Iodoethane → Methyl 2-(4-t-butylphenyl)butanoate

Participants

Reaction Steps

4 g (19 mmol) of methyl 4-tert-butylphenylacetate, 19.5 mL (1.0 M, 19 mmol) of lithium bis(trimethylsilyl)amide and 3.12 g (20 mmol) of ethyl iodide were reacted as described in Preparation 11 to yield 5.13 g of a brown oil. Chromatography, eluting with a gradient solvent of hexane to hexane/ethyl acetate 19:1 gave the title compound 2.35 g (53%).

Provenance & Evidence