What is the mechanism and equation (reaction scheme) for the cross-coupling reaction between 4-bromobenzoic acid and phenyl boronic acid is run entirely in water with palladium as the catalyst?
This is the description: The mechanism of the Suzuki-Miyaura cross-coupling reaction is outlined below. The ligands that coordinate to palladium throughout the reaction are omitted for clarity. An initiation step reduces the Pd(II) catalyst to Pd(0) species. In the first step of Suzuki cross-coupling, the electron rich Pd(0) catalyst breaks the carbon-bromine bond of the aryl bromide, providing an aryl Pd(II) intermediate. This first step is referred to as an oxidative addition because of the conversion of Pd(0) to Pd(II) (formally an oxidation or loss of electrons from the metal center). In the next step, the base (hydroxide in this case) displaces the bromide from palladium. This hydroxyl group, now attached to the palladium metal, can bind to the Lewis acidic aryl boronic acid, facilitating the transfer of the second aryl group to palladium. The process of transferring the aryl group from boron to palladium is referred to as transmetallation. In the final step, the two aryl groups form the key carbon-carbon bond and release the palladium catalyst. In this final step, Pd(II) regains two electrons, and the Pd(0) catalyst is regenerated. This final step is referred to as reductive elimination
Answer (1)
Overall reaction4-bromobenzoic acid + phenylboronic acid + hydroxide → 4-phenylbenzoic acid + boric acid + bromideMechanism (Suzuki–Miyaura cycle)1. Catalyst activationPd(II) precursor is reduced to Pd(0), the active species.2. Oxidative additionPd(0) + Ar–Br → Ar–Pd(II)–Br(Ar = 4-carboxyphenyl group)3. Base exchangeAr–Pd(II)–Br + OH⁻ → Ar–Pd(II)–OH + Br⁻4. TransmetallationPhenylboronic acid + OH⁻ → phenylboronateAr–Pd(II)–OH + phenylboronate → Ar–Pd(II)–Ph + B(OH)₃5. Reductive eliminationAr–Pd(II)–Ph → Ar–Ph (4-phenylbenzoic acid) + Pd(0)
