Methyl iodide → Ethyl 2-(2-naphthyl)propanoate

Participants

Reaction Steps

5 g (23 mmol) of ethyl 2-naphthylacetate, 23.3 inL (1.0 M, 23 mmol) of lithium bis(trimethylsilyl)amide, and 1.5 mL (24 mmol) of methyl iodide were reacted as described in Preparation 11 to yield 5.71 g of a dark oil. Chromatography eluting with a solvent gradient of hexane to hexane/ethyl acetate 19:1 gave the title compound 2.85 g (54%).

Provenance & Evidence