Formation of organolithium reagents from alkyl halides

by James

Description: Addition of lithium metal (2 equiv) to an alkyl or alkenyl halide results in formation of the organolithium reagent.
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Notes: X here is a halogen such as Cl, Br, or I (not F). Two equivalents of lithium are required. Note that the other equivalent of lithium metal makes the lithium halide salt. Ether here is “diethyl ether”, a common solvent for these reaction. Sometimes “pentane” is used as a solvent for these reactions.

Examples:

Notes: 2 equivalents of lithium is always required. Here, ether or pentane here are just solvents. They don’t actually participate in the reaction.

Mechanism: Lithium easily gives up its single valence electron, in this case to Br (Step 1, arrow A) to form the radical anion. Fragmentation of the radical anion (Step 2, arrow B) leads to brief formation of a free radical, which gains an addition electron from a second equivalent of Li (Step 3, arrow C).

Notes: Note that the byproduct here is lithium bromide (LiBr)

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{ 1 comment… read it below or add one }

Amitrajit July 6, 2012 at 6:41 am

what would be the reaction of R-Li with 4-Bromobutyryl chloride ??

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