Question

How does NMR spectroscopy work?

Answer 1

NMR is an analytical technique used in elucidating the structures of organic compounds.

Explanation:

it makes use of magnetic spins of the nuclei of atoms.

Answer 2

NMR spectroscopy directly interrogates `""^1H`, and `""^13C` nuclei on the basis of their symmetry. What follows is taken from several prior answers on the topic.

Explanation:

Of course, there are other nuclei whose NMR spectra we could assess, but in the modern organic and inorganic laboratories, `""^1H, ""^13C`, and `""^31P` `"nuclei"`, are the nuclei with which we typically do the experiments.

Any groups (of hydrogen, carbon etc.) that can be interchanged by a proper axis of rotation or a fast moving process are said to be `"equivalent"` (`"isochronous"`) in the NMR spectrum, and should give rise to the same chemical shift. How does this help? Well, you've got to look at the representation of your molecule (models help immensely) and recognize the equivalent, the symmetric hydrogens. Any groups of hydrogen that are interchanged by a mirror plane (and only by a mirror plane) will also give rise to the same absorption in the standard NMR experiment - such hydrogens are called `"enantiotopic"`.

If you knew nothing of NMR spectroscopy, you could look at a molecule of propane, `H_3C-CH_2-CH_3`. you could reasonably decide that the terminal methyl groups are equivalent. And, indeed they are. They should, in principle, give rise to a spectrum where there are 2 proton signals (or indeed two carbon signals), one for the terminal methyl protons, and one for the central methylene protons (for the moment forget coupling!). What about butane? Two signals again. What about pentane? What about hexane; what about ethane; what about 2-methylpropane?

And, conveniently, the area under the curve of these signals is proportional to the number of hydrogens; i.e. for `H_3C-CH_2-CH_3` we would see TWO absorptions in a `2:1` ratio. And since integration is a routine matter on the modern NMR spectrometer, the protons could be assigned directly and straightforwardly.

Protons and carbon nuclei exchanged (only) by a mirror plane, are said to be enantiotopic (i.e. interchanged by an improper axis of rotation). Enantiotopic nuclei are equivalent in the standard experiment, and should give rise to the same chemical shifts in all achiral NMR experiments. Should a homochiral NMR solvent be used (i.e. an optically active NMR solvent, which do exist), their interaction with the enantiotopic nuclei is diastereotopic, and the enantiotopic nuclei may be differentiated and become non-equivalent.

On other hand, there are protons that are constitutionally equivalent; i.e. they may be attached to the same carbon, and yet they CANNOT be interchanged by symmetry or a fast-moving process. Such protons are called `"diastereotopic"`, and I will give one example. Consider the dithioacetal of acetaldehyde, whose synthesis is given:

`H_3C-C(=O)H + 2HSCH_2CH_3 rarr (H_3C-CH_2S)_2C(CH_3)H`

Consider the methylene, the `CH_2` protons, bound to the `H_2CS` centre. Now these protons CANNOT be interchanged by any symmetry element, even tho' they are clearly constitutionally equivalent in that they are attached to the same carbon. Such `"diastereotopic"` protons should give rise in principle to 2 separate absorptions. (Of course the absorption is going to be a mess; they may be coupled to each other, and certainly to the methyl protons.)

I can't hope to teach you everything in a couple of paragraphs (for a start, I don't know everything). But consider a few simple molecules, and classify them on the basis of individual symmetries. Can the hydrogens be interchanged? If yes, 1 signal. If no, they should give rise to separate absorptions.