When 4‑chloro‑3‑methylbromobenzene (C6H4ClBrCH3) is treated with excess sodium amide (NaNH2) in liquid ammonia, which mechanism predominates? Outline the steps of the favoured mechanism and explain why it is preferred over the direct SNAr pathway.
Answer
Benzyne mechanism; deprotonation at the position ortho to chloride forms a carbanion, elimination of chloride gives benzyne, addition of NH2⁻ to the benzyne, protonation yields an aniline derivative, followed by deprotonation to give the final product.
Step-by-step solution
- Step 1: Identify the substituents on the aromatic ring – a chloro and a bromo group. Neither is a strong electron‑withdrawing group that would stabilize a σ‑complex for a direct SNAr.
- Step 2: In liquid ammonia, NaNH2 acts as a strong base and deprotonates the hydrogen ortho to the chloride (adjacent to the electron‑poor site).
- Step 3: Loss of the chloride ion from the carbanion generates a highly strained benzyne intermediate (triple bond between the two ortho carbons).
- Step 4: The nucleophile NH2⁻ adds to one of the carbons of the benzyne to form a carbanion intermediate (addition of the nucleophile to a 1,4‑addition pattern).
- Step 5: Protonation of the carbanion by ammonia (or the solvent) gives the substituted aniline derivative (3‑(amino)‑4‑chloro‑1‑methyl‑benzene).
- Step 6: The product is isolated after deprotonation and purification.
- Step 7: This benzyne route is preferred because the substrate lacks electron‑withdrawing groups required for the addition–elimination (SNAr) mechanism, making the formation of a Meisenheimer complex energetically unfavourable.
- Step 8: Thus the reaction proceeds via the benzyne mechanism, giving the substituted product with an amino group at the 3‑position.
Common mistake: Assuming direct SNAr occurs due to the presence of halogens
Concept tested: Benzyne mechanism in haloarene substitution
Published · How this solution was prepared and checked
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