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As the question says, I'm supposed to make cyclohex-4-ene-1,2-diol using cyclohexa-1,4-diene:

enter image description here

I can use any inorganic reactant, any solvent, and any organic molecule with 3 or fewer carbons. I am not sure how to go about this question, as I do not know how to use regional-specific reactions. Any help, even just mentioning a reaction mechanism I can use for this would be greatly appreciated.

I could put the cyclohexa-1,4-diene on $\ce{KMnO4}$ for a catalyst and $\ce{H2O}$ with $\ce{HO-}$ as solvents in order to break the double bond and replace it with the $\ce{HO-}$ molecules. Problem is that this would affect the molecule as a whole – not just a specific side which is what I need.

I also thought of using bromine as something temporary, since they are easy to remove however, again, this affects the whole molecule.

I also thought about Diels-Alder reaction, but we haven't learned anything about that so I do not know how to use it- much less on a cyclic molecule.

I have also seen the osmium tetroxide oxidation of alkenes. However, I still have the problem that I can't control this reaction to only affect one side.

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  • $\begingroup$ It doesn't matter you're not sure. Write some idea even if it wouldn't work. $\endgroup$
    – Mithoron
    May 20 at 0:13
  • $\begingroup$ I mean, I could put the cyclohexa-1,4-diene on $\ce{KMnO4}$ for a catalyst and $\ce{H2O}$ with $\ce{HO}$ as solvents in order to break the double bond and replace it with the OH molecules. Problem is that this would affect the molecule as a whole- not just a specific side which is what I need. I also thought of using Bromine as something temporary, since they are easy to remove however, again, this affects the whole molecule. I also thought about Diels-Alder, but we havent learned anything about that so I do not know how to use it- much less on a cyclic molecule. $\endgroup$ May 20 at 0:26
  • $\begingroup$ You should edit it into the question. $\endgroup$
    – Mithoron
    May 20 at 1:06
  • $\begingroup$ Ok, I added that to the question. $\endgroup$ May 20 at 1:20
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    $\begingroup$ I think if you use 1eq of KMnO4 you will still get some tetra-ol, the diol can and will react. The mixture will be separable and maybe this is how you have to do it. $\endgroup$
    – Waylander
    May 20 at 7:22
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I think the easier way of doing this is to epoxidise the diene with 1 eq of a peracid epoxidation with peracids . The standard reagent m-Chloroperoxybenzoic acid (MCPBA) breaks your rule of organic molecules with 3 carbon atoms or fewer, but you can also do it with peracetic or pertrifluoroacetic acid preparation of peracetic acid. You will inevitably get some di-epoxide, but this will be separable by chromatography. The mono-epoxide can then be reacted in mild aqueous acid to give the diol. You could also carry the mixture through to the aqueous acid treatment as the tetraol will be far more water soluble than the diol and can be removed by an aqueous wash.

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You mentioned the use of $\ce{OsO4}$ to form the enediol. The secret is not to perform the osmylation catalytically ($\ce{OsO4/HIO4}$) but rather stoichiometrically in ether. The initially formed osmate ester often precipitates from solution thereby suppressing overoxidation. The osmate ester may be decomposed with $\ce{H2S}$. The downside --- prohibitively expensive.

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  • $\begingroup$ In addition OSHA will be concerned about the toxicity of $\ce{OsO4}$ (reference). $\endgroup$
    – Buttonwood
    May 20 at 21:46
  • $\begingroup$ @Buttonwood: The OSHA concern goes without saying. A one gram ampoule of OsO4 is ~ 4 mmol. $\endgroup$
    – user55119
    May 20 at 21:54
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What is the desired stereochemical outcome of the product?

  1. For cis-diol, use 1/2 eq KMnO4/NaOH
  2. For trans-diol, use 1/2 eq MCPBA followed by catalytic H2SO4/H2O (excess H2SO4 will cause protonation of the π bond and lead to undesired polymerization and other unwanted side products)

Does this help? :)

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  • $\begingroup$ Item 1: What prevents overoxidation. One is bound to get tetraol and recovered diene. also. Item2: Why wouldn't protonation of the double bond then lead to hydration? $\endgroup$
    – user55119
    May 21 at 14:40

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