What are the products of the reaction of 1,3-butadiene with hot $\ce{KMnO4}$ (excess)?

I know that $\ce{KMnO4}$ will first cleave the double bonds and we should get $$\ce{CH2=CH-CH=CH2 ->[KMnO4] 2CH2=O + O=CH-CH=O}$$ Now the aldehydes will further oxidise to $$\ce{H2C=O <<=> H2C(OH)2 ->[KMnO4] O=C=O}$$ $$\ce{O=CH-CH=O ->[KMnO4] HOOC-COOH}$$

I don't understand how and why $\ce{HO2C-CO2H}$ would further oxidise to $2 \ \ce{CO2}$?

  • $\begingroup$ chemistry.stackexchange.com/questions/4310/… Related $\endgroup$ Jul 14 '20 at 6:33
  • $\begingroup$ @Safdar, the link just provides the balanced equation. But the question is why does it react in the first place? Also what is the mechanism? $\endgroup$ Jul 14 '20 at 6:37
  • $\begingroup$ So, acc. to book, answer is 2 + 2 = 4 CO2? $\endgroup$ Jul 14 '20 at 6:38
  • $\begingroup$ @RahulVerma, yes that is correct. This question is from Solomon's Organic Chemistry 11e, Problem 13.20 (f), (Chapter 13, pg no 619). $\endgroup$ Jul 14 '20 at 6:38
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    $\begingroup$ The KMnO4 oxidation of oxalic acid mechanism is highly complex and is under debate. This was the most recent paper I could find with possible mechanisms $\endgroup$ Jul 14 '20 at 6:57

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