= Solution
\b[The <malonate> <carbon> receives two <allyl groups>.] With $L=\mathrm{CH_2CH{=}CH_2}$, attached through its first <carbon>, the <product (chemistry)> is
$$
\boxed{D=\mathrm{EtO_2C{-}C(L)_2{-}CO_2Et}}.
$$
<sodium hydride> removes a methylene <proton> from <diethyl malonate>; <molecular hydrogen> formation accompanies production of an <enolate> stabilized by <resonance in chemistry>. Its <carbon> attacks the bromomethylene <carbon> of <allyl bromide> in an <SN2 reaction>, expelling <bromide>. The monoalkylated <malonate> still has one <hydrogen> between the <carbonyls>. Excess <base (chemistry)> removes that <hydrogen>, and a second equivalent of <allyl bromide> produces a second <carbon>–<carbon> bond. The final <malonate> <carbon> has no remaining <hydrogen> and cannot undergo a third analogous <deprotonation> there.
The two <alkenes> are preserved because the substitution occurs at the primary <carbon> bearing <bromine>. Counting the introduced groups gives \b[$\mathrm{C_{13}H_{20}O_4}$] for the diallylated diethyl <ester>. Its two identical <allyl groups> also mean that the central <carbon> is not a <stereocenter>. This stage is the <alkylation> portion of the <malonic ester synthesis>.
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