1
$\begingroup$

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 $2 \ \ce{CO2}$?

$\endgroup$
7
  • $\begingroup$ chemistry.stackexchange.com/questions/4310/… Related $\endgroup$ Jul 14, 2020 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, 2020 at 6:37
  • $\begingroup$ So, acc. to book, answer is 2 + 2 = 4 CO2? $\endgroup$ Jul 14, 2020 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, 2020 at 6:38
  • 1
    $\begingroup$ 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, 2020 at 6:57

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.