7: Hückel theory for polyatomic molecules

Transcription

1 7: ückel theory for polyatomic molecules Introduction Approximate treatment of π electron systems in organic molecules: 1 Approximations π and σ frameworks completely separated. Trial wavefunctions can be used of the form: Ψ = i c i p π i 3. ii = p i Ĥ p i = α same for all atoms 4. ij = p i Ĥ p j = β i bonded to j Note that β is negative. = 0 otherwise. 5. S ij = p i p j = δ ij, where δ ij = 1 when i = j and δ ij = 0 when i j.

2 Ethene Optimize orbital coefficients as before using the variation principle. Know already that this leads to secular equations ( α E β β α E )( c1 c ) So the energies are the solutions of the secular determinant α E β β α E = 0 E - = E + = = 0. Wavefunctions obtained by substituting the energies E ± back into secular equation: Ψ ± = 1 (p 1 ±p ) Thus, with respect to two non-bonded p π electrons, ethene has a π stabilization energy of β (i.e., β for each electron). See Sheet, question 1 for a problem involving the allyl radical.

3 Butadiene There is nothing in ückel theory that distinguishes cis from trans, or indeed linear, butadiene As before, the trial wavefunction, Ψ = c 1 p 1 +c p +c 3 p 3 +c 4 p 4, is optimized using the variation principle. The secular equation can be written down directly α E β 0 0 β α E β 0 0 β α E β 0 0 β α E c 1 c c 3 c 4 = 0. Multiplying out this leads to a quartic equation. Although this can be solved, here we will use of symmetry to simplify the algebra.

4 cont. Butadiene has a two-fold symmetry axis. Form symmetry adapted linear combinations (SALCs) of atomic orbitals χ 1 = 1 (p 1 +p 4 ) χ = 1 (p +p 3 ) χ 3 = 1 (p p 3 ) χ 4 = 1 (p 1 p 4 ) χ 1 and χ are symmetric (A), and χ 3 and χ 4 are antisymmetric (B) with respect to the C operation. Let Ψ = c 1 χ 1 +c χ +c 3 χ 3 +c 4 χ 4 ij and S ij vanish if i and j are of different symmetry: α E β 0 0 β α+β E α β E β 0 0 β α E c 1 c c 3 c 4 = 0. Matrix becomes block diagonal.

5 Look at some of the matrix elements 11 = χ 1 Ĥ χ 1 = 1 (p 1 +p 4 ) Ĥ 1 (p 1 +p 4 ) = 1 = α { p 1 Ĥ p 1 }{{} α + p 4 Ĥ p 4 }{{} α + p 1 Ĥ p 4 }{{} 0 14 = χ 1 Ĥ χ 4 = 1 (p 1 +p 4 ) Ĥ 1 (p 1 p 4 ) = 1 = 0 { p 1 Ĥ p 1 }{{} α p 4 Ĥ p 4 }{{} α p 1 Ĥ p 4 }{{} 0 33 = χ 3 Ĥ χ 3 = 1 (p p 3 ) Ĥ 1 (p p 3 ) = 1 { = α β p Ĥ p }{{} α + p 3 Ĥ p 3 }{{} α p Ĥ p 3 }{{} β Note that, as before, the overlap integrals in the SALC basis are + p 4 Ĥ p 1 }{{} 0 + p 4 Ĥ p 1 }{{} 0 p 3 Ĥ p }{{} β } } } cont. S ij = δ ij

6 cont. Now solve the upper determinant. (α E)(α+β E) β = 0 E ± = α+ 1± 5 β E 1 = α+1.6β E 3 = α 0.6β To get molecular orbital corresponding to E 1 substitute E 1 into secular equations 1.6βc 1 +βc = 0 c 1 = c 1.6

7 cont. But wavefunction needs to be normalized Ψ 1 = N [ ] 1 (p 1 +p 4 ) (p +p 3 ) Ψ 1 Ψ 1 = N [ ] = 1 So Ψ 1 = 0.37p p +0.60p p 4 Similarly, find MO Ψ 3 associated with energy level E 3 Ψ 3 = 0.60p p 0.37p p 4

