Why is methyl m-nitrobenzoate formed instead of the ortho or para isomers?

Answer 1

Hello, It can be answered if we take a look at the ressonance structures formed in each attack (ortho, meta and para). Considering methyl benzoate and NO2+ reactants, NO2+ (nitronium ion) formed by mixing HNO3 and H2SO4. If NO2+ attacks in ortho or para, three ressonance structures can be drawn, one containing a positive charge in the carbon linked to the carbon of -COOMe, which is electronally deficient (the oxygens, which are extremely electronegatives, pull electrons to them in order to complete their octet, leaving the carbon with a less dense electron cloud). This causes a slighly repulsion, disfavouring the formation of oand p isomers. If the attack takes place in meta, three ressonance structures can be drawn. However, differently from o and p, no repulsion is observed: the positive charge can circulate inside the ring without causing any repulsion. Notice that, besides the formation of o and p is not favoured, there's still formation o a very small quantity of them. Taking the discussion to an electron level, we are intented to remember Quantum Chemistry, and its probability theories. It is not because it is not favoured that it will never occur. This reaction is an eletrofilic attack and, for this type of reaction, -COOMe is a meta directioner. If we had -Cl or -Br in the place of -COOMe, it would be ortho/para directioner, because the positive charge created in the carbon linked to the halogen would be stabilized by the dense electron cloud of the halogens.

Answer 2

Methyl m-nitrobenzoate is formed because the nitro group is a strong deactivator and meta-director in electrophilic aromatic substitution reactions. This group directs the incoming electrophile to the meta-position relative to itself on the benzene ring. As a result, the meta isomer is preferentially formed over the ortho or para isomers.