MoleculeHere is the molecule. Clearly the 4th carbon is chiral. I have drawn the diagraph for it, but the exploratory pathways of the left and right chains are the same. So, the problem is that the two chains are actually different, but same in the diagraph. How to assign stereodiscriptors in such cases?

  • $\begingroup$ If you can visually tell that they're different, then the graphs you build is different. If you're getting different graphs, you're doing something incorrect in that respect. $\endgroup$ – Zhe Oct 16 '18 at 19:13
  • $\begingroup$ You mean "if you're getting same graphs"? The graphs are different, but the exploratory pathways are same, i.e. same set of atoms at any distance from the chiral carbon for both the chains $\endgroup$ – Shashank Oct 16 '18 at 19:19
  • $\begingroup$ it's in how do you hierarchically evaluate the paths' nodes … $\endgroup$ – mykhal Oct 16 '18 at 19:46
  • $\begingroup$ You just look at the sets of atoms at a distance of 1,2,... from the chiral carbon and wherever you find the first point of difference, then you can assign priorities. But, in this molecule the sets of atoms at all distances are same. So, you can't decide $\endgroup$ – Shashank Oct 16 '18 at 20:03
  • 1
    $\begingroup$ No, the connectivity matters. 2-propyl < 1-hydroxy-2-propyl You don't just evaluate the nodes at each distance. You need to evaluate the whole group, but closer atoms have more influence than atoms that are farther away. $\endgroup$ – Zhe Oct 16 '18 at 20:05

Not the proper, somewhat complicated, IUPAC ‘digraphs’… but might help.

Bold/colored bonds are the left and right branches' resulting paths taken when advancing to next sphere (because the decision could not be yet done) during determining a higher priority branch by finding a first difference of the compared atom set rankings. Sphere numbers a Roman. Grey digits are semi-conveniently arbitrarily chosen numbering just for reference, not nomenclature numbering nor priorities(!).

Fig. 1

Following list shows chosen highest priority atoms in comparison in each nth sphere and left/right branch (bold; also comparison with other sibling atoms in the same branch, if they are not just hydrogen), chosen according to the next bonded atoms from the (n+1)th sphere (in 1st braces; atom number in 2nd brace).

sphere № : [left] vs [right]
I: [ C(CCH) (1) ]   =   [ C(CCH) (4) ]
II: [ C(CCH) (2) > C(CHH) (3) ]   =   [ C(CCH) (5) > C(CHH) (6) ]
III: [ C(HHH) (7) = C(HHH) (8) ]   <   [ C(OHH) (9) > C(HHH) (10) ]

Please note that in the II. sphere, left, the carbon (3) leading to terminal -OH group could not be chosen, because it's C(CHH), compared to higher priority, chosen C(CCH) (2).

Also note that in sphere III, left, any of the two methyls' carbons could be chosen (equivalent).

In this III. sphere, the decision finally could be made. The right branch (R1) has higher priority – finally only hydrogens bonded at left: C(HHH) (7), vs oxygen and hydrogens bonded at right: C(OHH) (9).
(Next priority is left branch R2, then methyl, then H.)

enter image description here

So the compound configuration/name is

  • $\begingroup$ Oh! Ok, so you don't just keep on comparing sets of atoms for each of the chains at increasing distances from the chiral carbon and then try to get assign priorities. Instead, you see which branch to traverse in each chain, based on the priorities of atoms of each branch connected to the atom in the previous sphere. Am I right? $\endgroup$ – Shashank Oct 17 '18 at 13:44
  • $\begingroup$ @Shashank This is what I did not want,… to confuse. You just keep comparing, but your question looked like you thought that you traverse in each chain. I rather accidentally looks like it should be traversed, maybe the structure was intentionally designed such way (by our teacher?), or at least to be somewhat difficult. I'll try to improve the answer, mention some rules. $\endgroup$ – mykhal Oct 17 '18 at 13:52
  • $\begingroup$ Yes, I thought each and every branch of both the chains have to be traversed $\endgroup$ – Shashank Oct 17 '18 at 13:57
  • $\begingroup$ @Shashank … yes, it should be traversed (at least to the first atom in each branch☺) but with the evaluation, and finding the first difference. Your question looked like you are comparing sets of all possible traversal paths without evaluation/ranking of the attached atoms at every node. $\endgroup$ – mykhal Oct 17 '18 at 14:06
  • $\begingroup$ Yes, the was the mistake I made. Just to clarify, you didn't use the chain starting from C-3, leading to the terminal OH group. I thought, that will also have to be traversed $\endgroup$ – Shashank Oct 17 '18 at 14:12

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