Question: Assign the R/S configurations to the carbons which are attached to different rings of this molecule (carbons 1 and 3 according to IUPAC numbering): Question

My Solution

I first started out by numbering the carbons. The one at the top would be 1 and the bottom chiral carbon would be 3. Next, According to CIP rules, I have included the phantom atoms by replacing the double bonds while deciding the priority order. Moving on, at carbon 1, My priorities would be:

  1. $\ce{-OH}$
  2. Cyclopentene ring (on the left)
  3. Cyclopropene ring (on the right)
  4. Carbon 2

This gives me a counter-clockwise turn and since the least priority group is on the vertical line (away from us) the configuration I assign would be S.

Following a similar procedure for carbon 3, The priorities would be :

  1. Bottom 5-member ring (contaning $\ce{O-O}$ bond)
  2. 3 member ring on the right
  3. 5 member ring on the left
  4. Carbon 2

Again, this gives me a counter-clockwise turn and since the least priority group is away, I would assign the configuration: S.

The answer key, however, says that the answer is 1R, 3S.

Where am I going wrong?

  • 2
    $\begingroup$ Of the oxygen containing rings, doesn't the one on the bottom have the lowest priority because there is only one oxygen atom on the carbon attached to the rest of the molecule, whereas for the rings on the left and the right, there are two? $\endgroup$
    – Karsten
    Sep 4, 2020 at 1:12
  • 2
    $\begingroup$ I have included the phantom atoms by replacing the double bonds while deciding the priority order. - you did it wrong, as Karsten just mentioned $\endgroup$
    – Mithoron
    Sep 4, 2020 at 1:14
  • 2
    $\begingroup$ The configurations are 1R, 3S. For C1, the order is 1, 3, 2, 4 using your numbers. One switch gets you from S to R. For C3 the order is 2, 3, 1, 4. Two switches here keeps the S assignment. You were correct for the wrong reason. I'm preparing and answer but the drawing takes some time. $\endgroup$
    – user55119
    Sep 4, 2020 at 1:40
  • 1
    $\begingroup$ @sai-kartik: You may want to look at this link for help with digraphs: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… $\endgroup$
    – user55119
    Sep 4, 2020 at 11:45
  • 1
    $\begingroup$ Arpit Raj Choudhary: A real atom is the same as a duplicate atom as to the atomic number. The difference lies at the next level. In (1R, 5S, 6R)-5,6-Dimethylcyclohex-2-en-1-ol The real methyls have [H,H,H] attached while the duplicate methyls of the double bond have atoms of at. no. zero attached, i.e., [0,0,0]. Clearly, every real atom outranks a duplicate atom. (C) < C and [0,0,0] < [H,H,H]. I hope this clears things up and that you enjoyed the tour. $\endgroup$
    – user55119
    Aug 29, 2022 at 13:37

1 Answer 1


I have separated C1 and C3 into unique structures. For C1, it is clear that the oxygen of the hydroxyl group has the top priority while the methylene group has the lowest priority. A digraph is constructed such that we delineate two paths around each of the unsaturated rings. Thus, non-duplicate atoms C3 and C5 terminate with their respective duplicate atoms C(3) and C(5). The cyclopentene ring has locants in sphere 3 of {(7),7,H} which is a tie with {(4),4,H} in sphere 3 of the cyclopropene ring. But the locants {8,H,H} of the cyclopentene lose to {(3),3,H} of the cyclopropene ring. The priorities are O>cyclopropene>cyclopentene>methylene. C1 has the (R)-configuration. continued...

A different digraph for C3 is constructed by dissecting the three rings into a total of six pathways, two for each ring. The rings are designated with red, blue and black dots. The red and blue rings take precedent over the black ring because each one has locants {O,O,H} at the branchs while the black ring has {O,C,H}. To distinguish between blue and red, we move to the next atoms on the chain. Blue has two oxygens while red has two carbons. The priorities are blue>red>black>methylene. C3 has the (S)-configuration.

Related issues:

  1. What are the CIP rules for cyclic substituents?
  2. How to check for geometrical isomerism in cyclic compounds?
  3. More on Rings and Digraphs

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