= Solution
\b[The first <base (chemistry)> 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>:
$$
\mathrm{HO_2C{-}CH(L)_2\longrightarrow {}^-O_2C{-}CH(L)_2\longrightarrow {}^-O_2C{-}C^-(L)_2}.
$$
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 (chemistry)> workup reprotonates the <carboxylate>, giving \b[$F=\mathrm{HO_2C{-}C(Me)(L)_2}$]. 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 (chemistry)> has formula $\mathrm{C_9H_{14}O_2}$ and still has two identical <allyl groups>, so this methylation does not create a <stereocenter>.
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