Natural Science Forum / Chemistry / Organic Synthesis / October 2003

The lit prep breaks conjugation by having the (transient) carboxylate
anion go alpha to the nitrile and push the olefin into the ring.
Exo-methylenes are high energy compared to endocyclic oelfins.  Your
open chain won't have that driving force.  Any base sitting about will
catalyze reconjugation.

I'd do some small scale evaluations first using finely crushed (mortar
and pestle) soda lime (not Pyrex) glass.  Next shot would be some
anhydrous basic anion exchange resin, then basic alumina. If that
isn't going, try acid.  Maybe MgSO4(anh) for starters, then silica
gel, then juice it. If a calatyst does not dissolve in the reaction
medium it is best dispersed on an inert support.

Either way, you have to worry about catalyzed reconjugation.  If the
next step for the product is a base-catalyzed condensation, does it
make a difference where the double bond starts?

Uncle Al <UncleAl0@hate.spam.net> wrote in message news:<bii9p9$bu8@panther.Gsu.EDU>...

Thanks. The next step involves addition of trifluoromethyl radical to
the double bond, so it does matter where the double bond sits.

One way to make trifluoromethyl radicals is by removing one electron
from trifluoroacetate at a platinum anode.

Uncle already pointed out the trouble with possible product
re-conjugation and the effect of exo C=C on isomeration.

But if you are actualy observing some of the right product on NMR even
now, then there is hope. I would try to add a catalytic ammount of
oxalic acid - or pyridinium tosylate or a copper (I) oxide as Cu2O,
for example) - and heat the mix with a distilation column attached
(Vigreaux, 6-10 cm) at atmospheric or only slightly reduced pressure.
As you have seen, the volatility of the intermediate is already too
high and you cannot go high vacuum because you need the temperature
also.

Best luck. If everthing else fails, you can try vacuum flash pyrolysis
on the intermediate - this is supposed to be the ultimate 19th century
organiker's joy.

Btw. what is this stuff good for? Is a cyanide alkylation with
methallyl bromide unpleasant?

Uncle Al <UncleAl0@hate.spam.net> wrote in message news:<bii9p9$bu8@panther.Gsu.EDU>...

muhammar@hotmail.com (Muhammar) wrote in message news:<bj4og3$804@panther.Gsu.EDU>...

NMR after 2nd and 3rd redistillations don't look that different. The
unconjugated methallyl cyanide is more volatile and comes over as a
water white liquid. After time, reconjugation gives yellow colour. NMR
suggests that an equilibrium is reached between unconjugated and
reconjugated forms. Driving force towards reconjugation appears to be
not so big.

We're trying to make isoleucine labelled with a 13C trifluoromethyl
group in the 5 position. 13C chemistry is very expensive so we need to
start a dry run first with natural abundance.

That might work. There are several routes to methallyl cyanide.

My suggestion would be DMAP (N,N-Dimethylaminopyridine) or chlorid
ions under nucleophilic conditions (e.g. sodium chloride in
dimethylformamide with heating).

*CF3-CO2H -red-> *CF3-CH=O
*CF3-CH=O + CH3-NO2(exces) -NaOH(aq)/heat-> *CF3-CH=CH-NO2 + H2O
*CF3-CH=CH-NO2 -Zn/HCl-> *CF3-CH=CH-NH2 <==> *CF3-CH2-CH=NH
*CF3-CH2-CH=NH -H2O/oxyd-> *CF3-CH2-CO2H + NH3 + H2O
(*CF3-CH2-CO2)2Ca + (CH3-CO2)2Ca (exces)-pyrolyse/distillation->*CF3-CH2-CO-CH3 + CH3-CO-CH3
+ CaCO3
*CF3-CH2-CO-CH3 + HO-CH2-CH2-NO2 (exces) -NaOH(aq)/heat-> *CF3-CH2-C(CH3)=C(NO2)-CH2OH
Alternatively to HO-CH2-CH2-NO2, CH2=O + CH3-NO2 in exces might be used.
*CF3-CH2-C(CH3)=C(NO2)-CH2OH -reducer/H2-> *CF3-CH2-CH(CH3)-C(NH2)-CH2OH
*CF3-CH2-CH(CH3)-C(NH2)-CH2OH -oxydiser-> *CF3-CH2-CH(CH3)-C(NH2)-CO2H

PhZ