I think that the transition state shown in the mechanism
for the oxidation (p430) is slightly confusing. Intially, what happens is
that the attacking oxygen, of the hydroperoxide ion, donates both of its
electrons to the trialkyl borate. This gives the boron atom a negative
charge, and the reaction uses this negative charge to send one of the
adjacent R groups to give both of its electrons to the attacking oxygen
atom (to compensate for this negative charge). This, in turn, will give
that attacking oxygen (of the hydroperoxide ion) a positive charge and
allows for the hydroxyl leaving group to take the two electrons from their
bond to the oxygen and become and OH-. It may be helpful to use this
mechanism figure as a reference while reading this. Also keep in mind
that although I wrote this like a multi-step reaction, it is one
(near-simultanious) step, but sometimes I think it is easier to think of
some of these more difficult and involving mechanisms as a multi-step
process even though they may only comprise of a single step.
AL
On Tue, 3 Dec 1996 nguyentm@plu.edu wrote:
> On pg.430, the mechanism for the oxidation is shown. In the unstable
> intermediate state, it shows an alkyl group from the boron making a
> migration on to the adjacent oxygen. Whay forces are causing this
> migration, it appears as though the adjacent oxygen has no net charge?
>
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> tuan
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