Enantioselective catalysis is one of the important reactions in Organic chemistry which has huge applications in practical basis for synthesis of optically pure natural compounds. In this respect, I came across the following reaction of Aldehydes aimed towards converting them into Chiral alcohol with large enantiomeric excess with the help of (-)-3-exo-(dimethylamino)isoborneol [( -)-DAIB] catalyst.

The structure of (-)-DAIB is given as below,
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The reaction is given as,

enter image description here

If we slightly rotate the molecule and visualise, we will see that here the methyl group has come from the top side of the molecule,i.e. a si-face attack by methyl group on benzaldehyde has occured.
From the picture of the transition state also this picture become clear,( According to our reaction $\ce{R''}$ is methyl and $\ce{R'}$ is Phenyl in the picture)

enter image description here

But my question comes here, Why the orientation of the bigger group (here Phenyl) in the transition state is preferably axial? Shouldn't it be the opposite direction,i.e. shouldn't the preferred orientation be that transition state where the $\ce{R'}$ and $\ce{H}$ are just swapped with this picture ? In that case $\ce{R'}$ is farthest from $\ce{R''}$ and logically should be the most stable transition state. So, the major product will just be the enantiomer of the originally given product. I am unable to find any flaw in this logical approach. Then, why is the transition state is the given one not the proposed one ?

References: -

  • Soai, K.; Niwa, S. Enantioselective addition of organozinc reagents to aldehydes. Chem. Rev. 1992, 92, 833-856.;
  • Noyori, R.; Kitamura, M. Enantioselective Addition of Organometallic Reagents to Carbonyl Compounds: Chirality Transfer, Multiplication, and Amplification.Angew. Chem. Ini. Ed. Engl. 30 (1991) 49-69
  • Itsuno, S; and Frechet, J.M.J. Enantioselective addition of diethylzinc to aldehydes catalyzed by polymer-supported chiral amino alcohols. Evidence for a two zinc species mechanism. J. Org. Chem. 1987 52 (18), 4140-4142
  • $\begingroup$ R'' is equally close to R' and H. I'm more concerned about the proximity of R' and H to the proximal methyl group on the zinc. $\endgroup$ – Zhe Nov 30 '18 at 13:52
  • $\begingroup$ I guess you are misunderstanding the diagram. R'' is perpendicular to the plane of phenyl ring, hence doesn't make huge difference if it is flipped as far as R'' is concerned. But the interaction with the Methyl groups, Zn atom on the other side, if you have flipped would give rise to much more unfavourable interaction. $\endgroup$ – Esha Manideep Nov 30 '18 at 15:47

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