I would like to, theoretically, prepare oxalic acid from acetylene.

I think that using ozonolysis would be a good idea to do so. Upon ozonolysis of acetylene, we would get:


On oxidizing this using hydrogen peroxide, would I obtain the desired product?


Ozonolysis of acetylene won't produce oxalic acid. The triple bond holding the two acetylenic carbons together will be attacked as shown below. The reaction is complex and depending upon reaction conditions, different products can result (ref_1, ref_2).

enter image description here

Oxalic acid can be produced from acetylene by reacting acetylene with alkaline $\ce{KMnO4}$ (ref_3).

$\ce{H-C#C-H ->C[KMnO4/OH^-]\ HOOC-COOH}$

EDIT: Response to OPs comment Here is a rough idea of what is going on mechanistically:

1) the permanganate forms a cyclic intermediate

2) that produces a diol

3) since a double bond is still present the reaction can be repeated to yield 2 gem-diols that can equilibrate to an alpha-diketone

4) now the standard alkaline permanganate oxidation of a carbonyl occurs

5) hydroxide can add to a carbonyl and

6) the manganese ion complexes with the carbonyl oxygen

7) the alpha-proton donates its electrons to reform the carbonyl which is now part of a carboxylic acid

8) repeat steps 5-7 on the other carbonyl and you have oxalic acid

enter image description here

  • $\begingroup$ Isn't the reaction with hot alkaline $\ce{KMnO4}$ basically oxidative ozonolysis? $\endgroup$ Feb 16 '15 at 15:54
  • 1
    $\begingroup$ I've never heard it used that way. "Ozonolysis" implies a specific reaction involving ozone. $\endgroup$
    – ron
    Feb 16 '15 at 15:57
  • $\begingroup$ But the two will give different products. So it cannot be the same reaction. What is the mechanism behind the reaction that you have proposed? $\endgroup$ Feb 16 '15 at 16:20
  • $\begingroup$ Yes, two different reactions, two different mechanisms and two different sets of products. I've edited my answer to include a basic mechanism. $\endgroup$
    – ron
    Feb 16 '15 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.