Density functional calculations show that in the absence of Compound I, the primary oxidant species of P450, the precursor species, Compound 0 (FeOOH), can effect double bond activation of 5-methylenylcamphor ( 1). The mechanism is initiated by homolytic cleavage of the O–O bond and formation of an OH radical bound to the Compound II species by hydrogen bonding interactions. Subsequently, the so-formed OH radical can either activate the double bond of 1 or attack the meso position of the heme en route to heme degradation. The calculations show that double bond activation is preferred over attack on the heme. Past the double bond activation, the intermediate can either lead to epoxidation or to a glycol formation. The glycol formation is predicted to be preferred, although in the P450 cam pocket the competition may be closer. Therefore, in the absence of Compound I, Compound 0 will be capable of epoxidizing double bonds. Previous studies [E. Derat, D. Kumar, H. Hirao, S. Shaik, J. Am. Chem. Soc. 128 (2006) 473–484] showed that in the case of a substrate that can undergo only C–H activation, the bound OH prefers heme degradation over hydrogen abstraction. Since the epoxidation barrier for Compound I is much smaller than that of Compound 0 (12.8 vs. 18.9 kcal/mol), when Compound I is present in the cycle, Compound 0 will be silent. As such, our mechanism explains lucidly why T252A P450 cam can epoxidize olefins like 5-methylenylcamphor but is ineffective in camphor hydroxylation [S. Jin, T.M. Makris, T. A. Bryson, S.G. Sligar, J.H. Dawson, J. Am. Chem. Soc. 125 (2003) 3406–3407]. Our calculations show that the glycol formation is a marker reaction of Compound 0 with 5-methylenylcamphor. If this product can be found in T252A P450 cam or in similar mutants of other P450 isozymes, this will constitute a more definitive proof for the action of Cpd 0 in P450 enzymes.