The given reaction involves an aromatic compound where a chlorine atom (Cl) is present ortho to a bromine atom (Br). The reaction takes place with ammonia (NH₃) in ethanol (EtOH).
This is a classic example of a nucleophilic aromatic substitution reaction, also known as the SNAr mechanism. In such reactions, an electron-withdrawing group (like NO₂, CN, or halogens such as Br) facilitates the substitution of a halide (like Cl) by a nucleophile (here, NH₃).
In this case, the chlorine atom is ortho to the bromine atom. Although Br is less electronegative than NO₂ or CN, it still has a slight withdrawing effect via induction and a resonance donating effect. But in this structure, due to the presence of NH₃ in EtOH, the NH₃ acts as a nucleophile and attacks the aromatic ring where the Cl is attached, provided there is some activating influence nearby (as the Br gives partial stabilization).
Thus, the reaction proceeds via the SNAr mechanism. The Cl group is displaced by the NH₂ group, and the product formed is:
2-bromoaniline (that is, the amino group replaces Cl in the original position, and Br remains ortho to NH₂).
So the structure of the final compound is a benzene ring with NH₂ and Br in ortho positions.
Final Answer: The major product is 2-bromoaniline.
Correct Answer: Option C