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.
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The Correct Option isA
Solution and Explanation
Step 1: Understanding the Question:
Aniline has a very strongly activating -NH2 group. We want only one bromine atom, and it must go to the para position (4-position).
Step 2: The problem with direct bromination.
Bromine water reacts with aniline at once and gives 2,4,6-tribromoaniline as a white precipitate. The -NH2 group pushes electron density into the ring so strongly that all ortho and para positions are brominated. We cannot stop at one Br.
Step 3: The remedy.
We reduce the activating power of the -NH2 group by protecting it. Treating aniline with acetic anhydride or acetyl chloride gives acetanilide, \(\text{C}_6\text{H}_5\text{NHCOCH}_3\). The lone pair on N is now shared with the C=O group, so the ring is only moderately activated. The bulky acetyl group also blocks the ortho positions, so the para product is mainly formed.
Step 4: Remaining steps.
Acetanilide is brominated to give p-bromoacetanilide. Then acid or base hydrolysis removes the acetyl group and gives 4-bromoaniline. So the first step is acetylation.
Step 5: Check the options.
Option 1 is acetylation of the amino group, the correct first step. Option 2 makes a diazonium salt, which would lose the -NH2 group later, so it is wrong. Option 3, bromine water at room temperature, gives the tribromo product. Option 4, bromine with NaOH, is not a controlled ring bromination of an aromatic amine and does not give the 4-bromo product. So options 2, 3 and 4 are wrong.
Final Answer:
The first step is protection of the amino group by acetylation. This is option 1.
\[ \boxed{\text{Option 1: Acetylation of amino group}} \]
