Question

 

Hint:

In the Hell-Volhard-Zelinski reaction, the halogenation of carboxylic acids can lead to the substitution of the halogen atom at the alpha position to the carboxyl group. In this case, the final product undergoes diazotization followed by substitution, forming phenol.

Answer 1

Correct Answer: 2

Step 1: Analysis of Reactions.
- In the given reaction, the compound is a propan-2-one (acetone derivative), and the reagents indicate a multi-step reaction.

Step 2: Reaction (i) NH$_3$, \( \Delta \).
This step likely involves the formation of an imine (Schiff base) through nucleophilic attack by ammonia on the carbonyl carbon. However, this intermediate is not stable and will undergo further reaction in the subsequent steps.
Step 3: Reaction (ii) NaOH + Br$_2$.
This step indicates the Hell-Volhard-Zelinski reaction, where the carboxylic acid group undergoes halogenation, leading to a bromo-substituted intermediate, typically leading to an aromatic compound.
Step 4: Reaction (iii) HNO$_2$.
The treatment with nitrous acid leads to diazotization, which further results in the formation of a phenyl group.
Step 5: Reaction (iv) Ph–N$_2$Cl, \( \Delta \).
In this final step, the diazonium salt undergoes a substitution reaction, replacing the diazonium group with a hydroxyl group, yielding the final product, which is phenol (C$_6$H$_5$OH).
Step 6: Conclusion.
Thus, the major product of the given reactions is Phenol (C$_6$H$_5$OH), which corresponds to option (B).
Final Answer: (B) C$_6$H$_5$OH