Cahn-Ingold-Prelog Rules: More on Rings and Digraphs

How to Manipulate JSmol Structures

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This page describes how to treat small rings as substituents via the CIP method, how to convert cyclic compounds into acyclic digraphs and how to distinguish between duplicate and phantom atoms.

The 2013 update of the 1982 Cahn-Ingold-Prelog rules is located here.

 

(R)-Cyclopropyl-2-methylpropan-1-ol


The assignment of R/S-configurations to the enantiomers of 2-butanol is trivial by the CIP method. But how does one handle rings as substituents? The solution to the problem is to convert the structure to an acyclic version, a digraph as in 1a and 1b.

Using cyclopropyl alcohol 1 as an example, consider the path C4-C5-C6-C4 around the ring. The path C4-C5-C6-(C4), wherein the last term in the chain is a tetradentate, duplicate carbon attached to three atoms of atomic number zero (phantom atoms). The duplicate atoms, in this instance carbon, are enclosed in parentheses and still have their respective atomic number. The phantom atom ranks below hydrogen because they have atomic number zero.

In digraphs 1a and 1b the number of the duplicate atom is displayed as (4). Alternatively, the path around the ring could also be traversed via route C4-C6-C5-(C4).

The red numbers in digraph 1b or 1c signify the sphere or the distance the atoms are remote from the center of stereochemistry. The oxygen, C2, C4 and H are in the first sphere with oxygen having the top priority and the hydrogen the lowest. The priority of the C2 and C4 chains must be determined. No priority can be assigned at sphere 1 because O>C2= C4>H. C2 and C4 are equivalent [C,C,H] in sphere 2. Sphere 3 breaks the tie where C4 [C,H,H] supercedes C2[H,H,H]. The priorities for C1 are O>C4>C2>H. The stereocenter is of the (R)-configuration.

Notice that spheres 4 and 5 were not utilized. Sphere 4 contains a duplicate atom (4) while sphere 5 contains six phantom atoms of at. no. = 0. Most of the subsequent digraphs will not display hydrogens or phantom atoms but their presence will be implied.

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(S)-1-Cyclopropyl-2-propylpentan-1-ol

Cyclopropyl alcohol 2 is a homolog of cyclopropyl alcohol 1. The digraphs display a high degree of symmetry. C5 and C11 bear a total of three hydrogens each in sphere 5 because they are methyl groups whereas the two duplicate carbons (C6) bear "three atoms each of zero atomic number". The priorities are O>C2>C6>H. The stereocenter C1 is of the (S)-configuration.

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(3S,4S)-4-Methylhex-3-en-1-ol

The C4 (S)-configuration of allylic alcohol 3 is determined in the first sphere: C4{C3[O,C,H]>C5[C,C,H]>C7[H,H,H]>H}. The analysis of the C3 stereochemistry is not as obvious. The bold lines in digraph 3b distinguish betwee C2 and C4. In sphere 3 [6,H,H]>[(2),H,H] because in sphere 4 (not shown) the former is terminated with [H,H,H] while the latter has [0,0,0] as terminators. Therefore, C4>C2 because every other atom is of higher priority than one with an at. no. = 0. The order of priorities is O>C4>C2>H. The stereocenter C1 is of the (S)-configuration.

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(1R,5S,6R)-5,6-Dimethylcyclohex-2-en-1-ol

As was the case with allylic alcohol 3, allylic cyclohexenol 4 displays symmetry in its digraphs. The C5 and C6 assignments are resolved at the sphere 2 level: (S)-C5 {C6[C,C,H]>C4[C,H,H]>C7[H,H,H]} and (R)-C6 {C1[O,C,H]>C5[C,C,H]>C8[H,H,H]}.

C1's assignment is resolved in sphere 2 (digraph 4b). The C6 route has [C,C,H;8,5,H] while the C2 route has [C,(C),H;3,(3),H]. The order of priority for C1 is O>C6>C2>H. The stereocenter C1 is of the (R)-configuration.

