I would like to know how this reaction actually works. I think its an ring metathesis, espececially with the Grubbs catalysator II. But I dont know how to this reaction examole really works especially with respect to enantioselectivity. May anyone help? enter image description here

  • $\begingroup$ Which bit do you not understand. The alkylations or the metathesis? $\endgroup$ – Waylander Sep 1 '19 at 20:38
  • $\begingroup$ @Waylander seems to me that it's the diastereoselectivity of the alkylation(s). $\endgroup$ – orthocresol Sep 1 '19 at 20:51
  • $\begingroup$ As far as I can tell, the stereoselectivity is completely unimportant as you're doing a double allylation at the same position. $\endgroup$ – Zhe Sep 4 '19 at 15:53

I'm guessing this is a homework type question. Yet, I have some concern of this being a exam question, probably one already taken. So I give a step by step answer depicted in following diagram (courtesy of https://web.chemdoodle.com/demos/sketcher/):

Amino Acid B.jpg

  • 2
    $\begingroup$ I don't actually agree: in step (b) the reaction proceeds via an enolate, which destroys the preexisting stereochemistry, and the alkylation will again occur anti to the t-butyl group. So it is the allyl group and not the homoallyl group which ends up on the top face. $\endgroup$ – orthocresol Sep 1 '19 at 21:13
  • $\begingroup$ @ orthocresol: I agree with you 100%. However, OP didn't give final structure so I gave him what possible logical answer for him/her. I might have to use ChemDraw to do proper explanation, which I don't have an access during the weekend. :-) $\endgroup$ – Mathew Mahindaratne Sep 1 '19 at 21:25

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