Question

Please show the mechanism of Oxymercuration - demercuration ? Give an example to make it understand.

Answer 1

The reaction converts an alkene to an alcohol in a Markovnikov addition. The end result is just like acid-catalyzed hydration using `"H"_2"SO"_4//"H"_2"O"`.

The "oxymercuration" part is when the mercury complex adds on, and the "demercuration" part is when it comes off.

On an exam, you can feel free to write `"CH"_3"CO"_2^(-)` as `"AcO"^(-)`, as long as you leave enough detail that it can still be shown to make bonds and break bonds.

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Consider a general asymmetric alkene. The overall reaction is:

where each `R` can be either `"H"` or an alkyl group (`-"CH"_3`, `-"CH"_2"CH"_3`, etc).

If the reaction dissolves mercury(II) acetate in water, then THF (tetrahydrofuran) is also needed. `""^(-)"OAc"` is acetate, `""^(-)"O"_2"CCH"_3`. For simplicity, let's say `R_1 = "H"` and `R_2 = "H"`. Then...

Here's the mechanism:

1. The mercury(II) acetate complexes onto the alkene, just like in bromination (`"Br"_2` in `"CH"_2"Cl"_2` onto alkenes), forming a three-membered ring.

It's a two-way street - Here, mercury donates its `5d` electrons into the alkene carbon's antibonding `pi^"*"` orbital, and the alkene's `pi` orbital donates into mercury's `6s` orbital.

2. Nucleophilic attack occurs via the alcohol (`"ROH"`) or water (`"HOH"`) onto the less-substituted carbon. That forms a good leaving group, but we want it to stay.

3. Fortunately, this acid-base equilibrium favors the weak acetic acid over the alcohol (`"pK"_a` of `4.75` compared to about `16`), so this equilibrium favors the `"ROH"` staying by over `10^11`-fold.

4. This last step is the demercuration, i.e. the reduction of the `"Hg"` complex off of the substrate via sodium borohydride (`"NaBH"_4`) to generate the product (an alcohol if using water reagent, or an ether if using an alcohol reagent).