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(-)-(1S,5R)-α-Pinene

α-Pinene (5a) is a bicyclic monoterpene (C10H16) that occurs as both enantiomers in nature. [Note: The degree of unsaturation of α-pinene is 3, one double bond and two rings. Therefore, α-pinene is bicyclic.] The determination of the configuration at C1 and C5 is not trivial by just looking at structure 5 or 5a. Duplicate atoms are added to the double bond of 5 to form structure 5a. To assign the configuration at C1, digraph 5b is constructed. There are three carbon chains (C2,C6,C7) each of which splits into two additional pathways back to C1. Notice that there are six (1) terminations in the digraph. The six pathways are:

  1. 1-2-3-4-5-6-(1)
  2. 1-2-3-4-5-7-(1)
  3. 1-6-5-4-3-2-(1)
  4. 1-6-5-7-(1)
  5. 1-7-5-6-(1)
  6. 1-7-5-4-3-2-(1)

Hydrogen has the lowest priority and C7 (C,H,H) has the next lowest priority. Does C2 or C6 have the top priority? The distinction is made in sphere 2. C2 is bound to C3 , (C3) and C10 while C6 is bound to C5, C8 and C9. Only (C3) is monodentate and bound to 3 atoms with at. no. = 0. Therefore, given that O>C[H]>C(0)>H, the priority order for C1 is C6>C2>C7>H. The stereocenter C1 is of the (S)-configuration.

The configuration of C5 is detailed in digraph 5c. In sphere 2, C1, C8 and C9 are bound to C6, which takes top priority. The distinction between the C4 and C7 pathways is resolved in sphere 3. The C7 pathway displays C4, C6, and H [C,C,H] while the C4 route reveals C2, (C2), and H [C,(C),H] in sphere 3. The C7 path takes precedent over its C4 counterpart. The C5 priorities are C6>C7>C4>H. The stereocenter C5 is of the (S)-configuration.

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(1S,5S)-α-Pinene




(+)-(1R,4R)-Camphor

The assignment of configurations at C1 and C4 of camphor is easier that the case of α-pinene. Examination of structure 6 or digraph 6a reveals for C1 that the C10-methyl group has the lowest priority [H,H,H] while C2 has the highest priority [O,O,C]. There is no tie to break between C7 [C,C,C] and C6[C,H,H]. The priorities for C1 are C2>C7>C6>H; C1 has the (R)-configuration. This result is borne out in a more formal way in digraph 6b.

As with C1, C4 has no ties to break. In digraph 6c, sphere 1{C7 [C,C,C]}>sphere 2{C2[O,[O],C]}>sphere 3{C1[C,C,C]}>H. The priorities for C4 are C7>C3>C5>H; C4 has the (R)-configuration. For both C1 and C4 it is the spheres alone that determine priorities since there are no ties to be resolved.

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(+)-(1S,5R)-Myrtenal

Enantiomer 7 of myrtenal is a natural product and the oxidation product of (1S,5S)-α-pinene 5. In digraph 7b the hydrogen is the lowest priority group; C7 [C,H,H] precedes it. C2 and C6 are the same in sphere 1 because they have the equivalent patterns [C,C,(C); 3,10,(3)] and [C,C,C; 5,8,9], respectively, in sphere 2. The decision is made in sphere 3. The C2- C10 route displays {O,[O],H} while the C6 route reveals [C,C,H] in sphere 3. A single oxygen takes priority over any carbon. The priorities for C1 are C2>C6>C7>H. The stereocenter C1 is of the (S)-configuration.

For stereocenter C5, C6 takes top priority while the distinction between C4 and C7 is resolved in favor of C7's [C,C,H] over C4's [C,(C),H] in sphere 3. Notice that these determinations are made prior to sphere 5 where the oxygens appear. The priorities for C5 are C6>C7>C4>H.The stereocenter C5 is of the (R)-configuration.

The same conclusion for C5 is reached in the circular digraph 7d. Sphere 3 has a tie between C2 and C6 for the C7 --> C1 path versus C2 and (C2) in path C4 --> C3. The tie is broken in sphere 4 where the red ovals cancel each other but the green oval supercedes the phantom atoms in the blue oval. Duplicate atoms, which are in the same nth sphere as a non-duplicated atom, are equal to one another [C = (C)]. It is only in the (n+1)th sphere that a distinction is made.

