Calculation of the Number of cis-trans Isomers in a "Symmetric" Polyene Calculation of a formula for the number of cis-trans isomers in polyenes offers students a valuable exercise in dealing with stereochemistry and simple combinatorial analysis. Students readily understand tbat the number of isomers in an "asymmetric" hydrocarbon with n double bonds is 2", and they can be convinced that a "symmetric" polyene with n double bonds has fewer than 2" isomers, since some possible combinations are identical. (In this context, a "symmetric" polyene is one tbat, with respect to its single and double bonds, reads the same from left to right as it dces from right to left.) Thus, 2.4-hexadiene, a "symmetric" polyene, has only 3 geometric isomers: cis-cis, trans-trans, and cis-trans (trans-cis is identical to cis-trans.) We propose as an exercise for students the following problem. Correct analysis requires careful consideration of symmetry and combinatorial arguments. Problem: Derive a general expression far the number of cis-trans isomers in a "symmetric" straight chain polyene when there are n double bonds. (1) Let n he even (2) Let n be odd. Answer: (1) (2" + 2"12)/2 (2) [2" 2'" + " 9 / 2 Proof: Visualize each isomer as a one-dimensional array of single (S), cis-double (C), and trans-double (T) bonds. Define a balanced "symmetric" polyene as one that, with respect to its cis-trans bonds, reads the same from left to right as it does from right to left. For example, the balanced "symmetric" palyene
+
H
H
H
H
I l l
C-c.-c=c-C=C-C=C-c=C-C-C
H
H
I l l
I HI
H
is represented by SSCSTSTSCSS. An example of an unbalanced "symmetric" isomer is SSTSTSTSCSS. (1) Even number of double bonds. There are clearly 2"12 balanced configurations in all. The problem reduces to calculation of the number of unbalanced configurations. (a) There are exactly 2"-' different configurations among the first n - 1 bonds, excluding the right-most double bond. (b) There are exactly 2'"-2u2 balaneed configurations within the inner n - 2 double bonds. (c) The 2'"-2"2 inner balanced configurations yield 2(21"-2''2) = 2"12 halanced configurations because the outermost bonds may be either cis or trans. However, they yield only 2'"-2"2 = ( 2 " 9 / 2 unbalanced configurations since trans-balanced-cis is identical to eis-balaneed-trans. (d) When the inner n - 2 double bonds are not balanced, the compound must be unbalanced. Clearly, of the Zn-' configurations in the first n - 1 bonds, exactly 2(21"-2"2) are balanced after the first bond. Thus, (a) and ( c ) imply that of the 2"-' configurations in all, exactly 2"12 are balanced. (e) The total number of unbalanced configurations is then
+ 2"1 - 2"" = (2" - 2"'2)/2
2"'"'
(f) The total number of compounds is therefore 2"'?
+
(2"
-
2"'2)/2
= (2"
+ 2"'2)/2
(2) Odd number of double bonds. One can consider a compound with an odd number n of double bonds to be the result of inserting a double bond in the center of a compound with n - 1 double bonds. Since a cis or a trans in the center will yield different compounds, the number of configurations must be equal to twice the number of configurations in a hydrocarbon with n - 1double bonds. Thus, the answer is 2q.+1112
+
(2.
-
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