Question

When 3-methylpent-2-ene is treated with mercury(II) acetate in methanol and the resulting product isreacted with NaBH4, what is the primary organic compound which results

Answer 1

3-methylpentan-3-ol

Step-by-step explanation:

In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand how this molecule is produced we have to check the mechanism:

The mercury(II) acetate (
`Hg(OAC)_2`) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation
`HgOAc^+` and the anion
`AcO^-`. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the
`NaBH_4` would reduce the compound to produce the final alcohol.

See figure 1

I hope it helps!