Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: Correct statement about the given chemical reaction is

When we look at the nitration of aniline, we are dealing with a classic electrophilic aromatic substitution reaction. The reagents used are a mixture of concentrated nitric acid ($HN{O}_3$) and concentrated sulfuric acid ($H_{2}S{O}_4$) at $288\text{ K}$.

Normally, the amino group ($-N{H}_2$) is a strongly activating, ortho-para directing group due to the lone pair of electrons on the nitrogen atom. Based on this, one might intuitively expect the reaction to yield exclusively ortho and para products. However, chemistry is rarely that straightforward, and there is a significant catch in this specific reaction environment.

The nitrating mixture is strongly acidic. Aniline, being a Lewis base, readily accepts a proton ($H^{+}$) from the acidic medium. This protonation converts a substantial portion of the aniline molecules into the anilinium ion ($-\stackrel{+}{N}{H}_3$).

This transformation completely flips the script. The positively charged nitrogen atom in the anilinium ion can no longer donate electrons through resonance. Instead, it exerts a powerful electron-withdrawing inductive effect ($-I$ effect). This strong electron withdrawal deactivates the benzene ring and, crucially, directs the incoming nitronium ion ($N{O}_2^{+}$) to the meta position.

Because we have an equilibrium between the highly reactive, ortho-para directing unprotonated aniline and the deactivated, meta-directing anilinium ion, the final product mixture is quite unique.

The unprotonated aniline reacts rapidly to form p-nitroaniline (Compound A) as the major product, accounting for about $51\%$ of the yield. The steric hindrance from the amino group restricts the formation of the ortho product to a mere $2\%$.

Simultaneously, the anilinium ion undergoes nitration to form a surprisingly large amount of m-nitroaniline (Compound B), making up $47\%$ of the final mixture.

Comparing the yields, p-nitroaniline ($51\%$) narrowly edges out m-nitroaniline ($47\%$). Therefore, the reaction is entirely possible, and compound (A) remains the major product. This perfectly aligns with option (d).

As a bonus tip for your exams: if a chemist wants to synthesize exclusively p-nitroaniline without the massive meta impurity, they must first protect the amino group. This is typically done by reacting aniline with acetic anhydride to form acetanilide. The acetyl group reduces the basicity of the nitrogen, preventing protonation, while still remaining ortho-para directing. After nitration, the acetyl group is easily removed via hydrolysis.