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I was taught that according to CIP rules, we must check the priority order for the substituents. But how can we do the same for molecules with cyclic groups as substituents?

How do we compare the substituents for the following molecule?

Tetra-substituted ethene

How to determine EZ configuration for such molecules?

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  • $\begingroup$ Follow the same CIP as straight chains. $\endgroup$ – Mathew Mahindaratne May 10 at 2:18
  • $\begingroup$ How do I break the cyclic compound though? $\endgroup$ – Natru May 10 at 2:46
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    $\begingroup$ I feel more and more that many teachers don't really know what to teach. This exercises with no didacticak value resemble sudoku and staff like that. They should be reserved, quite paradoxically, to future chemists and later in the course of their studies and not part of a general teaching chemistry. Even a professional organic chemist dig into this just because a IUPAC acceptable name has to be given after isolation and in publication. But that is. $\endgroup$ – Alchimista May 10 at 7:48
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The compound is a tetra-substituted alkene with four different groups. Thus, it could be a (E)- or (Z)-ethene. To find the priority of the groups, you must follow CIP rules. The four groups are cyclobut-1-enyl, cyclobut-2-enyl, 2-methylcyclobutyl, and 1-methylcyclobutyl. Since it is an alkene (planner sp2–sp2 bond), we don't have to consider all four groups at once. What you have to do is consider two groups at once on each side of the double bond (say they are a and b):

CIP-Priority of tetra-substituted alkene

On side a the first carbon of each cyclobutyl group (cyclobut-1-enyl and 1-methylcyclobutyl) has equal priority ({C,C,C} and {C,C,C}) as indicated in the image. However, the second carbon of cyclobut-1-enyl group has the priority {C,C,H}, while the second carbon of 1-methylcyclobutyl group has the lower priority {C,H,H} compared to that in cyclobut-1-enyl group. Therefore, between these two groups, cyclobut-1-enyl group has the higher priority.

On side b the first two carbons of each cyclobutyl group (cyclobut-2-enyl and 2-methylcyclobutyl) have equal priority: First carbons of both rings have {C,C,H} and the second carbons of both rings have {C,C,H} as well (see the image). Yet, the third carbon of cyclobut-2-enyl group has the priority {C,C,H} while the that of 2-methylcyclobutyl group has the lower priority {C,H,H} compared to cyclobut-1-enyl group. Therefore, between these two groups, the cyclobut-2-enyl group has a higher priority.

Now, you see higher priority groups of a and b sides are oriented on the same side of the double bond. Therefore, the compound is (Z)-isomer. The name is (Z)-1-(cyclobut-1-enyl)-2-(cyclobut-2-enyl)-1-(1-methylcyclobutyl)-2-(2-methylcyclobutyl)ethene.

Also, note that this molecule also has three chiral centers as indicated in the image. However, their stereochemistry is not given, so I neglect them.

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    $\begingroup$ Complementary to this answer, see e.g., this and this example on chemistry.se. $\endgroup$ – Buttonwood May 10 at 5:10
  • $\begingroup$ "the second carbon of cyclobut-1-enyl group has the priority {C,C,H} while the second carbon of 1-methylcyclobutyl group has the lower priority ({C,H,H}) compared to that in cyclobut-1-enyl group" How did you find which side to go to for the "second" molecule, when the first molecules had the same priority. You went clockwise for the top case and to another substituent for the bottom one while comparing priorities for the second molecule. I don't understand on what basis you did those. $\endgroup$ – Natru May 10 at 8:44
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    $\begingroup$ @Natru You go both ways around, treating each direction as a separate branch, then make the comparison based on the higher-priority branch. I've tried to illustrate it with this image: the red atoms are the ones which distinguish the ligands, and the blue atoms are the ones which must be compared to find the difference. $\endgroup$ – LegionMammal978 May 10 at 12:18
  • $\begingroup$ @LegionMammal978 wow I found the image really helpful, great visualisation. Thanks! $\endgroup$ – Natru May 10 at 12:46
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    $\begingroup$ Natru: They are called digraphs. Look here: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… $\endgroup$ – user55119 May 10 at 17:14

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