Question

Identify the product Z in the following reaction
\[{C_6}{H_5}N{H_2} \overset{(Ac)_{2}O}{\rightarrow} X \overset{Br_{2}/CCl_{4}}{\rightarrow} Y \overset{HOH}{\rightarrow} Z\]
A. p-Bromoaniline
B. p-Bromoacetophenone
C. o-Bromoacetophenone
D. o-Bromoacetonilide

Answer 1

Hint: The reaction takes place in three steps. In the first step acetylation takes place, in the second step bromination takes place in presence of carbon tetrachloride and in the third step hydrolysis takes place to form the final product Z.

Complete Step by Step Solution:
In the first step, aniline reacts with acetic anhydride to form an acetylated compound known as N-phenylacetamide. Acetylation is a reaction where an acetyl group is introduced to the reactant. In the first step, the amine is getting converted to an amide.

In the second step, N-phenylacetamide reacts with bromine in presence of carbon tetrachloride to form N-(4-bromophenyl)acetamide. The reaction is known as a bromination reaction where bromine is added to the para position of the reactant.

In the third step, N-(4-bromophenyl)acetamide undergoes hydrolysis to form 4-bromobenzenamine or P-Bromoaniline. Hydrolysis is a reaction where one of the reactants. Here, the amide group is again converted to an amine group.

Image: Formation of P-Bromoaniline
Thus, X is N-phenylacetanamide, Y is N-(4-bromophenyl)acetamide, Z is p-bromoaniline.
Therefore the correct option is A.

Note: It should be noted that aniline is an o- and p- directing group. The lone pair of electrons present in the amine group is delocalised by the resonance with the benzene ring forming a negative charge at the ortho and para position in the resonating structure. Ortho and para directing groups are generally electrons donor whereas the meta directing groups are electron acceptors.