Reactivity Studies of Iridium(III) Porphyrins with Methanol in Alkaline Media

Abstract

Ir(ttp)Cl(CO) (1a; ttp = 5,10,15,20-tetrakis(p-tolyl)porphyrinato dianion) was found to cleave the C−O bond of CH3OH at 200 °C to give Ir(ttp)CH3 (3a). Addition of KOH promoted the reaction rate and gave a higher yield of Ir(ttp)CH3 in 70% yield in 1 day. Mechanistic studies suggest that, in the absence of KOH, Ir(ttp)Cl(CO) reacts with CH3OH initially to give Ir(ttp)OCH3, which then undergoes β-hydride elimination to produce Ir(ttp)H (4a). Ir(ttp)H further reacts slowly to cleave the C−O bond of CH3OH, likely via σ-bond metathesis, to give Ir(ttp)CH3. In the presence of KOH, Ir(ttp)Cl(CO) initially reacts with KOH more rapidly to give Ir(ttp)OH, which then cleaves the O−H bond of CH3OH by metathesis to give Ir(ttp)OCH3. Ir(ttp)OCH3 further isomerizes via β-hydride elimination/reinsertion to give Ir(ttp)CH2OH and concurrently undergoes base-assisted β-proton elimination to give Ir(ttp)−K+ (5a). Ir(ttp)CH2OH subsequently condenses with CH3OH to form Ir(ttp)CH2OCH3 (2). Finally, Ir(ttp)−K+ cleaves the C−O bond in CH3OH, most probably via nucleophilic substitution, to give Ir(ttp)CH3. Ir(ttp)CH2OCH3 also serves as the precursor of Ir(ttp)−K+ as it undergoes nucleophilic substitution by KOH to give Ir(ttp)−K+.