How to check for geometrical isomerism in cyclic compounds?

Question

I have no problem with identifying whether a given molecule has geometrical isomerism if it's not cyclic. But cyclic compounds are confusing for me. For example, how to check whether the following molecule posses geometrical isomerism?

For a molecule to show geometrical isomerism different molecules should be attached to the double bonded carbons. So, I'm sure that geometrical isomerism (if it is present) is along the double bonded carbons. But how to check whether different groups are attached to it when the double bonded carbons are the part of a cycle?

Answer 1

I hope the statutes of limitation haven't run out. Structure 1 is a (Z)-alkene. Using Cahn-Ingold-Prelog (CIP) protocol, the rings are deconstructed as shown in diagram 2. The larger red arrow marks the higher priority carbon (C,C,C). The smaller red arrow points to the lower priority[carbon (C,H,H). For the blue arrows, sulfur takes priority over carbon. [Addendum: Strictly speaking, the two blue arrows should point to the methylene groups next to the blue sp² carbon such that C(S,H,H) > C(C,H,H).] The two larger arrows establish a (Z)-configuration. See: What are the CIP rules for cyclic substituents?

Answer 2

Two ways to think about it:

• Intuitively: Take one side of the double bond (say, the sulfur-containing one) and imagine rotating it by 180° (so what was up is now down, in your drawing). Is this molecule the same? If it is, then you don't have isomerism. If the procedure gives a new, different molecule, then you have isomerism.
• More formal: use the Cahn–Ingold–Prelog rules to rank both sides of the two carbon atoms of the double bond. If priorities are different, then you have different isomers (and you can actually assign them a name.

In your particular example, the compound has another isomer, and the one you have drawn has the (E) configuration.