When an aryl halide is treated with the very strong basic amide ion in liquid ammonia, it is converted into aniline. The intermediate in the reaction is :

Question

Accepted Answer

The reaction in question involves the conversion of an aryl halide to aniline using a very strong basic amide ion in liquid ammonia. The correct intermediate in this reaction is benzyne, as indicated by Option 1. Here's a detailed explanation of why this is the case:

1. **Justification for Option 1 (Benzyne):**
   - The reaction mechanism involves the elimination-addition pathway, which is characteristic of nucleophilic aromatic substitution reactions involving aryl halides. In this mechanism, the strong base (amide ion) abstracts a proton adjacent to the halogen on the aryl halide, leading to the formation of a highly reactive and unstable intermediate known as benzyne.
   - Benzyne is a unique intermediate because it features a triple bond within the aromatic ring, which is not typical for stable aromatic compounds. This strained structure makes benzyne highly reactive and capable of undergoing further reactions, such as nucleophilic attack by the amide ion, ultimately leading to the formation of aniline.
   - The formation of benzyne is a well-documented and established step in the conversion of aryl halides to aniline under these conditions, making Option 1 the correct choice.

2. **Explanation for why the other options are incorrect:**

- **Option 2 (Phenyl carbocation):**
     - A phenyl carbocation is an unstable intermediate that is not typically formed in reactions involving aryl halides and strong bases. The aromatic ring's electron-rich nature makes it unlikely to stabilize a positive charge, which is why phenyl carbocations are not favored intermediates in this context.
     - Additionally, the reaction conditions (strong base in liquid ammonia) do not support the formation of a carbocation, as the base would more likely abstract a proton rather than facilitate carbocation formation.

- **Option 3 (Phenyl carbanion):**
     - A phenyl carbanion is also an unlikely intermediate in this reaction. While strong bases can generate carbanions, the aromatic ring's structure does not favor the stabilization of a negative charge on a carbon atom within the ring.
     - The mechanism involving benzyne does not require the formation of a carbanion, as the nucleophilic attack occurs on the benzyne intermediate itself.

- **Option 4 (Free radical):**
     - Free radicals are not typically involved in the nucleophilic aromatic substitution reactions of aryl halides with strong bases. The reaction conditions and mechanism do not support the formation of free radicals.
     - The elimination-addition mechanism leading to benzyne does not involve radical intermediates, making this option incorrect.

In conclusion, the correct intermediate in the conversion of aryl halides to aniline using a strong basic amide ion in liquid ammonia is benzyne, as supported by the elimination-addition mechanism. This justifies why Option 1 is the correct answer, while the other options do not align with the established reaction pathway.