Cahn-Ingold-Prelog Rules: Spiro Compounds

How to Manipulate JSmol Structures

[Powered by Jmol 14.24.1]

Index:



Fig. 1

1,7-Dioxaspiro[5.5]undecane

An asymmetric carbon of the type C(a,b,c,d), wherein the priorities are a>b>c>d, is readily assigned an R/S configuration by Rule 1 of the Cahn-Ingold-Prelog (CIP) protocol. The spiro ketals, Figs. 1 and 2, are chiral, enantiomers of one another. [Rotate the structures to align them as mirror images.] These compounds have only two identical groups proximate to the spiro carbon: oxygen and CH2. Each ring bears an O and CH2 with the O having higher priority over the CH2. The priority order is a>a'>b>b'. It is irrelevant as to which ring is a,b and which one is a',b'. Fig.1 is of the R-configuration. Using your right hand, point your thumb toward b' along the C-b' axis. Your fingers will point from a > a' > b. Fig. 2 will prove to be of the S-configuration. Since a' (O) and b' (CH2) have been switched relative to Fig. 1, the configuration must change.


Fig. 2


Fig. 3

1,6-Dioxaspiro[4.5]decane

Fig. 3 has the same spatial arrangement as Fig. 1 except that the left hand ring has been reduced to 5-membered. The same is true of the relationship between Figs. 2 and 4. Reducing the ring size changes the configuration without altering the bonds attached to the spiro carbon! These spiro ketals are no longer of the a, a', b, b' type but rather an a, b, c, d type. Structure 3b bearing black dots is adigraph that makes it easier to evaluate the priorities of the groups. By successively, and mentally, cutting each of the bonds attached to the spiro carbon in 3a, stretching them out and capping the ends with black dots (phantom spiro carbons), one can easily assign priorities to each of the chains. In digraph 3a the two chains labeled 1 and 2 are the top priorities (O > CH2) but which one is which? In chain 1 there is an extra CH2 group (the last one before the phantom group; red arrow) while chain 2 has one fewer CH2 groups, i. e., CH2 > phantom group. Chain 3 precedes chain 4 because, counting atoms outward from the spiro carbon, the fourth atom is oxygen (blue arrow) while chain 4 has a CH2 group as its fourth atom. That is, O > CH2. The spiro carbon in Fig. 3 is of the S-configuration. In Fig. 4 only the priority 1 and 4 switch positions. The spiro carbon in Fig. 4 is of the R-configuration.

Fig. 4



Fig. 5

2,6-Dichlorospiro[3.3]heptane

The 2,6-dichlorospiro[3.3]heptane in Fig. 5 has axial chirality like an allene. [Click on "show carbon #". Orient the structure as in 5a so that C2 is on the right,] The axis passes through C2, C4 and C6. The chirality of the molecule may be designated as 2M but IUPAC rules prefer an R/S assignment. A 3b type digraph may be employed wherein C2 or C6 are each displayed as in the left side of digraph 5b. This procedure may condensed to the analysis in 5a. Temporary assignments (R0/S0) are made at C2 and C6. With hydrogen as the lowest priority group and chlorine the highest. Pointing the thumb of your right hand to hydrogen, your fingers pass chlorine to C3, which received the designation R0. Likewise, your left hand assigns C1 as S0. The same procedure may be applied to C6. [Note that C2 and C6 are equivalent positions in that rotation of the structure 180o about an axis (z-axis) passing through C4 and perpendicular to the plane of the screen followed by a 90o rotation toward -z, reproduces the same structure.] Because R>S (R0>S0) and given the symmetry of the molecule, either R0 (C3 or C5) may be chosen as first and second priorities with the proviso that the third and fourth priorities be paired odd and even in separate rings. In 5a one has C5(1), C3(2),C7(3) and C1(4). The spiro carbon (C4) is of the R-configuration. Click on "show R/S" to confirm.

Click "show carbon #". Digraph 5b illustrates and configuration of C2. Carbon 4 is labelled r0 and s0, each one bears enantiomeric groups. The red dots are phantom groups of C2; the blue dots of C4. The r0 assignment at C4 follows the priorities C1>C6(R0)>S0(C6)>C3(phantom). Similarly, C4 may be labelled s0 in the lower portion of 5b. Because r0>s0, C2, as is C6, is of the S-configuration.

