Benzene: E. det E E 4 E 6 0.

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1 Benzene: We will solve Schodinger equation for this molecule by considering only porbitals of six carbons under the Huckel approximation. Huckel approximation, though quite crude, provides very useful results. Since bonding in planar molecules has different symmetry from the bonding, corresponding molecular orbitals separate in the Hamiltonian do not have cross offdiagonal element and thus can be solved separately. Besides, bonding is much stronger than bonding, and molecular orbitals lie within the gap. Thus HOMO and LUMO orbitals are either of, type or nonbonding ntype. All that allows considering only p z orbitals in conjugated molecules for treating their spectroscopic features, while s and p x,p y orbitals will be responsible primarily for bonding, i.e. molecule s shape. Huckel approximation can be summarized with following statements: ) p i p j ij all overlap integrals are zero 2) p i H p i diagonal elements equal atomic energies of carbon p z electrons 3) p i H p j neighbors the interaction energy is nonzero only for neighboring carbons. Both, and, are negative and can be parameterized for a typical conjugated carbon and carboncarbon bond. Such a parameterization can be also introduced for hetero atoms, S, O, N If we write the secular equation for benzene det E E E E E E E 2 2 E 4 E 0. The solutions can be found E,E 2,E,E 2. For an arbitrary cyclic molecule with identical conjugated bonds, the solutions can be written as E 2cos 2 n k and graphically represented as enegries horizontally leveled at the corners of the appropriate polygon symmetrically placed on one of its corners: 0;

2 "Graphical solution" to cyclic aromatics E = + 2 cos(2k/n) One can see that there are degenerate solutions but symmetries of the appropriate orbitals can be found by finding appropriate eigen functions. Let s do that analysis for benzene using group symmetry: Benzene belongs to the ooint group:d h When six p orbitals are taken as the basis set for a representation of this group we obtain Γ, The character matrix is also shown on the top. C 2 s and v s were chosen to go through carbon atoms and C 2 s and d s go in between. D h E 2C 2C 3 C C C 2 i 2S 3 2S h 3 d 3 v A g A 2g B g B 2g E g E 2g A u A 2u B u B 2u E u E 2u Γ Γ 3N

3 That representation can be reduced into irreducible for this point group: Γ B 2g E g A 2u E 2u A bit faster approach is to recognize that D h D C i and thus use a smaller group, D, instead: D E 2C 2C 3 C C C 2 A A 2 B B 2 E E Γ That representation can be reduced into irreducible for this point group: Γ B 2 E A 2 E 2 which can be easily assigned g and u subscript based on the required negative character with respect to operation h in D h. Thus Γ B 2g E g A 2u E 2u We have not found the eigen functions nor energies yet, just their symmetries. Now the task is easier than before. We have two options: a)trained eye can draw the nodal planes and assign the molecular orbitals by analysis of their characters with respect to appropriate symmetry operations as shown below.i prematuraly put energies next to the MO, but note that encreaing number of the nodal planes corresponds to increasing energy of MO. Also, a 2u with energy 2 is the lowest because both, and, are negative. The approach is handy but might be confusing when trying to find those c and c 2 coefficients to finish the construction. 3

4 c + + c c c + + c c c c 2 c + c + c c c c 2 c + c 2 c 2 c c c c c + c + c c c e g a 2u b) More straight forward (also a bit more tedious) approach is by using the projection operator: P R R O R Again, we ll resort to a smaller group, D, instead of D h. Then projection operators would look shorter :0). For example: P A 2 OE O C O C O C3 O C3 O C2 O C2 O C2 2 O C2 3 O C2 O Using which on p results in P A 2 p p p 2 p p 3 p 5 p 4 p p 5 p 3 p 2 p 4 p 2p p 2 p 3 p 4 p 5 After normalizing we obtain: 4 a 2u p p 2 p 3 p 4 p 5 p. The same way it can be dobe with others. Some attention is required for doubly degenerate species. The projection operator for E 2 : P E 2 2OE O C O C O C3 O C3 2O C2.when applied to say p gives: P E 2 p 2p p 2 p p 5 p 3 2p 4.That gives only one symmetry adapted orbital, 5 2p p 2 p p 5 p 3 2p 4 Another has to be found by generating a similar one after applying P E 2 say to p2 : P E 2 p2 2p 2 p 3 p p p 4 2p 5 and making it orthogonal to 5 : 2p 2 p 3 p p p 4 2p p 2 p 3 p p p 4 2p 5 2 2p p 2 p p 5 p 3 2p 4 2p 2 p 3 p p p 4 2p 5 3 p 2 2 p 3 p 5 p Analogously:P E 2OE O C O C O C3 O C3 2O C2 and 4

