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
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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.
Step 1: Understanding the Concept:
This is a multi-step synthesis problem involving electrophilic aromatic substitution on a substituted aniline derivative. We need to analyze each step to determine the structure of the intermediate A and the final product B. The key concepts are the directing effects of substituents on the benzene ring and protection of functional groups. Step 2: Detailed Explanation of Each Step: Step 1: Formation of Intermediate A
Starting Material: p-Toluidine (4-methylaniline)
Reagents: (CH\(_3\)CO)\(_2\)O, Pyridine (Acetic anhydride)
The starting material has a primary amino group (-NH\(_2\)) attached to the benzene ring. The amino group is a very strong activating group and is highly susceptible to oxidation. To control its reactivity and prevent side reactions during subsequent steps (like bromination), it is often "protected".
This reaction is the acetylation of the amino group. Acetic anhydride reacts with the amino group to form an amide. The pyridine acts as a base to neutralize the acetic acid byproduct.
\[ \text{p-CH}_3\text{-C}_6\text{H}_4\text{-NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{p-CH}_3\text{-C}_6\text{H}_4\text{-NHCOCH}_3 + \text{CH}_3\text{COOH} \]
The product, intermediate A, is N-(4-methylphenyl)acetamide or p-acetotoluidide. The acetamido group (-NHCOCH\(_3\)) is still an activating, ortho-, para-directing group, but it is much less activating than the original -NH\(_2\) group. This allows for more controlled substitution. Step 2: Formation of Final Product B
Starting Material: Intermediate A (N-(4-methylphenyl)acetamide)
Reagents: Br\(_2\), CH\(_3\)COOH (Bromine in acetic acid)
This is an electrophilic aromatic substitution (bromination) reaction. We need to determine where the bromine atom will attach. The substituents on the ring are:
- -CH\(_3\): An activating, ortho-, para-directing group.
- -NHCOCH\(_3\): An activating, ortho-, para-directing group. It is a stronger activating group than -CH\(_3\).
Both groups direct incoming electrophiles to their ortho and para positions.
- The position para to the -NHCOCH\(_3\) group is already occupied by the -CH\(_3\) group.
- The position para to the -CH\(_3\) group is already occupied by the -NHCOCH\(_3\) group.
So, we need to consider the ortho positions.
- The positions ortho to -NHCOCH\(_3\) are C2 and C6.
- The positions ortho to -CH\(_3\) are C3 and C5.
The powerful activating group, -NHCOCH\(_3\), will control the position of substitution. It will direct the incoming Br\(^+\) electrophile to its ortho position. Due to the steric hindrance from the bulky -NHCOCH\(_3\) group and the adjacent -CH\(_3\) group, substitution is most likely to occur at the position ortho to the -NHCOCH\(_3\) group and meta to the -CH\(_3\) group.
Let's number the ring starting from the carbon with the -NHCOCH\(_3\) group as C1. The -CH\(_3\) group is at C4. The available positions are C2, C3, C5, C6.
The -NHCOCH\(_3\) group directs to C2 and C6. The -CH\(_3\) group directs to C3 and C5.
The -NHCOCH\(_3\) is the stronger director. It directs the Br to its ortho position (C2 or C6, which are equivalent). The methyl group at C4 does not significantly hinder the position C2/C6.
So, bromine will substitute at the position ortho to the acetamido group.
The final product B is 2-bromo-4-methyl-N-phenylacetamide or N-(2-bromo-4-methylphenyl)acetamide. Let's find this structure in the options.
Looking at the options, structure (E) shows the Br atom ortho to the NHCOCH\(_3\) group and meta to the CH\(_3\) group. Step 3: Final Answer:
The final product B is N-(2-bromo-4-methylphenyl)acetamide. This corresponds to the structure in option (E).
