Activity 4 - Symmetry
Carbons in the same chemical environment have the same chemical shift. For example,
let’s think about benzene, C6H6. This molecule contains 6 carbons, but how many peaks
would we expect to see? One, that’s right, and this is because all of the carbons are in
exactly the same environment (you can put any of the carbons at the “top” of the molecule
(imagining you are drawing it on some paper) and it makes no difference to what is at each
point. Therefore the 13C spectrum of benzene would look like:
Figure 1. The
13
C spectrum of benzene showing a single peak.
Some molecule have symmetry, but not quite as much symmetry as benzene, therefore
some of the carbons are equivalent, but not all of them. A good example of this is the
molecule 1,4-dichlorobenzene (a benzene ring with one Cl attached at the top in place of an
H and another Cl attached opposite to this on the ring. If you call the carbon with the first Cl
attached C1 and then count around the ring you will find the second chlorine is attached at
the fourth carbon, hence 1,4-dichloro. In this case there are two mirror planes within the
molecule giving us two carbon environments (two different types of carbon); the one
attached to the Cl and the one attached at an H. This knowledge can be used to predict
what the spectrum would look like, keeping in mind everything we know about where the
peaks are likely to be found. Draw yourself a prediction and then see if it agrees with the
one below, why not also have a go at the isomers 1,2-dichlorobenzene and 1,3dichlorobenzene (use the same counting method to put the Cls on in the correct place).
Figure 2. Predicted spectrum for 1,4-dichlorobenzene
Figure 3. Predicted spectrum for 1,2-dichlorobenzene.
1,3-dichlorobenzene should have 4 peaks all in the same region.
You could also try 1,3,5-trichlorobenzene and 1-chloro-3-bromobenzene. Answers are on
the ‘Answers’ sheet.
Imagine I carried out a reaction and I knew that I had made butanol, but I did not know which
of the isomers I had made. Can you work out from the spectrum below which I must have?
(Hint – I would start off by drawing out the isomers you can think of, and covering up the
answer just below!).
Figure 4. The spectrum of an unknown sample of butanol (the peak at around 77 which looks
like it is made from three peaks very close together can be ignored as this comes from the
solvent my sample was dissolved in, in other figures this is labelled as CDCl3, the name of the
solvent. You can see from this example that the solvents used in NMR are often “deuterated”,
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i.e. they have deuterium or H where they would normally have H, this means that they appear
very differently in the spectrum so we know we can ignore them).
You can find the solution on the ‘Answers’ sheet to check your prediction.
Exactly the same ideas apply to 1H NMR: each distinct environment leads to a peak; again
we can break the spectrum down into general areas in which peaks of different types are
likely to be found. In this second case the regions are the same as with 13C NMR, only the
scale is different:
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Figure 5. Regions of the H spectrum.
The nice thing about 1H NMR, as well as it being very quick to carry out, is that we can get
more information from it than from 13C NMR. Here we find that the area under the peak tells
us the relative number of hydrogen atoms in the environment represented by that particular
peak. This is called the integral and is often given as an extended “s” shape by the side of
each peak. An example of this is shown below where if you measure the height of the
integrals (which corresponds to the area under the peaks) you should find a 1:3 ratio.
Figure 6. A spectrum showing integral lines.
Now have a go at these examples. You need to match the spectrum to the molecule.
You can find the solutions on the ‘Answers’ sheet.