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)?

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.

ill try to . no promises tho.

explanation:

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante