Pyrolysis of dimethyl peroxide

By using isobutane ( t ‐BuH) as a radical trapit has been possible to study the initial step in the decomposition of dimethyl peroxide (DMP) over the temperature range of 110–140°C in a static system. For low concentrations of DMP (2.5 × 10 −5 −10 −4 M ) and high pressures of t −BuH (∼0.9 atm) the first‐order homogeneous rate of formation of methanol (MeOH) is a direct measure of reaction (1): \documentclass{article}\pagestyle{empty}\begin{document}${\rm DMP}\mathop \to \limits^1 2{\rm Me}\mathop {\rm O}\limits^{\rm .},{\rm Me}\mathop {\rm O}\limits^{\rm .} + t{\rm - BuH}\mathop \to \limits^4 {\rm MeOH} + t{\rm -}\mathop {\rm B}\limits^{\rm .} {\rm u}$\end{document} . For complete decomposition of DMP in t ‐BuH, virtually all of the DMP is converted to MeOH. Thus DMP is a clean thermal source of Me \documentclass{article}\pagestyle{empty}\begin{document}$\mathop {\rm O}\limits^{\rm .}$\end{document} . In the decomposition of pure DMP complications arise due to the H‐abstraction reactions of Me \documentclass{article}\pagestyle{empty}\begin{document}$\mathop {\rm O}\limits^{\rm .}$\end{document} from DMP and the product CH 2 O. The rate constant for reaction (1) is given by k 1 = 10 15.5−37.0 /θ sec −1 , very similar to other dialkyl peroxides. The thermochemistry leads to the result D(MeOOMe) = 37.6 ± 0.2 kcal/mole and / H   ° f (Me \documentclass{article}\pagestyle{empty}\begin{document}$\mathop {\rm O}\limits^{\rm .}$\end{document} ) = 3.8 ± 0.2 kcal/mole. It is concluded that D(ROOR) and D(ROH) are unaffected by the nature of R. From Δ S   ° 1and A 1 , k 2 is calculated to be 10 10.3±0.5 M −1 · sec −1 : \documentclass{article}\pagestyle{empty}\begin{document}$2{\rm Me}\mathop {\rm O}\limits^{\rm .} \mathop \to \limits^2 {\rm DMP}$\end{document} . For complete reaction, trace amounts of t ‐BuOMe lead to the result k 2 ∼ 10 9 M −1 ·sec −1 : \documentclass{article}\pagestyle{empty}\begin{document}$2t{\rm - Bu}\mathop \to \limits^5$\end{document} products. From the relationship k 6 = 2( k 2 k 5a ) 1/2 and with k 5a = 10 8.4 M −1 · sec −1 , we arrive at the result k 6 = 10 9.7 M −1 · sec −1 : \documentclass{article}\pagestyle{empty}\begin{document}$2t{\rm - u}\mathop {\rm B}\limits^{\rm .} \to (t{\rm - Bu)}_{\rm 2}{\rm,}t{\rm -}\mathop {\rm B}\limits^{\rm .} {\rm u} + {\rm Me}\mathop {\rm O}\limits^{\rm .} \mathop \to \limits^6 t{\rm - BuOMe}$\end{document} .