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?


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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Not all reactions are driven by... wait, I think I already said that. $\endgroup$ – Ivan Neretin Dec 8 '16 at 6:09

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


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