Mechanism of Oxymercuration-Demercuration reaction

Question

I am trying to write a mechanism for OMDM reaction. I could write the mechanism for oxymercuration part of reaction but I couldn't write the mechanism for remaining part (demercuration) of the reaction. Could anybody help me to proceed forward.

Answer 1

Here is the mechanism I wrote out for you

Answer 2

Actually, the oxymercuration involves the formation of a cyclic carbocation. This is a non traditional carbocation, but the final reaction becomes non-stereospecific, as you will later find out.

1. First, the mercuric acetate breaks apart into ions, the acetate anion, and the mercuroacetic cation, which you have correctly depicted.

2. In the next step, the mercuroacetic cation gets added on to the double bond, which forms a cyclic cation like this: (Don't bother about the red arrow for now, we'll come to it in the next step.)

3. Now, water adds on to the carbocation formed from the other side, and this process follows Markownikoff's rule. Now, one $\ce{H+}$ leaves, and a result, we get exactly what you got in your last step. This is, as you called it, the end of the oxymercuration step.

4. Now, for the demercuration part, we can use any reducing agent that will be able to replace the mercuroacetic cation with $\ce{H}$. (E.g. $\ce{NaBH4}$). This essentially demercurates the product, forming an alcohol.

The entire process is shown here: .

Now, as it is written here, the addition of the mercuroacetic cation followed by water will be anti addition, but when the reducing agent is used, the hydrogen can be added from either side. As a result, we get a non-stereospecific product. In short, this can also be written as addition of just $\ce{H}$ and $\ce{OH}$ without rearrangements, which followed Markownikoff's rule. In fact, the formation of the cyclic carbocation prevents rearrangements, hence enabling chemists to get desired products without worrying about random stabilisations of the carbocations.