We are having some problems with the next reaction in our lab. As you can see, is the sulfonylation of an indole in which we use NaH (60% dispersion in mineral oil) as a base. This reaction goes with good yields when the NaH is a white solid. But lately, we bought a new one (also 60% dispersion in mineral oil) but in this case its colour is grey. With this one, the reaction never completes and it goes with lower yield. We bougth another one, because we thought that the other one came in bad conditions, and again, the product was grey. We have also tried to wash the NaH with hexane and dry it carefully but nothing changed. So my questions are: Do you know which is the difference between both of NaH? Where could we find the white one again? How can this happen? It's the same reactant but only with diferent colour. Thank you so much for your help!!

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    $\begingroup$ I'm used to NaH (60% in oil) being grey and distrusted the material that was white - so not really any help. $\endgroup$
    – Waylander
    Commented Feb 6 at 21:28
  • $\begingroup$ I don't particularly see why this reaction should be problematic (though as raptorlane pointed out, NaH in DMF should be avoided), but you can always consider alternatives. For example, you could skip NaH altogether by performing an N-trimethylsilylation and reacting that with the sulfonyl chloride. It's two steps, but may end up being a cleaner and higher-yielding process overall. $\endgroup$ Commented Feb 6 at 23:47
  • $\begingroup$ I've since learned a nice little trick. The reason NaH is gray is due to small amounts of Na metal left over from the industrial preparation. It's possible to purify the NaH by stirring it in dry THF, and adding small iodine crystals slowly. Initially the colour is quickly destroyed by the excess Na metal (forming soluble NaI), but eventually the colour persists when all the metal is consumed, much like a titration. Then you can filter the NaH under nitrogen, wash with fresh THF and dry under vacuum to obtain pure white NaH. Whether purification will help you, I don't know. $\endgroup$ Commented Apr 28 at 0:36

2 Answers 2


Sodium hydride on mineral oil is grey. When it reacts with water, for example from air humidity it reacts to NaOH which leads to discolouration which is white. One can wash NaH with hexane or THF to remove the oil and to increase reactivity, but unter practical consideration I did not have problems with using it directly. Pure NaH is available, but be careful when using it. Ideally this is handled only under protective atmosphere, because it is very reactive and will violently react to even traces of water. When the oil is removed, the neat hydride appears to be white. I honestly don't believe that the mineral oil is the problem for the described problem, because it is inert. More likely are:

  1. The solvent was not dry. This reduced the amount of available NaH. Increase equivalents and or additionally dry your solvent. For example with mol sieve. You could try to use not 1 eq but 3 and see whats the result.
  2. The atmosphere was not water free. Use nitrogen or argon with a schlenk line and dry your flasks beforehand reaction.
  3. Did you account for the 40 Percent oil? One can easily forget this and than you have 0.6 eq instead of 1 eq in your reaction. Consider this in your calculations.
  4. Try to increase the time for deprotonation and check if this makes a difference. You could try to prove full deprotonation by a reaction that consumes the nucleophile fully. Check afterwards by tlc or lcms.
  5. If possible use THF instead of DMF. I know, this is not always possible, but solvents like DMSO and DMF are not recommended with NaH, these mixtures can lead to a thermal runaway.

Explosion Hazards of Sodium Hydride in Dimethyl Sulfoxide, N,N-Dimethylformamide, and N,N-Dimethylacetamide



Washing in dry pentane was my first thought, too (already visible from the freely available abstract of eEROS).

Regarding the determination of the concentration of the active reagent: in the chapter about the selection of bases, Loewenthal's Guide for the perplexed chemist recommends to determine the active concentration of $\ce{NaH}$ by titration. So, rapidly weigh-in a sample of $\ce{NaH}$ to check in 2-propanol, intentionally add water to quench it and then titrate back with a standard solution of aqueous hydrochloric acid, all under with good swirling.

Else, if you have advanced access to a NMR spectrometer, Hoye et al. re-introduced No-D NMR spectroscopy to determine the active concentration of metal organic reagents including metal hydrides. For greater accuracy, you might want to modify (usually increase) the $t_1$ and $t_2$ times used to collect a routine $\ce{^1H}$-NMR spectrum which sometimes works easier on elder instruments (e.g., Varian200) than e.g., Bruker's Avance DRX. The specific times of course depend on the specific signal in the spectrum monitored. With 16 to 32 fid collected (classical/standard probe head) you easily collect the data you need within 10 min, then integrate reference and sample signal, etc.

Hoye, T. R.; Eklov, B. M.; Ryba, T. D.; Voloshin, M.; Yao, L. J. No-D NMR (No-Deuterium Proton NMR) Spectroscopy: A Simple Yet Powerful Method for Analyzing Reaction and Reagent Solutions. Org. Lett. 2004, 6, 953–956. https://doi.org/10.1021/ol049979+.

Hoye, T. R.; Aspaas, A. W.; Eklov, B. M.; Ryba, T. D. Reaction Titration: A Convenient Method for Titering Reactive Hydride Agents (Red-Al, $\ce{LiAlH4}$, DIBALH, L-Selectride, NaH, and KH) by No-D NMR Spectroscopy. Org. Lett. 2005, 7, 2205–2208. https://doi.org/10.1021/ol0506011.

Creary, X. No-Deuterium Proton (No-D) NMR as a Convenient Method for Analysis of Organic Solvents. J. Org. Chem. 2023, 88, 11545–11551. https://doi.org/10.1021/acs.joc.3c00807 (open access).


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