Question

In the reaction,

The major product $A$ is

Answer 1

4-methoxyaniline

Answer 2

The reaction involves 1-chloro-4-methoxybenzene (p-chloroanisole) reacting with sodium amide (NaNH₂) in liquid ammonia (liq. ammonia). This set of reagents is characteristic for a nucleophilic aromatic substitution reaction proceeding via the benzyne mechanism.

The mechanism involves three main steps:

1. Formation of the benzyne intermediate (Elimination): The strong base, amide ion (NH₂⁻), abstracts an acidic proton ortho to the halogen (Cl). There are two possible ortho positions to the chlorine atom (C1): C2 and C6. The methoxy group (-OCH₃) is at C4.

   • Abstraction of H from C2: This forms a carbanion at C2. C2 is meta to the -OCH₃ group. The resonance (+M) effect of -OCH₃ does not significantly affect the meta position, and its inductive (-I) effect (electron-withdrawing) can slightly stabilize the carbanion.
   • Abstraction of H from C6: This forms a carbanion at C6. C6 is ortho to the -OCH₃ group. The strong electron-donating resonance (+M) effect of -OCH₃ would destabilize a carbanion at the ortho position.

   Therefore, the abstraction of the proton from C2 is kinetically favored, leading to a more stable carbanion at C2. This carbanion then eliminates the chloride ion (Cl⁻) from C1 to form the benzyne intermediate with a triple bond between C1 and C2. This intermediate is 4-methoxybenzyne (considering OCH₃ at C4).

2. Nucleophilic attack on the benzyne (Addition): The amide ion (NH₂⁻) acts as a nucleophile and attacks one of the carbons of the benzyne triple bond.

   For 4-methoxybenzyne (triple bond between C1 and C2, OCH₃ at C4):

   • Attack at C1 (original Cl position): The triple bond shifts, forming a carbanion at C2. C2 is meta to the -OCH₃ group.
   • Attack at C2 (ortho to original Cl position): The triple bond shifts, forming a carbanion at C1. C1 is para to the -OCH₃ group.

   The -OCH₃ group is an electron-donating group by resonance (+M effect). It destabilizes negative charges at ortho and para positions. Thus, the carbanion formed at C2 (meta to -OCH₃) is more stable than the carbanion formed at C1 (para to -OCH₃).

   Therefore, the nucleophilic attack at C1 is preferred, leading to the carbanion at C2.

3. Protonation of the carbanion: The carbanion formed in the previous step (at C2) abstracts a proton from liquid ammonia (the solvent) to form the final product.

The final product, A, has the -NH₂ group at the original position of the chlorine atom (C1) and the -OCH₃ group at C4. This compound is 4-methoxyaniline (also known as p-anisidine).

Structure of Reactant (1-chloro-4-methoxybenzene):
      OCH3
        |
        C4
       /  \
      C3   C5
     /      \
    C2       C6
    |        |
    C1-------C
    |
    Cl

Structure of Major Product (4-methoxyaniline):
      OCH3
        |
        C4
       /  \
      C3   C5
     /      \
    C2       C6
    |        |
    C1-------C
    |
    NH2

The image provided in the question shows the major product A as:

      OCH3
        |
        C
       /  \
      C     C
     /       \
    C         C
    |         |
    C---------C
    |
    NH2

This corresponds to 4-methoxyaniline.

The final answer is $\boxed{\text{4-methoxyaniline}}$

Explanation of the solution:

The reaction of 1-chloro-4-methoxybenzene with NaNH₂/liq. ammonia proceeds via the benzyne mechanism. The strong base abstracts the ortho-proton at C2 (meta to -OCH₃) preferentially over C6 (ortho to -OCH₃) due to carbanion stability. This forms 4-methoxybenzyne. The nucleophile (NH₂⁻) then attacks C1 (para to -OCH₃) of the benzyne, as this leads to a more stable carbanion at C2 (meta to -OCH₃). Subsequent protonation yields 4-methoxyaniline as the major product.