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.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

personal, she chose to do it.