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What is / are the product(s) of ozonolysis of phenanthrene?

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I am not sure I understand how to proceed. Which of the $\ce{C=C}$ double bonds is most susceptible to this type of reaction?

I am confused given the resonance and aromatic character of the compound which doesn't make me feel comfortable doing ozonolysis the usual way.

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    $\begingroup$ In anthracene (an isomer of your compound), the central ring is less aromatic (and susceptible e.g., to the Diels-Alder reaction) than the two outer ones. Similar here, in phenanthrene, the double bond at position 9,10 is easier attacked by ozonolysis, than the other positions (relatively speaking). So start from there. You might infer from other resources as well, e.g. OrganicSynthesis, too. And it is much easier to keep track of the number of electrons with explicit dashes for single and double bonds, than with Thiele rings. $\endgroup$ – Buttonwood May 24 at 19:42
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    $\begingroup$ @Ashish In case of anthracene, Clar's rule is helpful. On occasion, Wikipedia sub-section's tagged "chemistry" (example) note general pattern. An identification of these may be helped by the structure explorer, too. $\endgroup$ – Buttonwood May 24 at 19:56
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Phenanthrene reacts quickly with one mole of ozone at the 9,10-double bond according to this paper here and the OrgSyn procedure here referred to above by @Buttonwood. In chloroform a polymeric mono-ozonide is obtained. In MeOH the reaction takes a different course resulting in the formation of a cyclic methoxy-hydroxy-peroxide and a cyclic dimethoxy-peroxide. Reductive work up (Zn/AcOH or Me2S) of these products gives the di-aldehyde.

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