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(+)-(1S,4S,5S,6R,8S)-Sativene

(arbitrary numbering system)

Sativene is a tricyclic sesquiterpene that has five stereocenters. Four of the assignments are straightforward, solvable at the first or second sphere level. However,the C5 stereocenter solution is a bit more complex. In partial digraph 8c the three carbons, C4, C6 and C8, contiguous with C5 are equivalent since they each protrude into sphere 2 as C4[3,11,H], C6[1,9,H] and C8[7,10,H], all of which satisfy [C,C,H]. These data are listed in the chart below. The blue pattern in the chart for sphere 3 assigns C4 to a priority just above the hydrogen in sphere 1. The pattern [14,15,H;C,C,H] in the path from C4 ranks lower than those from C6[2,7,12;C,C,C] and C8[1,13,(13);C,C,(C)], both of which are equal to each other. Sphere 4 resolves this tie. While C6[8,13,(13)] = C8[2,6,12] and C6[3,H,H] = C8[(7),H,H], C6[H,H,H] prevails over C8[0,0,0]. Thus the order of priorities is C6>C8>C4>H. The stereocenter C5 is of the (S)-configuration.

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(1S,5S)-5-Methylbicyclo[3.1.0]hex-2-ene

There are two stereocenters, C1 and C5, in the small, bicyclic alkene 9. Digraph 9b has hydrogen as the lowest priority group in sphere 1. Sphere 2 distinguishes among the remaining three groups. The 1--->5 path has [7,6,4;C,C,C] while the 1--->2 route shows [3,(3),H;C,(C),H and the 1--->6 path displays [5,H,H;C,H,H]. The order of priorities at C1 is C5>C2>C6>H. The stereocenter C1 is of the (S)-configuration.

The sphere 1 C7 methyl has the lowest priority for C5 in digraph 9c. The 5--->1 path in sphere 2 reveals [6,2,H;C,C,H]; route 5--->4 has [3,H,H;C,H,H]; and 5--->6 displays [1,H,H;C,H,H]. The tie between the last two paths is broken in sphere 3 as the 5--->6 route has [7,2,(5);C,C,(C)] priority over 5--->4 that has only [2,(2),H;C,(C),H]. The order of priorities at C5 is C1>C6>C4>C7. The stereocenter C5 is of the (S)-configuration.



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(1S,3s)-3-((R)-2-Hydroxy-3-((1r,3S)-3-hydroxycyclobutyl)propyl)cyclobutan-1-ol

(Note: The numbering scheme of 10 is arbitrary. Digraphs 10a and 10b are new on this page. For help, go here or here)

Carbons C3, C7 and C10 are stereogenic, chirotopic centers (R/S) and C1 and C5 are achirotopic (pseudoasymmetric), stereogenic (r/s) carbons. The descriptor is to be determined in digraph 10a. The rings are deconstructed such that the blue atoms are non-duplicate atoms and the red atoms are duplicated carbon atoms of atomic number six and terminated by three phantom atoms of atomic number zero. In the temporary assignments of Ro/So at C5, the longest path to the duplicate atom on the left side of the digraph will have the top priority. The assignments to the four C10 precedes So, pseudoasymmetric Carbons C1 are carbons follow the same methodology. Because Ro precedes So, pseudoasymmetric carbons C1 are assigned as lower case so. In the assignment of So to C3, the longer chain leading to duplicate atoms C1 will take priority over the shorter chain leading to duplicate atom C7.

Now R/S assignments may applied to non-duplicate atom C7. The SoSo pair (C5/C3) in the upper chain takes precedent over RoSo in the lower chain (C5/C3) because SoSo (like; lk) has precedent over RoSo (unlike; ul). [CIP Rule 4b] The order of precedent is O>SoSo>RoSo>H and C7 has the S-configuration.

The assignment of the configuration to C10 requires its own digraph 10b. After assigning temporary descriptors using the methods applied in digraph 11a. C10 is also of the S-configuration: O>RoRo>SoRo>H.

