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I was wondering what would be the product of the reaction of alkene with a diol (such as glycol) in alkoxymerucuration/demercuration, i.e. $\ce{(CF3CO2)2Hg/NaBH4}$?

I tried to look online but I couldn't land on a paper or website that goes through such reaction. I also tried Scifinder, I marked both compounds as reactants, but zero results stemmed from the reactants! Any input?

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I'm thinking if these possibilities (I have more possibilities in my mind, but for example I selected three) are structurally accurate and energetically stable to be produced in such reaction and which is favored?

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To classify them more ... to single and di addition, here is what I'm thinking:

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    $\begingroup$ Oxymercuration is often run in the presence of excess water or alcohol, but only one RO/OH adds. Therefore I presume in the presence of a diol you get mono-addition by one of the 2 available OHs, probably the least hindered. en.wikipedia.org/wiki/Oxymercuration_reaction $\endgroup$
    – Waylander
    Nov 8 '20 at 8:32
  • $\begingroup$ @Mithoron so Alkoxymercuration/demercuration (CF3CO2)2Hg/NaBH4) won't force this reaction? if it doesn't what can force this reaction? I want to see whether such reaction will add seperate glycol on each carbon of the double bond, or the glycol will form a cyclic with the two carbon of the double bonds? $\endgroup$
    – user99277
    Nov 8 '20 at 17:21
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    $\begingroup$ I think the top left option is most likely. I see no reason why you would get di-addition to the alkene $\endgroup$
    – Waylander
    Nov 8 '20 at 19:11
  • $\begingroup$ @Waylander I updated the question with single and di-addition possibilities, I'm not sure if excess of glycol or other factors will move the reaction toward single or di-addition or toward open or cyclic formation in either case. $\endgroup$
    – user99277
    Nov 8 '20 at 20:10
  • $\begingroup$ In your original scheme, the product in the upper left would likely have the hydroxyl attach to the benzylic position. Two of the remaining three products involve C-C bond formation. How is that accomplished with oxymercuration? $\endgroup$
    – user55119
    Nov 8 '20 at 20:11
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There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE.
To start, the alkene is the (E)-propenylbenzene 1 and the diol is butan-2,3-diol 3, devoid of stereochemistry. Formation of mercurinium ion 4 with mercuric trifluoroacetate (2) is drawn as unsymmetrical because in the transition state for addition the carbon bearing the greater positive charge is the site of SN2, C-O bond formation. Mercurial 5 is a racemate, ignoring any stereochemistry of 3, which is what is formed after the first reaction. The second reaction, reduction with NaBH4, proceeds via radical 6 to form alcohol 7. A minor product may be bis-ether 9 arising from alkene 1 and mercurial 5. For mercurial 5 to form 1,4-dioxane 8 requires an oxidation of the carbon bound to mercury and a reduction of Hg+2 to Hgo as shown by the red arrows in structure 5. How do we know that this process does not occur? Because if it did, NaBH4 would not be required! Consider the oxymercuration of alkene 1 in the presence of water. If what you are suggesting were true, then 1 would form a diol rather than the observed mono-ol prior to the participation of NaBH4.

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