[Roju (OP)]

Hello.
Can someone explain to me why this compound only has two stereoisomers? It follows Le Bel-van't Hoff rule, but I dont understand why the double bond marked on the picture doesn't give 2 more stereoisomers.

What about the double bonds in the cyclic structures?

I can't find any information online about stereoisomers in compounds with both double or tripple C-C bonds as well as chiral centra

Thanks in advance!  []
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[evan_au]

ChiralSPO or Bored Chemist could give an authoritative explanation, but I'll put in my small contribution:

I don't understand why the double bond marked on the picture doesn't give 2 more stereoisomers

It is a diagrammatic convention that if the (implicit) Carbon atoms aren't marked with a chemical group, it is (implicitly) assumed to be Hydrogen.

The two groups on the right-hand end of the double bond are identical methyl groups, so if you interchanged them, it would be indistinguishable from the original substance. So there aren't two different stereoisomers

What about the double bonds in the cyclic structures?

I assume that the ring on the left is a Benzene ring. This is illustrated with 6 carbon atoms linked by 3 double bonds and 3 single bonds. This looks like there could be 2 different benzene rings, with the double and single bonds swapped.

However, the Benzene ring actually has 6 identical bonds which are intermediate between a single and double bond - something like a "1.5 bond" (at 139pm bond length). So you can't swap the position of the single and double bonds, because it is just a diagrammatic convention.

See: http://en.wikipedia.org/wiki/Benzene#Ring_formula

[Roju (OP)]

Ah, i understand it now.
Correct me if my understanding is wrong.

The bond i marked has a carbon with two methyl groups, so it can't occur as stereoisomers.
The double bonds in the cyclic structures can only occur in that one configuration for the loop to be complete.

Therefore the only two isomers of the compound come from the chiral centre.

[puppypower]

Where you have the first circle on the left, connected to the -OH group, there is an implied fourth bond to hydrogen. Carbon needs four bonds. This junction can form a stereo isomer, since that center carbon has four different groups bonded to it.

[chiralSPO]

Ah, i understand it now.
Correct me if my understanding is wrong.

The bond i marked has a carbon with two methyl groups, so it can't occur as stereoisomers.
The double bonds in the cyclic structures can only occur in that one configuration for the loop to be complete.

Therefore the only two isomers of the compound come from the chiral centre.

That is correct.

        R1      R2
          \      /
The      C=C     structures can only lead to different isomers if R1 ≠ R3 AND R2 ≠ R4.
          /      \
        R3      R4

Additionally, if R1 and R2 (or R3 and R4) are connected in a ring that contains fewer than 8 consecutive bonds, there is only one  stereoisomer that is physically possible (R1 and R4 cannot be connected in a small ring).

[Roju (OP)]

That is correct.        R1      R2          \      /The      C=C     structures can only lead to different isomers if R1 ≠ R3 AND R2 ≠ R4.          /      \        R3      R4Additionally, if R1 and R2 (or R3 and R4) are connected in a ring that contains fewer than 8 consecutive bonds, there is only one  stereoisomer that is physically possible (R1 and R4 cannot be connected in a small ring).

Thank for you for broadening my understanding. Is the number 8 a set limit, or is it just a rule of thumb? I guess the limit has to do with the stereochemistry of the different bonds, lenghts and angles ?

[chiralSPO]

8 is a rule of thumb. I suppose it might be possible to have a 7-membered ring with some really long bonds in there...

The smallest trans-cycloalkene is trans-cyclooctene. If you have a molecular model kit, try to build a model of trans-cycloheptene, and you will see why the molecule has never been observed.