8 cont. From the lower determinant one obtains E = α+0.6β E 4 = α 1.6β with the normalized wavefunctions and Ψ = 0.60p p 0.37p p 4 Ψ 4 = 0.37p p +0.60p p 4 E 4 4 E 3 3 E E 1 1

9 Charge density on atoms Once the wavefunctions of the MOs are known the charge densities on each atom, q i, can be obtained from q i = k n k (c k i ), where the sum over k is over all occupied molecular orbitals, and n k is the occupation number of MO k (either 0, 1 or for non-degenerate orbitals). (c k i ) is the square of the coefficient in the kth orbital on the ith atom. For butadiene, this reads (for atoms 1 and ) q 1 = = 1 q = = 1 and similarly for atoms 3 and 4. That the charge densities on all the atoms are equal is a general property of the ground states of alternant hydrocarbons. These also have a pairing of the MO energies E = α±xβ. See MKT for more details.

10 Resonance stabilization energy In ückel theory the resonance stabilization energy provides an indication of the increased molecular stability that arises from π bonding. For the ground electronic state of butadiene the total electronic energy is E tot = (α+1.6β) + (α+0.6β) = 4α+4.48β Thus the resonance stabilization energy is E = E tot 4α = 4.48β This stabilization can be compared with that of 4β that would be generated from two isolated π bonds (i.e. in two isolated ethene molecules). E deloc = E 4β = 0.48β. This energy provides a measure of the delocalization energy, the extra stabilization that arises from delocalization of the electrons over the π system.

11 Bond order Provides an indication of the strength of the π bond between adjacent atoms i and j ρ ij = k n k c k i ck j. The sum is over MO s k, with n k the occupation number of those orbitals. For butadiene ρ 1 = ( ) + ( ) = 0.89 ρ 3 = ( ) + ( ) = 0.45 Suggests a partial π bond between the central atoms, so the terminal bonds might be shorter than the central bond. Bond lengths in excited states E 4 Similar treatment for excited electronic state. ρ 1 = ( ) + ( ) + ( ) = 0.45 ρ 3 = ( ) + ( ) + ( ) = 0.7 E 3 E E 1 Outer bonds expand, inner bond contracts on excitation to the excited electronic state. (See Spectroscopy lectures, and the Franck-Condon principle - vibrational excitation accompanies electronic transition.)

12 ESR spectrum of the radical cation a a 1 E 4 E 3 E E 1 The strength of the hyperfine coupling to the protons is proportional to the unpaired electron density on the adjacent carbon atom. For the butadiene cation, [C 4 6 ] +, a 1 = a(c 1 ) = a 0.60 = 0.36a a = a(c ) = a 0.37 = 0.14a where a is a universal constant. The following spectrum is therefore predicted: 0.36a 0.14a electron coupling to 4 almost equivalent s electron coupling to equivalent s

13 Susceptibility to electrophilic attack Rate of reaction is proportional to the barrier height. Estimate relative barriers for different pathways. E = E σ + E π # + or E E + { Suppose that E σ is insensitive to position of attack. E E π = E π E butadiene π E butadiene π Attack at central carbon = change in resonance stabilization = 4α+4.48β reactants + E + products E π(1) α+β a bit like ethene Attack at terminal carbon E π (1) = α.48β E π() α+ β a bit like allyl cation E π () = α 1.65β Attack at terminal atom preferred. Many approximate methods for calculating the localization energy have been devised (e.g., frontier orbitals) - see recommended texts.

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17 8: Applications of ückel Theory Aromaticity π MOs of a cyclic polyene (N carbon atoms) 3 4 N= α E β β β α E β β α E β β α E β β α E β β β α E c 1 c c 3 c 4 c 5 c 6 x x x x x x c 1 c c 3 c 4 c 5 c 6 = 0, with x = (α E)/β. A general row of the secular equations gives c n 1 + xc n + c n+1 = 0 For even membered rings (N = l where l is an integer) c l+n = c n

18 cont. ence try: c k n = e ikπn/l coefficient on n th atom for k th MO where k = 0,±1,±,,+l Substitute c k n back into equation for coefficients or or e ikπ(n 1)/l +xe ikπn/l +e ikπ(n+1)/l = 0 ) (e ikπ/l +x+e ikπ/l e ikπn/l = 0 x = cos kπ l. Therefore E k = α+βcos kπ l k = 0,±1,±,+l

