In nucleophilic aromatic substitution, what leaving group and substituent pattern typically enable the reaction?

• A. A leaving group on an unactivated ring with an electron-donating group.
• B. The reaction occurs with alkanes.
• C. A leaving group on an activated ring (e.g., halide) with an electron-withdrawing group such as nitro substituent.
• D. A strong base is required rather than a leaving group.

Answer: (C)

Nucleophilic aromatic substitution works best when the ring is activated toward nucleophilic attack and there is a good leaving group on the ring. The activation comes from an electron-withdrawing substituent that lies ortho or para to the leaving group, which stabilizes the negative charge that develops in the intermediate (the Meisenheimer complex) when the nucleophile adds. This stabilization makes it possible for the leaving group to depart, giving the substituted aromatic product. An activated ring with a leaving group such as a halide next to an electron-withdrawing group like nitro is the classic, reliable pattern for SNAr.

If the ring isn’t activated by an electron-withdrawing group, the intermediate is not stabilized and the reaction doesn’t proceed readily. Reactions on alkanes aren’t SNAr because there’s no aromatic ring to form the Meisenheimer intermediate. And while a strong base can drive other pathways (like benzyne formation), SNAr itself relies on that leaving group plus ring activation to proceed via the addition–elimination mechanism, not solely on base strength.