Task: Explain the reaction mechanism of 1,4-Dioxane in terms of how you get peroxides. Start here with the autoxidation with one radical X.

My explanation:

in the first step 1,4 Dioxan reacts with Dioxygen, so one gets an initiatorradical and and HOO* (* symbolises here a free electron, a radical). In the second step the initiatorradical reacts with oxygen and one receives a peroxyradical. In the end the peroxyradical reacts with an hydrogen atom and one comes to hydroperoxide. Is that right?enter image description here

  • 2
    $\begingroup$ You can't have free hydrogen atoms floating around in solution. The requisite hydrogen atom comes from dioxane itself. $\endgroup$ May 8, 2019 at 19:58
  • $\begingroup$ Doesn't the dioxane get the hydrogen atom from another substance? And is the rest correct? $\endgroup$ May 8, 2019 at 20:03
  • $\begingroup$ Right, i got it. It comes from the 1,4-Dioxane. Thanks! But what would be the Initation here? the first step, which I draw? and the second line would be the propagation? Is there an termination step, too? $\endgroup$ May 8, 2019 at 20:10

2 Answers 2


I could not find a definitive mechanism for 1,4-dioxane, but this is the mechanism for the formation of THF peroxide is likely very similar

enter image description here

diagram from here

  • $\begingroup$ So, the initiation is always that R* (here an hydrogen-atom) will be separated? Could the molecule do not react with oxygen at first, so that you get an peroxyradical and in the end this radical reacts with a hydrogen-atom to hydroperoxide? $\endgroup$ May 8, 2019 at 19:26
  • $\begingroup$ I've edited the question, Would it look like this? $\endgroup$ May 8, 2019 at 19:43
  • $\begingroup$ & is it right that 1,4-dioxane prefers to do autoxidation because it is an ether (it contains oxygen-atoms in it, connected to two different alkyl rests.). Or are there any other reasons, why it does like to do autoxidation? $\endgroup$ May 8, 2019 at 21:08
  • $\begingroup$ The radical is stablised by the oxygen. This is why all ether-based solvents have a warning for peroxide formation. $\endgroup$
    – Waylander
    May 9, 2019 at 8:34
  • $\begingroup$ But why do cyclic ethers tend more to peroxides than acyclic ethers? $\endgroup$ May 9, 2019 at 12:14

Liquid-Phase Photolysis of Dioxane has been studied in several occasions in the past. One of these studies, several neat aliphatic ethers including dioxane were irradiated for $\ce{24 h}$ with a medium-pressure mercury lamp (Ref.1). As a result, the neat liquid-phase dioxane has been reported to yield gaseous, liquid, and solid products after the irradiation. Among the products, the major gaseous products were identified as $\ce{H2, CO, CH4, C2H4,}$ and $\ce{C2H6}$, but the liquid products have not been analyzed. However, the solid products were shown to be the racemic and meso forms of dioxyldioxane (see total reaction illustrated in the diagram to give solid product):

Dioxyldioxane formation

Pfordte rationalized the formation of dioxyldioxane as a simple dimerization of dioxyl radical formed by photoinitiated hydrogen abstraction from dioxane (Ref.1, 2):


Thus, it is safe to conclude that dioxane is susceptible to sun light and/or a foreign radical from any other source. Thus, after a long time exposure to sunlight or a in the presence of a foreign radical, the formation of dioxyl radicals is possible and exposure to environment would supply oxygen (also dissolved $\ce{O2}$) to further react as shown in the propagation step as follows:



  1. Von Klaus Pfordte, “Intermolekulare Dehydrierungen durch UV-Licht, IV – Photoreaktionen der Äther,” Justus Liebigs Ann. Chem. 1959, 625(1), 30–33 (in German) (https://doi.org/10.1002/jlac.19596250104).
  2. P. H. Mazzocchi, M. W. Bowen, “Photolysis of dioxane,” J. Org. Chem. 1975, 40(18), 2689–2690 (DOI: 10.1021/jo00906a029).

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