Date: Thu Mar 27 1997 - 02:59:00 EST
I'm interested in the reaction of Cp*2Sm with t-BuCl which was reported
by Evans et al.
The reaction proceeds smoothly to produce Cp*2SmCl in high yield, along
with isobutene and isobutane as by-products.
My question is the reaction mechanism. Cp*2Sm is well-known as
an one-electron reducting agent; that reacts with variety of alkyl
halide(RX) to generate Cp*2SmCl and Cp*2SmR as follows;
Cp*2Sm + RX --> Cp*2SmCl + Cp*2SmR(very instable and reacts further)
[Cp*2SmR + RX -->-->--> Cp*R + SmX3]
(I don't show equations by using precise ratio for each compounds.
Please keep this in mind when seeing the equations.)
Applying the above scheme, the reaction involving t-BuCl should occur
Cp*2Sm + t-BuCl --> Cp*2SmCl + Cp*2Sm-t-Bu
Cp*2Sm-t-Bu is, somebody around me said, very sterically unfavored, so
the reaction must occur like this;
Cp*2Sm + t-BuCl --> Cp*2SmCl + t-Bu(radical)
t-Bu(radical) ---> disproportion to give isobutane and isobutene
This seems rational, but I cannot fully accept this theory.
Radical may really be involved, I agree. But if that is the cause of
hydrogen abstraction from another t-Bu radical, the solvent (toluene)
also should be involved. This should be noteworthy that the reaction
proceeds really smoothly and rapidly under mild conditions.
Would anybody please suggest me the explanation for the reaction
Thanx in advance.
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