 Solution

The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving $F=\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{Me})(\mathrm{L}{)}_2$. The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula ${\mathrm{C}}_9{\mathrm{H}}_{14}{\mathrm{O}}_2$ and still has two identical allyl groups, so this methylation does not create a stereocenter.