Question

The major product (F) in the following reaction is

• A. o-Anisidine
• B. m-Anisidine
• C. p-Anisidine
• D. p-Chloro aniline

Hint:

Using a simple frame or just bolding for the box Key Points: Reaction of aryl halides lacking strong EWGs with NaNH$_2$/liq NH$_3$ proceeds via benzyne mechanism (Elimination-Addition). Benzyne formation involves deprotonation ortho to the leaving group, followed by elimination of the leaving group. Nucleophilic addition to the benzyne is regioselective, often guided by the inductive effects of substituents. Electron-withdrawing groups (like -OCH$_3$ via -I effect) direct the incoming nucleophile meta to themselves to best stabilize the intermediate carbanion. The product is m-anisidine.

Answer 1

The Correct Option is B

This reaction utilizes sodium amide (NaNH₂) in liquid ammonia, conditions suitable for nucleophilic aromatic substitution via an elimination-addition (benzyne) mechanism.
(A) Elimination (Benzyne Formation): The strong base, amide ion (NH₂^-), removes a proton from a carbon next to the carbon with the leaving group (Cl). In m-chloroanisole, protons are available ortho to Cl (at C2 and C4 relative to OCH₃). Both pathways form benzyne intermediates where the triple bond involves C3. Example (deprotonation at C4): OCH₃- --> Benzyne
(D) Addition (Nucleophile Attack): The nucleophile (NH₂^-) attacks a carbon of the benzyne triple bond. The substituent (-OCH₃) influences the addition direction. The methoxy group (-OCH₃) has a substantial inductive electron-withdrawing effect (-I), which directs addition to benzyne. This effect stabilizes the negative charge when distant from the electronegative oxygen. The nucleophile (NH₂^-) preferentially adds to the carbon atom meta to the OCH₃ group (C3 in both benzyne possibilities), placing the carbanion at C2 or C4. 
(G) Protonation: The aryl anion is rapidly protonated by the solvent (liquid NH₃) to yield the final product. Addition of NH₂^- at C3 (meta to OCH₃) results in m-anisidine (3-methoxyaniline) formation.