Why is S-Ethyl 3-hydroxybutanoate SO MUCH more stable than its R enantiomer?

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The boiling points of R-ethyl 3-hydroxybutanoate and S-ethyl 3-hydroxybutanoate differ significantly, with R at about 75°C and S at around 180°C, raising questions about stability and enantiomer behavior. Enantiomers typically have the same boiling points under normal conditions, suggesting that the observed differences may stem from experimental conditions or measurement errors. The discussion highlights the role of enzyme active site symmetry in determining enantiomer preference during reactions, particularly in the case of alcohol dehydrogenase. The predominance of the S enantiomer in reactions is attributed to the enzyme's ability to interact more favorably with it, while the R enantiomer is produced in smaller amounts due to induced fit mechanisms. Overall, the conversation clarifies the complexities of enantiomer behavior in chemical reactions and the factors influencing their stability.
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The boiling point for R-ethyl 3-hyrdoxybutanoate is about 75 C
The boiling point for S-ethyl 3-hyrdoxybutanoate is about 180 C

How!?
How is the same molecule that much more stable just by rearrangement of the hydroxy group. I'm trying to explain how some R enantiomer may boil off along with ether if heated enough but I'd like to explain why it is that the R enantiomer is less stable
 
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This must be wrong (the spelling of the substance name is also wrong by the way). Enantiomers have the same boiling point under normal conditions. Were it true you'd have a wonderful business opportunity separating the enantiomes and selling them. I checked the data at Sigma Aldrich and I suspect that is where you have got your data because I find the same odd difference. The racemate is stated to have a bp of 170 oC so the higher value is probably true.
 
Woops! I meant hydroxy. Hm, maybe I read this was under pressure. My lab book says 180-182 for the s enantiomer. I'm still confused as to why the re face of ethyl acetoacetate is so much more favorable in my reduction reaction using alcohol dehydrogenase. Any clue?
 
Yes, with the dehydrogenase it has nothing to do with stability. It is related to the lack of symmetry in the enzyme active site. If I remember correctly there are some rules of thumb to predict which enantiomer that predominates for a new substrate but the thumb of those rules tend to be rather thick so often you get what you get.
At pdb there are lots of structures here is a relevant one showing an inhibitor bound to the active site: http://www.rcsb.org/pdb/explore/explore.do?structureId=1AXE
 
cool, that helps. Basically, I've taken it that the reason why I got 92 percent S in this reaction of yeast alcohol dehydrogenase and acetoacetate is because NADH can only donate to the acetoacetate from one side that fits (the S). And the small amount of R would likely be from induced fit model meaning that the enzyme changes to fit some si faced acetoacetates making R enantiomers. This however must take longer than the S and thus is less. Does that sound right, to your knowledge?
Edit: Fits from one side (re face creating an S)
 
Exactly!
 
Thank you so much. I've been trying to explain this lab for the entire day and couldn't quite get it. I appreciate the help!
 
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