I am struggling with a problem regarding the enantioselectivity of BINAL-H for quite some time now. This reaction scheme

The reaction scheme shows the reaction with (S)-BINAL-H as it is observed plus the transition state (green) preferred to avoid repulsion of lone-pair of the oxygen atom at the aluminum and the alkyne function. However I found an alternative transition state (red) that yields the opposite enantiomer. What am I doing wrong?

  • $\begingroup$ On paper drawing, you may think you have a different transition state (TS). But in reality, both TSs you have drawn are the same. Try to get both views using models. You'd realize then. $\endgroup$ May 1, 2022 at 21:39
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    $\begingroup$ To understand the concept better, read this original article. $\endgroup$ May 1, 2022 at 21:40
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    $\begingroup$ Thanks a lot, Mathew Mahindaratne. Due to the oxygen in my modeling kit being maximally two-valent I totally overlooked the effect of its substituent. Case closed! BUT: The transition states are not the same. Its even shown in the article that they are in fact different. However the red one is not realized due to steric hindrance between O-Et and the binaphtyl. $\endgroup$
    – marsem
    May 1, 2022 at 22:05
  • $\begingroup$ True to your statement. Of cause, I overlooked the two transition states. Actually, the red one is derived from R-BINAL.-H. Look at stereochemistry at Al atom (H on Green is going into the plane of paper and H on Red is coming iout of the plane of paper). Thus, Green one is (S) and Red one is (R). $\endgroup$ May 1, 2022 at 22:20
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    $\begingroup$ @Mathew: I took my comment down as soon as I saw your response. I'll revise and repost so your response has an antecedent. The question arises, why isn't the alkynyl group in the TS axial having a smaller A-value than the methyl. Switch the methyl and alkynyl group in the red structure. Both reagents (green and red) are S-(P)-BINOLs having a two fold axis of symmetry. After all, these TS's are rationalizations. See BINOLs here. $\endgroup$
    – user55119
    May 1, 2022 at 22:47

1 Answer 1


I think our readers need a good explanation for this question based on user55119's comment "The question arises, why isn't the alkynyl group in the TS axial having a smaller A-value than the methyl. Switch the methyl and alkynyl group in the red structure. Both reagents (green and red) are S-(P)-BINOLs having a two fold axis of symmetry. After all, these TSs are rationalizations."

OP's original question was:

I found an alternative transition state (red) that yields the opposite enantiomer. What am I doing wrong?

First, I need to clarify that the reducing reagent we are talking about is $\ce{BINAL-OEt}$ $(\ce{LiAlH(Binol)OEt})$, which gives pretty high enantioselectivity (than when used $\ce{LiAlH2(Binol)}$) for reduction of ketones that have a $\pi$-system directly attached on one side of the carbonyl group.

The chiral $\ce{BINAL-OEt}$ can be prepared by using enanthiomerically pure BINOL $(\ce{Binap(OH)2^{*}})$. The stereoisomer considering here is 2,2'-dihydroxy 1,1'-binaphthyl (BINOL), which shows atropisomerism (Ref.1) defined by $M$ and $P$ notations (alternatively, $R$ and $S$ notations, respectively). Atropisomers are stereoisomers arising because of hindered rotation about a single bond:


For BINOL, $\ce{A = A' = OH}$ (the highest priority group) and $\ce{B = B' = benzo}$ function (the lowest priority group). Thus, stereochemistry of BINOL group of both Red and Green transition states are $P$ (or $S$).

It is known fact that these atropisomers can transforem chirality to substrates it attached to. For example, the aluminum atom in $\ce{BINAL-OEt}$ is a chiral center based on the use of $M$- or $P$-BINOL. Thus, when enanthiomerically pure $P$-BINOL is reacted with equimolar amounts of $\ce{LiAlH4}$ and ethanol would result a diastereomeric mixture:

$$\ce{LiAlH4 + Binap(OH)2^{*} + EtOH -> LiAl^sHBinap(O2)^{*}OEt + LiAl^rHBinap(O2)^{*}OEt}$$

As I'd explain later in the text (vide infra), each of these two diastereomers would give opposite enanthiomer (when used in pure isomer form) in the reduction reaction in hand. Based in these information, as I pointed out in my comment (and as OP agrees on), the two Green and Red Zimmerman-Traxler Chair-Like Transition States (Ref.2) are not the same:

Zimmerman-Traxler Chair-Like Transition States

If you look closely, you would realize that the $\ce{Al}$ stereocenter in each of two Green and Red Transition States are not not identical (c.f., the orentation of napthyhyl ring the bottom box, which is identical in both Green and Red TSs). I assigned the on in Green TS as $s$ while that in Red TS as $r$. To my knowledge, these two TSs are arisen from the reaction of same starting material with each of diastereomers (the optical active reducing reagents) I mensioned earlier:

$$\ce{LiAl^sHBinap(O2)^{*}OEt + LiAl^rHBinap(O2)^{*}OEt}$$

But, it is not possible becase only one of these two diastereomers should be using in the reaction. Therefore, one of these two transition states should be negated.

