 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.The carbon end of this enolate attacks the primary carbon bearing iodine in $\mathrm{IC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OMe}$. The C–I bond breaks to iodide in an SN2 reaction, and carbonyl reformation givesThus 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.