If we split the signal into a doublet and then split each of the two peaks into a quartet, we will get the same result.
If the peaks don't overlap nicely, it can be a difficult signal to analyze.
A multiplet is a type of signal.
An example of a multiplet is shown.
We saw examples of complex splitting in the previous section.
In this section, we will look at cases where there is no splitting at all.
Richard A. Tomasi provided the spectrum of the protons.
The spectrum shows the characteristic signals of an ethyl group, which is a quartet with an integration of 2 and a triplet.
There is another signal at 2.2 parts per million.
It is difficult to predict where a signal between 2 and 5 ppm will appear.
The hydroxyl protons are not split into a triplet from the neighboring methylene group.
The process of transferring protons occurs at a faster rate than the timescale of an NMR spectrometer, which results in a blurring effect that averages out any possible splitting effect.
It is possible to slow down the rate of transfer by removing trace amounts of acid and base.
There is a common example of neighboring protons that do not split.
In the previous sections of this chapter, we have learned all of the individual tools that you need for analyzing a proton NMR spectrum, considering the number of signals, analyzing chemical shifts, assessing integration values, interpreting the multiplicity of each signal, pattern recognition, etc.
We are about to put all of these together.
There is one more important tool that you will need, and we will cover it in this section.
Imagine that you have a compound with a formula.
There are many constitutional isomers.
A careful analysis of the formula can provide clues about the structure of the compound.
Let's begin by analyzing the formula of several alkanes.
Paying special attention to the number of hydrogen atoms attached to each carbon atom is how to compare the structures of the following alkanes.
There are two hydrogen atoms on the ends of the structures, and there are two hydrogen atoms on every carbon atom.
All of the compounds have a formula of C H.
They have the maximum number of hydrogen atoms possible compared to the number of carbon atoms present.
A compound with a double or triple bond has fewer hydrogen atoms than a compound with a single bond.
A compound with a ring has fewer hydrogen atoms than a compound with a double bond.
A compound has one degree of unsaturation for every two hydrogen atoms that are missing.
There are some possible constitutional isomers for C H. Let's expand our skills set with this in mind.
Chlorine takes the place of a hydrogen atom.
If you want to calculate the HDI, treat a halogen as if it were a hydrogen atom.
C H should have the same HDI.
The oxygen atom does not affect the number of hydrogen atoms.
If an oxygen atom appears in the formula, it should be ignored.
The number of expected hydrogen atoms is affected by the presence of a nitrogen atom.
C H N should have the same HDI.
The rules allow you to determine the HDI for simple compounds.
HDI is the number of carbon atoms, N is the number of nitrogen atoms, H is the number of hydrogen atoms, and X is the number of halogens.
All compounds containing C, H, N, O, and/or halogens will work with this formula.
Calculating the HDI gives clues about the structural features of the compound.
The compound cannot have any rings or p bonds if the HDI is zero.
It is very useful information when trying to determine the structure of a compound, and it is easy to get by simply analyzing the formula.
If the HDI is two, there are several possibilities: two rings, two double bonds, or one ring and one double bond.
HDI analysis for an unknown compound can be useful, but only if the formula is known.
This technique will be used in the next section.
The skill of calculating and interpreting the HDI of an unknown compound is developed through the following exercises.
The compound has the same HDI as the one with the formula C H.
This compound has one degree of unsaturation because two hydrogen atoms are missing.
HDI is 1.
The compound has to contain one ring or one double bond, but not both.
Two degrees of unsaturation is required for a triple bond.
If the formula is given, inspect it as it provides useful information.
Important clues about the structure of the compound can be provided by calculating the hydrogen deficiency index.
The compound does not have any rings or p bonds.
The compound has either one ring or one p bond according to the HDI.
The number of signals and integration of each signal give clues about the symmetry of the compound.
Draw fragments consistent with each signal after analyzing it.
Our puzzle pieces must be assembled to produce a structure.
The following exercise shows how this is done.
The calculation of the HDI is the first step.
The formula shows 9 carbon atoms and 20 hydrogen atoms in order to be saturated.
The HDI is 5 because there are only 10 hydrogen atoms.
It wouldn't be efficient to think about all the ways of having five degrees of unsaturation.
We should be on the lookout for an aromatic ring when we encounter an HDI of 4 or more.
When analyzing the spectrum, we must keep this in mind.
An aromatic ring with one other degree of unsaturation is what we should expect.
The number of signals and integration value are the second step.
There are four signals in this spectrum.
The ratio is 2 to 1.
Look at the formula.
The relative integration values represent the actual number of protons in the compound.
The analysis of each signal is the next step.
There are two triplets with an integration of 2.
One or more factors is shifting these signals downfield because they don't appear at 1.2 where methylene groups are expected.
Our structure needs to take that into account.
After analyzing the HDI, we suspected that the next signal was aromatic protons and was just above 7 ppm.
Useful information is rarely given by the multiplicity of aromatic protons.
A messy multiplet of signals is often observed.
Important information is given by the integration value.
The aromatic ring is monosubstituted because there are five aromatic protons.
The last signal has a singlet at 10 parts per million with an integration of 1.
This is similar to an aldehydic proton.
The last step is to assemble the fragments.
There is only one way to assemble these puzzle pieces.
Each methylene group is being shifted downfield by one or more factors.
The structure explains the chemical shifts.
One methylene group is shifted by the carbonyl group and slightly by the aromatic ring.
The aromatic ring and carbonyl group are shifting the methylene group.
A structure for a compound with the formula C H that is consistent with the following spectrum is proposed.
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