Formation of diazonium salt in the presence of \(\text{NaNO}_2\) and HCl.
Reaction with Bromine water at room temperature.
Reaction with Bromine in the presence of NaOH.
Show Solution
The Correct Option isA
Solution and Explanation
Step 1: Spot the target.
The product has Br para to NH2. Only one halogen is added, at one position. This needs a controlled reaction.
Step 2: Why not brominate directly?
The nitrogen lone pair in aniline donates into the benzene ring. This makes the ortho and para carbons so electron rich that bromine water substitutes all three positions at room temperature. The result is 2,4,6-tribromoaniline, not 4-bromoaniline.
Step 3: Tone down the activation.
Convert aniline to acetanilide by reaction with acetyl chloride or acetic anhydride. In acetanilide the nitrogen lone pair is also pulled towards the carbonyl, so less of it reaches the ring. The ring is now mildly activated and the large NHCOCH3 group hinders the ortho sites.
Step 4: Finish the sequence.
Br2 in ethanoic acid gives p-bromoacetanilide as the major product. Hydrolysis with aqueous acid or alkali removes the acetyl group and gives 4-bromoaniline.
Step 5: Match with the options.
The step that comes first is the protection step, which is acetylation. A diazonium salt (option 2) would later be replaced by another group, so the amino group would be lost. Options 3 and 4 do not protect the amine.
Final Answer:
Acetylation of the amino group comes first, which is option 1.
\[ \boxed{\text{Acetylation of amino group (option 1)}} \]
