I am teaching myself organic chemistry and am very frustrated that I cannot get a complete mechanism for the oxymercuration-reduction of an alkene to an alcohol. The reagents are an alkene, mercury(II)acetate, and water for the first step... the step I understand. It ends by forming organomercurial alcohol. This is then reacted in a second step with sodium borohydride in a solution of sodium hydroxide. I have searched for a mechanism for this step (specifically the reduction portion at the very end) but I cannot find one anywhere. The closest I have come to an answer is a source that says it probably had to do with radicals and inorganic chemistry. The products are an alcohol with the C-Hg(OAc) bond transplanted with a C-H bond, tetrahydroxyborate, reduced mercury (standard conditions), and OAc anion. Thanks any help is appreciated

Update:my attempt at a mechanism in which the mechanism runs a total of 4 times I am attaching my attempt at a mechanism and would like someone to verify if it is the correct mechanism or tell me why it would not work. Thanks


The direct displacement of the mercury acetate by borohydride, as prooposed by @xavier_fakerat, looks attractive because it (1) makes some chemical sense and (2) gives the correct products. However, that mechanism can be disproven by looking at the stereochemical outcome of the reduction step.

Pasto and Gontarz did just that (Pasto JACS 1969). When cyclopentene was subjected to oxymercuration, followed by reduction with sodium borodeuteride, the product observed was the trans-configuration of the hydroxyl and deutero groups. The replacement of mercury for deuterium went with retention. This rules out direct displacement (SN2 like mechanism) which would lead to inversion of configuration.

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When cis- and trans-butene were subjected to the same process in independent experiments, in both cases the result was a 50:50 mixture of syn- and anti-stereochemistries. This indicates that there is cleavage of the carbon-mercury bond and sufficient lifetime for the resulting intermediate to undergo bond rotation before carbon-hydrogen bond formation. For other substrates, C-C bond rearrangements were observed.

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The authors propose that hydride displaces acetate at mercury. This is followed by homolytic C-Hg bond cleavage, leaving a carbon centered radical. Finally, hydrogen atom abstraction from the mercury-hydride gives the reduced product and elemental mercury.

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  • $\begingroup$ Thank you for this. I wanted to make sure that in the last step, is the carbon radical abstracting the hydrogen from mercury so that the D-Hg bond is split homolytically and the D with its one electron then combines with the C radical? And in my example, the hydride is coming from the BH4+ but how is the B(OH)4 formed? Thanks $\endgroup$ – Joe Jun 7 '17 at 1:34
  • $\begingroup$ @Joe After borohydride delivers hydride, a trivalent boron species will be formed. This is Lewis acidic and can accept electrons from Lewis basic water. As boron continues to release hydrides, additional water molecules will add to boron, eventually getting to the maximum of four. If this is unclear, I can add it to my answer. $\endgroup$ – jerepierre Jun 14 '17 at 18:16
  • $\begingroup$ @Joe with respect to the final deuteration step, you are interpreting that correctly. $\endgroup$ – jerepierre Jun 14 '17 at 18:17

I managed to do a scribble for the reation mechanism.

This is an example of an alkene I took from my book, to explain:

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The mechanism of the oxymercuration of an alkene to yield an alcohol comprises 3 main steps.

The first two steps you're familiar with, as you said in question but here are just for reference.

  • In (1), the electrophilic addition of $\ce{Hg2+}$ gives a mercurium ion.

  • In the second step, the $\ce{H2O}$ molecule attacks the more substituted carbon (bearing a greater share of the positive charge) forming an organomercury compound

  • In step (3), $\ce{NaBH4}$ cleaves the C-Hg bond. I believe this occurs through a hydride shift $\ce{:H-}$

Hope this helps

  • $\begingroup$ This may be close but I am certain the mechanism involves a radical intermediate. Thanks $\endgroup$ – Joe Jun 6 '17 at 21:14

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