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For the H NMR of the compound I am wondering why peak (A) is the farthest out of all of them? I think that it is due to the electro negativity of oxygen. Am I correct? If so, why would the presence of oxygen push the peak to the left?enter image description here

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The further to the left a peak in an NMR spectrum is, the more deshielded it is. Deshielding or shielding are terms to describe how much a certain nucleus is shielded from the outer magnetic field that is applied during NMR measurements. Shielded nuclei will experience less of a magnetic field and thus resonate at a lower frequency meaning lower ppm-values; and vice-versa. In general, shielding can be explained by more electrons being in the vicinity of an atom because the electrons — in layman’s terms — can be thought to create an electric current induced by the outer magnetic field, which in turn generates an opposing magnetic field and lowers the effective magnetic field at a nucleus.

If we have an electronegative atom close by, that atom will draw electrons towards it, reducing the electron density around the nucleus we are observing and thus deshielding it. That not only works for electronegative atoms alone, but also for electron-withdrawing groups like the carbonyl group.

In your specific example we have two electron-withdrawing groups that will deshield neighbouring protons. However, the deshielding that the acetyl group is responsible for is stronger than that of the chlorine atom. So both pairs of protons are moved downfield (to higher ppm values) when compared to toluene or para-xylol, but the protons closer to the acetyl group are deshielded more.

It ultimately boils down to oxygen’s electronegativity in combination with an $\ce{sp^2}$ carbon that is not attached to any hydrogen, but it may be easier to think of carbonyl groups being electron-withdrawing per se.

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Building on Jan's very useful answer, it is often easy to consider these substituted benzenes by understanding the chemistry associated with the substituents; both $\ce{-Cl}$ and $\ce{-C(O)CH3}$ are electron withdrawing groups, and it is this property rather than straight electronegativity that controls the chemical shifts in this case. Halides are weakly electron withdrawing, and acetyl groups are reasonably strongly electron withdrawing. If we consider benzene at $\delta = 7.34$ as our starting point, we can see that both sets of aromatic signals have a deshielding response, but one more so than the other.

EW subsituents on a benzene ring will have the greatest influence on the protons on the ortho position and para position, with the least influence on the meta position. This can be justified by drawing out resonance structures:

enter image description here

I'll leave you to work out the effects of an electron donating group. Often, a good sign for working out whether you have an electron withdrawing or electron donating group attached to your ring is to note the relative shifts for the ortho and meta protons; ortho are upfield of the meta for EDG and downfield for EWG.

Now, you have two subtituents which are para-substituted. Knowing that the meta position is the least affected by substituents (in most cases — there are exceptions), it can be easy to rationalise the relative effects of the deshielding influence by considering the relative strengths of the electron withdrawing groups. An acetyl group has much greater electron withdrawing potential than chlorine, and hence the protons ortho to this group experience greater deshielding, and have a larger chemical shift.

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