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

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.

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).