Assuming a downward magnetic field is applied and the nuclear spin of two equivalent protons in $\ce{H2}$ are all in ground state $\downarrow\downarrow$ initially. Would the required energy of one of the protons excited to spin up state $\uparrow\downarrow$ be different from the energy of proton excited form state $\uparrow\downarrow$ to state $\uparrow\uparrow$ ?

And if it's true why only one peak is observed in NMR spectrum?


2 Answers 2


For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, the energy difference can be calculated to be $0.21~\mathrm{J \cdot mol^{-1}}$ which corresponds to a population ratio as follows:

$$\frac{n_\mathrm{ex}}{n_\mathrm{g}} = \exp \left(\frac{\Delta E}{RT}\right) = 0.999916$$

Next, consider the extremely high symmetry of the $\ce{H2}$ molecule: It has the point group $D_{\infty\mathrm h}$, one of the most symmetric ones around. Due to this high symmetry, we cannot distinguish between either proton at all — both can be transformed into each other by a rotation as defined by the point group. Thus, the two nuclei are magnetically equivalent — not only must their chemical shift be identical but also the coupling to the other one.

Think of it this way: Assume, that the spin transition of one proton happened at a slightly different frequency than the other’s. That would mean that one proton had to be different from the other in some way (a different transition energy is equivalent to a different chemical shift is equivalent to a different environment). But how would you explain a different environment on one side of the $\ce{H2}$ molecule compared to the other? Correct: You cannot.

Now we could still assume that the energy required to excite a nuclear spin whose neighbour is parallel be different from one whose neighbour is antiparallel. However, no matter how we excite, the transition is always $\ce{parallel <=> antiparallel}$. And since we cannot distinguish between the hydrogens, we also cannot distinguish which one pointed in which direction, so those two energies must be identical. Therefore, we only see one peak and no $^1J_\ce{HH}$ coupling.

  • $\begingroup$ Well, if no magnetic field applied, would the energy level of parallel states $\uparrow\uparrow or \downarrow\downarrow$ be different from the energy level of antiparallel states $\uparrow\downarrow or \downarrow\uparrow$? I think there should be a energy different because of the spin-spin coupling, and when magnetic field applied, the two degenerate parallel states should split equally in energy difference. Therefore, the transition energy between antiparallel states and two distinct parallel states should be different. Am I thinking right? $\endgroup$
    – Junbo
    Commented Dec 29, 2015 at 16:58
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    $\begingroup$ You are correct in that ↓↓ and ↑↑ are different from ↓↑ and ↑↓. However, each pair of them is identical. However, ↑↑ is still identical to ↑↑ and the transition between antiparallel and parallel is identical no matter what the different ones are. $\endgroup$
    – Jan
    Commented Dec 29, 2015 at 17:47
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    $\begingroup$ Technically, for two magnetically equivalent spins, the transitions are no longer between αα/αβ/βα/ββ but rather between linear combinations of them. This is explained more thoroughly in the answers to chemistry.stackexchange.com/q/53478/16683. $\endgroup$ Commented Mar 12, 2019 at 23:19
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    $\begingroup$ It’s also slightly inaccurate to say that αα and ββ are the same, as the α and β spin states are no longer degenerate in the presence of an external magnetic field. The correct point is that the transition (to αβ, or βα, or more accurately, (2^-1/2)(αβ+βα)) occurs at the same energy. $\endgroup$ Commented Mar 12, 2019 at 23:32

If by referring to peak you mean a single chemical shift — in low-resolution H-NMR — the number of peaks available on the spectrum would depend on the type of hydrogen environments there are. For example, ethanol ($\ce{C2H6O}$) has three different H environments: $\ce{C\mathbf{H}_3-C\mathbf{H}_2-O\mathbf{H}}$.

ethanol low-res H NMR
(source: rsc-cdn.org)

The H's in the two methyl groups are of different environments because one methyl group has its electrons pulled on stronger by the oxygen in the hydroxide functional group ($\ce{OH}$).

In that case, $\ce{H2}$ only has one hydrogen environment, with both hydrogen atoms in the same environment, resulting in a single chemical shift signal.

However, you could have meant the spin-spin coupling determined by high-resolution H-NMR — the one where the tinier individual peaks within a certain signal are visible. The number of tinier peaks that show up for a particular chemical shift is dependent on the number of hydrogen atoms on a neighboring carbon atom.

ethanol high-res H NMR spectrum
(source: alevelchem.com)

In ethanol, the hydrogen environment of the central carbon atom ($\ce{CH3-C\mathbf{H}_2-OH}$) would have a quartet (4 peaked) signal because the neighboring carbon has 3 hydrogen atoms. The number of peaks due to spin-spin coupling is = number of neighboring hydrogen atoms + 1

With respect to $\ce{H2}$, the hydrogen atoms have no neighboring hydrogen atoms — there is no carbon in $\ce{H2}$ — so a singlet or single peak would apeak above the chemical shift for $\ce{H2}$. Even if a compound contained carbon atoms, like the H-NMR standard tetramethylsilane $\ce{Si(CH3)4}$ only a singlet is produced because all hydrogen atoms are in the same environment of a methyl group attached to a silicon atom.

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    $\begingroup$ This is quite false. The “n+1 rule” is a very basic simplification which assumes that only three-bond couplings contribute to the multiplet shape. While couplings over three bonds are the most commonly seen ones, it is not correct to imply that they are the only types of coupling. One-bond coupling is quite a real thing, it’s just that we cannot observe it due to the equivalence of the spins. $\endgroup$ Commented Mar 12, 2019 at 23:23

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