I'm unable to understand the enantioselectivity of Noyori's hydrogenation reaction, and would appreciate any help with the same:

Here's a mechanism I found on Wikipedia:

enter image description here

but, to my disappointment, the highlighted statement, i.e. The (R)-BINAP-Ru catalyze the synthesis the (R)-Product, and the (S)-BINAP Ru catalyze the synthesis the (S)-product with high ee. contradicts the attached paper about Noyori's hydrogenation.

enter image description here

The last two entries of the table mention formation of S products from R-BINAP.

Also, I feel it's not really possible to determine the absolute configuration beforehand, i.e. without knowledge of surrounding groups.

It'd be great if someone could describe the mechanism based enantioselectivity of this reaction in detail, so that correct predictions about the stereochemistry of the product can be made, given the catalyst (R or S BINAP) and the substrate.

Thanks in advance!

  • $\begingroup$ Quoting from Reinhard Brückner's book "Organic Mechanisms: Reactions, Stereochemistry and Synthesis": "Sadly, there is no vivid explanation why the enantioselective addition to the C=O double bond of the subtrate occurs only if it is flanked by one conjugated substituent (Aryl, C=C)." (translated by me). I should note that in the book's example, Aluminium was used instead of Ruthenium. $\endgroup$ Feb 18, 2020 at 0:02
  • $\begingroup$ I've learned to be extra wary of Wikipedia on highly specific matters like this. For example, my adviser made it a point to include his lecture that the entry on kinetic isotope effects was written by someone who didn't understand kinetic isotope effects. $\endgroup$
    – Zhe
    Feb 18, 2020 at 13:37

1 Answer 1


I think this is easy to answer than you think. Let's look at the stereochemistry of compounds in the question, 1g and 1h (I also put the common product, 1a, for convenience):

beta--Hydroxy Carb acid

The stereospecific reduction is on carbon-3 ($\ce{C}$3) in all cases. Accordingly (see diagram above), if you apply Cahn–Ingold–Prelog (CIP) sequence rules for each, you'd find (3R)- for 1a (methyl group is priority 3 based on $\ce{CHH}$ vs $\ce{HHH}$), (3S)- for 1g (isopropyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCH}$), and (3S)- for 1h (phenyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCC}$). All these assignments are $\ce{C}$2 versus $\ce{C}$4 since priory 1 for all three groups on $\ce{C}$3 is $\ce{OH}$.

Thus, we can conclude that the change of stereochemistry is mainly due to the priority rules (not due to the change in stereochemistry on transition states or intermediates). This is also supported by the example given in the same reference (Ref.1) in the box in depicted diagram. Even when bulkiness at $\ce{C}$2 was increased by adding a methyl group, the reaction (with R-BINAP) has undergone same manner to give (R)-configuration to $\ce{C}$3 center.


  1. Ryoji Noyori, Takeshi Ohkuma, Masato Kitamura, Hidemasa Takaya, Noboru Sayo, Hidenori Kumobayashi, Susumu Akutagawa, "Asymmetric hydrogenation of $\beta$-keto carboxylic esters. A practical, purely chemical access to $\beta$-hydroxy esters in high enantiomeric purity," J. Am. Chem. Soc. 1987, 109(19), 5856-5858 (https://doi.org/10.1021/ja00253a051).

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