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What is the mechanism of the reductive elimination of mercury metal using hydrides that is carried out after oxymercuration of the alkene? I have come across a free radical mechanism, but that suggests loss of the stereospecificity of the product, while the reaction is stereospecific in reality. Also, why isn't an acid sufficient to remove the mercury as it is with alkynes? Why is a reducing agent necessary at all?

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The reducing agent NaBH4 is used for this reaction because it is highly selective. If H2SO4 were used in place of NaBH4 in this reaction the molecule would be dehydrated and we would probably see some HgSO4 form as well. The main thing to remember here is that a reducing agent has much more selectivity than an acid solution.

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  • $\begingroup$ Do organomercury compounds in general give alkanes with acids (save the case when alcohol groups are also present in the compound, or any such groups which might also react with the acid)? Grignard's reagents are also organometallics, and they do. $\endgroup$ – Charles Oct 5 '15 at 15:18

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