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From the above mechanism it is clear to see that the electrons from the double bond attack one of the terminal oxygens in ozone, thus meaning it's an electrophile. How can this be? If you draw the resonance structures, the negative charge is concentrated on the terminal oxygens too. I am OK with the rest of the mechanism but I would never think to draw oxygen as an electrophile.

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    $\begingroup$ This type of reaction is more appropriately described as a pericyclic reaction. $\endgroup$ – Ben Norris Feb 14 '15 at 23:10
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Electrons are concentrated on the two terminal oxygens in ozone but remember, ozone exhibits resonance. The electrons can be delocalized across each oxygen through p-orbitals.

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So yes, at first if you think about the reaction in purely Coulombic terms then it doesn't make much sense. Why would electrons go to a region that already has a relative surfeit of electrons?

But what happens is that by attacking the terminal oxygen, the double bond effectively displaces electrons to the central, electron-deficient oxygen atom.

I'm sure someone can explain this in much greater detail with reference to molecular orbital theory but I'll leave that to someone else.

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As Ben Norris mentioned in his comment, ozonolysis is better described as a pericyclic reaction. Pericyclic reactions fall outside of the usual nucleophile-electrophile paradigm. This reaction is an example of a dipolar cycloaddition, because ozone is a 1,3-dipole. Chemists have observed that electron-rich alkenes react faster with ozone than electron-poor alkenes. This implies that the alkene is providing the HOMO to the reaction and ozone is providing the LUMO. In nucleophile-electrophile reactions, the nucleophile reacts through its HOMO, and the electrophile reacts through its LUMO. So if you want to make an analogy between a dipolar cycloaddition and a nucleophile-electrophile reaction, you could say that the alkene is the 'nucleophile' and ozone is the 'electrophile.'

It's not always obvious that the 1,3-dipole is going to provide the LUMO in a dipolar cycloaddition, but we can speculate as to why ozone does. Oxygen is a very electronegative atom, so its atomic orbitals are lower energy than carbons. Since ozone is made up of three oxygens (with low energy atomic orbitals orbitals), its pi-system will be low energy with respect to an alkene's pi-system. Furthermore, the pi-system will be even lower energy due to placing a cation on an electronegative atom. These factors contribute to the LUMO being low energy relative to an alkene.

To a first approximation, if there's a positively charged oxygen in a compound, it can be assumed that compound will be electrophilic.

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    $\begingroup$ I'm sure someone on this site could run calculations to support these rationalizations. $\endgroup$ – jerepierre Feb 18 '15 at 1:17

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