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Advanced Organic Chemistry: H-1 NMR spectrum of ethylbenzene

The H-1 hydrogen-1 (proton) NMR spectrum of ethylbenzene

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK IB KS5 A/AS GCE advanced A level organic chemistry students US K12 grade 11 grade 12 organic chemistry courses involving molecular spectroscopy analysing H-1 NMR spectra of ethylbenzene

C8H10 low and high resolution H-1 proton nmr spectrum of ethylbenzene analysis interpretation of chemical shifts ppm spin spin line splitting diagram H1 1-H nmr for ethylbenzene doc brown's advanced organic chemistry revision notes

TMS is the acronym for tetramethylsilane, formula Si(CH3)4, whose protons are arbitrarily given a chemical shift of 0.0 ppm. This is the 'standard' in 1H NMR spectroscopy and all other proton shifts, called chemical shifts, depend on the individual (electronic) chemical environment of the hydrogen atoms in an organic molecule - ethylbenzene here.

The chemical shifts quoted in ppm on the diagram of the H-1 NMR spectrum of ethylbenzene represent the peaks of the intensity of the chemical shifts of (which are often groups of split lines at high resolution) AND the relative integrated areas under the peaks gives you the ratio of protons in the different chemical environments of the ethylbenzene molecule.

Ethylbenzene, C8H10 , C6H5CH2CH3 , (c) doc b , (c) doc b

Interpreting the H-1 NMR spectrum of ethylbenzene

For relatively simple molecules, the low resolution H-1 NMR spectrum of ethylbenzene is NOT a good starting point.

At low resolution, the protons appear to be in a 5:2:3 ratio, however with very high resolution you can demonstrate that the hydrogen atoms (protons) of ethylbenzene occupy 5 different chemical environments - the proton ratio from chemical shifts (a) to (e) is 1:2:2:2:3.

Considering in detail the chemical shifts (a) to (e) on the H-1 NMR spectrum diagram for ethylbenzene.

Although there are 10 hydrogen atoms in the molecule, there are 5 possible different chemical environments for the hydrogen atoms in ethylbenzene molecule.

The high resolution H-1 NMR spectrum of ethylbenzene

The ppm quoted on the diagram represent the peak of resonance intensity for a particular proton group in the molecule of ethylbenzene - since the peak' is at the apex of a band of H-1 NMR resonances due to spin - spin coupling field splitting effects - see high resolution notes on ethylbenzene below.

(a) to (c) 1H Chemical shifts of 7.0 to 7.45

These correspond to three groups of protons of the benzene ring protons on carbon atom 4, atoms 3 = 5 and atoms 2 = 6 (conventional numbering).

All five benzene ring protons of ethylbenzene are in a very similar chemical environment - hence the similarity of chemical shifts.

This theoretical gives proton ratios of 1:2:2, but the proton resonances, and their splittings, are so close as to appear as a ratio group of 5.

I couldn't find very high H-1 NMR resolution data for ethylbenzene from the internet.

However, you can readily apply the n+1 rule to the side-chain alkyl group of ethylbenzene and make some predictions and deductions.

(d) 1H Chemical shift of 2.63 ppm.

This resonance corresponds to the protons of the CH2 group.

It is split into a 1:3:3:1 quartet by the protons of the CH3 group.

Evidence for the presence of a CH3 group in the molecule of ethylbenzene

(e) 1H Chemical shift of 1.22 ppm

This resonance corresponds to the protons of the CH3 group.

It is split into a 1:2:1 triplet by the protons of the CH2 group.

Evidence for the presence of a CH2 group in the molecule of ethylbenzene

In the absence of other similar chemical shifts, (d) and (e) provide evidence of an ethyl group in the ethylbenzene molecule.

Also, note the significant difference in the chemical shifts of alkyl protons compared to those of the aryl (benzene ring) protons in the H-1 NMR spectrum of ethylbenzene.


Number of directly adjacent protons 1H causing splitting Splitting pattern produced from the n+1 rule on spin-spin coupling and the theoretical ratio of line intensities
0 means no splitting             1            
1 creates a doublet           1   1          
2 creates a triplet         1   2   1        
3 creates a quartet       1   3   3   1      
4 creates a quintet     1   4   6   4   1    
5 creates a sextet   1   5   10   10   5   1  
6 creates a septet 1   6   15   20   15   6   1

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