The pathway for the third dehydrogenation reaction is shown in
Fig. 4 and the details are shown in Figs. S8 and S9 (Supporting information). The reaction leading to the IrC
5H
6+ + H
2 products is predicted to proceed
via oxidative addition, reductive elimination, ring-forming and dehydrogenation processes. If the reaction begins with the singlet state
1Ir(
η2-C
4H
4)
+ + CH
4 or
1Ir(
η4-C
4H
4)
+ + CH
4 as the ground reactants, the efficient dehydrogenation at the thermal energies cannot occur because the energy of transition state is higher than the reactants (Figs. S6 and S7), this is consistent of experimental results that no product is observed when ions are cooled by Ar gas. The experimental results can be explained that the part of the release energy in the reactions of IrC
2H
2+ or Ir
+ with C
2H
4 is not transferred away and the product IrC
4H
4+ owns the extra energy to overcome the barrier. So the reaction can begin with the triplet state
3Ir(
η2-C
4H
4)
+ + CH
4 as ground reactants or singlet state with the extra energy [
54]. Along the triplet state surface, the ethene adsorption complex,
3I1 lies 0.47 eV lower in energy than the ground state reactants. When the activation of the first C-H bond of CH
4 occurs, the second transfer of hydrogen atom leads to the intermediate
3I3 by the transition state
3TS2 which lies 0.05 eV above the reactants. Because of this, efficient dehydrogenation at thermal energies cannot occur along the triplet state surface, so further calculations along the triplet state surface are not conducted. The triplet state Ir(
η2-C
4H
4)
+ is directly coordinated to CH
4 for forming the ethene adduct,
1I1, lies 1.13 eV lower in energy than the ground state reactants. From
1I1, the first oxidative addition process with the transfer of the first H atom from CH
4 to the Ir
+ occurs through
1TS1 (−0.11 eV) resulting in the formation of the intermediate
1I2 (−0.64 eV). The first reductive elimination process, with the transfer of H atom from Ir
+ to adjacent C atom leads to intermediate
1I3 (−1.74 eV) by a transition state
1TS2 (−0.24 eV). The second oxidative addition process with the transfer of the H atom from -CH
3 to the Ir
+ occurs through
1TS3 (−0.89 eV) resulting in the formation of the intermediate
1I4 (−1.36 eV). Then the first ring-forming reaction with the coupling of -C
4H
5 and -CH
2 from the activated product of reactant CH
4 forms a metallacycle structure
1I5 (−1.54 eV) through a transition state
1TS4 (−0.41 eV). The third oxidative addition process, with the transfer of H atom from -CH
2 (the activated product of reactant CH
4) to Ir
+ atom leads to intermediate
1I6 (−1.20 eV) by a transition state
1TS5 (−0.87 eV). Then the H atom is transferred between the two C atoms to form intermediate
1I7 (−1.65 eV) by a transition state
1TS6 (−0.43 eV). The second ring-forming reaction with the coupling of two C atoms adjacent to Ir
+ through
1TS7 (−0.30 eV) results in the formation of the intermediate
1I8 (−0.43 eV) with a like-half-sandwich structure that IrH
2+ connects to the five-membered ring -C
5H
6. The final oxidative addition process, with the transfer of the H atom from -CH
2 to the Ir
+ center takes place to form the intermediate
1I9, in which H
2 is adsorbed by the metal Ir through weak interactions. Finally, the half-sandwich structure IrHCp
+ is generated with the reductive elimination process of H
2, which is consistent with the experimental results. The generated pathways of isomers Ir(
η5-C
5H
6)
+ and IrH(
η2-C
5H
5)
+ are exothermic processes which lie 0.98 eV and 0.96 eV below ground state reactants. But the energies of Ir(
η5-C
5H
6)
+ and IrH(
η2-C
5H
5)
+ are higher than the IrHCp
+, Therefore, the IrHCp
+ is major product, Ir (
η5-C
5H
6)
+ and IrH (
η2-C
5H
5)
+ may be also coexisting, which is similar with the isomers of IrC
4H
4+. The IRC calculations of the pathways
1I3–
1TS3–
1I4 and
1I4–
1TS5–
1I6 are performed and the results are shown in Figs. S10 and S11 (Supporting information).