LCAO APPROXIMATIONS OF ORGANIC

Transcription

1 Principles of Organic hemistry lecture 19, page 1 LAO APPROXIMATIONS OF ORGANI Pi MO SYSTEMS BENZENE MOs o ollecting all the molecular orbitals for benzene and arranging them according to energy and putting in the electrons gives the following energy diagram What is the resonance energy of the benzene ring? o Imagine putting three walls between the double bonds to isolate them. Aromaticity o 4N+2 What would be the pi-energy of this version of benzene? E = 6 β. Where does this result come from? The pi-energy of benzene is E = 2( 2 β ) + 4( β ) = 8 β The difference, and therefore, the resonance energy of benzene is 2β. o filled atomic orbitals paired

2 Principles of Organic hemistry lecture 19, page 2 lead to stability o aromatic species Note the electronic configuration of 6 6. It is like electronic configuration of the mono-atomic noble gasses. o filled molecular orbitals need to be cyclic and fully conjugated 1,3,5-hexatriene above: is fully conjugated but acyclic; and 1,3,5- cycloheptatriene is cyclic but not totally conjugated. Frontier Molecular Orbital Theory o reactivity and coefficients There is a methylene unit insulating the p orbitals at 1 and 6 from parallel interaction. these are important for anions and cations o kinetic results versus thermodynamic results ow does this relate to MO calculations? Electronic changes are instantaneous compared to changes in atomic displacement. Let s talk about the protonation of 1,3-butadiene o electron density functions as c 2 so.60 2 /.37 2 = 2.6.

3 Principles of Organic hemistry lecture 19, page 3 Frost-ückel diagrams The electron density is heavily weighted toward the termini. Where is the molecule going to like protonate in the first elemental event? What should we consider if we are going to talk about the stability of species on the reaction pathway in the protonation of l,3-butadiere? o Only valid for fully conjugated systems. o 2βcos(m2π/n) = E m, n = # atoms, m (integer) = 0, 1,... n/2 for cyclic conjugated molecules for benzene n=6 2π/3 π/3 for cyclobutadiene for linear conjugated molecules for allyl θ for butadiene

4 Principles of Organic hemistry lecture 19, page 4 o The frost method gives the energies straight away! θ o 2βcos(mπ/[(n+1)) = E m, n = # atoms, m (integer) = 1,... n IN BOT fully conjugated cyclic systems and linear systems we are looking form the projection of the radius of the circle onto the energy axis. Don t commit these equations to memory. Establish an operational definition and work them out every time. o an you envision how to use these with the secular determinants to solve more complex systems? You can get the energies with these Frost diagrams and then use the method of expansion by cofactors to get the MO coefficients! The cofactor process is simplified by the fact that you are not looking for roots of polynomials! You are looking for ratios of expressions as a function of energies. The energies are calculated directly above for many interesting systems. Odd/Even Alternate and non-alternate hydrocarbons. o For the operational definition, in conjugated molecules, mark the conjugated atoms so that the number of marked atoms is maximized. An alternate fully conjugated molecule is one in which alternate atoms can be marked with no marks adjacent to each other. Examples.

5 Principles of Organic hemistry lecture 19, page 5 The fully conjugated molecule is non-alternate if two un-marked atoms or two marked atoms are adjacent after the making operation. so azulene at right is non-alternate. some more examples of non-alternate conjugated hydrocarbons: All linear and branched structures are by definition alternate. Even alternate conjugated molecules have the same number of marked versus unmarked atoms. Odd alternate conjugated molecules have an unequal number of marked versus unmarked atoms. example and see the benzyl cation above. Now here is the utility in all this: Large structures lead to difficult secular determinants. owever the following rules allow some prediction regarding the characteristics of the OMOs of these orbitals. o Alternate, conjugated molecules dispose orbital energies symmetrically about zero.

6 Principles of Organic hemistry lecture 19, page 6 The non-branched acyclic even alternate conjugated molecules dispose orbital energies above and below zero, but not at zero. The odd alternate, conjugated molecules dispose orbital energies such that at least one orbital is non-bonding and has value 0 β. We would have had a good head start on allyl with this knowledge. o The non-alternate simple cyclic systems like cyclopropenyl, cyclopentenyl, cycloheptenyl dispose orbital energy dissymmetrically about zero Non-alternate bicyclic systems and non-alternate cyclic branched systems loose this property. Symmetrical systems of this variety can dispose orbital energies symmetrically about the zero line. What can we say about trimethylenemethane? o Is it alternate? If it is alternate, is it odd or even? o We want to maximize the number of marks on the structure. When we maximize the number of marks, the nodes in the NBOs (non-bonding orbitals) occur at the unmarked atoms. The sum of the coefficients of the atomic orbitals on the starred atoms directly linked to ONE unmarked atom is zero. Example = 0 3 ow should the orbitals of trimethylenemethane disposed in energy? ow many molecular orbitals are there? (four) It is an odd alternate conjugated system. After you put one molecular orbital at the zero line you have three left. o There is no way to symmetrically dispose these remaining three orbitals about zero without putting another at 0β. o If you carefully workout the secular determinant you will find roots 0, 0, SQRT(3), SQRT(3) an you do this?

7 Principles of Organic hemistry lecture 19, page 7 There are two unique ways to symmetrically satisfy = 0. o 2 = 3 and 2 = 3. Think about how this relates to our solution of the benzene molecule when we used mirror planes to simplify the secular determinant. These two solutions lead to 4 = 0 and 4 = 2 2. After you normalize you get: Ψ 2 = (1/SQRT(6))χ 2 + (1/SQRT(6))χ 3 (2/SQRT(6))χ 4. Ψ 3 = (1/SQRT(2))χ 2 (1/SQRT(2))χ 3. o This structure may look so esoteric that you think it is not important to consider it. If we add to electrons to trimethylenemethane and change the atoms to N atoms we get guanidinium cation. If we change the in the middle to a boron atom and the atoms connected to oxygen atoms we can think about an inorganic structure. N N N O B O the pi systems in these structure are isoelectronic. O o Their MO s will be similar with perturbations by the electronegativity of the N, B and O atoms. o Your author does the benzyl example for you. You should read it, page I don t agree with the analysis that resonance structures predict +2/5 on 1. You need atomic coefficients to predict charge.

8 Principles of Organic hemistry lecture 19, page 8 I don t agree that the benzyl system was the only system for which we get different results when we do ückel MO theory versus resonance theory. Remember we found bond order differences in simpler allyl system.