So in order to predict NMR spectra, we should count protons with different magnetic environment. Protons may differ in chemical and magnetic equivalence due to their difference in both attachment and arrangement in the space. Here we will see few examples and how we can determine the possible number of peaks in proton NMR for each compound.
Organic compounds mainly contain different protons attached at different locations to the parent chain. All these protons are chemically non-equivalent and therefore give different signals NMR spectroscopy. Protons with different linkage give different NMR signals. Initially observe the structure for protons with different linkage.
Protons with same linkage are considered as similar type. Here all protons are equivalent as they are similarly connected to carbon. Hence the number of NMR signals is 1. In aliphatic compounds protons attached to same carbon are completely equivalent and therefore give single NMR signal. Then what about the solvent used in NMR? They may have protons which may give their own peaks in NMR spectra.
Practically this can be achieved by deuterated solvents where all the hydrogens are replaced by deuterium. In the above example, all the six protons are equivalent as they are connected in similar way to carbon which is connected to another carbon. Here all these protons are primary protons.
Here we can observe two types of protons viz. Hence the number of NMR signals is 2. Since these protons are chemically not equivalent, they will show different chemical shifts chemical shifts in NMR spectra. Sometimes we can easily identify the type of proton based on their chemical shift by comparing the value in NMR spectrum table. In the above example, if a proton shows a chemical shift of 0. The problem seems to be that the position of the -OH peak varies dramatically depending on the conditions - for example, what solvent is used, the concentration, and the purity of the alcohol - especially on whether or not it is totally dry.
Do you need to worry about this? Not really - you can assume that in an exam question, any NMR spectrum will be consistent with the chemical shift data you are given. If you measure an NMR spectrum for an alcohol like ethanol, and then add a few drops of deuterium oxide, D 2 O, to the solution, allow it to settle and then re-measure the spectrum, the -OH peak disappears!
By comparing the two spectra, you can tell immediately which peak was due to the -OH group. Note: Deuterium oxide sometimes called "heavy water" is simply water in which all the normal hydrogen-1 atoms are replaced by its isotope, hydrogen-2 or deuterium. The reason for the loss of the peak lies in the interaction between the deuterium oxide and the alcohol. All alcohols, such as ethanol, are very, very slightly acidic. The hydrogen on the -OH group transfers to one of the lone pairs on the oxygen of the water molecule.
The fact that here we've got "heavy water" makes no difference to that. The negative ion formed is most likely to bump into a simple deuterium oxide molecule to regenerate the alcohol - except that now the -OH group has turned into an -OD group. Deuterium atoms don't produce peaks in the same region of an NMR spectrum as ordinary hydrogen atoms, and so the peak disappears.
You might wonder what happens to the positive ion in the first equation and the OD - in the second one. These get lost into the normal equilibrium which exists wherever you have water molecules - heavy or otherwise.
Unless the alcohol is absolutely free of any water, the hydrogen on the -OH group and any hydrogens on the next door carbon don't interact to produce any splitting. The -OH peak is a singlet and you don't have to worry about its effect on the next door hydrogens. The left-hand cluster of peaks is due to the CH 2 group. It is a quartet because of the 3 hydrogens on the next door CH 3 group. You can ignore the effect of the -OH hydrogen. Similarly, the -OH peak in the middle of the spectrum is a singlet.
It hasn't turned into a triplet because of the influence of the CH 2 group. Note: The reason for this is quite complex, and certainly goes beyond A'level. It lies in the very rapid interchange that occurs between the hydrogen atoms on the -OH group and either water molecules or other alcohol molecules. To find out about it you will have to read either a degree level organic chemistry book or one specifically about NMR. For A'level purposes just accept the fact that -OH produces a singlet and has no effect on neighbouring groups!
Don't expect these to be easy reading though - this is university level stuff. Hydrogen atoms attached to the same carbon atom are said to be equivalent. Equivalent hydrogen atoms have no effect on each other - so that one hydrogen atom in a CH 2 group doesn't cause any splitting in the spectrum of the other one.
But hydrogen atoms on neighbouring carbon atoms can also be equivalent if they are in exactly the same environment. These four hydrogens are all exactly equivalent.
You would get a single peak with no splitting at all. Because the molecule now contains different atoms at each end, the hydrogens are no longer all in the same environment. This compound would give two separate peaks on a low resolution NMR spectrum.
The high resolution spectrum would show that both peaks subdivided into triplets - because each is next door to a differently placed CH 2 group. At this introductory level, all you can safely say about hydrogens attached to a benzene ring is how many of them there are.
If you have a molecular formula which has 6 or more carbon atoms in it, then it could well contain a benzene ring. Look for NMR peaks in the 6. If you are given a number like 5 or 4 alongside that peak, this just tells you how many hydrogen atoms are attached to the ring.
If there are 5 hydrogens attached to the ring, then there is only one group substituted into the ring. If there are 4 hydrogens attached, then there are two separate groups substituted in, and so on. There should always be a total of 6 things attached to the ring. Every hydrogen atom that is missing has been replaced by something else.
Splitting patterns involving benzene rings are far too complicated for this level, generally producing complicated patterns of splitting called multiplets. Note: If you are unfortunate enough to have examiners who ask about spitting in benzene rings at this level for 16 - 18 year old chemistry students , look carefully at the question and mark scheme so that you can see exactly what they want, and just learn that. But it may well be that all they want is for you to notice the number of hydrogen atoms involved in the ring see above.
If this is the first set of questions you have done, please read the introductory page before you start. The difference between high and low resolution spectra What a low resolution NMR spectrum tells you Remember: The number of peaks tells you the number of different environments the hydrogen atoms are in.
High resolution NMR spectra In a high resolution spectrum, you find that many of what looked like single peaks in the low resolution spectrum are split into clusters of peaks.
0コメント