Animated Solution for Chemistry - Organic Compounds Containing Nitrogen: The major product $Z$ obtained in the following reaction scheme is

Organic chemistry is often like a strategic game of chess, where each move sets up the board for the next. In this problem, we are presented with a three-step reaction sequence starting from $m$-bromoaniline. Our goal is to trace the transformations and predict the final major product, $Z$. Let's embark on this molecular journey step by step.

The Diazotization Reaction

Our starting material, $m$-bromoaniline, features a primary aromatic amine group ($-N{H}_2$). The first set of reagents is a mixture of sodium nitrite (${\text{NaNO}}_2$) and hydrochloric acid ($\text{HCl}$) kept at a chilly $273-278\text{ K}$ ($0-5^{\circ }\text{C}$).

This is the classic recipe for diazotization. The nitrous acid generated in situ reacts with the primary amine, converting it into a highly reactive diazonium salt. The resulting intermediate $X$ is $m$-bromobenzenediazonium chloride. The diazonium group ($-N_2^{+}C{l}^{-}$) is an exceptionally good leaving group, setting the stage perfectly for the next transformation.

The Sandmeyer Reaction

Next, intermediate $X$ is treated with cuprous bromide (${\text{Cu}}_2{\text{Br}}_2$). This triggers the famous Sandmeyer reaction.

Through a radical mechanism facilitated by the copper catalyst, the diazonium group is expelled as nitrogen gas (${\text{N}}_2$), and a bromine atom takes its place on the benzene ring. Our new intermediate $Y$ is $1,3$-dibromobenzene. We have successfully replaced the amine group with a second halogen.

Electrophilic Aromatic Substitution (Nitration)

Now comes the climax of our synthesis. Intermediate $Y$ is subjected to a nitrating mixture of concentrated nitric acid (${\text{HNO}}_3$) and sulfuric acid (${\text{H}}_2{\text{SO}}_4$). This mixture generates the powerful nitronium ion (${\text{NO}}_2^{+}$), an electrophile hungry for the electron-rich $\pi$-cloud of the benzene ring.

To predict where the nitronium ion will attack, we must analyze the directing effects of the two bromine atoms already present. Halogens are unique in electrophilic aromatic substitution: while they deactivate the ring overall via their electron-withdrawing inductive effect ($-I$), they are ortho/para directing due to their electron-donating resonance effect ($+R$).

Both bromine atoms in $1,3$-dibromobenzene will direct the incoming electrophile to their respective ortho and para positions. Let's evaluate the available spots:

1. Position 2: This spot is ortho to both bromine atoms. Electronically, it is highly activated. However, it is sandwiched directly between two large, bulky bromine atoms. The resulting steric hindrance makes it incredibly difficult for the nitronium ion to squeeze in. Attack here is highly disfavored. 2. Position 4 (and the equivalent Position 6): This spot is ortho to one bromine atom and para to the other. It is electronically activated by both halogens and, crucially, it is much less sterically hindered than position 2. 3. Position 5: This spot is meta to both bromine atoms and is not activated by either.

Because molecules always seek the path of least resistance, the electrophile will bypass the crowded position 2 and attack the accessible position 4.

Therefore, the nitro group attaches at position 4, yielding $1,3$-dibromo-$4$-nitrobenzene as the major product $Z$. This perfectly matches option (c), concluding our elegant synthetic sequence!