1H nmr of Substituted Benzenes
A Worksheet

Remember, the chemical environment of a proton determines where it will resonate when placed in a

magnetic field, i.e. determines its chemical shift. Charge distribution in aromatic rings is reflected in the

chemical shifts of protons attached to the rings. Charge distribution is another way of discussing electron

density. The pi-electron distribution in aromatic rings is strongly affected by electron donation or

withdrawal by conjugated substituents through resonance. So charge distribution or electron density can

be predicted by drawing out all the resonance structures which are possible. Large electron density (or

negative charge) shields the proton and thus shifts it upfield. Small electron density (or positive charge)

deshields the proton and shifts it downfield.

There are two things to keep in mind when analyzing the 1H nmr spectrum of protons attached to

an aromatic ring (also called aromatic protons).

1. Electron density around the proton of interest determines its chemical shift.

One normally speaks of chemical shift difference from benzene protons (7.24 ppm). Consider

aniline for example. Would we expect the chemical shift of the ortho protons to be shifted upfield from

benzene (< 7.24 ppm) or downfield from benzene (> 7.24 ppm)? To answer this question we need to draw

the various resonance structures of aniline. These structures demonstrate the strong electron donation into

the ring by the nitrogen substituent.

NH2

H

H

HH
Notice in the first resonance structure there is a negative charge located on the carbon to which an ortho

H

H

NH2

-

••

H

H

H

H

This proton has lots of
electron density around it

+

proton is attached. This means it is experiencing greater electron density than it would if there were no

nitrogen attached to the ring, therefore we would expect it to be shifted upfield with respect to benzene. In

fact, it comes at 6.5 ppm. Draw the other resonance structures of aniline in the space provided and predict

the relative chemical shift of the meta and para protons.

2. Splitting by adjacent protons only occurs if the chemical shift of the two protons are significantly

different.

An example of where this plays a significant role in the appearance of the aromatic region in the 1H
nmr spectrum is toluene.

CH3

H

H

H

H

H

para

meta

ortho

There are obviously 3different types of aromatic protons, however, there is only one signal in the 1H nmr

spectrum in the aromatic region. This signal is a singletat 7.10 ppm. Although methyl is an electron

donating substituent, as evidenced by the observed chemical shift being upfield from benzene, it is only

weakly so. It does not donate enough electron density to cause a magnetic non-equivalence at 60 MHz.

Since the protons have the same chemical shift they do not split each other and therefore a singlet is

observed.

Splitting is observed when the chemical shifts of aromatic protons are significantly different. One

way to get such differences in the chemical shift is to have both an electron donating group and an electron

withdrawing group on the same ring. An example is p-nitroaniline, draw all the resonance forms, then

predict the chemicals shifts (relative to benzene) and the splitting pattern one would observe.

NH2

H

H

N

H

H

O

O

-

+

The only way to learn the process of predicting the chemical shifts of aromatic protons on substituted

benzenes by resonance theory is by practice. The following examples will help accomplish this.

Write the resonance structures for each of thefollowing compounds based on the charge

distribution in each structure which protons will be shielded (come upfield) and deshielded (come

downfield) with respect to the protons in benzene?

O

CH3

H

H

H

H

H

C

H

H

H

H

H

O

H

O

CH3

H

H

NH2

H

H

The last example contains chlorine, a halogen. We have not discussed halogens in the context of nmr yet.
However, remember how it affected the aromatic ring towards electrophilic aromatic substitution? Was it
an electron donator or an electron withdrawer? The same principles hold here. Look on page two of your
other handout for the actual 1H nmr of this molecule. Chlorine has essentially no observable effect on the
chemical shift of the protons on the ring because the electron donation (by resonance) is canceled out by its
electron withdrawing ability (by induction). This is also true of the other halogens.

Cl

H

H

NH2

H

H