[nitinrao]

Hello everyone!

I'm new to the forums, but from the look of these expansive forums, I can tell that this is a very vivid community.

I'm writing because of a mechanism I could not grasp from a paper. The image shows the schematic of this substitution mechanism....could someone please explain each step of this mechanism to me?

This is what I understand so far: the carbanion undergoes a Sn2 reaction with the least substituted carbon in the epoxide ring (being in basic conditions), and opening the ring. Since that carbon's antibonding orbital is planar, it is involved the nucleophile is the plane....but then again, why do we see CN as a racemixture (squiggly line)? This is in equilibrium with the epoxide reformation, where the oxygen attacks the C-Cl bond and displaces it. But then, how is the bottom two products form? and their stereochemistry? I don't understand how the product on the is formed either.



(My undergraduate class has thus far covered substitution, elimination, proton transfer, and stereochemistry involved in these)

Thanks,

nitinrao

[sjb]

Depending on how you formed the carbanion in the first instance, I'd think that this is a racemic mixture - don't forget that carbanions don't (readily) invert, so that the alkoxide is also a mixture of isomers at that centre. Once you have the alkoxide, you can either reform the epoxide by direct displacement of the chloride, and then re-deprotonate at the benzylic position which then attacks the secondary centre on the epoxide, leading to the bottom products. Or you can transfer the proton from the benzylic position to the alkoxide which then displaces the chloride

[nitinrao]

I really don't understand how the stereochemistry changes in each reaction step....could you please help me out with this?

[movies]

I would guess that the epoxide starting material (epichlorohydrin) is enantioenriched in this case.

So in the attack at C-3 you generate a stereocenter on the nucleophile part of the product.  This is a mixture of stereoisomers since there is essentially no control over which face of the nucleophile reacts.  There is no change in the C-2 stereocenter.  The nucleophilic attack leading to the cyclobutanol comes from deprotonation adjacent to the CN group and intramolecular alkylation.  The stereochemical magic that seems to take place is really just because the isomers of the cyclobutanol are not chiral due to a plane of symmetry.  There are two isomers – one with the CN and OH syn and one with them anti.  These two arise from the different faces of the deprotonated alkyl nitrile again.  The title of the scheme is really incorrect since the molecules are not chiral they can not be "(±)-."

In the other pathway, the closure to form the epoxide (the equilibrium step) does not generate any new stereocenters.  The next step to form the cyclopropane again comes from deprotonation of the C adjacent to the CN followed by attack at the internal C of the epoxide.  This will occur with inversion of the C–O stereocenter.  However, a new stereocenter is generation at the nucleophilic C and therefore you get a mixture of diastereomers as show.

Note that it is incorrect to say that a single stereocenter in a molecule with multiple stereocenters is "racemic."  Racemic means a mixture containing equal amounts of two enantiomers.  In this case, the product of the C-3 attack is a mixture of diastereomers since the epichlorohydrin is enantioenriched.  Racemic can only refer to a complete structure, not just a portion of the molecule like a single stereocenter.

I hope that is helpful!

BTW, would you mind posting the literature citation for this paper?