= Solution
The hydroxyl group first binds <aluminium>, normally by <deprotonation> to an <aluminium> <alkoxide> with release of <hydrogen>. Coordination also brings <aluminium> close to the <alkyne>. \b[<directed trans-hydroalumination>] then puts <hydride> on the <alkyne> <carbon> proximal to CH2O and <aluminium> on the distal <carbon>.
One useful mechanistic description has external <hydride> attacking the coordinated <alkyne> opposite the aluminium-containing face. The resulting <vinylaluminate> is stabilized by a five-membered O–Al–C–C–CH2 chelate. This is not the ordinary undirected syn addition of a neutral dialkylaluminium <hydride>. Propargylic-alcohol-directed trans-hydroalumination with <Red-Al> gives <vinylaluminate> intermediates that can be intercepted by <electrophiles>.<trans-Hydroalumination and alkylation by Neil F. Langille and Timothy F. Jamison>[Abstract, stereoselective vinylalanate formation from propargylic alcohols]{ref}
<iodine> replaces the distal C–Al bond with C–I without scrambling the <alkene> geometry; aqueous work-up releases the <alcohol>. <hydride> and <iodine> are consequently \b[trans] across the <alkene>, while <iodine> and the CH2OH group are on the same side.
By <Cahn-Ingold-Prelog priority rules> priority the product is therefore \b[$Z$]: CH2OH outranks H at one <alkene> <carbon> and I outranks CH2R' at the other. Describing the two <carbon> chains as trans would not make this an $E$ <alkene>, because the <iodine> changes which substituent has higher priority.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2021/paper-1-b4-hydroalumination.png]
{title=Directed anti hydroalumination, stereoretentive iodination and the Z priority assignment}
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