Question

In the following reaction, the final product B is:
C\(_6\)H\(_5\)NH\(_2\) + (CH\(_3\)CO)\(_2\)O (Pyridine) \(\rightarrow\) A \(\xrightarrow{Br, CH_3COOH}\) B

• A. A
• B. B
• C. C
• D. D
• E. E

Hint:

In electrophilic aromatic substitution reactions, the electron-withdrawing groups (like \(-COCH_3\)) direct substitution to the meta position, while electron-donating groups would direct it to the ortho/para positions.

Answer 1

Answer: (E)

Step 1: Analyze the starting reaction.
The starting material is aniline (\(C_6H_5NH_2\)) reacting with acetic anhydride \((CH_3CO)_2O\) in pyridine. This is a typical acetylation reaction, where the amino group (\(NH_2\)) reacts with the acetic anhydride to form an amide group \(-NHCOCH_3\). Thus, the intermediate product A will be acetylated aniline, where the amino group is now \(NHCOCH_3\).

Step 2: Analyze the second reaction step.
In the second step, the reaction is carried out with bromine \((Br)\) and acetic acid \((CH_3COOH)\). Bromination typically occurs at the position para to the \(NHCOCH_3\) group because of the electron-withdrawing nature of the amide group, which directs substitution to the meta positions in electrophilic aromatic substitution reactions.

Step 3: Determine the final product.
In the final product, we expect bromination at the para position to the amide group. This results in the bromo-substituted acetylated aniline. Therefore, the final product B corresponds to option \((5)\).

Step 4: Verify other options.
Other options show different positions for the substituents, but they do not match the expected para substitution pattern from the bromination.

Step 5: Final answer.
The final product B is:
\[ \boxed{\text{Option (5)}} \]