6
$\begingroup$

I'm having a hard time naming these molecule myself.

For the first molecule I start with one of the stereocenters, give priority "4" hydrogen and "3" to the attached CH3. How are the other two attached nodes given priority "1" and "2" in an anticlockwise fashion, like in the graph below?
For the second one I have the analogous problem, who is "1" and "2" in each stereocenter?

enter image description here

an

$\endgroup$
3

1 Answer 1

4
$\begingroup$

The Cahn-Ingold-Prelog (CIP) assignments of the configurations of the two tetramethylcyclobutanes above are accomplished as follows:

1) Tetramethylcyclobutane 1: This isomer has all of the methyl groups on the same face of the ring. There are two mirror planes passing through C1-C3 and C2-C4. Numbering the cis-isomer is arbitrary in that there is a 4-fold axis of symmetry present. The common methods for R/S-configurations do not apply. Because the ring carbons lie in symmetry planes, they are stereogenic and achirotopic requiring r/s-assignments. When an assignment is made to one carbon, the same assignment may be applied to the remaining three.

Temporary assignments for 1a are made to C2, C3 and C4 (r0/s0). Priorities are assigned at each of these carbons such that C1 is the highest priority group. Returning to C1 the priorities are r0>s0>CH3>H. C1 is of the s-configuration. Digraph 1b illustrates the assignment to C1 in a different way. The horizontal line represents the plane of the ring with C2, C3 and C4 above the plane of the ring and the dotted line (C1) signifies the carbon under assignment. Since r0 precedes s0, C4 precedes C2. C1 is of the s-configuration.

2) Tetramethylcyclobutane 2: This stereoisomer has ring carbons that are both stereogenic and chirotopic. Here R/S assignments apply. There is a plane of symmetry passing through the C1-4 and C2-3 bonds while a two-fold axis of rotation exists between the C1-2 and C3-4 bonds. Applying the methodology of 1a to 2a, C1 is assigned the R-configuration. Referring to digraph 2b, R0R0 precedes R0S0 [like(lk)>unlike(ul)]. Thus, C2 precedes C4. C1 is of the R-configuration.

One may apply this method to the remaining three ring carbons but there is a shortcut. Because C1 mirrors C4, C4 is of the S-configuration. The rotational axis shows that both C1 and C2 are of the R-configuration while C3 and C4 are both of the S-configuration. The numbering system requires that the lowest numbered atoms, if possible, be labeled R.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.