Structure & Reactivity
Nuclear Magnetic Resonance Spectroscopy
NMR13. Combined Structure Determination
There are a many ways we can use NMR spectroscopy to analyse compounds. One common application is in determination of an unknown structure. Given the MS, IR, ^{13}C and ^{1}H NMR spectra, what might be the structure of an unknown sample?
It is often easiest to start with the IR spectrum.
Table NMR13.1. An IR data table waiting to be filled in.
cm^{1} 
asst 
For example, a student might obtain the following IR spectrum.
Figure NMR13.1. An infrared spectrum of an unknown sample.^{1}
From that information, she constructs the following table. She might even write this table, by hand, directly on her spectrum. She makes useful notes on the edges, and might even include some guesses, which she later crosses out, but does not erase. She is assisted in this task by consulting an IR table, that suggests what some of these peaks might mean.
Table NMR13.2. An IR data table for an unknown sample.
Remember:
Look at the ^{13}C spectrum.
Table NMR13.3. A ^{13}C NMR data table waiting to be filled in.
ppm 
asst 
For example, a student might obtain the following ^{13}C NMR spectrum:
Figure NMR13.2. A ^{13}C NMR spectrum of an unknown sample.^{1}
From that information, she puts together the following table:
Table NMR13.4. A ^{13}C NMR data table for an unknown sample.
Remember:
As in ^{13}C NMR, you should be able to assign all peaks in the ^{1}H NMR spectrum. You may be able to do so by making notes on the spectrum. If you think you know the structure, you may be able to draw it and note which peak belongs with which proton.
A formal proof of structure might require a table of assignments.
Table NMR13.5. A ^{1}H NMR data table waiting to be filled in.
ppm 
int 
mult 
partial structure 
assignment 














An example of a spectrum and its accompanying data table is given below. Here is the spectrum:
Figure NMR13.3. A ^{1}H NMR spectrum of an unknown sample.^{1}
Here is a data table:
Table NMR13.6. A ^{1}H NMR idata tablee for an unknown sample.
Things to note:
There are a couple of additional tools that can help to confirm the structure at this point. Alternatively, if the structure is still elusive, these tools might help to produce some ideas.
The first tool is the formula. Once we have NMR tables, we can begin guessing at the numbers of carbons and hydrogens in the structure. With the addition of an IR table, we can begin guessing at the presence of other atoms, such as oxygen or maybe nitrogen.
For example, in the ^{13}C NMR table above, there were seven peaks. That means there are probably at least seven carbons. We can start off the molecular formula as C_{7}. However, there may be additional carbons if there is some symmetry. There may also be a few carbons that do not show up very well in the spectrum. If you have ever obtained a real ^{13}C NMR spectrum, you will know that carbonyl peaks can be hard to find, especially if there are no hydrogens attached to the carbonyl. In the table above, it looked like there was a benzene, so maybe there werereally six carbons in the aromatic region, and not just four carbons. That would mean the formula, so far, is C_{9}.
In the ^{1}H NMR table, the integrals added up to a total of 10H. So, maybe the formula is C_{9}H_{10}.
Furthermore, the IR table suggested the possible presence of two different oxygen atoms. The formula may actually be C_{9}H_{10}O_{2}.
Once we have a formula, we actually get a great deal of information automatically. One of the most important pieces is "units of unsaturation" or "degrees of unsaturation" (DU). The DU is the result of a formal comparison of the C/H ratio in the compound to that in a normal alkane. In a normal alkane, the formula is always C_{n}H_{2n+2}. If you picture a long hydrocarbon chain, there will be two hydrogens on each carbon along the chain, plus one more hydrogen at either end of the chain. However, an alkene contains one pi bond, and at the site of that pi bond there are two hydrogen atoms missing from that alkane formula. A simple alkene always has the formula C_{n}H_{2n}. That missing pair of hydrogens in the formula is called a degree of unsturation.
Figure NMR13.4. Examples of degrees of unsaturation in hydrocarbons.
The same thing also happens to the formula if there is a ring present. One DU can correspond to the presence of a double bond or a ring. If DU=2, there may be two double bonds, two rings, or one of each.
If there are oxygen atoms present in the formula, we can just ignore them and pay attention to the hydrocarbon part. Conceptually, because oxygen forms two bonds, we can think of it as squeezing in between any two atoms in a hydrocarbon structure to form a new compound. The ratio of carbon to hydrogen is unchanged. If there is a degree of unsturation in a formula containing oxygen, it simply suggests the presence of a ring or a double bond, just like in a hydrocarbon.
Figure NMR13.5. Examples of degrees of unsaturation in oxygencontaining compounds.
Sometimes, if there are other atoms present, we need to adjust the formula to take them into account. For example, any time a halogen is found in the structure, it conceptually replaces a hydrogen atom. In order for a halogen to be found in the structure, there would have to be one fewer hydrogen atoms in order to open up a spot for the halogen. To adjust for the presence of a halogen, we need to add one hydrogen into the formula, then compare it to the standard alkane formula.
Figure NMR13.6. Examples of degrees of unsaturation in amines and haloalkanes.
Nitrogen, on the other hand, has three bonds. Unlike oxygen, if we squeeze it in between two other atoms, it still needs one extra bond. It always brings an extra hydrogen into the formula. To adjust for the presence of nitrogen, we need to subtract one H from the formula, then compare it to the standard alkane formula.
In the formula we just calculated, we have C_{9}H_{10}O_{2}. We can ignore the oxygens and look at the C9H10. If this were a saturated hydrocarbon with nine carbons, its formula would be C_{9}H_{20} (since 2 x 9 + 2 = 20). We are missing five pairs of hydrogens, so DU = 5. That is a lot. However, if we have one benzene in the structure, that would account for three double bonds and one ring all at once. That four degrees of unsaturation. An additional carbonyl would bring the number up to the required five. If we had not yet arrived at the idea of a benzene ring, this comparison might make us think of it. Alternatively, if we knew about the benzene but hadn't yet spotted the carbonyl, we might be on the lookout for it now.
Once we have a possible formula, another useful tool is mass spectrometry (MS). Even if you don't know much about mass spectrometry, the basic idea is simple. A mass spectrometer takes a molecules and bashes it into little pieces, then measures the molecular weights of each of those fragments. If you are lucky when you run the experiment, some of the molecules are left intact, and you get the molecular weight of the entire molecule, too.
Figure NMR13.7. A cartoon of a mass spectrum.
If we calculate the molecular weight based on the formula and compare it to the possible molecular weight from the mass spectrum, we might get confirmation that we are on the right track. Alternatively, maybe our calculated molecular weight will come up short. If we are off by 16, maybe we have missed an oxygen atom somewhere. If we are off by 14, maybe we have missed a carbon and a pair of hydrogens. This information might help us to correct some mistakes.
In the above example, the formula leads to a molecular weight of 150 g/mol. If the mass spectrum did not match, we would want to check our work to see if we overlooked something.
Problem NMR13.1.
Using the approach outlined above, build a case for the structure of the compound represented by the data below.
Problem NMR13.2.
Using the approach outlined above, build a case for the structure of the compound represented by the simulated data below.
Ref. 1. SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology of Japan, 14 July 2008)
This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted). It is freely available for educational use.
Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons AttributionNonCommercial 3.0 Unported License.
Send corrections to cschaller@csbsju.edu
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