5 P E p 2p p 2 p p 5 p 3 2p 4 ; 2 2p p 2 p 3 2p 4 p 5 p P E p2 2p 2 p 3 p p p 4 2p 5 and 3 2p 2 p 3 p p p 4 2p 5 3 p p 3 p p 2 3 p 2 2 p 3 p 5 p B 2g p p 2 p 3 p 4 p 5 p E g 2 2p p 2 p 3 2p 4 p 5 p 3 p 2 2 p 3 p 5 p A 2u 4 p p 2 p 3 p 4 p 5 p E 2u 5 2p p 2 p 3 2p 4 p 5 p p 2 2 p 3 p 5 p 2 2p p 2 p 3 2p 4 p 5 p So we found the eigen functions which graphically are represented below with shaded areas corresponding to a negative sign and the size of each circle resembling the value of coefficient in the eigen function. e g a 2u The energies of these MO can be found by evaluating appropriate i H i : E( ) H p p 2 p 3 p 4 p 5 p H p p 2 p 3 p 4 p 5 p 5

6 p H p p 2 p 3 p 4 p p 2 H p p 2 p 3 p 4 p p 3 H p p 2 p 3 p 4 p 5 p p 4 H p p 2 p 3 p 4 p 5 p p 5 H p p 2 p 3 p 4 p 5 p p H p p 2 p 3 p 4 p 5 p matrix element for highlighted orbitals are zero p H p p 2 p p 2 H p p 2 p 3 p 3 H p 2 p 3 p 4 p 4 H p 3 p 4 p 5 H p 4 p 5 p p H p p 5 p 2 2 Energies of other orbitals are calculated the same way: E p p 2 p 3 p 4 p 5 p Ee g 2 2p p 2 p 3 2p 4 p 5 p 3 p 2 2 p 3 p 5 p Ea 2u 4 p p 2 p 3 p 4 p 5 p E 5 2p p 2 p 3 2p 4 p 5 p p 2 2 p 3 p 5 p (p p 2 p 3 p 4 p ) () 0.289(2p p 2 p 3 p 4 p ); (2) 0.5( p 2 p 3 p ) e g e g () 0.289(2p p 2 p 3 p 4 p ); e g (2) 0.5( p 2 p 3 p ) a u 2a u 0.408(p p 2 p 3 p 4 p ) We see that MO energies are symmetric with respect to value of. All planar conjugated hydrocarbons with such property are called alternant. Alternant hydrocarbons can be recognized by their ability to have all carbons labeled intwo colors alternatingly (thus the name), i.e. when each neighbor has a different color. The ground state configuration is a 2u 2 e g 4 X A g. Note that this ground state energy is lower than that of cyclohexatriene (benzene with three localized double bonds). The effect of sharing electrons, the socalled delocalization or resonance energy, equals 2. The lowest excited state configuration is a 2u 2 e g 3 (with energy 2 above the

7 ground state) from which following states can be constructed: e g B u B 2u E u.each of which can be either a singlet or a triplet, i.e.: e g a u e g a u e g a u X: A g B 2u, B u, E u 3 B 2u, 3 B u, 3 E u At this level of comlexity we cannot choose which excited state is the lowest energy, B u, B 2u, or E u (all we know is that they each have corresponding triplet of lower energy 3 B u B u, 3 B 2u B 2u, and 3 E u E u ). We have to analyse other higher energy oneelectron excited state configurations and find if there are any of the same symmetry as the three under consideration. Configurational interaction between configurations of the same symmetry lowers the appropriate lower energy state while increasing the corresponding high energy state energy. The next one electron excited state configurations (with energy 3) are:a 2u 2 e g 3 and a 2u e g 4. Neither of them has a proper symmetry. The former makes: e g e 2g, while the latter: a 2u e 2g E 2g E 2g B 2u We shall include even higher one electron excited state configuration with energy 4. There is only one, a 2u e g 4, which produces a 2u b 2u. Thus, out of the three states, B u, B 2u, and E u, only B 2u has two configurations. Consequently, it will be the lowest energy state. 7

