# How do you determine non equivalent hydrogens?

Sep 17, 2015

$\text{Symmetry}$ is key to the process.

#### Explanation:

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 $\text{equivalent}$ ($\text{isochronous}$) 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 $\text{enantiotopic}$.

If you knew nothing of NMR spectroscopy, you could look at a molecule of propane, ${H}_{3} C - C {H}_{2} - C {H}_{3}$. you could reasonably infer 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, 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? Three signals are observed. 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 ${H}_{3} C - C {H}_{2} - C {H}_{3}$ 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.

On the 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 $\text{diastereotopic}$, and I will give one example. Consider the dithioacetal of acetaldehyde, whose synthesis is given:

${H}_{3} C - C \left(= O\right) H + 2 H S C {H}_{2} C {H}_{3} \rightarrow {\underbrace{{\left({H}_{3} C - C {H}_{2} S\right)}_{2} C \left(C {H}_{3}\right) H}}_{\text{dithioacetal of acetaldehyde}}$

Consider the methylene, the $C {H}_{2}$ protons, bound to the ${H}_{2} C S$ centre. Now these protons CANNOT be interchanged by any symmetry element, even tho' they are clearly constitutionally equivalent in that they are bound to the same carbon. Such $\text{diastereotopic}$ 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.