I read somewhere that a silane group will react with hydroxyl groups.

Does anyone know why this would happen?


It is well known that silicon forms very strong bonds with oxygen (think sand, $\ce{SiO2}$). Hence, any time one can convert a $\ce{Si-H}$ bond into $\ce{Si-OH}$, it will happen.

I believe that silanes are extremely air sensitive due to the moisture content.

  • 1
    $\begingroup$ I don't know about H-Si groups, but alkoxysilanes are surprisingly unreactive in the short term. I suspect there may be some sort of kinetic barrier in place. If you mix a methoxy/ethoxy silane 1:1 with 10M NaOH, it takes a good minute for the reaction to take place. Once it does, though, it's incredibly incredibly exothermic. You still wouldn't want to leave it out uncapped for long though. Source: cleaned the results off of my fume hood ceiling about 10 hours ago. $\endgroup$
    – chipbuster
    Nov 20 '13 at 8:43
  • $\begingroup$ @chipbuster Interesting, seems to be some sort of significant kinetic barrier. I wonder why that is. $\endgroup$ Nov 20 '13 at 11:11
  • $\begingroup$ @NicolauSakerNeto A kinetic barrier must be at least as high as the energy of the endothermic (up-hill) product. $\endgroup$
    – Eric Brown
    Nov 20 '13 at 12:38

Hydroxyl radicals are strong proton acceptors, due to an oxygen with additional electrons, and will therefore abstract hydrogens from anything that has hydrogens, e.g., silane or methane. This attack can be enhanced, depending on the electronic properties of the atom sitting at the center (silicon or carbon). The more stable the atom, the more stable the 3-hydrogen radical will be, the more thermodynamically favorable the hydrogen abstraction reaction will be.

In the case of silane combustion, this leads to the formation of $\ce{^.SiH3}$ radicals, which react with molecular oxygen ($\ce{O2}$) in a chain-branching step. Depending on whether the temperature is low or high, this step can occur along various different pathways. It generally occurs following the reaction:

$$\ce{^.SiH3 + O2 -> Si(O)OH + H + H}$$

but the intermediates change depending on temperature and the energetics of the reaction.

You can find more information in a couple of papers:

  • Babushok, Tsang, Burgess Jr., Zachariah, "Numerical study of low- and high-temperature silane combustion." Twenty-Seventh International Symposium on Combustion (1998) pp. 2431-2439.
  • Kondo et al., "Ab Initio Energetic Calculations of Elementary Reactions Relevant to Low-Temperature Silane Oxidation by Gaussian-2 Theory." Journal of Physical Chemistry A (1997) pp. 6015–6022
  • 1
    $\begingroup$ Just to be clear for future readers, hydroxyl radicals and hydroxyl groups are quite different things. Your answer is certainly correct (and we are very appreciative of sourced answers!), but it may not directly answer the OP. I suspect they were talking about hydroxyl groups in molecules, in which case Eric's answer (though rather short) is correct thermodynamically, and kinetically it probably has to do with silicon's larger size which would stabilize a five-coordinated transition state where the hydroxyl group attacks the silicon atom in a silane directly. $\endgroup$ Nov 20 '13 at 0:19
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    $\begingroup$ Excellent point. I pounced on "hydroxyl" because silane so readily combusts, and hydroxyl radicals play an important role in any combustion process, but I neglected "groups." Oops! $\endgroup$ Nov 20 '13 at 3:49

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