Metaldehyde MA-1:

Metaldehyde (MA; 2,4,6,8-tetramethyl-1,3,5,7-tetraoxocane) is the tetrameric acetal of acetaldehyde. Four stereoisomers of MA are possible: MA-1, MA-2, MA-3 and MA-4. MA-1, which has the four methyl groups on one side of the ring, has planes of symmetry orthogonal to the tetraoxocane ring passing through C₁-C₃ and C₂-C₄. The two planes of symmetry through alternate oxygens is irrelevant. Carbons C_1-4 are all stereogenic but achirotopic. Therefore, the CIP designation will be r/s and not R/S.

To determine the r/s assigment for MA-1, a digraph of the structure is made as shown in Figs. 1. In Fig. 1a two pathways around the ring from C1 and back to a duplicate C1 are created. To the right one has 1-O-2-O-3-O-4-O-(1) and to the left 1-O-4-O-3-O-2-O-(1). The red dot at C1 is the carbon under analysis and it is the non-duplicate atom. The red dots at the end of the chains are duplicate C1 atoms of atomic number six and they are enclosed in parentheses. Each of these atoms is attached to three phantom atoms of atomic number zero. Atoms C2-4 are now assigned temporary chiralities, Ro/So. Consider C2 on the right side of Fig. 1a designated by the blue arrow. Clearly, hydrogen has the lowest priority and the methyl group has the next lowest priority. The longer oxygen chain has priority over the shorter one. Longer chain: O-1-O-4-O-3-O-2-O-(1) Shorter chain: O-3-O-4-O-(1) C3 in the longer chain is attached to O, C, H {8,6,1} while C(1) in the shorter chain is attached to three atoms of mass zero {0,0,0}. Therefore, the priority at C2 is O (longer chain) > O (shorter chain) > CH3 > H. Consequently, C2 at the blue arrow in the right hand chain is of the So configuration. When the same methodology is applied to C3 and C4 in the right hand chain, both of these carbons are also of the So configuration. Return to the top of the page.

Careful examination of Fig. 1a shows that the left hand and right hand chains are mirror images of one another. Therefore, C_2-4 of the left hand chain must be of the R_o configuration. You can test this claim at C₄ by the green arrow. Now the configuration at C₁ can be determined. Because the left hand chain has all R_o and the right hand chain has all S_o, one simply follows CIP Rule 5, R_o > S_o > CH₃ > H. C₁ has the s-configuration. Fig. 1b is another digraph representation. The horizontal line contains the ring atoms and the methyl groups are above this line. As noted early on in this discussion, all four ring carbons have the same s-configuration. Select the R/S;r/s button in the JSmol structure below for MA-1.

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Metaldehyde MA-2:

MA-2 has three methyl groups on one side of the ring that necessarily must be contiguous. A plane of symmetry passes through C₁-C₃ which are stereogenic and achirotopic (r/s) while C₂ and C₄ are both stereogenic and chirotopic (R/S). The configuration of C₁ in MA-2 above is easily determined without the aid of a digraph. Ignoring the oxygens and the employing the methodology from MA-1, the pathway C₂ -C₃ -C₄ is all R_o while the pathway C₄ -C₃-C₂ is all S_o. Therefore, C₁ is of the r-configuration. Verify this result with the JSmol version of MA-2.

Once you become more familiar with digraphs, you will recognize that you can take shortcuts as shown in Fig. 2a for C1 of MA-1. When the configuration of C2-4 (Ro) in the counterclockwise direction need only point back to the non-duplicate atom C1 which is the start of the longer chain. In the clockwise direction the C2-4 configurations are all So. [See the discussion here related to the inositols.] Return to the top of the page.

Now that the s-configuration of C1 has been determined for MA-2, let's deal with the other stereogenic, achirotopic atom at C3 in Fig. 2b. Once the temporary assignments have been made on one side of the plane, the other side of the plane have the opposite configurations. The assignment of the configuration at C3 will depend on CIP Rule 5 where Ro > So > CH3 > H. C3 has the s-configuration. Return to the top of the page.

Thus far the discussions have dealt with stereogenic, achirotopic carbon atoms. However, C₂ and C₄ are stereogenic and chirotopic. These atoms do not lie in a mirror plane. They will qualify for R/S designation. The temporary descriptors are assigned in the usual way. In Fig. 3b two descriptors are selected on either side of C₂. In past examples only one descriptor was utilized. The reason is that the previous examples were defaults utilizing CIP Rule 5 whereas CIP Rule 4b has precedence over Rule 5. Rule 4b states that RR/SS > RS/SR, which also applies to temporary assignments. Comparing C₁ and C₄ on both sides of C₂ shows R_oR_o> R_oS_o. Therefore, the preferences are R_oR_o> R_oS_o > CH₃ > H. C₂ has the S-configuration. If this comparison were to fail, then C₁ and C₃ are compared and, if needed, C₄ is compared with C₃. If these options all fail, then Rule 5 is invoked.

Verify this result with the JSmol version of MA-2 above. The analysis of C₄ of MA-2 is left to you.

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Metaldehyde MA-3:

MA-3 has alternating stereochemistries around the ring with two planes of symmetry through C₁-C₃ and C₂-C₄. As was the case with MA-1, solving the stereochemistry for one center applies to all four centers. Rotation in the plane of the page by 180^o or a 180^o rotation about a vertical axis followed by a 90^o in page rotation both lead to the same structure. The digraphs Fig. 4a and 4b both reveal the symmetry on both sides of the C₁-C₃ plane. CIP Rule 5 applies and C_1-4 all have the s-configuration.

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Metaldehyde MA-4:

MA-4 is the only one of the four stereoisomers that does not have any planes of symmetry passing through carbon atoms. Thus, carbons C_1-4 are stereogenic and chirotopic (R/S). The planes of symmetry through the oxygens have no bearing on the assignments. Figs. 5a/5b show that the two sides of the digraph are not mirror images which calls into play CIP Rule 4b. Because S_oS_o> S_oR_o, the priority order is S_oS_o > S_oR_o > CH₃ > H. C₁ has the S-configuration.

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