I am working on a practice problem and the answer key shows one major product instead of two. My question is why does the product below form as opposed to a racemic mixture since two methoxy groups are blocking the incoming alkylation from the top and bottom face? Wouldn't the enolate not have a preference when attacking the methyl iodide?
The pyrrolidine portion of the substrate amide is a chiral auxiliary, a single enantiomer with the (R,R)-configuration. Both of these centers are equivalent. The $\ce{-CH2OCH3}$ groups are designed to act as coordinating ligands. Lithium disopropylamine (LDA) exists as a dimer, but for convenience, it is shown as the monomer in amide 1 chelated to both the amide oxygen and methoxy group. Either of the methoxy groups may play the role of chelating agent.
Deprotonation of the amide may occur intramolecularly (dimeric LDA?) or intermolecularly. What is critical is the conformation of the carbon chain. Conformation 1a (transoid) places the chain remote from the non-chelating $\ce{-CH2OCH3}$ groups while conformation 1b (cisoid) seemingly has a steric interation between the chain and the proximate $\ce{-CH2OCH3}$ group. Conformation 1a leads to the (Z)-enolate 2 while conformation 1b affords the (E)-enolate 4. Enolate 2 has the (Re)-face exposed to methylation while enolate 4 has the (Si)-face exposed. The non-chelating $\ce{-CH2OCH3}$ chain is remote from the site of alkylation. In the event, enolate 2 leads to the major (R,R,S)-3 diastereomer while enolate 4 yields the minor diasterereomer (R,R,R)-3. The diastereoselectivity of the reaction is not reported but this analysis favors the diastereomer (R,R,S)-3
