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Chemical and Magnetic Equivalence
Understanding whether two atoms in a molecule are equivalent or inequivalent is important in
understanding the NMR spectra produced. If two atoms are equivalent they give one signal in an
NMR spectrum. Conversely if two atoms are inequivalent they give two different signals in the
NMR spectrum. This idea can be extended to more than two atoms, and the concept of “atoms”
can also be replaced with “functional groups”. The discussion will focus on protons (hydrogen)
and carbon, but can be applied to any atom in a molecule.
The concept of equivalence can be approached in two ways: look at the structure, replace one
atom with a different atom, and examine the results or look for a symmetry element in the molecule. The first example is propane. Replace any one of the six protons on the end of the molecule with a bromine atom, the result is the same—1-bromopropane. These six protons are thus
chemically equivalent and give one signal in a 1H NMR spectrum. The protons are also said to be
chemical shift equivalent, another (possibly better) word for this is homotopic.
Replacement of either of the two protons on C2 (the middle carbon) with a bromine atom also
gives the same result—2-bromopropane. These two protons are chemically equivalent, chemical
shift equivalent, or homotopic.
Now compare the result of replacing a proton on the end of propane with replacing a proton in
the middle. 1-Bromopropane and 2-bromopropane are different compounds, and the protons on
the ends of propane are NOT chemically equivalent to the protons in the middle of propane. The
protons can also be described as heterotopic. The 1H NMR spectrum for propane has two signals;
a six-proton signal at 0.90 ppm and a two-proton signal at 1.34 ppm.1
We can get the same results using symmetry elements in the molecule. First examine a methyl
group, and recall that there is free rotation around the single bond attaching the methyl group
to the rest of the molecule. We can move the protons around this rotation axis and get the same
arrangement every time. The three protons of a methyl group are chemically equivalent by symmetry.
Whether propane is rotated 180° around the middle carbon, or reflected in a mirror through the
middle carbon (perpendicular to the C–C bonds), the two methyl groups are chemically equivalent and thus homotopic. Rotation around the same axis as above, or a reflection through a mirror
plane containing all the C–C bonds, shows the two protons on the middle carbon are chemically
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equivalent. No symmetry operation will make the methyl groups equivalent to the protons on
C2, and thus we get the same result as above.
Coupling between protons now must be considered. Do chemically equivalent protons couple?
The answer is yes, BUT the coupling is not observable. The full explanation involves long journeys into the quantum mechanics of NMR, so we’ll leave it for the time being (or for all time …).
The result is that a group of homotopic protons uncoupled to any other protons will give a singlet. Examples here would be a methyl group attached to an aromatic ring, or a methylene group
in an acetal (–O–CH2–O–).
Protons that are not chemically equivalent (i.e. heterotopic)
will couple if they are close enough in bonding distance—2
or 3 bonds if through bonds, possibly longer if  bonds
are present. For propane, the CH2 protons are not chemically equivalent to the two CH3 groups, coupling occurs,
and the 1H NMR spectrum of propane1 (right) has a septet
for the CH2 signal and a triplet for the CH3 signal (look
closely).
Aromatic compounds present interesting situations because of this long-range coupling. Consider 4-chlorotoluene as the example. The symmetry in this para-disubstituted molecule should be
evident. Thus the two protons next to the methyl group are chemically equivalent, as are the two
protons next to the chlorine.
Considering a proton ortho to the methyl group (Ha), we would
initially expect a doublet for that signal—the 3J coupling between
the closest protons should be about 8 Hz, and the 5J coupling is
less than 1 Hz and not easily observed. Therein, however, lies the
problem—the Ha proton has a different coupling constant to two
chemically equivalent protons. The protons are said to be magnetically inequivalent. Note that from the point of view of one Ha proton the two Hb protons are different. The most noticeable effect in
the spectrum is the presence of “extra” resonances of low intensity on either side of the strong
doublet. The exact position and intensity of each of the lines can be calculated using the secular
determinants (quantum mechanical stuff again) but this is beyond the scope of this course. The
spectrum is shown at the top of the next page.
The correct designation for this system is an [AX]2 spin system, the designation AA’XX’ is seen
in older texts. The signal may not be described as a “doublet”. It is an interesting example where
the usual first order approximation is not valid, and an increase in the spectrometer frequency
will not change this observation (most signals become more “first order” as the spectrometer
frequency increases). The spectra may not be simplistically analyzed. While the chemical shifts
are the mid-points of the patterns, the coupling constants are not necessarily equal to the spacing
between the most intense signals. Thus reporting a range of  values for the signal is preferred.
You should consider the following style for reporting the analysis:
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
Multiplicity
Integration
J
Assignment & Comment
7.24–7.18
m
2H
~ 8.0 Hz
Hb protons of [AX]2 spectrum. The
reported J is the approximate coupling constant based on the most
intense lines.
7.15–7.00
m
2H
~ 8.0 Hz.
Ha protons of [AX]2 spectrum. The
reported J is the approximate coupling constant based on the most
intense lines.
The same effect occurs in ortho disubstituted compounds when the
substituents are the same, as in o-xylene. The symmetry and chemical equivalence of the protons is fairly evident. The initial expectation is a doublet of doublets for each signal. Consider either Ha—
this proton should have a 3J coupling to the nearest Hb proton and
a 4J coupling to the other Hb proton. Again, the proton being observed has a different coupling constants to two chemically equivalent
protons. Magnetic inequivalence again. The spectrum of o-xylene
has other complications and is not shown.
There is a curious difference in symmetry for the last two examples. For the para disubstituted example, different substituents were necessary, otherwise all the aromatic protons are homotopic.
For the ortho-xylene example, the same substituent was necessary, otherwise all the aromatic
protons would be heterotopic. Curious indeed!
Examples also exist for aliphatic compounds, usually where there is
symmetry in the molecule. Consider 1,4-dibromobutane. There are two
chemically equivalent methylene groups next to the bromine atoms (C1
and C4) and two chemically equivalent methylene groups in the middle (C2 and C3). One would
hope for a nice-and-easy pair of triplets. But remember that chemically equivalent protons do
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couple with themselves—we just don’t observe the coupling in a normal 1H NMR spectrum. So
the protons on C2 couple to the protons on C3. From the point of view of the C1 methyl group, it
couples to the protons on C2, but not to the protons on C3. We again have an example of magnetic
inequivalence, in that the C1 methyl group has different coupling constants to chemically equivalent protons. Note that it doesn’t matter that one of the coupling constants is zero. The signals are
more complex that expected. This effect is called “virtual coupling” in other sources.
Does magnetic inequivalence cause complexity in 13C spectra? It could, and it should, but we
don’t see it. Remember that when 13C spectra are recorded, the protons are decoupled, and no
effect of 13C-1H coupling are observed. The natural abundance of carbon-13 takes care of any 13C13C coupling, as it is rare that a small molecule will have two 13C atoms in it.
Lastly, consider 1,1-difluoroethene. It’s clear from the symmetry that the two
protons are chemically equivalent, as are the two fluorine atoms. But because
of the fixed stereochemistry of the molecule, the protons and fluorines are magnetically inequivalent. This is observed in both the 1H and 19F spectra.
Reference
1.
SBDS (spectral database) #9726: http://riodb01.ibase.aist.go.jp/sdbs/cgi-bin/cre_index.
cgi?lang=eng.
© Peter Marrs, University of Victoria Chemistry 363 NMR Notes, September 2014