AndersHoveland - 1-4-2011 at 23:32
Is there someone that can provide a good explanation about the shape of this molecule:
http://pubs.acs.org/doi/abs/10.1021/ol200121f
More specifically, excactly how the presence of a trifluoromethyl group keeps the bicylobutane in a strained (more than usual obviously) configuration
?
The author wrote some informal comments about the reaction here:
http://newreactions.wordpress.com/
(it is the second reaction from the bottom)
As a sidenote, I also found this from one of the links in the site:
"Bubbling HCl gas (made by adding conc. sulfuric acid to sodium chloride) through a methanol solution of a carboxylic acid for 10 minutes gave 100%
yield of the methyl ester"
Seems like an interesting procedure for making esters.
[Edited on 2-4-2011 by AndersHoveland]
ckellz - 2-4-2011 at 04:53
Hi Anders,
My theory as to how the CF3 group keeps it stable is via the perfluoroalkyl effect. It was discovered by Lemal and coworkers in the late 70s but
essentially what it does is it provides a high kinetic barrier to prevent bond cleavage. The way I like to think about it is that the bond is so
strongly electron withdrawn (and since that brigde head bond is more double bond like in character) that it can't react with many electrophiles (or
even radicalize) hence it just sits there. And we still have a sample of it and its still just as good as when I made it a year ago. If you have more
questions about the chemistry just let me know, I'd be happy to help!
-Ckellz
AndersHoveland - 2-4-2011 at 06:14
If not for the trifluoromethyl group, could the bicyclobutane (in the strained configuration) convert to its more stable shape without bond cleavage,
or would it decompose breaking the rings? The interesting thing is not that bicyclobutane exists ( I have once even prepared a derivitive of it), but
rather the unusual shape, and thus added strain, beyond that of the plain bicyclobutane adjoining rings.
[Edited on 2-4-2011 by AndersHoveland]
ckellz - 2-4-2011 at 06:55
hmm interesting question, I know other derivatives have been prepared (cyano, carbonyls) and I believe they have similar conformations. But if that
isnt the case, I would argue that the path on how we formed the bicyclobutane proceeded through such a mechanism that lead to the shown conformation
(ie through a W conformation y-silyl elmination) which should give that confirmation. I think that in order to convert to the more typical
conformation, you would actually require bond breakage and since the CF3 group greatly dimishes that possiblity, we get stuck in a kinetic well that
prevents isomerization. Interestingly Lemal noted the same sort of thing when working with perfluoroalkyl benzenes. He found that he was able to form
the much less stable dewar isomer with the hexa(trifluoromethyl) analog and it was stable under ambient conditions but the other hexamethyl isomer
readily isomerized. Its crazy stuff but I believe thats whats occuring here!
bicyclopentane
AndersHoveland - 4-4-2011 at 13:59
This bicyclobutane compound would be expected to have a high heat of formation caused by the added strain from the stereoisomerism, in addition to the
usual bicyclopentane ring strain.
Is it be possible that bicyclopentane may exhibit analogous stereoisomerism when a CF3 group is added ?
Here is the normal structure of bicyclopentane (this derivitive has a hydroxyl group instead of a trifluoromethyl):
http://upload.wikimedia.org/wikipedia/commons/5/5c/1-Bicyclo...
Such bicyclopentane derivitives could potentially be useful as energetic compounds.
It may be possible that a dinitrofluoromethyl group could be used instead of a trifluoromethyl group (trinitromethyl is not thermally stable).
Because of the high degree of strain, geminal (on same carbon atom) or vicinal (on adjacent carbon atoms) nitro groups would be expected to be much
more thermally stable attached directly onto the bicyclopentane structure (another example of ring strain adding thermal stability to two geminal
nitro groups is TNAZ).
[Edited on 4-4-2011 by AndersHoveland]