About user55119's comment: Zimmerman and Traxler have suggested these Chair-Like Transition States to explain the resulting stereochemistry of a reaction, which are as suggested by user55119, in deed rationalizations. Besides, having the alkynyl group in the axial position in the TS is not just because steric reason (even if having a smaller A-value than the methyl). There are other factors such as $n,\pi$-repulsion forces (Ref.1, and Ref.3-4):

Electronic effect

As shown in the above diagram, favorism to the accepted TS is govenrned by a repulsive interaction between the $\pi$ system of the ketone (here it is alkynyl group) and the lone pair of an oxygen atom of the $\ce{BINAL-OEt}$.


  1. Jean Michel Brunel, "BINOL:  A Versatile Chiral Reagent," Chem. Rev. 2005, 105(3), 857–898 (DOI: https://doi.org/10.1021/cr040079g).
  2. Howard E. Zimmerman and Marjorie D. Traxler, "The Stereochemistry of the Ivanov and Reformatsky Reactions. I," J. Am. Chem. Soc. 1957, 79(8), 1920–1923 (DOI: https://doi.org/10.1021/ja01565a041).
  3. R. Noyori, I. Tomino, Y. Tanimoto, and M. Nishizawa, "Asymmetric synthesis via axially dissymmetric molecules. 6. Rational designing of efficient chiral reducing agents. Highly enantioselective reduction of aromatic ketones by binaphthol-modified lithium aluminum hydride reagents," J. Am. Chem. Soc. 1984, 106(22), 6709–6716 (DOI: https://doi.org/10.1021/ja00334a041).
  4. R. Noyori, "Centenary Lecture. Chemical multiplication of chirality: science and applications," Chem. Soc. Rev. 1989, 18, 187-208 (DOI: https://doi.org/10.1039/CS9891800187).
  • $\begingroup$ Thank you for the effort on this detailed answer. You are right on the electronic repulsion and the favorism that derives regarding axial and equatorial position of the alkyne function. However I want to focus on what you mentioned above. There are no diastereomers of the reducing agent. It is C2-symmetrical. Rotation around the axis yields same position of BINOL-Substituent, but switches the positions of OEt and H in the plane. Therefore both TS, green and red could exist. However the steric hindrance of OEt and Naphthylring disfavors the red one. See TS 9&10 here: doi.org/10.1021/ja00334a041 $\endgroup$
    – marsem
    May 4, 2022 at 16:47
  • $\begingroup$ In short: The transition states are diastereomeric, but the reducing agent itself is not. $\endgroup$
    – marsem
    May 4, 2022 at 16:49
  • $\begingroup$ @marsem: I stand by my analogy in this case. If you look at the ring systems of BINOL residue, configurations of Al-O of front naphthyl (dark filling) and back naphthyl (no filling) stays same on both Green and Red TSs. Thus, if you switch the OEt and H groups on Al, you get diastereomers. Yet, TS 9 and TS 10 have different configuration on front and back naphthyl groups (O-Al is axial on front naphthyl in TS 9 and it is axial on back naphthyl in TS 10) thus, switching of OEt and H is allowed without changing the stereochemistry of whole system. $\endgroup$ May 4, 2022 at 18:11
  • $\begingroup$ When regarding the transition states, they are diasteromeric, as the tetrahedral aluminum is fixed in the drawn structures. But the single molecule BINAL-H itself only shows axial chirality of the BINOL-substituents. When switching OEt and H in the molecule it is the same molecule rotated by 180° (C2 symmetry). However in the transition states drawn it is then fixed and no rotation of the BINAL-H is possible, thus the TS are diastereomeric, but not the single BINAL-H molecule. $\endgroup$
    – marsem
    May 4, 2022 at 19:19
  • $\begingroup$ If you do not trust me, then build the models of the "two" molecules and you will realize that they are in fact superimposable with each other. $\endgroup$
    – marsem
    May 4, 2022 at 19:23

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