Atomic oxygen rate-determining formation

The necessary condition is that the rates of formation of the products Pi and Pi per oxygen atom reacting with A1 and 42, respectively, are independent of whether only one or both A1 and A 2 are present. This condition is satisfied at sufficiently high pressures (where pressure-dependent fragmentation is suppressed) by most of the products of addition of oxygen atoms to olefins. If the initial concentrations (vl,) and (A2)j are sufficiently high so that they remain to a good approximation constant in the course of the reaction, the ratio of the rate constants is determined from the expression... 

The kinetics of the liquid-state reaction are also first order. For both the solid and liquid the activation enthalpies and entropies are the same, AS = 8 J deg1, indicating that the rate-determining step is the breaking of a Cl-O bond, with the formation of atomic oxygen... 

No thermodynamic values have been measured. Decomposition kinetics have been studied in some detail,77-80 with some of the studies being concerned with the effect of additives and irradiation on the course of decomposition. Various equations were used to represent different stages of the reaction. The activation energy is 209 kJ, consistent with the breaking of an 1-0 bond and the formation of atomic oxygen as the rate-determining step. 

Steps 1,3, and 5 cannot be slow as they are just proton transfers between oxygen atoms (Chapter 13). That leaves only steps 2 and 4 as possible rate-determining steps. The bimolecular addition of the weak nucleophile water to the low concentration of protonated ester (step 2) is the most attractive candidate, as step 4—the unimolecular loss of ethanol and re-formation of the carbonyl group—should be fast. What p value would be expected for the reaction if step 2 were the rate-determining step It would be made up of two parts. There would be an equilibrium p value for the protonation and a reaction p value for the addition of water. Step 1 involves electrons flowing out of the molecule and step 2 involves electrons flowing in so the p values for these two steps would have opposite charges. We know that the p value for step 2 would be about +2.5 and a value of about -2.5 for the equilibrium protonation is reasonable. This is indeed the explanation step 2 is the rate-deter-... 

The evidence for the proposed mechanism comes from kinetic, spectroscopic (multinuclear NMR), X-ray structure, and theoretical calculations. The kinetic rate law under optimum catalytic conditions is very complex. Under pseudo-first-order conditions, where the concentrations of both 9.35 and the hydroperoxide are much greater than that of allyl alcohol, the rate expression 9.5 is obeyed. In this expression the inhibitor alcohol is an inert alcohol such as isopropanol or f-butanol that is deliberately added to slow down the reaction for convenient rate measurements. The inert alcohol acts as an inhibitor, since it competes with both hydroperoxide and allyl alcohol for coordination to the Ti center. Note that expression 9.5 is consistent with the formation of an intermediate like 9.36, before the rate-determining oxygen atom transfer step. 

Hence, barring the formation of EO(RSH) complex in a rate-determining step, the nucleophilic attack by sulfur at EO oxygen, or transfer of hydrogen atom from S—H, does not appear to be the preferred reaction pathway, Further, the reduction of Compound I by azide does not conform to the stoichiometric mechanism [Eq. (27)] (143,175). 

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