19 cont. ence, the resulting energy levels can be displayed diagrammatically E k = α+βcos kπ l α+βcosθ k = 0,±1,±,+l, where θ = kπ/l correspond to the angles subtended by the vertices of a regular polyhedron. E = 4α+4β two ethenes. No extra resonance stabilization due to delocalization. Note unpaired electrons and degenerate ground state. E = 6α+8β, i.e. β lower than three ethenes - extra resonance stabilization due to delocalization. E = 8α+4(1+ )β, i.e. 4( 1)β lower than four ethenes - extra resonance stabilization but with unpaired electrons and degenerate ground state.

20 For odd membered rings (N = l +1) cont. ence Therefore c k n = e ikπn/(l+1) c l+1+n = c n k = 0,±1,±,,±l E k = α+βcos πk l+1 k = 0,±1,±,±l Origin of the 4n + π-electron rule for stability of aromatic molecules.

21 Perturbation theory and steric interactions Find approximately the properties of a system from those of a system with a slightly different (simpler) amiltonian. General result - just quote First order energy E k = Ψ k Ĥ Ψ k E k change in energy of the k th orbital on perturbation Ψ k orbital k of the original (unperturbed) problem Ĥ difference between the original amiltonian and the full one (i.e. the perturbation)

22 Example 1 Find the π MO energies of having solved the butadiene problem O The only major difference is in the energy of the oxygen atom p π orbital with respect to that of carbon, i.e. p 4 Ĥ p 4 = α O < α C owever, we can estimate the energy lowering ( 0.6β) and can write α O = α C +0.6β Now use butadiene MOs in E k = Ψ k Ĥ Ψ k = i c k i p i Ĥ j c k j p j = i,j c k i ck j p i Ĥ p j (1)

23 cont. But the only matrix element which is affected is p 4 Ĥ p 4 such that All other terms in Eqn.(1) are zero. E k = (c k 4 ) p 4 Ĥ p 4 = (c k 4 ) 0.6β. (-0.37) x x 0.6 (-0.60) x x 0.6 O

24 Example In butadiene the π bond order between the central (-3) atoms is less than the terminal (1-, 3-4) bonds. Matrix element 3 is less negative than 1 and 34. Let 1 = 34 = β 3 = β +δ, where β is negative (as usual), but δ is positive. r = 1.3A r = 1.4A Find shift in the MO energies due to this effect. Perturbation to amiltonian only involves p Ĥ p 3, therefore E k = (c k c k 3 +c k 3c k ) p Ĥ p 3 = c k c k 3δ. -x0.60 x0.37 -x0.37 x0.60 Tending to the energy level pattern for two ethenes.

25 Non-crossing rule Consider mixing of two orbitals φ 1 and φ Ψ = c 1 φ 1 +c φ. The orbitals φ i could be the MOs that we have just been considering for butadiene. We know that the energies of the mixed orbitals will be the solutions of the secular equations (see Problem sheet 1) E ± = 1 { } ( AA + BB )± ( AA BB )+4AB. So E + = E only if AB = 0. In general, AB = 0 only if the states have different symmetry (otherwise in region of crossing there will always be some small term in the amiltonian which mixes the two orbitals). BB E - AB AA E + Bond length or reaction coordinate States of the same symmetry cannot cross.

26 Correlation diagrams Non-crossing rule is helpful in determining reaction pathways based on orbital symmetry arguments. Consider the ring opening of cyclobutene. B A D C? A B C D? A B D C Two categories of ring-opening. B A D C conrotatory A B C D A or B C v A B C D disrotatory A B D C + or -

27 cont. The ground electronic state of cyclobutene (σ π ) smoothly correlates with the ground state (Ψ 1 Ψ ) of butadiene in the conrotatory mode, giving for thermal reaction. B C A 4 butadiene conrotatory A B cyclobutene disrotatory - - D butadiene 4 3 B A A B B A From the first excited electronic state (σ ππ ) of cyclobutene there is a smooth correlation with the excited state (Ψ 1 Ψ Ψ 3 ) of butadiene in the disrotatory mode, giving A for photon-initiated reaction. This is the basis of Woodward-offmann rules. B D C