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I am studying additional reactions in organic chemistry and I am confused as to why ozone cleaves double bonds? I already know the mechanism of what happens but can anyone give a more conceptual understanding of why this occurs?

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closed as unclear what you're asking by Klaus-Dieter Warzecha, Jan, bon, Todd Minehardt, Wildcat Dec 9 '16 at 22:12

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    $\begingroup$ The mechanism actually tells you why this occurs. It's a step-by-step description. The overall reaction is favored. Each step is favored or provides easy access to a favored step. $\endgroup$ – Zhe Dec 8 '16 at 3:17
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    $\begingroup$ Not all reactions are driven by... wait, I think I already said that. $\endgroup$ – Ivan Neretin Dec 8 '16 at 6:09
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Analysis of the frontier molecular orbitals of the alkene and ozone molecules easily and elegantly predicts the occurrence of the first step of this reaction:

The actual cleavage step forms a carbonyl and carbonyl oxide—a pair of molecules that are electronically identical to the starting reagents, but are thermodynamically favored over their regenration:

$$ \small \begin{array}{lcc} \hline & \text{Bond Dissociation} & & \text{Bond Dissociation}\\ \text{Bond} & \text{Enthalpy}\ \mathrm{(kJ\ mol^{-1})} & \text{Bond} & \text{Enthalpy}\ \mathrm{(kJ\ mol^{-1})} \\ \hline \ce{C-C} & 347 &\ce{C=C} & 614\\ \ce{O-O} & 146 & \ce{O=O} & 495\\ \ce{C-O} & 358 & \ce{C=O} & 745^{[1]}\\ \hline \end{array} $$

Formation of the secondary ozonide is also thermodynamically favored:


$^{[1]}$Bond Dissociation Enthalpy Tables

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