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

When an aryl halide is treated with a very strong basic amide ion in liquid ammonia, it undergoes a reaction known as the elimination-addition mechanism, which involves the formation of an intermediate called benzyne. Here's a detailed explanation supporting why Option 1 is correct and why the other options are incorrect:

1. **Benzyne (Option 1)**: 
   - In the reaction of aryl halides with strong bases like the amide ion in liquid ammonia, the aryl halide first undergoes an elimination step. This step involves the removal of a halide ion and a hydrogen atom from adjacent carbon atoms on the aromatic ring, resulting in the formation of a highly reactive intermediate known as benzyne. Benzyne is characterized by a triple bond within the aromatic ring, which is not a typical feature of stable aromatic compounds. This intermediate is crucial for the subsequent steps of the reaction, where the amide ion attacks the benzyne to form aniline. Thus, benzyne is the correct intermediate in this reaction mechanism.

2. **Phenyl carbocation (Option 2)**:
   - A phenyl carbocation is a positively charged species where the positive charge is localized on the carbon atom of the aromatic ring. However, in the context of this reaction, the conditions do not favor the formation of a carbocation. The strong basic conditions and the nature of the aryl halide do not support the stabilization of a carbocation intermediate. Therefore, a phenyl carbocation is not formed in this reaction.

3. **Phenyl carbanion (Option 3)**:
   - A phenyl carbanion is a negatively charged species with the negative charge localized on the carbon atom of the aromatic ring. The formation of a carbanion is unlikely in this reaction because the aromatic ring is not conducive to stabilizing a negative charge due to its electron-rich nature. Additionally, the reaction mechanism involves the formation of a benzyne intermediate, not a carbanion.

4. **Free radical (Option 4)**:
   - A free radical is a species with an unpaired electron. The reaction conditions and mechanism do not involve the formation of free radicals. The elimination-addition mechanism specifically involves the formation of benzyne, which is a distinct intermediate with a triple bond, rather than a free radical species.

In summary, the correct intermediate in the reaction of aryl halides with strong basic amide ions in liquid ammonia is benzyne, as it is the key species that facilitates the conversion to aniline. The other options do not align with the reaction mechanism and conditions, making them incorrect in this context.