8 SinceintheD h group the dipole moment components behave like E u (x,y) and A 2u (z), the only dipole allowed transition is C E u X A g which indeed shows a high extinction coefficient, while the lower energy transitions, B u and B 2u,are electronically forbidden. They are observed anyway but due to so call vibronic coupling. Before that let s consider electric dipole allowed transitions first. The matrix element for the transition moment is: M e v e"v" e v e v e e v v M e e" v v v2 v2... The selection rules in this case are defined by the selection rules for electronic part M e e" and the FranckCondon factor, v v 2, for vibrational part. The latter would be nonzero if that integrand has the totally symmetric species, which means that for totally symmetric vibrations all vibrations are allowed: Δv 0,,2,3,... For all other vibrations the totally symmetric integrand appears only for even difference in vibrational number, i.e. when: 8

9 Δv 0,2,4,... In cases like the lowest energy transition in benzene, which is electronically forbidden, the only way the matrix element for the transition moment would be nonzero is tostep back from the BornOppenheimer approximation and consider mixing nuclear (vibrational) and electronic coordinates, so called vibronic coupling. The electronic Hamiltonian parametrically depends on on vibrational coordinates: H e H 0 H e e i Q i Q i.. H 0 e H The excited state wavefunction 0 f becomes accordinly mixed with other zeroorder electronic states : e 0 0 f k c k k where the degree of mixing, c k k 0 H 0 f E 0 f E0, depends on both, the vibronic coupling k element 0 k H 0 f and the energy separation between the states E 0 f E 0 k.the electronic transition moment becomes: M e e" e e 0 0 f e k c k 0 0 k e While the first term (00 transition) is zero, the other terms might be not. Then the intensity of transition would become nonzero, or as it is often referred to as intensity is borrowed from a neigboring transition. That is why B u transition, beeing closer to the allowed E u, is more intense than B 2u. The energy factor E f 0 E k 0 is not the only one responsible for nonzero c k, the other part, k 0 H f 0, enforces additional selection rules: k 0 H f 0 H e Q i k 0 Q i 0 f Qi 0 or Δv 3,5,... for vibronicaly active mode, the mode for which the symmetry of Γ k 0 Γ Q i matches that of the ground state X. The rules for other vibrational states would follow the rules of ordinary FranckCondon factor described above. In case of benzene, vibronically allowed transition should be a 2u and e u.forthea B 2u state it translates to a reqire for either b g b g B 2u B 2u or e 2g e 2g B 2u E u. There are 2330 vibrational modes in benzene, which for the ground state would be realized in 20 fundamental frequencies: 2a g a 2g a 2u 2b u 2 2b 2u e g 3e u 4e 2g 2. From these modes only e 2g satisfy the selection rules ( 5 e 2g through 8 e 2g ) for vibronic transition. By symmetry all four of them can contribute and those that have the highest value of H e Q i contribute the most. CH vibrations do not affect the energies of electrons very much, it that should be a mode that shakes the molecular sceleton associated with system. A likely candidate is CCC in plane bending, 8 e 2g, with energy ca cm (0 cm in the ground state and 522 cm in the electronically first excited state). Thus the excited vibronic transitions that should be observed are: A B 2u 8 0 transition from vibrational ground state in the X A g to A B 2u at vibrationally excited state with " at 8 e 2g. Another possibility is to observe a hot A B 2u 8 0 transition from the vibrationally excited with at 8 e 2g X A g to the A B 2u and "0 at 8 e 2g. The latter transition is very temperature dependent (hot). What about combinational transitions? Again, the other vibrations have to be involved in ring distortion vibrations (preferably CC stretching since it affects the Hamiltonian the most) and posses appropriate symmetry, which 9

10 can be achived by combining 8 e 2g with totally symmetric combinations of the ring distortion vibrations such as 2 a g (992 cm in the ground state and 923 cm in the electronically first excited state). As a result, a series of vibronic transitions A B2u 2 0 m 8 0 appears as a famous benzene spectrum near 20 nm. 0