`C_(6) H_5 NH_2 overset(CH_3 -CH=C=CH_2 //H^(+) )to A,` compound 'A' will be

Options

A. para-isopropyl aniline
B. ortho-isopropyl aniline
C. orthop-n-propylaniline
D. reaction does not occur

Answer

To solve the question, we need to analyze the reaction of aniline (C₆H₅NH₂) with the given reagent (CH₃-CH=CH₂ in the presence of H⁺).

### Step-by-step Solution:

1. **Identify the Reactants**:
- The reactant is aniline (C₆H₅NH₂), which is an aromatic amine.
- The other reactant is CH₃-CH=CH₂, which is an allyl group, and H⁺ indicates an acidic condition.

2. **Protonation of Aniline**:
- Under acidic conditions (presence of H⁺), the lone pair of electrons on the nitrogen atom of aniline can be protonated.
- This leads to the formation of an anilinium ion (C₆H₅NH₃⁺).

3. **Formation of Anilinium Ion**:
- The reaction can be represented as:
\[ C₆H₅NH₂ + H^+ \rightarrow C₆H₅NH₃^+ \]
- The anilinium ion is a positively charged species where the nitrogen is bonded to three hydrogen atoms and has a positive charge.

4. **Deactivation of the Aromatic Ring**:
- The formation of the anilinium ion deactivates the aromatic ring towards electrophilic substitution reactions because the positive charge on nitrogen withdraws electron density from the ring.
- This means that the aromatic ring is less reactive towards further electrophilic attack.

5. **Conclusion**:
- Since the anilinium ion is formed and the aromatic ring is deactivated, no further reaction occurs with the allyl group (CH₃-CH=CH₂).
- Therefore, the final compound A is simply the anilinium ion, and no substitution or addition reaction occurs.

### Final Answer:
The compound A will be the anilinium ion (C₆H₅NH₃⁺), and no further reaction occurs.