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CanaDAM 2013 Memorial University St, John’s Newfoundland

CanaDAM 2013 Memorial University St, John’s Newfoundland. Just how are the Clar and Fries structures of a fullerene related ? Jack Graver Syracuse University This is joint work with Elizabeth Hartung.

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CanaDAM 2013 Memorial University St, John’s Newfoundland

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  1. CanaDAM 2013Memorial UniversitySt, John’s Newfoundland Just how are the Clar and Fries structures of a fullerene related? Jack Graver Syracuse University This is joint work with Elizabeth Hartung

  2. Suppose that we have a fullerene with Clar structure (C,A) and that the Clar chains are non-interfering. Then there are just 3 types: • Type 1 Type 2 Type 3

  3. Given the Clarstructure (C,A), we have the formula: |C|= |V|/6 -|A|/3 • Given any Kekule structure K, we let B3 denote the number of benzene rings, A1 denote the set of edges of K that belong to exactly 1 benzene ring and A0the set of edges of K that belong to no benzene ring. • For such a Kekule structure K, we then have the analogous formula: |B3|= |V|/3 – (2|A0|+|A1|)/3 Proof: Double counting the edges of K, we have 3|B3|+|A1|+2|A0|=2|K|=|V|

  4. In |C|= |V|/6 -|A|/3,we identify |A|as the Clar deficit and to compute the Clar number of the fullerene, we seek to minimize |A|over all Clarstructures for the fullerene. In |B3|= |V|/3 – (2|A0|+|A1|)/3, we identify 2|A0|+|A1| as the Fries deficit and to compute the Fries number of the fullerene, we seek to minimize 2|A0|+|A1| over all Kekulestructures for the fullerene.

  5. Given the Clar structure (C,A), we have the induced improper face 3-coloring with the Clar faces forming one color class – the yellow faces in this case. • To obtain a Kekule structure for this Clar structure, we must select an orientation of the double bonds for each face in C. • We have just two choices for each Clar face: • We denote them by RV for “red void” and BV for “blue void”

  6. The interface between the RV and BV Clar faces may be identified with a degree 2 subgraph of the dual graph. • Hence, this interface corresponds to a collection of disjoint circuits in the dual graph. • Each dual-vertex/face in these circuits contributes 1 or 2 to the Fries deficit. • However, these contribution can be reduced by shrinking these circuits until they surround individual chains • These contributions can then incorporated into the Fries deficits for each individual chain.

  7. We conclude that associated with each Clar structure there are two natural Kekule structures that give a minimal Fries deficit. • They are obtained by choosing all Clar faces to be of type RV or all to be of type BV and make adjustments in the neighborhoods • Will we always get THE Kekule structure that gives THE Fries number from some Clar structure in this way? • The answer is yes!

  8. Let K be THE Kekule structure that gives THE Fries number. • Let C denote a maximum independent set of benzene rings for this Kekule structure. • The set C along with A, the set of edges in K that do not lie on any face in C, form a Clar structure (C,A). • It is not too difficult to see that K differs from one of the two Kelule structures associated with (C,A) only in regions around the chains.

  9. Actually, we don’t need a Clar structure to get a chain decomposition or chain pairing. • Given a fullerene and any Kekule structure K and any coupling of exiting vertices across the faces, we get a set of closed chains and six open chains pairing the pentagons. • Ignoring the closed chains, a set of open chains of shortest total length over all Kekule structures & all possible couplings is a shortest chain paring.

  10. We will assume • that we are given a fullerene with a unique shortest chain pairing and • that these chains are “widely separated.” We then construct and fix an improper face 3-coloring. We will identify each chain by its Coxeter coordinates and its color code.

  11. W w We will identify each chain by its color code and its Coxeter coordinates: this is a B-Y (6,3) chain. Note that the longer chain is a B-R chain. We will use the shorter chain color code in identifying the pairing even if the longer chain is used in the Clar or Fries decomposition.

  12. By folding we may construct equivalent chains of the same length:

  13. Given our widely separated chain decomposition with the induced face 3-coloring, we have a basic choices to make: • Which of the 3 color classes will be the void faces. • For each of these 3 options, we will compute the contribution to the Fries and Clar deficit of a B-Y (m,n) chain and illustrate it with our B-Y (6,3) example.

  14. Let the blue faces be void. The contributions to the Fries and Clar deficits are: • 0 to the Fries deficit; • m to the red Clardeficit; • m+n to the yellow Clardeficit. • For the red Clar, the white faces are yellow; for the yellow Clar, the white faces are red.

  15. Let the yellow faces be void. The contributions to the Fries and Clar deficits are: • 4m+2to the Fries deficit; • m to the red Clardeficit; • 3m+2to the blue Clardeficit. • A0 consists of the m red edges, these edges belong to both Clar chains; A1 consists of the 2m+2 light blue edges and belong to the blue Clar chain. Hence the Fries deficit is the sum of the two Clar deficits.

  16. When the red faces are void there are two different “fixes.” • The fix based on the m chain gives the correct blue Clar deficit but Fries and yellow Clar deficits that are too large. • The fix based on the m+n chain gives the correct Fries and yellow Clar deficits but a blue Clar deficit that is too large. • Hence, in this case the Fries number and the blue Clar number will not be given by the same Kekule structure.