Digraph 10c allows assignment of the configuration to C3. Because of the symmetrical nature of the digraph, the pseudoasymmetric carbons come into play. Since ro>so [CIP Rule 4c], C3 has the R-configuration.

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(1R,1'S,3r,3'r)-3,3'-(2-hydroxypropane-1,3-diyl)bis(cyclobutan-1-ol)

Whereas the previous example 10 contains a cis and trans 1,3-disubstituted cyclobutane ring, stereoisomer 11 has two trans 1,3-disubstituted cyclobutane rings. This change renders structure 11 achiral with C3 both achiral and non-stereogenic. The assignment of the "r" configuration seems intuitively reasonable for an achiral compound but yet C7 and C10 have the S- and R-configuration, respectively, even though both rings are identical and they are not mirror images. However, using the guidelines of the 2013 IUPAC Blue Book CIP Rules, these assignments apply.

The R-configuration for C10 arises from digraph 11a. [For the basics of these digraphs go here.] In deconstructing the rings the red dots are duplicate atoms of the proximate blue dot atoms. Duplicate atoms in the nth sphere terminate in the (n+1)th sphere with three atoms of atomic number zero. Therefore, a duplicate atom will always be of lower priority than a "real" atom in the same sphere.

For example, the assignment of Ro to C1 (red circle) in the upper chain of digraph 11a is an issue of how far removed are the duplicate atoms from C1. Apart from the hydrogen, there are three methylene groups in the first sphere. The most remote duplicate atom from C1 is the duplicate atom C4 (red arrow, sphere 8). It's chain receives top priority. The next highest priority chain terminates at the duplicate C10 in the lower chain (green arrow, sphere 6). The remaining chain, just below C1, is duplicate atom C10 (blue arrow, sphere 2). Applying the right hand rule to the C1 atom in question, temporary assignment Ro is made.

The remaining centers in digraph 11a are assigned in a similar fashion. Note that C3 is assigned temporary descriptor Ro yet, ultimately, it will not receive a descriptor. [Not surprisingly C3 in digraph 11b is So.] Pseudoasymmetric center C5 is given the ro descriptor [CIP Rule 4c] because C4-chain> Ro>So>H.

C10 (digraph 11a) has the R-configuration based on CIP Rule 4b because of the priorities O>RoRo>SoRo>H. Similarly, C7 (digraph 11b) has the S-configuration with priorities O>SoSo>RoSo>H.

Digraph 11c illustrates that C3 is both non-stereogenic and achirotopic. Both chains have identical temporary descriptors.

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((1s,3s)-3-Methylcyclobutyl)((1'r,3'R)-3-methylcyclobutyl)((1"r,3"S)-3-methylcyclobutyl)methane

(Note: The numbering scheme of 12 in the right panel (left side) is arbitrary. Digraphs 12a-d are new on this page. For help, go here or here.)

Achiral alkane 12 bears three cyclobutane rings one of which is cis-substituted and two that are trans-substituted. The structure has a plane of symmetry that passes through C1, C3 and C13 rendering these atoms stereogenic but achirotopic.

Digraph 12a permits an assignment to C13. The priorities at C13 are Ro>So>CH3>H and are determined to be of the s-configuration according to CIP Rule 5. This rule postcedes Rule 4c, which states r>s (ro>so). In this example, both pseudoasymmetric carbons, C2 and C4, are designated as ro. Thus, no designation can be made by Rule 4c and default to Rule 5 is required.

Digraph 12b determines the configuration of C7. In this example, The assignment ro at C2 has precedent over so at C3 (Rule 4c), bypassing the designations at C4 (Rule 5). It is the order of the CIP Rules that are determinant and not that the Ro/So descriptors at C4 are proximate to C7. C7 has the S-configuration.

In interpreting digraph 12c, the top priority chain is the one containing C2 (Ro) and C4 (ro). But it is the latter designation that prevails because it invokes Rule 4c, while the former arises from subsequent Rule 5. Consequently, C10 has the R-configuration.

The configuration of the three pseudoasymmetric carbons C2, C3 and C4 are determined in digraph 12d by Rule 5.

