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+.

User-friendly aerobic reductive alkylation of iridium(iii) porphyrin chloride with potassium hydroxide: scope and mechanism

“…These Ir(ttp)-alkyls are five-coordinated complexes in a square pyramidal geometry and the Ir-C α bond lengths (2.057-2.154 Å) and the average lengths of Ir-N (2.011-2.024 Å) are consistent with the reported data of analogous Ir(ttp)R (R = CH 2 OMe, Bn-p-Me). 15,23 The Ir-C α bond length of Ir(ttp)-adamantyl (3k) (2.154 Å) is similar to that of Ir(ttp)Ph (2.158 Å). 24 The porphyrin planes of 3a, 3e, 3i•CH 2 Cl 2 and 3k display a slight saddle form.…”

Base-Promoted C–O Bond Cleavage of Primary Alcohols by Iridium(III) Porphyrin Chloride

“…8 We previously have reported the direct cleavage of the C−OH bond in methanol by Ir(ttp)(CO)Cl to give Ir(ttp)Me under solvent-free and basic conditions, in which Ir(ttp)H was proposed as the active species to cleave the C−O bond likely via σ-bond metathesis to give Ir(ttp)Me. 9 In this work, we have successfully expanded the scope of the alcohols to benzyl alcohols and other aliphatic alcohols with β hydrogens. Ir(ttp)(CO)Cl can efficiently cleave the benzylic C−OH bond in toluene solvent or aliphatic C−OH bond in benzonitrile solvent to give Ir(ttp)Bn or Ir(ttp)-alkyls, respectively.…”

“…To our delight, the reaction was complete in 2 h and gave Ir(ttp)Me (3f) in 88% yield (Table 5, entry 1), which was higher than that under the previously reported solventless conditions giving a 70% yield. 9 Using these conditions, other alkyl alcohols with β-hydrogens, such as EtOH, n-PrOH, n-BuOH, n-pentanol, and n-octanol, were also screened, and moderate to good yields of Ir(ttp)-alkyls 3g−l were obtained in 24 h (Table 5, entries 2−6). The yields slightly decreased with increasing alkyl chain length, possibly due to product decomposition by the facile β-hydride elimination.…”

“…3,12a In addition, [Ir(ttp)] 2 has been ruled out to be an intermediate in the C−O cleavage of MeOH by Ir(ttp)(CO)Cl in alkaline media. 9 Therefore, we proposed that similar mechanistic pathways operate in benzene or benzonitrile solvent. Therefore, the reactivities of only Ir(ttp) − and Ir(ttp)H as the probable intermediates toward C−O bond cleavage of benzyl alcohol were examined individually.…”

“…Ir(ttp) − was then generated in situ from the deprotonation of Ir(ttp)H (1b) (the pK a of Ir(ttp)H is 15) 9 with KOH in C 6 D 6 and reacted with 30 equiv of benzyl alcohol to form the C−O cleavage product Ir(ttp)Bn in 78% yield in 2 h (Scheme 3,B). These outcomes imply two mechanistic possibilities: (1) the Ir(ttp) − anion directly cleaves the C−O bond of benzyl alcohol possibly by an S N 2-like pathway and (2) KOH is essential and plays a promoting role in the C−O cleavage reaction with Ir(ttp)H. We support the second mechanistic proposal because hydroxide is a very poor leaving group under basic conditions.…”