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(3S,5R)-3-ethyl-5-methylcyclohex-1-ene

(3S,5R)-3-ethyl-5-methylcyclohex-1-ene

Does anyone know how one reaches the conclusion that the 5C is R? I tried using the Cahn–Ingold–Prelog priority rules but kept getting confused.


From chem.libretexts.org:

When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

From Wikipedia:

To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.

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You don't need a digraph for this one. Your structure is A. Remove the double bond and add duplicate carbon atoms to each of the double bond carbons (B). These duplicate carbon atoms each bear three phantom atoms of atomic number zero. Clearly at C5 hydrogen has the lowest priority and methyl is the next lowest. The methylene groups at C4 and C6 are a tie. [There is no need to consider C2 since it comes into play for both routes. It is a wash.] The red carbon of the ethyl group bears C,H,H while the red duplicate carbon at C1 bears 0,0,0. Place your right hand on C5 and point your thumb in the direction of the hydrogen. Now your remaining fingers will point as follows: C4>C6>CH3>H. C5 is of the R-configuration. See, Twisting stereoisomers with rings to determine R/S.

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The most difficult thing in the presented problem is to determine the priority of the substituents at C5. User55119 explained the priority well but ended everything again with the notorious "hand rule", which is no less catastrophic than molecular turns. Because it is difficult to understand how to orient the hand. Knowing the priority of the substituents and the direction of their decrease AS IT SEEN on a picture (CW or CCW), you just need to assign a configuration. In the above example, the priority decreases counterclockwise, and the configuration should be S, but the lowest priority group (H) is CLOSER to us, so we have to change the configuration to the OPPOSITE and get 5R.

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    $\begingroup$ Your suggestion is fine as long as the lowest priority group is above or below the plane of the other three atoms. But how convenient is it if the lowest priority group lies in the plane? As to the notoriety of the hand rule, to each his own. After all, chiral is derived from the Greek for hand. $\endgroup$
    – user55119
    Oct 17 at 15:27
  • $\begingroup$ Good question! The solution is very simple - swap any two substitutes twice so that the younger group is behind the plane of the screen (sheet, board, etc.). orgchem.tsu.ru/ENGLISH/R_S_swap.PNG $\endgroup$ Nov 12 at 2:45
  • $\begingroup$ In VERY RARE cases of cyclic compounds the lowest priority group is placed in the ring, and swapping would be confusing. $\endgroup$ Nov 12 at 2:58

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