IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

rags

O-methoxybromobenzene is treated with sodamide and liquid ammonia.

What is the product.THis was a question in my IIT test.

magiclko

this is an example of birch-reduction..

The Birch Reduction offers access to substituted 1,4-cyclohexadienes.... in this question... The effect of electron-withdrawing substituents is to be seen... the free radical formed will be stabilised. due to the electron with drwaing nature of OCH3 and Br group... so it will be formed nearest to them.... this will give the product as shown in attached figure!!!

khinart

As the magicko has given the answer it is the best.On further simple explanation I would say that such type of reactions follow a benzyne mechanism.Now it would be clear to U that the attacking nucleophile in the question is NH2 negative.So it will deliberately create a charge separation and then would form an intermediate benzyne(U must be knowing the structure I believe).The benzyne then breaks up and develops a + charge at ortho and meta positions giving rise to two products viz o-toluidine and p- toluidine.And as we know the OCH3 group is a -I group and hence stabilizes the ortho position so ortho product will be major.

arun-rashi

this reaction follow benzyne mechanism so this reaction go through elimination sustitution mechanism.in my view benzyne formation and -I effect of -OCH3 group major product is meta-bromoaniline and minor product is ortho bromo aniline.for detail machanism you consult I.L. FINAR.you must check the answer.

chem4884

i also think it is an elimination-addition reaction.......proceeding through benzyne intermediate....first the removel of HBr & then addition of HNH2...giving o-& p- substituted mothoxy benzene

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