Question

Write the structures of A, B and C in the following reaction:

Hint:

Acetylation is the standard "protection" strategy in aromatic synthesis to prevent over-substitution and oxidation of the sensitive \( -NH_{2} \) group.

Answer 1

Step 1: Understand why aniline needs to be protected first.
The amino group on aniline pushes so much electron density into the ring that direct bromination cannot be controlled, it happily substitutes at all three activated positions at once, giving the tribromo product. So before bromination, the amine is toned down by acetylation with acetic anhydride, giving acetanilide, compound A, $C_6H_5NHCOCH_3$.
Step 2: Brominate the tamer molecule.
The $-NHCOCH_3$ group is still ortho, para directing but far less powerfully activating, and it is also bulky. That bulk pushes the incoming bromine mostly to the less crowded para position, giving compound B, p-bromoacetanilide.
Step 3: Remove the protecting group.
Once the bromine is safely in place, acidic hydrolysis strips off the acetyl group and regenerates the free amine, giving compound C, p-bromoaniline.
\[ \text{Aniline} \xrightarrow{(CH_3CO)_2O} A \xrightarrow{Br_2/CH_3COOH} B \xrightarrow{H_3O^+} C \]
\[ \boxed{\text{A = Acetanilide, B = p-bromoacetanilide, C = p-bromoaniline}} \]