Major Product of Diazotization and Intramolecular Coupling

To determine the major product of the given reaction sequence, let us break down the reaction step-by-step:

Step 1: Diazotization

When 2-amino-3'-hydroxybiphenyl is treated with sodium nitrite ($\text{NaNO}_2$) and hydrochloric acid ($\text{HCl}$) at $0^\circ\text{C}$, the primary aromatic amine ($-\text{NH}_2$) group undergoes diazotization to form an aromatic diazonium salt:

$\text{Ar-NH}_2 \xrightarrow{\text{NaNO}_2, \text{HCl}, 0^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^-$

Step 2: Deprotonation in Alkaline Medium

Treatment with aqueous sodium hydroxide ($\text{aq. NaOH}$) deprotonates the phenolic hydroxyl group ($-\text{OH}$) on the second ring to form a phenoxide ion ($-\text{O}^-$). The phenoxide ion is a strongly activating group that directs electrophilic aromatic substitution ($EAS$) primarily to the para-position (and secondarily to the ortho-position).

Step 3: Intramolecular Azo Coupling

The diazonium cation ($-\text{N}_2^+$) serves as a weak electrophile and undergoes an intramolecular azo coupling with the activated phenoxide ring:

1. Regioselectivity: The attack occurs preferentially at position $6'$ of the second ring, which is para to the phenoxide group (located at position $3'$) and ortho to the inter-ring $\text{C1-C1'}$ single bond.
2. Ring Stability: Coupling at position $6'$ leads to the formation of a stable, planar, conjugated 6-membered azo ring (a derivative of benzo[$c$]cinnoline).

Since the phenoxide group ($-\text{O}^-$) is located para to the site of coupling (position $6'$), protonation during workup yields the 6-membered fused heterocyclic compound with the hydroxyl group ($-\text{OH}$) at the position para to the newly formed $\text{C-N}$ bond.

This matches the structure shown in Option [C].

Correct Answer: [C]