Animated Solution for Chemistry - Organic Chemistry: The product(s) of the following reaction sequence is(are)

Imagine you are a chemist tasked with adding exactly one bromine atom to an aniline molecule. It sounds simple, right? However, aniline is a notoriously over-enthusiastic molecule. The $-N{H}_2$ group is a powerful activating group due to the $+M$ (mesomeric) effect of the nitrogen's lone pair. If you simply toss aniline into a flask with bromine water, it won't stop at one bromine. It will rapidly brominate at all available ortho and para positions, leaving you with a white precipitate of 2,4,6-tribromoaniline.

To achieve a controlled, mono-bromination, we must first "tame" the aniline. We need to temporarily reduce its reactivity.

Acylation

Our first move is to protect the amino group. By reacting aniline with acetic anhydride ($A{c}_{2}O$) in the presence of pyridine, we convert it into acetanilide.

Why does this help? In acetanilide, the lone pair of electrons on the nitrogen atom is no longer fully dedicated to the benzene ring. Instead, it is caught in a resonance tug-of-war with the adjacent carbonyl group ($-C=O$). This cross-conjugation significantly decreases the electron density pushed into the benzene ring, effectively lowering its reactivity. The ring is now primed for a controlled, single electrophilic attack.

Para-Bromination

With the ring's reactivity dialed down, we introduce our brominating agents: $KBr{O}_3$ and $HBr$. These reagents react in situ to generate bromine ($B{r}_2$) gas steadily and safely.

Now, where will the incoming bromine electrophile ($B{r}^{+}$) attack? The acetamido group ($-NHCOC{H}_3$) is quite bulky. This steric hindrance makes the ortho positions highly inaccessible. Consequently, the electrophile is forced to attack the less hindered para position, yielding p-bromoacetanilide as the major product.

Our bromine atom is successfully installed at the para position. The protecting group has served its purpose, and it is time to remove it. We perform an acidic hydrolysis using $H_3{O}^{+}$ and heat. This cleaves the amide bond, releasing acetic acid and unmasking our original amino group to give p-bromoaniline.

Next, we prepare the molecule for its final transformation. We treat the p-bromoaniline with sodium nitrite ($NaN{O}_2$) and hydrochloric acid ($HCl$) at ice-cold temperatures ($0-5^{\circ }C$). This classic reaction, known as diazotization, converts the primary amine into a diazonium salt. The $-N{H}_2$ group is transformed into the $-N_2^{+}C{l}^{-}$ group, which contains one of the best leaving groups in organic chemistry: nitrogen gas ($N_2$).

Sandmeyer Reaction

In the grand finale, we employ the Sandmeyer reaction. By treating our diazonium salt with copper and hydrobromic acid ($Cu/HBr$), a radical-nucleophilic aromatic substitution occurs. The diazonium group departs as stable nitrogen gas, and a bromide ion seamlessly takes its place.

The result? A perfectly synthesized molecule of 1,4-dibromobenzene (also known as p-dibromobenzene). By carefully orchestrating protection, directed substitution, deprotection, and functional group transformation, we successfully bypassed aniline's hyper-reactivity to achieve our exact target.