I'm having a doubt on the mechanism of oxymercuration Demercuration reaction. How does the reaction, initially heading towards anti addition(after cyclic transition state formation) ,get spoiled by $\ce{NaBH4}$ and ends up being stereorandom? Can anyone point out the importance of using mercury as well? The explanation would be greatly helpful.

Thank you

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    – M.A.R.
    Apr 13, 2015 at 17:55

1 Answer 1


The stereorandomization occurs in the formation of the cyclic mercurinium ion intermediate.

formations of chiral mercurinium ion

When the mercury acetate approaches the alkene, it can do so from either the top face:

mercury approach from the top face

or the bottom face:

mercury approach from the bottom face

Since the initial alkene is achiral and the steric factors are the same for both approaches, we get both ions, which are enantiomers.

Each enantiomer reacts stereospecifically with water to produce a chiral product. Note that the position where the mercury sits is still a chirality center in this example, but in many simpler cases it is not.

attack of water on one enantiomer

attach of water on other enantiomer

Finally, $\ce{NaBH4}$ is added to this compound as a second step to reduce the $\ce{C-Hg}$ bond to a $\ce{C-H}$ bond. In most cases, if the mercury was at a chirality center, that position no longer is one because we have two $\ce{C-H}$ bonds present.

reduction of C-Hg bond to produce racemic alcohol

Why do we use mercury anyway? It works. It works to put the OH on the more substituted position without rearrangements typical of carbocations.

However, there is ongoing research to replace toxic mercury.

  • $\begingroup$ But isn't the formation of cyclic mercurinium ion lead to anti addition products and not cause stereorandomness? So far, I've learned that cyclic transition state leads to anti addition. $\endgroup$
    – user73157
    Apr 13, 2015 at 18:36
  • 2
    $\begingroup$ True - the cyclic intermediate (it is not a transition state) leads to anti addition, but a racemic mixture of cyclic mercurium ions is formed - both of which ring-open stereospecificaly top produce a racemic mixture of anti addition products. The reduction step removes the mercury group (and often the second chirality center), but does not affect the chirality center where the alcohol is. $\endgroup$
    – Ben Norris
    Apr 14, 2015 at 0:39

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