Question

How can NMR be used to distinguish the ring substitution at meta position instead of ortho or para nitronium attack on methyl benzoate?

Answer 1

Nitration is one of the most important examples of electrophilic aromatic substitution. Aromatic nitro compounds are used in products ranging from explosives to pharmaceutical synthetic intermediates. The electrophile in nitration is the nitronium ion (NO2 + ). The nitronium ion is generated from nitric acid by protonation and loss of water, using sulfuric acid as the dehydrating agent. The reaction is given as -

Electron‐withdrawing substituents are usually meta‐directors. Ortho‐para attack on a deactivated ring would place a positive charge on the carbon that bears the any substituent X. However, when X is an electron‐withdrawing substituent, the electronic effects of the substituent destabilize the positive charge. Consequently, meta attack is favored because all resonance forms, avoid this unfavorable electronic interaction.

For 3-nitro methyl benzoate, you'd expect the following signals on a 1H NMR spectrum:
the 3 hydrogens in the methyl group would appear as 1 singlet at around 4 ppm.
for the 4 other hydrogens, on the benzene ring, they'll appear around 9-7 ppm, the aromatic area. you'd have that the H on position 2 (between the carbmethoxy and the nitro groups) would also apear as a singlet, and, being alpha to two electron withdrawing groups, you can expect it to be the signal at the highest ppm. the H on positions 4 and 6 are alpha to 1 of the groups, and would appear somewhere on the middle, and the H on position 5 would appear at the lowest ppm of them all (always in the area between 9 and 7).
Now, if you look at your spectrum:
the peak at 4 ppm integrates for 3, as we predicted.
also, there's a singlet at 9 ppm which integrates for 1 (H on C_ring-2), a triplet at 7.6 ppm, which also integrates for 1 (H on C_ring-5. the fact that its a triplet means that the couplings with its neigbouring H atoms are not too different from eachother). and finally, at 8.2 ppm you have two signals, doublets, really close to eachother. they integrate for 2 and correspond to the remaining H atoms.