Question

In the reaction: \(\text{Aniline} \xrightarrow{\text{NaNO}_2/\text{dil. HCl}} P \xrightarrow{\text{Phenol/NaOH}} Q\)

• A. $C_6​H_5​N_2​Cl$
• B. Ortho-hydroxyazobenzene
• C. Para-hydroxyazobenzene
• D. Meta-hydroxyazobenzene

Hint:

Aniline diazotizes to benzene diazonium chloride (P), which reacts with phenol to form para-hydroxyazobenzene (Q).

Answer 1

The Correct Option is C

To solve the given reaction sequence, let's break it down into steps: 

1. **Formation of Diazonium Ion (P):** 
   Aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)) reacts with sodium nitrite (\( \text{NaNO}_2 \)) and dilute hydrochloric acid (\( \text{HCl} \)) to form benzene diazonium chloride (\( \text{C}_6\text{H}_5\text{N}_2\text{Cl} \)). This reaction occurs through the following steps: 
   \(\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{NaCl} + \text{H}_2\text{O}\)
2. **Coupling Reaction with Phenol in NaOH (Q):** 
   The benzene diazonium chloride (P) undergoes a coupling reaction with phenol in the presence of NaOH to form an azo compound. The coupling primarily occurs at the para position of the phenol ring, leading to the formation of para-hydroxyazobenzene.

 

The end product (Q) is para-hydroxyazobenzene. The reason para substitution is favored is due to steric and electronic considerations, which make the para position more accessible and reactive in phenolic compounds during coupling reactions.

Therefore, the correct answer is Para-hydroxyazobenzene.