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. $\endgroup$ Apr 9, 2016 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? $\endgroup$ Apr 9, 2016 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. $\endgroup$ Apr 9, 2016 at 10:17
  • $\begingroup$ @DGS Thanks! I've uploaded a different set of sketches. Do you feel this set conveys the appropriate sterochemistry? $\endgroup$ Apr 9, 2016 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. $\endgroup$ Apr 9, 2016 at 10:28

1 Answer 1


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? $\endgroup$ Apr 9, 2016 at 23:00
  • 1
    $\begingroup$ No, they are diastereomers. $\endgroup$
    – Lighthart
    Apr 10, 2016 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. $\endgroup$ Apr 10, 2016 at 1:29

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