Studies in the synthesis of n-butyl strontium

Strontium does not easily react with n-butyl bromide in ether or benzene to produce an organostrontium compound. Strontium amalgam reacts to give a trace of the butyl mercury compound, perhaps through the formation of an organostrontium compound as intermediate.

The conversion, 2BuLi + SrCl₂ = 2LiCl + Bu₂Sr, seems to proceed predominantly to the left, perhaps because of relative solubilities of the lithium and strontium halides in the solvents used.

The conversion, Bu₂Hg + Sr = Hg + Bu₂Sr, appears to proceed to a slight extent to the right. So far, this reaction seems to offer the most promise in synthesizing an organostrontium compound.