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When o-bromoanisole is treated with nanh2 in nh3 (l), a reaction proceeding via a benzyne intermediate, what two isomeric products are possible? (check two products)?

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When orthobromoanisole is treated with sodamide (NaNH₂) and ammonia (NH₃), benzyne is formed and the tripple bond is formed between C₂-C₃ carbon atoms of benzene ring. Then NH₂⁻ can attack at ortho position as well as meta position.

The reaction takes place in three steps. Step-1:The amide NH₂⁻ ion attacks the H -atom that is ortho to C3 , generating a carbanion. Step-2:Loss of Br - to form a benzyne intermediate.Step-3:Addition of NH₂⁻ ion. The strain caused by a triple bond in a benzene ring can be relieved by a nucleophilic addition of NH₂⁻.

Two isomeric products are formed: (1) o-methoxyaniline and (2) m-methoxyaniline. The structures of both the isomeric products are given in diagram.

When o-bromoanisole is treated with nanh2 in nh3 (l), a reaction proceeding via a-example-1
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