Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Imagine you are looking at a molecular battlefield. We start with a biphenyl system—two benzene rings connected by a single bond. On the left ring, we have a primary aromatic amine ($-N{H}_2$) sitting comfortably at the ortho position relative to the biphenyl linkage. On the right ring, we have a hydroxyl group ($-OH$) positioned meta to the linkage.

This molecule is a ticking time bomb of reactivity because it contains both a potential electrophile precursor (the amine) and a potential nucleophile (the phenol). All it needs are the right reagents to set off a spectacular intramolecular reaction.

Diazotization

The first step introduces our classic diazotization mixture: sodium nitrite ($NaN{O}_2$) and hydrochloric acid ($HCl$) at a chilly $0^{\circ }C$. This mixture generates nitrous acid ($HN{O}_2$) in situ, which acts as a molecular sniper, specifically targeting the primary amine.

The amine undergoes a series of protonations and dehydrations, ultimately transforming into a highly reactive diazonium salt ($-N_2^{+}C{l}^{-}$). The diazonium group is an exceptional electrophile—it is electron-deficient and desperate to find a source of electron density. Meanwhile, the phenol on the other ring remains untouched in this acidic environment.

Phenoxide Formation

Next, we change the rules of the game by adding aqueous sodium hydroxide ($NaOH$). This basic environment immediately deprotonates the acidic phenol, converting it into a phenoxide ion ($-O^{-}$).

Why is this crucial? A neutral phenol is a decent nucleophile, but a phenoxide ion is an absolute powerhouse. The negative charge on the oxygen is delocalized into the aromatic ring, making the ring incredibly electron-rich and highly activated for electrophilic aromatic substitution.

The Geometric Masterpiece

Now we have the perfect storm: a hungry electrophile (the diazonium group) and a wealthy nucleophile (the phenoxide ring) tethered together in the same molecule. They are destined to couple, but where exactly will the attack happen?

The phenoxide oxygen directs incoming electrophiles to its ortho and para positions. Let's evaluate the options on the right ring. The ortho position (closest to the biphenyl bond) is sandwiched right between the bulky biphenyl linkage and the oxygen atom itself. It suffers from severe steric hindrance. Trying to force the large diazonium group into that tight space is like trying to park a bus in a bicycle lane.

However, the para position is wide open. But it gets even better. If we trace the path from the diazonium nitrogen to the para carbon, we realize that coupling here forms a brand new ring. Let's count the atoms: two carbons from the left ring, two nitrogens, and two carbons from the right ring. It forms a six-membered ring! Because all the atoms involved are $s{p}^2$ hybridized, the geometry naturally folds into a perfect, strain-free regular hexagon with $120^{\circ }$ internal angles.

Driven by this geometric perfection and lack of steric hindrance, the electrophilic nitrogen attacks the para carbon. After the loss of a proton to restore aromaticity, the azo linkage ($-N=N-$) is permanently forged.

The final product is a beautiful fused tricyclic system. If you look closely at the correct option, you will see the azo bond connecting the two lower vertices of the biphenyl system, while the $-OH$ group remains exactly where it started—perfectly positioned para to the newly formed azo bond.