What is the product when naphthalene is oxidized by alkaline (or acidic) solutions of $\ce{KMnO4}$?

Some possible reactions show up in a google search, but they have no references:

  1. Naphthalene to phthalic acid by alkaline $\ce{KMnO4}$:
  2. Another product on oxidation of napthalene by alkaline $\ce{KMnO4}$:

If possible, please include the reaction mechanisms in the answer as well.

  • 4
    $\begingroup$ Quick search on Reaxys shows that both the products you've already listed have been made, also plus 2-formylbenzoic acid. Some are from 19th-century Chem. Ber. articles, which I'm not inclined to look up. $\endgroup$ Commented Jan 19, 2018 at 21:45
  • 1
    $\begingroup$ Here is a reference: sciencedirect.com/science/article/abs/pii/S0045653510001906 $\endgroup$
    – Karsten
    Commented Apr 18, 2021 at 18:24
  • 1
    $\begingroup$ The ketoacid in 2 is oxidized to phthalic acid by high valence manganese. Derivatize ArCOCO2H as ArCOCO2Mn(+7). Move electrons from the sigma C-C bond toward Mn+7.which forms ArCO+ , CO2 and Mn+5. The acylium species is captured by base. Alternatively, base attacks the keto group of the derivatized Mn+7 species, which upon collapse, moves electrons toward Mn+7 forming o-phthalic acid. $\endgroup$
    – user55119
    Commented Apr 18, 2021 at 18:32
  • 1
    $\begingroup$ jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf. This link has a mechanism for the first reaction. However, the conditions are not clear. As for the second product, is can be observed in multiple sources (but no mechanism): orgsyn.org/demo.aspx?prep=cv2p0523 and pubs.acs.org/doi/pdf/10.1021/j150083a001. From what I am getting, it is likely the first product is in acidic conditions while the second one is in basic conditions. The mechanism for the second one isn't too clear though. $\endgroup$
    – M.L
    Commented Apr 19, 2021 at 23:07
  • 1
    $\begingroup$ @ShoubhikRMaiti For #1 in basic media, there are old references like 1907JPhysChem93. For running the reaction in acidic media, you find via google e.g. this, too. $\endgroup$
    – Buttonwood
    Commented Apr 23, 2021 at 20:45

1 Answer 1


I already made a comment about some of what I am about to say but I will provide a partial answer. I say partially because I could not find any mechanism for the second product. However, from literature, what I found was that in acidic conditions, $\ce{KMnO4}$ will oxidize naphthalene into the first product, phthalic acid1. In basic/alkaline conditions, $\ce{KMnO4}$ will oxidize naphthalene into Phthalonic Acid2. The results also agree from other papers. The proposed mechanism for the acidic solution reaction is here (taken from the paper in the first link): enter image description here

It is essentially the same mechanism for the cleavage of a double bond but it happens in two positions in naphthalene. Note this is the mechanism for $\ce{KMnO4}$ oxidative cleavage:

enter image description here

Also note that the mechanism in the first image was based on a 2017 paper studying the kinetics of the reaction and then formulating a mechanism. So this reaction is still being studied. Along the same lines, I could not find the mechanism for the reaction of $\ce{KMnO4}$ and naphthalene in alkaline conditions. I would assume that its mechanism is still being researched.


  1. https://www.jchps.com/issues/Volume%2010_Issue%201/40-0581116.pdf
  2. The Oxidation of Naphthalene to Phthalonic Acid by Alkaline Solutions of Permanganate, R. Arthur Daly, The Journal of Physical Chemistry 1907 11 (2), 93-106, DOI: 10.1021/j150083a001
  3. http://www.orgsyn.org/demo.aspx?prep=cv2p0523
  4. https://www.acros.com/_rainbow/pdf/oxidation_brochure_manga.pdf
  5. http://studymaterial.unipune.ac.in:8080/jspui/bitstream/123456789/8059/1/POLYNUCLEAR%20HYDROCARBONS.pdf

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