  17. Let the red faces be void. The contributions are: • 8m+8to the Fries deficit; • (A0 - the 3m+2 red and light blue edges. A1- the 2m+4 green edges.) • 3m+2 to the blue Clardeficit (red and light blue edges); • 5m+6 to the yellow Clardeficit (red, light blue and light green edges) • Again 8m+8 = 3m+2 + 5m+6.

  18. Let the red faces be void. The contributions are: • 4(m+n)+2to the Fries deficit; (A0 – the m+n yellow edges. A1- the 2(m+n)+2 green edges.) • 3(m+n)+2 to the blue Clardeficit (yellow and light green edges); • m+nto the yellow Clardeficit (yellow edges) • Again 4(m+n)+2 = 3(m+n)+2 + m+n.

  19. Permuting the colors in this case enables us to make a table of the Fries and Clar deficits for all X-Y (m,n) chains. • This, in turn, enables us to get exact formulas for the Fries and Clar numbers for many families of fullerenes with widely separated pairing.

  20. Consider the fullerene with this structure graph: • If we take r to be a multiple of 3 and s a large enough integer not a multiple of 3, this will be a fullerene with widely separated paired pentagons – the red and purple edges.

  21. Hence, we may give formulas the the Fries and Clar numbers in terms of the variables p, rand s: • If s is congruent to 1 mod 3, the chains alternate Blue-Red, Yellow-Red, Blue-Red, … and we have: Fries # = 8r2/3 + 4s2 + 8rs + 4(p-1)(r+2s) - 2 Clar # = 4r2/3 + 2s2+ 4rs + 2(p-1)(r+2s) • If s is congruent to 2 mod 3, the chains alternate Blue-Red, Red-Blue, Blue-Red, … and we have: Fries # = 8r2/3 + 4s2 + 8rs + 4p(r+2s) - 4(r+s) - 2 Clar # = 4r2/3 + 2s2 + 4rs + 2p(r+2s) – 2(r+s) • In both cases, the same Kekule structure gives both the Fries and Clar numbers.

  22. As we saw, there are cases where different “fixes” are needed to the Fries # and Clar #:

  23. However, this slight difference over a few chains does not seem to explain all the cases in which different Kekule structures are required for the Fries and Clar numbers. • There are two other possible explanations: • Could the Fries and ClarKekule structures require different pairings of the pentagons? • Could they require entirely different Kekule structures, that is different choices for the void faces?

  24. We do not know the answer to the first question. • However, we have a construction for a family of fullerenes in which the Kekulestructure that gives the Fries number is virtually disjoint from the Kekule structure that gives the Clar number. • The plan is to construct a fullerene for which one must choose the red faces to be void for the Fries number and the blue faces to be void for the Clar number.

  25. We start with 3 widely separated chains: one Red-Blue (m,n) chain, one Yellow-Red (p,q) chain and one Blue-Yellow (u,v) chain. One may then patch these together to get a nanotube cap as pictured on the next slide with a Red-Blue (7,4) chain, a Yellow-Red (9,3) chain and a Blue-Yellow (7,1) chain. An equatorial chain is indicated in green and a half-turn about an axis through the light blue points, gives a duplicate bottom cap with blue and yellow interchanged.

  26. NOTE: to keep the model “small”, these chains are NOT widely separated. In the general case, we now have a Red-Blue & a Red-Yellow (m,n) chain, a Yellow-Red & a Blue-Red (p,q) chain, a Blue-Yellow & Yellow-Blue (u,v) chain

  27. If we choose the red faces to be void, the Fries deficit will be 8(p+u+v+1) If we choose the yellow or blue faces to be void, the Fries deficit will be 4(2m+n+p+q+u+2) The inequality 8(p+u+v+1)<4(2m+n+p+q+u+2) simplifies to p+u+2v < 2m+n+q and, if it is satisfied, the Fries set giving the Fries number requires the the red faces be void.

  28. If we choose the red faces to be void, the Blue or Yellow Clardeficit will be 2m+n+4p+4u+v+4 If we choose the yellow or blue faces to be void, the Red Clardeficit will be 6m+2p+2q+2u+4 and the Blue or Yellow Clar deficit will again be 2m+n+4p+4u+v+4 The inequality 6m+2p+2q+2u+4<2m+n+4p+4u+v+4 simplifies to 4m+2q < n+2p+2u+vand, if it is satisfied, the Clarset giving the Clarnumber requires the the blue or yellow faces be void.

  29. Hence, any fullerene in this class that satisfies p+u+2v<2m+n+q & 4m+2q<n+2p+2m+v along with the mod 3 congruence conditions will obtain its Fries number with the red face void and its Clar number with Clarred faces when either the blue or yellow faces void. These inequalities are satisfied by our example, (m,n)=(7,4), (p,q)=(9,3) and (u,v)=(7,1), and give a Fries deficit of 144 (red faces void) and a Clar deficit of 84 (Clar faces red, yellow or blue void).

  30. This model has 1728 atoms and 872 faces. • Since these chains are not widely separated we cannot be certain that this is the correct chain decomposition. Assuming that it is, we have that: • When the red faces are void, there are 528 benzene faces giving Fries # = 528 and there are 259 blue (or yellow) benzene faces. • When the blue (yellow) faces are void there are only 524 benzene faces; but, there are 260 red benzene faces and only 259 yellow (blue) benzene faces giving:Clar # = 260.

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