I wanted to analyse and suggest the peaks for the NMR spectrum and ran into a problem.

I knew that the methyl $\ce{CH3}$ produces a singlet peak with integration $3$ but when I ran into the benzene protons, I thought that the two equivalent protons (one on each side) would split into a doublet (considering if coupling occurs through 3 bonds or fewer only).

So this was my solution to the problem:

  • s, $\ce{3H}$, integrated $3$, ppm $2.0$ due to $\ce{CH3}$ group.
  • d, $\ce{3H}$, integrated $2$, ppm $7.2$ due to $\ce{H}$ on benzene furthest from $\ce{NO2}$.
  • d, $\ce{3H}$, integrated $2$, ppm $7.9$ due to $\ce{H}$ on benzene closest to $\ce{NO2}$.

However, the spectrum answer given by the mark scheme was as such:

enter image description here

Where have I made a mistake?


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