= Solution
Only one bromine is replaced by <hydrogen>. Writing the unchanged protected chain as $R=\mathrm{THPOCH_2CH(Me)}$, the product is \b[$R\mathrm{CH=C(Me)CH=CHBr}$], retaining the original internal <alkene> geometry and forming predominantly the \b[$Z$ terminal bromoalkene]: bromine and the <carbon> chain lie on the same side of that new stereodefined double bond. The formula is consequently $\mathrm{C_{13}H_{21}BrO_2}$.
Phosphine dissociation exposes Pd(0), which undergoes preferential <oxidative addition> into the less hindered C–Br bond, the bromine trans to the <carbon> chain in the gem-dibromoalkene. <hydride> transfer from <tributyltin hydride> replaces the <palladium> bromide ligand, producing tributyltin bromide and an alkenyl–Pd–H intermediate. <reductive elimination> joins the alkenyl <carbon> and <hydride>, regenerating Pd(0). Retention of the alkenyl geometry through this sequence puts the surviving bromine cis to the chain. The conjugated <carbon> skeleton and <protecting group> remain intact; this is not addition of tin <hydride> across either <alkene>. The stereoselective monohydrogenolysis is established for this reagent combination.<A general and convenient synthetic method of geometrically pure (Z)-1-bromo-1-alkenes by Junichi Uenishi and others>[Tetrahedron Letters 37, pp. 6759–6762: palladium/tributyltin-hydride monoreduction]{ref}
\Image[/past-exam-of-the-chemistry-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-3-organometallic.png]
{title=Original product and metallacycle connectivity drawings for the organometallic transformations}
Back to article page
