1
$\begingroup$

Are (S,E)-4-isopropoxypent-3-en-2-ol and (R,Z)-4-isopropoxypent-3-en-2-ol both enantiomers and diastereomers? I am a bit confused because they would be diastereomers when only looking at the configuration of the double bond or enantiomers when just looking at the chiral center. However, it doesn't make sense that molecules can be diastereomers and enantiomers at the same time. Could someone explain? Thanks.

$\endgroup$
1
  • $\begingroup$ You don't separate the stereochemical entities. Do they have the same atom connectivity? YES! Are they enantiomers? NO! Then they are diastereomers. $\endgroup$
    – user55119
    Commented Nov 24, 2023 at 20:52

1 Answer 1

3
$\begingroup$

Diastereomers, by definition, are stereoisomers that are not enantiomers, so you're right they can't be both.

I think you might be confused because you are trying to apply a simplified heuristic (opposite configuration of double bond = diastereomer, opposite configuration at chiral center = enantiomer) outside of its proper scope. This works well for molecules with a single stereocenter, but for slightly more complex cases you have to step back and think about the meaning of the terms for a moment.

Two molecules are enantiomeric when they are mirror images of one another, that is, if they are identical in every respect, except the configuration around their chiral center(s). (R)-4-isopropoxypentan-2-ol and (S)-4-isopropoxypentan-2-ol are enantiomers, because their single chiral center contains the exact same set of four substituents (H, OH, methyl and 2-isopropoxypropyl), just in reverse orders. However, your pair of molecules have only three substituents (H, OH and methyl) in common; the fourth substituent at C2 is (E)-2-isopropoxypropen-1-yl in the first compound and (Z)-2-isopropoxypropen-1-yl in the second, which are isomeric but clearly distinguishable groups. (R)-4-isopropoxypentan-2-ol and (S)-4-ethoxypentan-2-ol aren't enantiomers, and neither are these molecules.

They are, however, still stereoisomers of one another, so they must be diastereomers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.