= Solution
<Lithium diisopropylamide> removes an alpha <proton> from the ethyl side of the <ketone>. The C–H <electrons> form the C=C part of a <lithium> <enolate>, while the C=O pi <electrons> move to <oxygen>. The phenyl side has no competing alpha C–H site. Diisopropylamine is the conjugate-acid byproduct.
$$
\mathrm{PhCOCH_2Me}\xrightarrow{\mathrm{LDA}}\mathrm{PhC(O^-Li^+){=}CHMe}.
$$
The <carbon> end of this <enolate> attacks the primary <carbon> bearing iodine in $\mathrm{ICH_2CH_2CH_2OMe}$. The C–I bond breaks to <iodide> in an <SN2 reaction>, and <carbonyl> reformation gives
$$
\boxed{\mathrm{PhCOCH(Me)CH_2CH_2CH_2OMe}.}
$$
Thus \b[a three-<carbon> methoxypropyl chain is attached at the <ketone> alpha <carbon>]. The remote methyl <ether> survives. The hindered <amide> base is chosen to generate the <enolate> rather than to add to the <carbonyl>. No absolute stereochemical preference is specified for the new <stereocenter> under these <achiral> conditions.
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