3‑Chloro‑4‑methylbenzene is reacted with sodium amide ($\mathrm{NaN}{\mathrm{H}}_2$) in liquid ammonia. (a) Does the substitution proceed via the direct SNAr mechanism or via the benzyne mechanism? (b) What is the major organic product formed? (c) Outline the steps of the favored mechanism.

Step-by-step solution

1. 1. Examine the aromatic substrate: chlorobenzene has a $\mathrm{Cl}$ at the meta position relative to the methyl group; no strong electron‑withdrawing group is present.
2. 2. Direct SNAr (addition‑elimination) requires an electron‑withdrawing group ortho or para to the leaving group to stabilise the Meisenheimer complex. Without such a group, the SNAr pathway is highly unfavourable.
3. 3. The absence of an EWG makes the benzyne mechanism the dominant pathway for this substrate.
4. 4. Benzyne mechanism involves:
5. (i) Deprotonation of the $alpha$‑C (C2) adjacent to the chloride by $\mathrm{N}{\mathrm{H}}_2^{-}$ to generate a carbanion.
6. (ii) Elimination of $\mathrm{C}{\mathrm{l}}^{-}$ gives a highly reactive benzyne intermediate (C2–C3 triple bond).
7. (iii) Nucleophilic attack of $\mathrm{N}{\mathrm{H}}_2^{-}$ at the electron‑poor carbon of the benzyne (C3).
8. (iv) Protonation of the anionic intermediate yields the amino substituted product.
9. 5. Thus the chloride at C3 is replaced by an amino group while the methyl group remains at C4, giving 4‑methyl‑3‑aminobenzene as the major product.

Concept tested: Benzyne mechanism in aromatic substitution

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked