enter image description here

The above are two compounds.
The objective is to find if the compounds are identical/enantiomers/diastereomers

I can see a plane of symmetry if we keep the $\ce{H-C-CH3}$ in the plane. So, concluding it as identical looks great.

but, if so, how can one transform the first compound to the second?

Applying, CIP rules, we might invalidate the P.O.S. The solution given for this question is "identical".


I have somehow manipulated the bonds such that I can prove the identity. Even though it is extremely-sterically hindered as indicated in the comments section (Now moved to chat). The forceful manipulations are done keeping in mind, not to break any bonds and only .

enter image description here

However this again proves that the molecules are not identical.

Now, since it is proposed that these molecules are different, can I get a lab or computer synthesized version of both molecules? Upon many search attempts, I could get this document which discusses about the same cyclohexene ring but Deuterium instead of the $\ce{CH3}$ at the bridge head and substituted oxygen ions. enter image description here

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – user7951
    Aug 25, 2020 at 21:13

1 Answer 1


7-Methylbicyclo[2.2.1]hex-2-ene exists as two diastereomers, anti-1 and syn-1, which differ by the relationship of the methyl group and the double bond. Both compounds are meso with the bridgehead carbons stereogenic and chirotopic (R/S) while C7 is stereogenic but achirotopic (r/s). The plane of symmetry in each one does not produce an enantiomer of itself. Stereoisomers have the same molecular formula and the same atom connectivity. You probably know that the stereoisomers (R,R)- and (S,S)-tartaric acid are enantiomers and bear a diastereomeric relationship with (R,S)-tartaric acid, which is a meso-compound. A pair of meso-compounds with the same molecular formula and the same atom connectivity may also be diastereomers. Just like structures 1, the central carbons (C3) of ribitol and xylitol are stereogenic and achirotopic. The properties of these two pentaols are different as are the properties of anti-1 and syn-1.


Your Answer

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