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In my lab course I had to synthesize 9,10-phenanthrenequinone (phenanthrene-9,10-dione) from phenanthrene and chromium(VI) oxide. The synthesis was made as explained in Organic Syntheses, Coll. Vol. 4, p.757 (1963)

The reactants are: phenanthrene (1.00 eq), chromium(VI) oxide (7.50 eq), sulfuric acid (15.0 eq), water (83.3 eq)

I now have to explain the reaction mechanism, but unfortunately I don't find it in any of my textbooks, neither on the internet.

Here is my assumption:

  • The chromium(VI) oxide reacts with water and forms chromic acid.
  • The chromic acid can oxidize an alcohol to a ketone (Jones-Oxidation)

So what misses is the alcohol. Somehow I have to get the 9,10-phenanthrenediol, which could react as alcohol in the Jones-Oxidation.

How could this possibly work. Remember, I don't have $\ce{OsO4}$, which could easily dihydroxilate an alkene.

EDIT 1: As an answer to the proposed mechanism by @Zhe:

Is this ok in your eyes?

Mechanism

EDIT 2: An other idea for the first step (dihydroxilation)

A person of the scientific staff of my university told me, that the $\ce{CrO3}$ should react in a very similar way as $\ce{OsO4}$ does. As I'm using 7.50 eq of chromium(VI) oxide, there is enough for the dihydroxilation and the Jones-Oxidation. He said that a radical mechanism is rather unlikely, because I'm working with water as solvent and water is an inhibitor for radical reactions. This is the mechanism I came up with, regarding these new information.

Mechanism 2

What do you think about this possibility?

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  • $\begingroup$ This probably happens via a single electron oxidation of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone. I can suggest a more complete mechanism later if I time, but try that and see where you can take it. $\endgroup$
    – Zhe
    Commented Jun 25, 2017 at 19:09
  • $\begingroup$ @Zhe Thanks for your comment. Would you please explain this a bit more? I only know the single electron oxidation for carbonyl compounds and I don't find anything about alkenes. Thanks in advance! $\endgroup$
    – Sam
    Commented Jun 26, 2017 at 9:00
  • $\begingroup$ That's not a bad idea. Pretty similar to how $\ce{OsO4}$ works. Again, without more mechanistic evidence, it's hard to know for sure, but I find this quite plausible. $\endgroup$
    – Zhe
    Commented Jun 27, 2017 at 15:45

1 Answer 1

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Here's what I was thinking.

This probably happens via a single electron oxidation of that double bond to a radical cation as a first step. The cation can trap water and then we can oxidize the radical to a cation. Eventually, you get to a diol which you can oxidize to a diketone.

Mechanism

I ommited the steps for chromium oxidation from the alcohol to the ketone. Sounds like you are already familiar with those.

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  • $\begingroup$ Thank you very much. How is the first step induced? How do you get the cation and the radical? $\endgroup$
    – Sam
    Commented Jun 26, 2017 at 18:55
  • $\begingroup$ Chromium(VI) is a strong oxidant, so I assumed it can just pick up an electron. Granted, that is not how the oxidation of alcohols to ketones works, so I may want to come up with a mechanism that involves a $\pi$-complex with the chromium. $\endgroup$
    – Zhe
    Commented Jun 26, 2017 at 19:04
  • $\begingroup$ I edited my question. It would be great if you'd have a look at it. $\endgroup$
    – Sam
    Commented Jun 26, 2017 at 20:38
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    $\begingroup$ Not sure that the SET would behave as drawn. There may be some significant changes regarding the oxo groups on chromium. Again, this is just a guess, and I think it's reasonable, but I wouldn't declare it correct since I don't have any mechanistic data. I'm also considering using the olefin a $\eta^{2}$ ligand, but I don't have a productive way of eliminating the chromium that I like. That's not to say that that mechanism isn't better. The best bet would be to look at the literature more thoroughly for mechanistic studies. Sadly I don't have journal access anymore to help there... $\endgroup$
    – Zhe
    Commented Jun 26, 2017 at 20:50
  • $\begingroup$ As an undergraduate student I don't have any access to those either :( And in my textbooks, there is nothing that seems to help with this. $\endgroup$
    – Sam
    Commented Jun 26, 2017 at 22:18

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