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Cahn–Ingold–Prelog Rules, cyclic structures, and breaking ties

Twisting stereoisomers with rings to determine R/SHow are non-carbon stereogenic centers named (S/R)?How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenesWhich has higher priority (Cahn-Ingold-Prelog)?Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl?How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistryWhich alkene has a higher Cahn-Ingold-Prelog priority in (2Z,5E)-hepta-2,5-dien-4-ol?Cahn–Ingold–Prelog priority of carboxylic group versus ortho carboxylic acidWhat are the CIP rules for cyclic substituents?Cahn–Ingold–Prelog priority of CN vs CH2NH2Which one has higher priority according to Cahn–Ingold–Prelog rules?

3

$begingroup$

(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.

nomenclature stereochemistry chirality

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edited Jun 10 at 17:15

andselisk♦

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3

$begingroup$

(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.

nomenclature stereochemistry chirality

share|improve this question

edited Jun 10 at 17:15

andselisk♦

21.4k775143

asked Jun 10 at 16:19

SarahSarah

162

New contributor

Sarah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

(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.

nomenclature stereochemistry chirality

share|improve this question

edited Jun 10 at 17:15

andselisk♦

21.4k775143

asked Jun 10 at 16:19

SarahSarah

162

New contributor

Sarah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited Jun 10 at 17:15

andselisk♦

21.4k775143

asked Jun 10 at 16:19

SarahSarah

162

New contributor

Sarah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question

edited Jun 10 at 17:15

andselisk♦

21.4k775143

asked Jun 10 at 16:19

SarahSarah

162

New contributor

Sarah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Jun 10 at 17:15

andselisk♦

21.4k775143

edited Jun 10 at 17:15

andselisk♦

21.4k775143

asked Jun 10 at 16:19

SarahSarah

162

New contributor

Sarah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 10 at 16:19

SarahSarah

162

1 Answer
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4

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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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edited Jun 10 at 18:04

answered Jun 10 at 17:35

user55119user55119

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4

$begingroup$

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.

.

share|improve this answer

edited Jun 10 at 18:04

answered Jun 10 at 17:35

user55119user55119

4,87511242

$endgroup$

add a comment | 

4

$begingroup$

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.

.

share|improve this answer

edited Jun 10 at 18:04

answered Jun 10 at 17:35

user55119user55119

4,87511242

$endgroup$

add a comment | 

$begingroup$

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.

.

share|improve this answer

edited Jun 10 at 18:04

answered Jun 10 at 17:35

user55119user55119

4,87511242

$endgroup$

share|improve this answer

edited Jun 10 at 18:04

answered Jun 10 at 17:35

user55119user55119

4,87511242

answered Jun 10 at 17:35

user55119user55119

4,87511242

answered Jun 10 at 17:35

user55119user55119

4,87511242