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Alkene Reactions: Oxymercuration of Alkenes using Hg(OAc)2, H2O, and NaBH4

Mercuric acetate [Hg(OAc)2], water (H2O), and sodium borohydride (NaBH4) react together with alkenes to form alcohols (R-OH):

Alkene Reactions: Oxymercuration of Alkenes using Hg(OAc)2, H2O, and NaBH4 image3.png

The water molecule binds in Markovnikov addition, bonding to the most substituted carbon atom (red). Unlike other alkene-alcohol reactions, no rearrangement occurs with this reaction (compared to H3O+ or H2SO4). If both carbons on the alkene bond have the same degree of substitution, a mixture of molecules will be formed where OH is added to both carbons. This reaction may also result in the formation of a stereocenter however, there is no preference to syn or anti addition, so a mixture of enantiomers or diastereomers occurs :

No Rearrangement, Addition to both carbons

Alkene Reactions: Oxymercuration of Alkenes using Hg(OAc)2, H2O, and NaBH4 image1.png

Syn and Anti addition (enantiomers)

Alkene Reactions: Oxymercuration of Alkenes using Hg(OAc)2, H2O, and NaBH4 image2.png

The reaction mechanism is depicted below:

Alkene Reactions: Oxymercuration of Alkenes using Hg(OAc)2, H2O, and NaBH4 image4.png

In the first step, electrons from the alkene bond react with the mercury from the mercuric acetate [Hg(OAc)2] forming a ring structure as an intermediate (mercurinium ion) and displacing when of the acetate groups (OAc-).

In the second step, the water molecule attacks the more substituted carbon in a Markovikov addition.

In the third step, the hydrogen atom that was part of the water group is removed by the free, negatively charged acetate group (OAc-).

In the fourth step, sodium borohydride (NaBH4) removes the mercury atom from the alkane, resulting in a newly formed alcohol molecule. For more information on how sodium borohydride removes atoms, visit our Reaction Library.

The reagent used for this reaction is Hg(OAc)2, H2O, and NaBH4.