Ortho Meta Para Nmr Shifts

[Alice Palecek]

Protonsdirectly attached to an aromatic ring, commonly called aryl protons, show up about 6.5-8.0 PPM. This range is typically called the aromatic region of an 1H NMR spectrum. Protons on carbons directly bonded to an aromatic ring, called benzylic protons, show up about 2.0-3.0 PPM.

For the 1H NMR spectrum of p-bromotoluene, the absorption for the benzylic protons appears as a strong singlet at 2.28 ppm. Due to the molecule's symmetry, the aryl protons appear as two doublets at 6.96 & 7.29 ppm.

Carbons in an aromatic ring absorb in the range of 120-150 ppm in a 13C NMR spectrum. This is virtually the same range as nonaromatic alkenes (110-150 ppm) so peaks in this region are not definitive proof of a molecule's aromaticity.

Due to the decoupling in 13C NMR, the number of absorptions due to aromatic carbons can easily be observed. This can be used to determine the relative positions (ortho, meta, or para) for di-substituted benzenes. Peak assignments can be simplified by noting that 13C peaks tend to be larger if two carbons contribute to the absorption.

Benzenes substituted with two identical groups have a relatively high amount of symmetry. In all three configurations, there is a plane of symmetry which reduces the number of distinct aryl carbon absorptions to less than six. The ortho configuration has a plane of symmetry which mirrors each carbon in the benzene ring causing only three 13C aryl absorptions to occur. The meta configuration's plane of symmetry mirrors two carbons of the benzene ring allowing for four aryl absorptions to occur. The para configuration actually has two planes of symmetry (one vertical and one horizontal on the structure below) through the benzene ring, which only allows two distinct aryl absorptions to occur.

The lack of a plane of symmetry in asymmetrical di-substituted benzenes makes each carbon in the ortho and meta configuration unique. Consequently, their 13C NMR spectra show six arene absorptions. However, the para configuration has a plane of symmetry drawn through the two substituents which mirrors two carbons of the benzene ring causing only 4 arene absorptions to appear.

15.7: Spectroscopy of Aromatic Compounds is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Steven Farmer, Dietmar Kennepohl, Layne Morsch, Tim Soderberg, & Tim Soderberg.

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Monosubstituted rings will have 5 protons in the region 6.5-8.5 ppm; disubstituted rings will have 4 protons; trisubstituted rings will have 3 protons (and so on). Examples of the NMR of aromatics of mono-, di-, and tri-substituted aromatics are shown below. When interpreting the spectrum of an aromatic compound, remember to count the number of protons in the aromatic region to determine how many times the ring is substituted.

If an aromatic ring has more than one substituent, careful analysis of the shifts and splitting pattern of the protons in the aromatic region reveals the positions on the ring of the different substituents. However, the the shifts depend on the substituents and the splitting is not really first order. Most sophomore-level courses do not cover NMR spectroscopy in the depth required to analyze the aromatic region. Therefore, you will not be able to designate the exact ring substition in many cases. In the specific case of disubstituted aromatic rings, para-substituted rings usually show two symmetric sets of peaks that look like doublets. The para-substitution NMR aromatic region pattern usually looks quite different than the patterns for both ortho- and meta-substituted aromatic rings.

Examples of ortho, meta, and para substitution are illustrated in the NMR spectra of different isomers of chloronitrobenzene, below. The CDCl3 peak is pointed out in each spectrum. (The samples were run using CDCl3 as the solvent, and a small contaminant of this deuterated solvent is CHCl3, which shows up at 7.24 ppm. This is used to calibrate the spectrum.) Note the symmetry of the para substituted chloronitro benzene. (Click on each full-size image to view details of the region from 6.5-8.5 ppm.)

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.

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!)

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.

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.

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