In the structures below (2) vs (3) seem obvious optical isomers to me.

But what about (1)? Is that just a stuctural isomer (e.g. Butane vs Isobutane) or is there a more specific way to describe the relation of (1) vs (2) / (3)?

enter image description here

Alternative sketch of (2) & (3) based on the comment by @DGS below:

enter image description here

enter image description here

  • $\begingroup$ (2) and (3) look like the same molecule to me. I see the wedged and dashed bonds, but those should be up and down, and the way to flip that spot is to switch which is up and which is down. $\endgroup$ – SendersReagent Apr 9 '16 at 9:03
  • $\begingroup$ @DGS You mean there's an error in the sketch I've used? Or you mean beta vs epi-beta are the same molecule? $\endgroup$ – curious_cat Apr 9 '16 at 9:29
  • $\begingroup$ I mean I don't think the wedge and dash is the appropriate way to show stereochemistry from that view. There are two different molecules, I just don't think the image properly represents them both. $\endgroup$ – SendersReagent Apr 9 '16 at 10:17
  • $\begingroup$ @DGS Thanks! I've uploaded a different set of sketches. Do you feel this set conveys the appropriate sterochemistry? $\endgroup$ – curious_cat Apr 9 '16 at 10:24
  • $\begingroup$ Hmm... I don't think any of these are optical isomers of each other now that I look at them again. 2 and 3 both have three chiral centers and only one flips. $\endgroup$ – SendersReagent Apr 9 '16 at 10:28

There is no exocyclic methylene in structure 1. Is it not an optical isomer of 2 and 3.

Likely there is a pericyclic rearrangement between 1 and 2/3 which would formally render them tautomers.

2 and 3 are diastereomers.

  • $\begingroup$ So do you say 2 and 3 are optical isomers? $\endgroup$ – SendersReagent Apr 9 '16 at 23:00
  • 1
    $\begingroup$ No, they are diastereomers. $\endgroup$ – Lighthart Apr 10 '16 at 1:25
  • 1
    $\begingroup$ Okay. Just wanted to make sure because you hadn't made it real clear. The "epi" was a good indication, but the first time I looked at it, I missed the two carbons attached to the bridging carbon. $\endgroup$ – SendersReagent Apr 10 '16 at 1:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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