In the resting state, the LPMO active site contains Cu(II). It is now firmly established that reduction is necessary to form the catalytically active Cu(I). However, events after this initial reduction have remained unclear. Putative mechanisms from experimental as well as theoretical investigations are shown in Fig. . In the left path, O 2 takes the role of the co-substrate and binds Cu(I) and is thereby activated. Notably, this activation requires external electrons (in addition to the priming electron) and protons to form intermediates 3 and 4 with several different possibilities depending on the time of proton/electron Figure : Active site of the LsAA9 LPMO complexed with a Glc3 substrate; the histidine brace is comprised of His1 and His78. The proposed reaction mechanisms (with numbered intermediates) from O 2 and H 2 O 2 as co-substrates, respectively, are shown below. Residue numbers refer to the PDB 5ACF. Oxidation states are shown only for the Cu ion in moieties where they are unambiguous. delivery (3a-3b and 4a-4c). As seen from Fig. , the substrate blocks the entry to the active site, so the delivery-mechanism of these electrons has been a conundrum for the O 2driven mechanism since the LPMOs are free of typical electron transfer co-factors, although electron-transfer chains based on amino acid residues have been suggested. Initial studies supported the route depending on O 2 , 10,30 but many investigations have now established that the H 2 O 2 -driven pathway (right path in Fig. ) is orders of magnitudes faster. In fact, some LPMOs exclusively use H 2 O 2 as co-substrate. The H 2 O 2 -driven pathway does not require external electrons and protons, and theoretical investigations have shown how H 2 O 2 can be placed close to Cu(I) by hydrogen binding to certain second-sphere residues, forming 5a in Fig. . Several of these residues have been shown through mutation studies to have an impact on the LPMO mechanism. The oxidation of the key C-H bond in the polysaccharide substrate requires a very strong oxidant and the intermediates in Fig. with an intact [CuO 2 ] n+ moiety are not sufficiently reactive. Therefore, both pathways (O 2 and H 2 O 2 in Fig. ) lead to a putative Cu-oxyl (6b) intermediate that performs the initial C-H abstraction from the substrate. In the O 2 -pathway, the 6b intermediate can be formed from dissociation of the O-O bond in 4b or 4c, whereas in the H 2 O 2 -driven pathway, the O-O bond is broken by first forming a so-called "caged" OH • radical (5b) with a [CuOH] + moiety loosely bound to a OH • radical. The OH • radical abstracts a hydrogen from [CuOH] + , forming 6b.