# What are all products of an ozonolysis of alkenes using zinc and water?

I'm studying ozonolysis reactions in alkenes, and saw that in one of them there is the formation of $\ce{H2O2}$ (Can anyone explain from where?) that I presumed that reacts with $\ce{Zn}$, one of the reagents, as the following: \begin{align} \ce{2H2O2 &-> 2H2O + O2} \\ \ce{2H2O + Zn &-> Zn(OH)2 + H2} \\ \hline \ce{H2O2 + Zn &-> Zn(OH)2} \end{align}

The general reaction is \begin{align} \ce{Unsaturated~compound (-C=C-) + O3 &-> Ozonide\\\\ Ozonide + Zn &-> R^1-(C=O)-R^2 + R^3-(C=O)-R^4 + X} \end{align}

But one of my books shows that this $\ce{X}$ is $\ce{Zn(OH)2}$ only and other $\ce{ZnO}$ only. What about this? What am I doing wrong?

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.We prefer not to use markup in the title field, see here for details. – Martin - マーチン Jan 4 '16 at 6:37
• Your second equation is not balanced: $$\ce{2H2O + Zn -> Zn(OH)2 + H2}$$ Hence $$\ce{H2O2 + Zn -> Zn(OH)2}.$$ – Martin - マーチン Jan 4 '16 at 6:44
• @Martin Ooops! Thank you! But $\ce{H2}$ and $\ce{O2}$ will not form water? – Vinicius ACP Jan 4 '16 at 7:01
• Well I personally would assume that hydrogen peroxide directly reacts with zinc. So my estimate would be that there is no hydrogen or oxygen formation. It probably depends a lot on the mechanism and your $\ce{X}$ will ultimately depend on how much water and what pH is present. – Martin - マーチン Jan 4 '16 at 7:08