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I came across a question in which the major product was asked for Kolbe's electrolysis of potassium salt of 2,3-dimethylbutane-1,4-dioic acid. The answer was given as trans-(but-2-ene). I couldn't understand how does the trans isomer form as the specific major product from this reaction, following free radical mechanism. Would someone please help me with this. Thanks!

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  • $\begingroup$ As to your concern about the lack of references to the Kolbe process, try: J. Meyers, et al., Intramolecular Biradical Recombination of Dicarboxylic Acids to Unsaturated Compounds: A New Approach to an Old Kolbe Reaction, ChemElectroChem, 2020, 7 (24), 4873-4878. $\endgroup$
    – user55119
    Jul 18 at 15:17

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It has little to do with mechanism and more to do with the fact that stable products are, in general, the major products in most of the chemical reactions.

In this case, trans-(but-2-ene) is more stable than cis-(but-2-ene), due to steric reasons.

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Generally during electrolysis of dicarboxylic acid we will get a few different products but the main product is unsaturated hydrocarbon. Maybe at this point, based on ChemElectroChem, 2020, 7, 4874 - down of page, the mechanism reaction is like in drawing:

enter image description here

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  • $\begingroup$ Thanks for the answer. My concern with the mechanism shown is that according to this article, the syn-conformer is more stable than the anti-conformer that you have drawn here. $\endgroup$
    – ananta
    Jul 19 at 16:46
  • $\begingroup$ @ananta: Just because the syn-clinal conformation may be more stable in solution, does not imply what goes on when radicals are formed. $\endgroup$
    – user55119
    Jul 19 at 23:14
  • $\begingroup$ @user55119 under kinetic conditions (for example, high volatage), the syn-conformation may not to turn into the anti-conformation in time. We need to know how much of the cis- and trans-isomers are formed vs the voltage, temperature, pressure graphs. Do some metallic surfaces promote the formation of cis-conformer? Most studies are carried over a pristine platinum surface. Surely, there are unanswered questions here, and an incompletely understood mechanism. $\endgroup$
    – ananta
    Jul 20 at 1:29

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