Fig. 6 is the enantiomer of Fig. 5. Compare the two enantiomers.

Fig. 6


Fig. 7

7-Methylspiro[3.5]nonan-2-ol

Figs. 7 and 8 are enantiomeric analogs of Figs. 5 and 6 with two differences. The substituents are not identical, which is not and issue but the rings are of different sizes, a situation that causes some changes in priorities. Click on "show R/S". Note that the spiro carbon is of the S-configuration while the spiro carbon in Fig. 5 has an R-configuration. Take note that the R0/S0 assignments are the same in both as are the configurations at the carbons bearing substituents. Why the difference? Ring sizes make a difference! Notice that the priorities are different in 7a compared with 5a. Starting at the spiro carbon C4 and counting out two ring atoms in either ring, the 4-membered ring reaches C2 bearing a hydroxyl group, while the 6-membered ring has a methylene group at either C6 or C8 (Click on "show carbon #"; Rule 1). The cyclobutane ring has the top two priorities (R0 > S0; Rule 5) and the 6-membered ring has the two lowest priorities (R0 > S0). The spiro carbon is of the S-configuration with priorities C1 > C3 > C5 > C9.

In digraph 7b the chains bearing the red dot are the lowest priority with the remaining three ligands of C4 are R0 > S0 > -CH2CH2-. Because C4 bears two enantiotopic groups, it bears the r0/s0 designation. The configuration of C7 is S; r0 > s0 > -CH3, H.

Digraph 7c at C4 has the following priorities: -CH4CH(OH)-, R0 > S0 > -CH2-red dot. The priorities at C2 are: OH > r0 > s0 > H. Carbon C2 has the S-configuration.

Convince yourself that the enantiomer represented by Fig. 8 has the R, R, R-configuration.

Fig. 8


Fig. 9

1,5,8,11-Tetraoxatetraspiro[2.0.24.0.27.0.210.03]dodecane

Figure 9 has three epoxide rings with the oxygens on the same side of the cyclobutane ring at C3, C4 and C10. The fourth epoxide ring at C4 has the methylene group on the same face of the cyclobutane ring as the oxygens in the other three epoxides. This compound is achiral with a plane of symmetry passing through C4 and C10. This plane will cause the configuration at C3 to be the opposite of C7. The compound is a cyclobutane analog of an inositol, which has six oxygens attached to each carbon atom of a cyclohexane. The assignment of configurations to all the inositols has been described in detail. Nonetheless, the basics will be reviewed here. Each stereocenter bears an oxygen atom and methylene group in the epoxide ring. The oxygen has the highest and the methylene the lowest priority. But how does one determine the second and third priorities? The digraphs for the four centers are on the right. Consider the top digraph. The carbon under consideration is C3 (in red). The cyclobutane carbons to the left of C3 are successively C10, C7 and C4 while to the right they are C4, C7 and C10. The vertical lines below the atom numbers indicate the location of the oxygen atoms, either above or below the plane of the ring, which is the horizontal line. To assign temporary configurations for C3, one moves successively to the left around the ring from C10 --> C7 --> C4 and, to the right, from C4 --> C7 --> C10. At each carbon the configuration is determined assuming the target atom, in this example C3, has the second priority. The descriptors for the atoms on the left of the digraph mirror their partners on the right side. But the descriptors themselves do not mirror one another! Assignments are made by 1,2 comparisons to find like pairs (Rule 4). If none is found then a 1,3-comparison and, if needed, 2,3-comparison. S0S0 is a like (lk) pair while S0R0 is unlike (ul). Now lk > ul. Therefore for C3, C10 has second priority while C4 has third priority. Carbon 3 has an R-configuration.

The configuration of C7 can be solved using the digraph but, as mentioned above, C3 and C7 are mirror images of one another. Hence, C7 is S.

The digraphs for stereotopic, achirotopic C4 and C10 display the symmetry in both atoms and descriptors. There is no possibility of a lk/ul comparison. Default is made to Rule 4, r > s (r0 > s0).