[raney nickel]

Only a small blurb was left in my text and I'd like some more info if its convenient.  It basically said you can make ethers by adding another alcohol to the solvent (instead of H2O as in oxymercruation).
Does this work for most alcohols, or does steric hindrance play a big part in which ones will work?  For example can I use a tertiary ROH like tert butyl and come out with a reasonable amount of ether product?

What about the history of this rxn? Is this primarily a product of academic research or are there actual industrial applications for it? I would think a williamson, etc would be more useful, instead of playing with Hg all day, but thats just me.

[movies]

I don't know of an example where t-butanol has been used, but I am pretty sure that isopropanol works just fine.  I did find an example where a tertiary alcohol was used in an intramolecular reaction.

Industry will never use this reaction because it requires a full equivalent of Hg.  It does still see occassional use in academic labs, however.  This reaction has been known for a long time and the alternatives to achieving this transformation are not really that great.  To do a Williamson ether synthesis you would first have to convert the alkene to an alcohol (note that hydroboration gives the opposite regioselectivity from Hg so you have to use something else) and then you have the second step of the Williamson synthesis which requires the use of an alkyl halide which might be sensitive and difficult to handle.

Also for anyone looking at the Wikipedia Oxymercuration-Reduction page, the mechanism for demercuration is wrong!!  The mechanism actually proceeds through radical intermediates.  I'm too lazy to fix it there, however.

[raney nickel]

Thanks for your reply. I appreciate the extra info. 

I checked out the demercuration page and yeah its written strange.  The writer had this to say:

Nevertheless, the mechanism that I have drawn also involves a radical pathway (notice the half arrow I drew ). The hydride bridges both the boron and the mercury atom by an 'electron deficient bond.' Thus, the lines connecting the boron and the mercury to the hydride only represent one electron rather than the classical two. These types of bonds behave like radicals. Sorry about the confusion. I hoped that helped. --Nemsen 19:26, 6 January 2007 (UTC)

I think that means he's too lazy to fix it, too.