In the Meisenheimer complex (carbanion intermediate) formed during the reaction of 1-chloro-4-nitrob — Aromatic Hydrocarbons Chemistry Question
Question
In the Meisenheimer complex (carbanion intermediate) formed during the $S_NAr$ reaction of 1-chloro-4-nitrobenzene with hydroxide ion, how many carbon atoms strictly within the six-membered aromatic ring remain $sp^2$ hybridized?
💡 Solution & Explanation
During the initial addition step of the $S_NAr$ mechanism, the incoming nucleophile attacks the carbon bearing the leaving group. This specific carbon atom changes its hybridization from $sp^2$ to $sp^3$ to accommodate both groups simultaneously. The remaining 5 carbon atoms in the six-membered ring remain $sp^2$ hybridized to allow for the resonance delocalization of the negative charge.