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(R,z)-4-Ethylidenecyclohexane-1-carboxylic acid 13

and

(S,e)-4-Ethylidenecyclohexane-1-carboxylic acid 14

Carboxylic acids 13 and 14 are enantiomers and each is chiral. This fact may be better appreciated if one imagines that the cyclohexane ring is a double bond, in which case, one has a chiral allene.The double bond is enantiomorphic, which is reflected in the lower case z/e designation. The method for assigning configurations and double bond geometries for enantiomer 13 are illustrated in digraphs 13a and 13b. In 13a, the double bond is partitioned into its Z and E components. Because Z>E, the priorities are CO2H>Z>E>H. C1 has the R-configuration.

The geometry of the double bond is illustrated in digraph 13b. Here the carboxyl group is partitioned with the duplicate (C4) groups (red dots). Temporary descriptors Ro and So are assigned on the basis that CO2H>C4>(C4)>H. One end of the double bond has CH3>H while the other end has Ro>So. Therefore, the double bond is of the z-configuration.

The same methodology can be applied to the enantiomer 14 using digraphs 14a and 14b.

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(1E,4E)-1,4-Diethylidenecyclohexane 15

and

(1Z,4Z)-1,4-Diethylidenecyclohexane 16

One might instinctively assign dienes 15 and 16 as having the Z- and E- configuration, respectively, from the position of the methyl groups on each double bond. While each diene would have the correct configuration, the reasoning would be wrong.

Whatever the configuration of one of the double bonds in either diene is, each of them will have the same configuration as the other one owing to the 2-fold axis of symmetry in 15 and the plane of symmetry in 16.

The digraphs 15a and 16a clarify the respective assignments. Because the priorities about the double bonds are CH3>H and Z>E, diene 15 has the the E-configuration and diene 16 has the Z-configuration.

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(1R,2S,z)-4-Ethylidene-1,2-dimethylcyclopentane 17

and

(1R,2S,e)-4-Ethylidene-1,2-dimethylcyclopentane 18

The cyclopentanes 17a and 18a are enantiomers and consequently have a double bond that is enantiomorphic and utilize the descriptors e/z (lower case). While the IUPAC Blue Book 2013 does not assign descriptors to such a double bond, the assignment allows one to know which enantiomer is which. Because R>S (Rule 5), 17a is assigned the z-configuration shown in red. Enantiomer 18a follows the same logic. Structures 17b and 18b are digraphic representations of 17a and 18a, respectively.

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(1R,2S,4R)-2--Bromo-2-methylbicyclo[2.2.2]octane


2-Bromo-2-methylbicyclo[2.2.2]octane

Bromide 20a has the (S)-configuration at C2 because Br>C1>C3>C9 (CIP rule 1). Structure 20b is structure 20a from a different perspective. To determine the configurations of the bridgehead carbons, C1 and C4, a digraph is the method of choice. In the "Digraph of C1" one starts at the non-duplicate atom C1 and traces the six possible paths back to duplicate atoms C(1) that are each attached to three phantom atoms of atomic number zero. The vertical chain has the top priority while the hydrogen (not shown) at C1 has the lowest priority. The two horizontal chains are mirror images and they must be assigned to the second and third priorities. In these chains C2 has been predetermined as having the (S)-configuration. Locate C4 in the left chain. It has the (R)-configuration because C3>C8>C5. Locate these atoms in structure 20b and convince yourself. [Use your hands. Point your right thumb from C4 to H and your fingers will point from C3 to C8 to C5]. In the right hand chain C4 has the (S)-configuration because C3>C5>C8. Check these priorities in structure 20b.
Now the left hand chain is designated as RS and the right hand chain as SS. Since RR/SS>RS/SR (CIP rule 4b), the order of carbon atoms attached to non-duplicate C1 is C2>C6>C7. Locate these atoms in structure 20b and convince yourself that C1 does indeed have the (R)-configuration! Now that you are versed in the method, the priorities around non-duplicate C4 is C3>C8>C5 because C3-chain>SS>RS>H. C4 also is of the (R)-configuration!
Bromide 21a is the enantiomer of bromide 20a. Determine the configurations for yourself.

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(1S,2R,4S)-2--Bromo-2-methylbicyclo[2.2.2]octane