Animated Solution for Chemistry - Organic Chemistry: Aniline reacts with mixed acid (conc. ${\text{HNO}}_3$ and conc. ${\text{H}}_2{\text{SO}}_4$) at $288\text{ K}$ to give P ($51\%$), Q ($47\%$) and R ($2\%$). The major product(s) the following reaction sequence is (are) :-

Our journey begins with the nitration of aniline using a mixed acid ($conc.{\text{HNO}}_3$ and $conc.{\text{H}}_2{\text{SO}}_4$). In a strongly acidic medium, aniline is protonated to form the anilinium ion, which is strongly deactivating and meta-directing. This is why we get a surprisingly high yield of the meta product ($47\%$). However, the problem directs our attention to the minor $2\%$ product, R. This minor product is $o$-nitroaniline.

The Synthesis of Intermediate S

We take $o$-nitroaniline and subject it to a carefully orchestrated sequence of reactions.

First, we react it with acetic anhydride (${\text{Ac}}_2\text{O}$) in pyridine. This step protects the highly reactive amine group, converting it into an acetanilide ($-\text{NHAc}$). Why is this necessary? Because a free amine group is so strongly activating that it would lead to uncontrollable polysubstitution in the next step.

Next, we introduce bromine in acetic acid. The protected amine group is ortho-para directing, while the nitro group is meta-directing. The position para to the $-\text{NHAc}$ group is sterically free, whereas the ortho position is crowded by the adjacent nitro group. Thus, the electrophile attacks the para position, yielding 4-bromo-2-nitroacetanilide.

Time to remove the protection! Acidic hydrolysis (${\text{H}}_3{\text{O}}^{+}$) strips away the acetyl group, giving us back the primary amine. Immediately after, we treat it with sodium nitrite and $\text{HCl}$ at $0^{\circ }\text{C}$. This classic reaction converts the amine into a highly unstable diazonium salt ($-{\text{N}}_2^{+}{\text{Cl}}^{-}$).

Finally, heating this diazonium salt with ethanol ($\text{EtOH}$) reduces it. The entire nitrogen group leaves as a gas, replaced by a simple hydrogen atom. We have successfully synthesized our intermediate S, which is 3-bromonitrobenzene.

The Final Transformation

Moving to the second reaction sequence, we treat intermediate S with tin and $\text{HCl}$ ($\text{Sn}/\text{HCl}$). This is a standard reduction that converts the nitro group into a primary amine, yielding 3-bromoaniline.

Here is the crucial step: we add excess bromine water. The newly formed amine group is strongly activating and, unlike before, it has no protection. It aggressively directs bromine to all available ortho and para positions simultaneously. The positions ortho and para to the $-{\text{NH}}_2$ group are rapidly brominated, resulting in a heavily substituted ring.

Now, we need to remove this amine group to reach our final product. We diazotize it again using ${\text{NaNO}}_2$ and $\text{HCl}$, converting it back into a diazonium salt. Immediately after, we reduce it using hypophosphorous acid (${\text{H}}_3{\text{PO}}_2$). This completely removes the nitrogen group, leaving a hydrogen atom in its place.

Look at our final molecule. We have four bromine atoms on the benzene ring. According to IUPAC nomenclature rules, we must number the carbons to give the lowest possible locant set to the substituents. Starting from one bromine and numbering around the ring yields the positions 1, 2, 3, and 5.

Thus, our final product is 1,2,3,5-tetrabromobenzene, which perfectly matches option (D).