# Why does the Oxymercuration Demercuration mechanism lead to stereorandomization?

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

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

or 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.

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.

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

• 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. – user73157 Apr 13 '15 at 18:36
• 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. – Ben Norris Apr 14 '15 at 0:39