Take, for example, methanol ($\ce{CH3OH}$). The hydroxyl proton ($\ce{-O$\color{red}{\ce{H}}$}$) is three bonds away from three more protons ($\ce{-C$\color{blue}{\ce{H}}$_3}$), which aren't equivalent to it, so why would this not produce a quartet using the $n+1$ rule?


4 Answers 4


You are both right and wrong.

At low temperatures, the NMR spectrum will indeed be as you predicted: the $\ce{H}$ from $\ce{OH}$ will produce a quartet and likewise the $\ce{H}$ from $\ce{CH_3}$ will produce a doublet. Note that at low temperatures the formation of $\ce{H}$ bonds is favored, hence leading to more stable components and significantly decreasing proton exchange.

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At higher temperatures however, the following spectrum will be observed:

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Due to the increase in temperature, the $\ce{H}$ bonds are weakened, and hence shields its signal. Most importantly, the intermolecular proton exchange, non negligible at those temperatures, will render the interactions between the two groups of $\ce{H}$ undetectable via NMR. Hence why you get two singlets.

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    $\begingroup$ Wouldn't the H-bonding cause shielding, not deshielding, based on chemical shift here? I would expect higher electron density on the hydoxyl proton during hydrogen bonding, too, so shielding makes sense to me both theoretically and experimentally here. $\endgroup$ Commented Mar 21, 2016 at 4:23
  • $\begingroup$ @DGS this and that source among others indicate that H bonding display resonance signals down-field of their usual signal. "This downfield shift is a result of a number of not yet fully understood and partially competing factors, including a decrease in the electron density around the hydrogen nucleus and deshielding effects from the electronic currents of the acceptor atom." $\endgroup$ Commented Mar 21, 2016 at 16:08
  • $\begingroup$ But that contradicts the data you presesented. Unless there is actually increased hydrogen bonding at low temperature, which I would believe. But then that would negate the idea that H-bonding reduces J-coupling. $\endgroup$ Commented Mar 21, 2016 at 16:12
  • $\begingroup$ Could it be that at lower temperature, there is more H-bonding but less exchange, so you see an downfield shift but still see J-coupling? $\endgroup$ Commented Mar 21, 2016 at 16:18
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    $\begingroup$ @DGS You're totally right, I mixed up the two cases in my answer. Amended. Please do tell me if there's still something that seems wrong. $\endgroup$ Commented Mar 21, 2016 at 16:19

This is provided as a comment as much as anything - but a lengthy one, in order to address some concerns I have with the accepted answer - not that it is entirely incorrect. There is much too much emphasis on H-bonding, and not enough about exchange in order to justify whether one observes coupling or not.

The loss of coupling in this instance is a classic example of exchange decoupling. The reason that the coupling disappears is due to exchange processes. The important part in factoring whether coupling will be observed relates to whether there are any mechanisms that will promote or inhibit this exchange. H-bonding may be a mechanism to reduce exchange at low temperature, however it is also a well-established mechanism that promotes exchange especially at higher concentrations. The original paper by Kivelson and Kivelson in 1958 clearly demonstrated that the observation of coupling in methanol was dependent on reduced concentrations which lower the probability of H-bonding. The opposite could be argued for ethanol, and Shoolery (also in 1958) showed that higher concentrations of ethanol actually give rise to observation of coupling between alcohol and CH2. Neat ethanol clearly shows this coupling at room temperature. This is also explained by H-bonding mechanisms giving rise to stable dimers and higher order oligomers, which in fact undergo very fast signal averaging (exchange). So, H-bonding is not the answer for whether coupling is observed - it is purely a factor to consider in determining whether active exchange pathways exist for chemical exchange.

In a nutshell, a dilute sample of methanol in acetone will show splitting quite clearly at RT, but a concentrated sample does not. S.Burt quite rightly points out that using a dry solvent will improve your chances of observing coupling. Choice of solvent is critical, and aprotic and nonpolar solvents, removed of any acidic impurities help to promote observation of coupling. I use dry pyridine-d5 with a number of sugars to demonstrate coupling through to alcohol protons, and then add a drop of D2O to make them disappear. Dry CCl4 is another great solvent, and will easily allow the observation of couplings in a dilute sample of methanol. Methanol is used as a low temperature thermometer and low temperature calibration standard down to about 180K, and coupling is never observed in these samples.


What you propose actually happens if your solvent is very dry. A colleague of mine works with sugar chemistry and when his students run very dry samples, they not only see every alcohol on the sugar, but the alcohol is split by the vicinal protons (and those protons are split by the OH) even in CDCl3!

Typically, there is enough water in typical samples (or the solvents themselves!) to ensure that the hydroxyl proton is rapidly exchanging and this effectively decouples the spin from its neighbors.

If you're in a position to make a sample yourself - get a stable organic solid with a hydroxyl that you can dry in a drying oven (or lyopholize it if you have access to a lyopholizer) and prepare a sample in dry DMSO-d6. You almost always see the splitting to/from hydroxyl peaks in DMSO.

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    $\begingroup$ Good answer. Using the right solvent also is a great help for observing couplings to alcohol protons. Pyridine-d5 is quite an easy solvent to get dry, and I use this for running sugar samples to see couplings to the alcohol protons. $\endgroup$
    – long
    Commented Mar 21, 2016 at 21:15
  • $\begingroup$ You don’t even necessarily need very dry solvents. I have had my fair share of couplings to hydroxy protons in 2-hydroxycarbonyl compounds where an intramolecular hydrogen bond makes hydrogen exchange less favourable. $\endgroup$
    – Jan
    Commented Apr 18, 2016 at 17:37

have a look at the solution of this problem which has been published some decades ago:

D. Gerritzen, H. H. Limbach, Ber. Bunsenges. Phys. Chem., 1981, 85, 527-535. http://dx.doi.org/10.1002/bbpc.19810850702 Kinetic and Equilibrium Isotope Effects of Proton Exchange and Autoprotolysis of Pure Methanol Studied by Dynamic NMR Spectroscopy.

H. H. Limbach, J. Magn. Reson., 1979, 36, 287-300. http://dx.doi.org/10.1016/0022-2364(79)90106-9 NMR Lineshape Theory of Superimposed Intermolecular Spin Exchange Reactions and Its Application to the System Acetic Acid/Methanol/Tetrahydrofuran-d8.

H. H. Limbach, W. Seiffert, J. Am. Chem. Soc., 1980, 102, 538-542. http://dx.doi.org/ 1H NMR Spectroscopic Study of Cyclic Proton Exchange between Acetic Acid and Methanol in Tetrahydrofuran-d8.

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    $\begingroup$ Welcome to Chemistry SE! If you haven't yet, take the tour and visit the help center. While we definitely want well sourced answers on the site, we generally prefer that the answer doesn't just consist of sources or links to material. Could you offer a brief summary of the key points from those papers that apply to this question? @Hans-HeinrichLimbach $\endgroup$
    – Tyberius
    Commented Jun 27, 2018 at 15:02

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