[Marcus]

Here is the NMR spectrum of the following compound:





Moving upfield to downfield, the first and second singlets correspond to the methyl group of the acetamide and the methyl group para to it, respectively.

Next, there is a doublet and singlet which I am confused about. I know that these two correspond to the hydrogens ortho and meta to the acetamide.

Thirdly, there is the second doublet. This must correspond to the proton ortho to the nitro and methyl groups.

Finally, the small singlet far downfield would seem to be the nitrogen's proton in the acetamide.

My confusion arises between the two remaining protons between the acetamide and the methyl group. The NMR spectrum of 4-methylacetanilide reveals that the chemical shift of the proton ortho to the methyl group is smaller than that of the proton ortho to the acetamide. So...

Introduction of a nitro group is going to heavily deshield its ortho proton, explaining the furthest doublet downfield. But its effect on its para and meta protons is unclear.

I'm trying to understand the change in chemical shifts in terms of deshielding through introduction of another electron withdrawing group, but I think I'm just confusing myself.

[Smrt guy]

The nitro group is a strongly electron withdrawing group both in terms of resonance and inductive effects.  Overall, all protons should be deshielded by the nitro group, but the o/p- positions should be most deshielded.  Also, remember and acetamide is resonance donating.

[orgopete]

If I name the compound as 4-methyl-2-nitroacetanilide, then the assignment of H3, H5 and H6 must be as follows:
H6 must be the hydrogen at 8.5 ppm as it has an ortho-hydrogen.
H3 must be at 7.9 as it does not have an ortho-hydrogen.
H5 must be at 7.4 as it has an ortho and meta-hydrogen.

ChemDraw calculates a similar assignment.

[pacifyer]

Another approach, is to take advantage of the J values: large (close to 10 Hz) for ortho coupling, small for meta coupling... and nil or close to zero for para couplings... that allows the assignment in this case, too.
The signal at 8.5 should only have an o-coupling, the one at about 8, only a small, m-coupling, and the last, at 7.4, is a dd...
Hope it helps, too..