is treated with bromine in the presence of an aqueous or ethanolic solution of sodium hydroxide ($\text{NaOH}$), it undergoes the Hofmann bromamide degradation reaction. In this reaction, the carbonyl group is removed as carbonate, yielding a primary amine containing one carbon atom fewer than the starting amide:

$\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$

The primary product formed is aniline ($\text{C}_6\text{H}_5\text{NH}_2$), and not benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$). Therefore, Statement I is false.

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Analysis of Statement II: Direct nitration of aniline using a mixture of concentrated $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ at $288\text{ K}$ leads to protonation of aniline in the strongly acidic medium, forming the anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$).

Since the anilinium ion is a strong $m$-directing group, a substantial amount of $m$-nitroaniline is formed alongside $o$- and $p$- isomers. The percentage yield of the products obtained is:

• $p$-nitroaniline: $51\%$
• $m$-nitroaniline: $47\%$
• $o$-nitroaniline: $2\%$

Since $47\% > 2\%$, $m$-nitroaniline is formed in a significantly higher amount than $o$-nitroaniline. Therefore, Statement II is true.

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Conclusion:

• Statement I is false, but Statement II is true.

Thus, the correct choice is Option D.