According to Wikipedia, $\ce{C2H5ONa}$ melts at 260°C, but no boiling point is given. What will happen if $\ce{C2H5ONa}$ is heated further?

  • 1
    $\begingroup$ I don't know but my instinct is that it decomposes perhaps with loss of ethene $\endgroup$
    – Waylander
    Commented Jul 4, 2018 at 12:08
  • $\begingroup$ Yeah, it decomposes. $\endgroup$ Commented Jul 4, 2018 at 17:26
  • $\begingroup$ Thanks. Do you have a reference, and do you know at what temperature it decomposes? $\endgroup$
    – Martin
    Commented Jul 4, 2018 at 21:59

1 Answer 1


Yes, sodium ethoxide starts to decompose on heating and thus its boiling point cannot be determined. According to [1]:

The decomposition of sodium ethoxide is observed to start above 573 K agreeing well with that reported by Blanchard et al. The gaseous products formed on decomposition were mainly methane and ethylene with minor quantities of ethane, propylene and butylene. Solid residue was found to be a mixture of sodium carbonate, sodium hydroxide and amorphous carbon. However, in an earlier study on decomposition of sodium ethoxide, by collecting and condensing the evolved gases in liquid nitrogen and analysis by gas chromatography, the authors have reported the gaseous products to be mainly of ethylene and minor quantities of propane, heptane, ethanol and water. The authors have also stated that the non-condensable gas at liquid nitrogen temperature could be hydrogen or methane. The TG trace for decomposition of sodium ethoxide in their study indicated a slow and steady decomposition right from 273 K with a sharp weight change above 573 K which implied that the sodium ethoxide could contain some amount of ethanol and moisture that evaporated on heating.[...]


  1. Chandran, K.; Kamruddin, M.; Ajikumar, P.; Gopalan, A.; Ganesan, V. Kinetics of thermal decomposition of sodium methoxide and ethoxide. J. Nucl. Mater. 2006, 358 (2-3), 111–128. DOI: 10.1016/j.jnucmat.2006.07.003.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.