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I have the following reaction: A peroxide reacting with a sulfide to form anthracene, sulfoxide and water or sulfoxide and water depending on the reactant. The reaction is from a paper discussing the oxygen indused oxidation of bitumen (asphalt binder).

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The paper cites another paper for the reaction mechanism which is of interest to me and is shown below:


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In my opinion the two reactions differ greatly in their products and the mechanism doesn't make sense.
1. In the first step the why would the sulphur atom attack the negative oxygen atom?
2. Is the reaction even possible with hydrogen peroxide?


Edit: I have also found another paper which somewhat changes the initial from the sulphur attacking the oxygen to the oxygen attacking the sulphur. I assume either the oxygen is attacking or the sulphur is attacking. Who could be right?
J. Am. Chem. Soc. 2004, 126, 3, 900–908

1 Claine Petersen, J. “A Dual, Sequential Mechanism for the Oxidation of Petroleum Asphalts.” Petroleum Science and Technology, vol. 16, no. 9–10, 1998, pp. 1023–1059, https://doi.org/10.1080/10916469808949823.

2 Hargrave, R. “Oxidation of Organic Sulphides - VI. Interaction of Hydroperoxides with Unsaturated Sulphides.” Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences, vol. 235, no. 1200, 1956, pp. 55–67, https://doi.org/10.1098/rspa.1956.0064.

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    $\begingroup$ The mechanism invokes the use of acid, H-X. R2S attacks the terminal oxygen to gives R2SO and an alcohol. Protonation of the hydroxyl group of the alcohol causes lose of water and the formation of anthracene. The mechanism shown does not elucidate the individual steps. Think of one of the peroxide oxygens as negative and the other one being positive, the site of nucleophilic attack. The reaction occurs with $\ce{H2O2}$ and ozonides. $\endgroup$
    – user55119
    Commented May 20, 2023 at 21:33
  • $\begingroup$ @user55119 what could be possible individual steps? If another H atom would protonate the OH group it would form water but leave the R group positively charged. $\endgroup$
    – bobsburger
    Commented May 22, 2023 at 20:16
  • $\begingroup$ I believe I told you the individual steps. What I think you are suggesting is to protonate the terminal oxygen, lose water, and form anthrone, not anthracene. That doesn't get you where you want to go! $\endgroup$
    – user55119
    Commented May 22, 2023 at 22:01
  • $\begingroup$ @user55119 "Protonation of the hydroxyl group of the alcohol causes lose of water" yes but also the formation of a carbocation instead of anthracene. Or is the step from carbocation to anthracene trivial? $\endgroup$
    – bobsburger
    Commented May 23, 2023 at 22:54
  • $\begingroup$ Loss of water would give a positive charge on the oxygen that is still attached to the carbon framework. Loss of a proton would give anthrone. There would be no positive charge on carbon. $\endgroup$
    – user55119
    Commented May 24, 2023 at 1:12

1 Answer 1

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Bobsburger: If I am understanding you correctly, you are proposing a reaction scheme wherein the hydroperoxide 1 forms anthracene via the loss of water. Pathway 1 $\rightarrow$ 2 $\rightarrow$ 3 $\rightarrow$ etc. does not provide anthracene but rather anthrone 4. There is no role for dimethyl sulfide (DMS) because hydroperoxide 1 and anthrone 4 are of the same oxidation level. This process is merely an acid-catalyzed rearrangement and not a redox reaction.
On the other hand and following the red arrows, attack of DMS on the terminal oxygen of hydroperoxide 1 affords an intermediates alkoxide 5 and the protonated form of dimethyl sulfoxide (DMSO) 6. A neutralization of these two species provides alcohol 7 and DMSO. This alcohol in acid medium forms carbocation 9 via the protonated alcohol 8. Loss of a proton from carbocation 9 provides anthracene 10. Accordingly, hydroperoxide 1 is reduced by two electrons and DMS is oxidized by two electrons.
Now, if this reaction were run in some sort of aqueous acid medium, then one might argue that protonation of the inner oxygen of 1 leads to a solvolysis to alcohol 7 via carbocation 8, both of which lead to anthracene 10. With or without water, hydrogen peroxide would also be formed. If this pathway were viable, then there is no need for the presence of DMS.

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  • $\begingroup$ yes I was referring to reaction pathway 1-5-7-8-9-10 in your scheme. The last step, loss of a proton is still unclear to me. How does that happen? $\endgroup$
    – bobsburger
    Commented May 25, 2023 at 7:13
  • $\begingroup$ 9 is a strong acid. Anthracene is its conjugate base. Any convenient base such as water will effect the deprotonation. $\endgroup$
    – user55119
    Commented May 25, 2023 at 13:26
  • $\begingroup$ I have found in a different paper that O attacks S in the initial attack instead of the other way around. what would be the explanation for this? $\endgroup$
    – bobsburger
    Commented May 28, 2023 at 21:41
  • $\begingroup$ What is the paper? Reference? The O-O bond is electron deficient and S is a good nucleophile. $\endgroup$
    – user55119
    Commented May 28, 2023 at 23:53
  • $\begingroup$ J. Am. Chem. Soc. 2004, 126, 3, 900–908. I’ve also put it in the question $\endgroup$
    – bobsburger
    Commented May 29, 2023 at 12:03

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