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As the question says, I'm having trouble understanding where does magnetic inequivalence in $\ce{^1H}$ NMR come from. All the sources I've found say that two protons are magnetically equivalent if all $J$ values to all other spins in the molecule are identical. So for example, in 1,2-dibromobenzene, the $\ce{H^3}$ and $\ce{H^6}$ are not magnetically equivalent because $J(\ce{H^3, H^4}) \neq J(\ce{H^6, H^4})$ and $J(\ce{H^3, H^5}) \neq J(\ce{H^5, H^6})$. I understand how to apply the rule, but I don't understand why does this affect the spectra. For example, assuming a not unusual $J_\mathrm{ortho} = \pu{7 Hz}$ and $J_\mathrm{meta} = \pu{2 Hz}$, then: $J(\ce{H^3, H^4}) = \pu{7 Hz} = J(\ce{H^5, H^6})$, and $J(\ce{H^3, H^4}) = \pu{2 Hz} = J(\ce{H^5, H^6})$. So they are both split the same way, but in reality they also split each other because of their magnetical inequivalence. This is what I find hard to grasp: if they are being in overall affected the same way, by the same type of atoms, in the same positions, how do they "know" that they are, in fact, different.

I guess related to this would be, why do all protons in ethene give the same signal? If they are also related differently between each other? ($\ce{H^1}$ and $\ce{H^{$1'$}}$ have different coupling constants to each $\ce{H^2}$ and $\ce{H^{$2'$}}$, cis and trans, respectively).

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Firstly, thank you for asking a very interesting question that certainly pushed my limits of knowledge (and others' as well). I apologise if I have made mistakes, it is a real possibility. All references are to Levitt's Spin Dynamics, 2nd ed. (an excellent book, by the way).

From a quantum mechanical point of view, the complexity in the spectrum of magnetically inequivalent nuclei should be considered the "norm". Fundamentally, in any strongly coupled system where the difference in Larmor frequency between two spins $\Delta \omega$ is much less than the coupling constant $J$, we expect to see complex spectra (see second half of this answer for more discussion).

Magnetic equivalence is a special case which leads to simplifications in the Hamiltonian and hence in the spectrum. As I mentioned in a comment, it can be shown that any coupling between magnetically equivalent nuclei "disappear". The word "disappear" merits further explanation, though. It is not true that the coupling constant $J$ goes to zero, for example. Neither is it true that it is averaged to zero by some sort of physical process, e.g. molecular rotation. It only means that the coupling term in the Hamiltonian has no impact on the expectation value of any quantum mechanical observable, or its evolution. (Formally, this is because the coupling Hamiltonian between two magnetically equivalent nuclei

$$H = 2\pi J_{ij}(I_{ix}I_{jx} + I_{iy}I_{jy} + I_{iz}I_{jz})$$

commutes with every other relevant Hamiltonian—supposedly a 'well-known' fact, according to the famous product operator paper: Prog. Nucl. Magn. Reson. Spectrosc. 1984, 16, 163–192.) Anyway, since an NMR spectrum is simply us humans reading out some expectation values, the NMR spectrum itself is therefore unaffected by the coupling term between magnetically equivalent nuclei.

Having established that it is magnetic equivalence, not inequivalence, that is the special case, I would like to rephrase your question to: "Is there a physical explanation for why magnetic equivalence leads to simplification of the spectrum?"

The mathematical reason is given in Appendix A.9 of Levitt. I won't reproduce it here, so I will just mention it has to do with some operators which commute with others, and these only commute under very specific circumstances, i.e. the rules for magnetic equivalence that you have surely been taught (they must be chemically equivalent, plus either they have the same coupling constant to all other nuclei, or there are no other spins in the molecule).

Given the (relative) complexity of the mathematical explanation, I suspected that a physical explanation would not be particularly instructive. Levitt, though, does offer us something (p 461):

The apparent disappearance of J-couplings between magnetically equivalent spins is quite surprising. A physical explanation runs as follows. For magnetically equivalent spins, the local magnetic fields, whatever their source, are exactly the same on the two spins. It follows that the motion of the two spin polarizations is identical. Whatever happens, the polarizations of the two spins are locked in the same relative orientation. There is, therefore, no need to incorporate a term in the spin Hamiltonian that takes into account the dependence of the energy on this relative spin orientation.

What this means is: it does not matter what the relative orientations of the spins are. For example, in ethene, it does not matter whether two spins are up and two are down, or whether all four are down, or whether all four are somewhere between up and down (i.e. a superposition of $\alpha$ and $\beta$, which is in fact the most likely case by far). The relative orientations can be anything, but the point is that the time evolutions of all four spins are synchronised. Therefore, the coupling terms between them – which are terms, scalar products, that depend on the relative orientation of spin vectors – may be omitted from the Hamiltonian without any impact on the observables of the system.

Returning to a mathematical interpretation, another sign of magnetic equivalence between nuclei $i$ and $j$ is when the labels $i$ and $j$ can be permuted in the spin Hamiltonian without changing it. This implies that $[\hat{\vec{I}}_i,\hat{H}] = [\hat{\vec{I}}_j,\hat{H}]$. From the standard formula

$$\frac{\mathrm d}{\mathrm dt}\langle Q \rangle = -\frac{\mathrm i}{\hbar}\langle[\hat{Q},\hat{H}]\rangle$$

it is immediately obvious that $\mathrm d\langle\vec{I}_i\rangle/\mathrm dt = \mathrm d\langle\vec{I}_j\rangle/\mathrm dt$, thus proving that the spins $i$ and $j$ move in concert.

Regarding ethene: you are correct that the cis and trans (and vicinal) couplings in ethene are different. However, all four of the protons in ethene are chemically equivalent and must be considered together as a group. It is not correct to look at two out of the four protons and call them magnetically inequivalent, without also taking into account the remaining two. In this case, it turns out that if you do the maths, the relevant commutators don't cancel out if you only consider two out of the four nuclei. However, when you consider all four nuclei together, the relevant operators fully commute.

Now, none of the simplifications above apply to magnetically inequivalent nuclei. These experience very tiny instantaneous differences in magnetic fields – hence the name – and therefore all the couplings in the system end up being reflected in the spectrum, as we should have expected a priori. Returning to your analysis ("the coupling constants are the same, so shouldn't they be the same?"), the issue is that the coupling constant $J_\ce{AX}$ is only one part of the coupling term in the Hamiltonian, which is $2\pi J_\ce{AX}(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$. Although $J_\ce{AX}$ is indeed equal to $J_\ce{A'X'}$, the dot product $(\hat{\vec{I}}_\ce{A}\cdot\hat{\vec{I}}_\ce{X})$ is in general not equal to $(\hat{\vec{I}}_\ce{A'}\cdot\hat{\vec{I}}_\ce{X'})$.

Warning: as with the ethene case, one cannot look at individual A and A' nuclei, because they are symmetry-related. You have to look at both of them together as a group. Once you do that, though, the spectra become possible to analyse.

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What ethene misses compared to a benzene derivative, is a ring current. There would be no one in the ground state of an isolated molecule, but it will arise in magnetic field. The ring current in the molecule with (artificially numbered) atoms can have one of the two directions, which makes, for instance, H4 and H5 magnetically inequivalent.
In other words, ethene in magnetic field retains it point group symmetry (D2h), while 1,2-dibromobenzene loses their C2v because of asymmetric look of the ring current.

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  • $\begingroup$ I like this idea. But couldn't ethene might as well have a left or right direction of the electrons on the $\pi$ bond? $\endgroup$
    – ralk912
    Commented Dec 31, 2017 at 0:27

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