The major product (F) in the following reaction is
This reaction involves sodium amide (NaNH2) in liquid ammonia, which are conditions for nucleophilic aromatic substitution proceeding via an elimination-addition (benzyne) mechanism.
(A) Elimination (Formation of Benzyne): The strong base, amide ion (NH2-), removes a proton from a carbon adjacent to the carbon bearing the leaving group (Cl). In m-chloroanisole, protons are available ortho to Cl (at C2 and C4 relative to OCH3).Both pathways lead to benzyne intermediates where the triple bond involves C3. Example (deprotonation at C4): OCH3- --> Benzyne
(D) Addition (Attack by Nucleophile): The nucleophile (NH2-) attacks one of the carbons of the benzyne triple bond. The direction of addition is influenced by the substituent (-OCH3).The methoxy group (-OCH3) has a significant inductive electron-withdrawing effect (-I) which dominates in directing the addition to benzyne. This effect stabilizes the negative charge when it is further away from the electronegative oxygen. Therefore, the nucleophile (NH2-) preferentially adds to the carbon atom meta to the OCH3 group (C3 in both benzyne possibilities), placing the resulting carbanion at C2 or C4.
(G) Protonation: The resulting aryl anion is quickly protonated by the solvent (liquid NH3) to give the final product. Addition of NH2- at the C3 position (meta to OCH3) leads to the formation of m-anisidine (3-methoxyaniline).
The acid formed when propyl magnesium bromide is treated with CO_2 followed by acid hydrolysis is:
The best reagent for converting propanamide into propanamine is: