Reduction of unsymmetrical ketones using BINAL-H (Enantioselectivity)

I am struggling with a problem regarding the enantioselectivity of BINAL-H for quite some time now.
[![This reaction scheme](https://i.stack.imgur.com/prtPw.png)](https://i.stack.imgur.com/prtPw.png)


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?




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:


[![Atropisomerism](https://i.stack.imgur.com/SPSwn.jpg)](https://i.stack.imgur.com/SPSwn.jpg)


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](https://i.stack.imgur.com/kKV11.jpg)](https://i.stack.imgur.com/kKV11.jpg)


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](https://i.stack.imgur.com/q73BN.jpg)](https://i.stack.imgur.com/q73BN.jpg)


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 $\ce{BINAL-OEt}$.




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**References:**


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>).


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