# What happens to chemical shifts of protons close to aromatic systems in ¹H-NMR?

I have two questions regarding chemical shifts on aromatic systems.

First, what does being near an aromatic system do to protons? If we look at the spectra of 1-napthalenemethanol and then 9-anthracenemethanol, the peak of the two aliphatic protons of the benzylic methylene group get shifted way up. Can someone explain why adding in that third aromatic ring does this? The spectra:

I was also under the impression that protons near electronegative elements would be deshielded and be shifted higher. However in the spectrum of Benzil, the protons para to the carbonyl group are higher shifted than the protons meta to the carbonyl. Why is this? Is it to do with conjugation of the carbonyl with the aromatic ring? I'm having trouble finding a proton nmr in the literature, so there's nothing to compare it to.

• @Martin Thanks for your effort. It is much more easy to read now and hence more likely to get an answer.
– Jori
Apr 13 '15 at 8:16

The aromaticity found in certain ring systems (e.g. benzene, see Hückel's rule) causes a magnetic field because of the circular motion of the electrons in the ring (see picture).

(Source: Clayden et al. Organic Chemistry 2nd edition, p. 277)

This additional field enforces the magnetic field of the NMR machine inside the ring, while countering the field outside the ring, causing the signals of the outer hydrogens to shift downfield (i.e. to the larger shifts).

• First, what does being near an aromatic system do to protons? If we look at the spectra of 1-napthalenemethanol and then 9-anthracenemethanol, the peak of the two aliphatic protons of the benzylic methylene group get shifted way up. Can someone explain why adding in that third aromatic ring does this?

Lookig at the difference in shift between the benzylic methylene protons of 1-Naphthalenemethanol and 9-Anthracenemethanol we must conclude that the induced magnetic field of the second is stronger at that position than that of the former, as there is no other appreciable difference. Intuitively this seems to make sense as the field of the 9-Anthracenemethanol is more uniform at and closer to (with respect to the "right" side of the molecule as in the pictures) the discussed proton. Of course this could/should be confirmed by quantum mechanical calculations, but I do not have the tools or the knowledge to do that (if someone has, please feel free to expand this answer!)

• I was also under the impression that protons near electronegative elements would be deshielded and be shifted higher. However in the spectrum of Benzil, the protons para to the carbonyl group are higher shifted than the protons meta to the carbonyl. Why is this? Is it to do with conjugation of the carbonyl with the aromatic ring? I'm having trouble finding a proton nmr in the literature, so there's nothing to compare it to.

Yes, conjugating effects and inductive effects can counter each other, but there is no easy general way to say which will win out in a particular situation, although usually it is the conjugating effect.

Jori’s answer does a nice job in explaining why aromatic rings generally shift proton shifts downfield, and why this is not restricted to the aromatic protons themselves; see the pictures at the beginning of his post for explanation (and replace the benzene proton with a $\ce{CH2}$ group in the right-hand side picture for further clarity).

The difference in chemical shift $\Delta \delta(\ce{C\mathbf{H}_2OH}) = 0.47~\mathrm{ppm}$ which is a very small, almost neglegible difference. Compare:

• the difference between naphthalene-1-ylmethanol and benzyl alcohol: $\Delta \delta(\ce{C\mathbf{H}_2OH}) = 0.41~\mathrm{ppm}$

• the difference between benzyl alcohol and methanol: $\Delta \delta(\ce{C\mathbf{H}_{$n$}OH}) = 1.12~\mathrm{ppm}$

(Note that we cannot strictly compare the chemical shift difference between naphthalene-1-ylmethanol and anthracene-9-ylmethanol since the latter has been measured in $\ce{(CD3)2SO}$ rather than $\ce{CDCl3}$. However, the general trend works)

The big difference in shifts occur when we add an aromatic ring to our methyl alcohol; the number of six-membered aromatic rings does not really matter. This is because the first ring is added in close vicinity to the centre in question while the other rings are further away — we might compare this to the difference in chemical shifts between ethanol and 2-phenylethanol: $\Delta \delta (\ce{C\mathbf{H}_2OH}) = 0.09~\mathrm{ppm}$ (even smaller). However, this last comparison is not strictly valid due to the flexibility of 2-phenylethanol being larger than that of the polyaromatic systems.

Summed up in a nutshell: The chemical shift difference between naphthalene-1-ylmethanol and anthracene-9-ylmethanol can be attributed to the presence of a third aromatic ring in relative proximity of the group in question.

Your second (unrelated, and probably should have been posted separately) question deals with how groups outside of a phenyl ring influence the chemical shifts of protons belonging to that ring.

In the case of benzil, we are dealing with an electron-withdrawing carbonyl group, and we might as well choose the simpler benzaldehyde system for comparison. You can picture the electron-withdrawing effect of the carbonyl group by a set of enolate resonance structures including charge separation: We can relocate the positive charge to the ortho- and para-positions of the phenyl ring, but not to the meta-positions, resulting in a lower electron density at ortho and para and thus greater deshielding and a downfield shift.

Similarly for electron-donating groups such as the methoxy group in anisole: The ortho- and para-positions are shifted upfield because the electron-donating group supplies their positions with a greater electron density and thus more shielding. The meta-groups are rather unaffected by that.

This can be generalised to the (not necessarily always true!) statement:

Groups in benzylic positions generally leave meta-positions unaffected and affect mainly the chemical shifts of ortho- and para-positions.

Therefore we expect a downfield shift for ortho- and para-protons in benzil and no shift change for the meta ones.