Perturbation Theory and Symmetry - Lecture Notes | CHEM 6311, Study notes of Mechanics

Download Perturbation Theory and Symmetry - Lecture Notes | CHEM 6311 and more Study notes Mechanics in PDF only on Docsity! 136 VIII. Perturbation Theory and Symmetry A. Interaction between orbitals 1. Degenerate case (both orbitals are at the same energy) ∆E2 ∆E1 φa φb ψ1 ψ2 8.1 a. Molecular orbitals: i) ψ1 = clφa + clφb cl≠c2 (cl < c2 ) ii) ψ2 = c2φa - c2 φb b. Energy: I∆E1I < I∆E21 i) ∴ 2e- → net bonding ( attraction ) ii) 4e- → net antibonding (destabilization) iii) ∆E1 and ∆E2 ∝ Sab (overlap between φa and φb) c. Overlap: i) This depends on both orbita1 type and geometry. In general: σ > π > δ ii) Some common cases for the dependence of Sab on geometry changes: 137
 
 r r r r θ θ = 90ο Sab φa φb r Sab φa φb r Sab φa φb 1.0 -1.0 1.0 0.0 0.0 Sab φa φb1.0 0.0 0.0 Smax 8.2 0o 45o 90o 
Sab1.00.0Smax 0o 45o 90o 8.30 note at θ=90° bonding and antibonding cancel, Sab = 0 note that at φ=45° only 30% of the overlap is lost 2. Orbitals at different energies 
  ∆E2 ∆E1 φa φb ψ1 ψ2 8.4a. Molecular orbitals:i) ψ1 = clφa + c2φb cl > c2 ii) ψ2 = c3φa - c4φb c3< c4 iii) In other words, the resultant MO most strongly resembles that starting orbital closest to it in energy. 140 i) Take linear combinations of the set of s orbitals of the H atoms, so they are either S or A with respect to all symmetry elements in the molecule. ii) Interact these combinations with the AO’s of the central atom, with the provision that only orbitals of the same symmetry can interact. b. AH2:

  A H H m1 m2 C2 C2 m1 m2 C2 m1 m2 SSS SSS SSS ASA AAS AAS (a1) (a1) (a1) (b2) (b2) (b1) ψ1 ψ2 ψ3 ψ4 ψ5 ψ6 8.11 Notice that these are just the orbitals of H2. They are not much split in energy because the H---H distance is long. i) The forms of ψ2, ψ4 and ψ5 are easy to see: ii) ψ1, ψ3 and ψ6 show pattern typical of mixing three orbitals: iii) Notice that we started with 6 atomic orbitals ψ2 ψ4ψ5 8.12 
 ψ1 ψ3 ψ6 = = = + + + + + + bonding nonbonding antibonding 8.13 141 (or orbital combinations) and we formed 6 MO’s. The number of orbitals is always conserved. iv) These MO’s are all symmetry correct for H2O. v) The electrons are distributed in bonding and non-bonding orbitals. (a) There are 2 valence electrons from the 2 H’s and 6 electrons from O. (b) This gives us 8 valence e¯ s which will be placed in the 4 lowest MO’s.   ψ2 ψ4 ψ1 ψ3 ionization potential (eV) 12.6 13.8 17.2 33.2 lone pairs on oxygen O-H σ bonds 8.14 (c) There are now two distinctly different O-H s bonds and two different lone pairs. If one used sp3 hybrids at oxygen, then this should generate two identical sp3 hybrids for the lone pairs and two sp3-Hs σ bonds. The LCAO and VB approaches are clearly different. Which is right (most correct)? (d) A way to tell the difference is by photoelectron spectroscopy which measures the ionization poten- tials associated with the MOs. The basic operation is that a photon beam at constant energy, hν, ionizes electrons ejecting them from the MOs with a kinetic energy, KE. Then IP = hν - KE and -ei = IP (Koopmanns’ theorem). 142 (d) The PE spectrum of H2O is shown below. There are three ionizations (the fourth is at lower energy) and, therefore, the LCAO model appears to be much more appropriate. (e) The splitting between ψ3 and ψ4 is quite large: 1.2eV = ~28 kcal/ mol! (I eV = 23.06 kcal/mol) There is appreciable difference between these “lone pair” orbitals. c. Let us now turn to a planar AH3 molecule i) In this case the orbitals are easy to construct: 145 e. We can do exactly the same thing for the H-A mol- ecule: 146 4. The bond orbital approach to forming symmetry cor- rect orbitals. a. There are three steps i) Start from a VB (localized) description of the bonding, nonbonding and antibonding orbitals. ii) Take linear combinations of the VB orbitals to make them conform to the symmetry of the mol- ecule (a) if the molecule has a mirror plane of symmetry or C2 axis, take + and - combinations, i. e. for χa and χb (two equivalent VB orbitals) ψ+ = χa + χb ψ- =χa - χb (b) If there are three equivalent VB orbitals and the molecule has a C3 axis, then for χa, χb. χc ψ1 = χa + χb + χc ψ2 = 2χa -χb -χc ψ3 = χb - χc iii) Simplify the linear combinations and order them in the correct energy b. Example one, H2O: - recall we have done this before via the LCAO ap- proach for AH2 }the degenerate combinations O H H O H H two equivalent lone pairs two equivalent σ bonds 8.23 m1 m2 C2 147 i) Before you continue, write down the atomic orbitals composing each hybrid orbital, then recom- bine the AO’s and try to envision how they gener- ate the new, symmetrized orbitals. Compare the AO’s used here to the ones used in the LCAO approach. ii) Notice that these orbitals generated from B.O.s exactly match those obtained from the LCAO approach, except for ψ3. ψ1 and ψ5 in the B.O. approach also mix into ψ3, however this is a sec- ondary effect, and is not so important here for what we want to do with the MOs  
  
  
ψ1 ψ5 ψ2,ψ3 ψ6,ψ7 a1’ e’ a1’ e’ ψ4 a2" 8.26 150 i) The essence of the bonding interaction in ethane is, therefore, simply given by the interaction be- tween the two nonbonding orbitals of the CH3 units to form the C - C σ and σ* orbital as shown below The essential fragment bonding interaction ii) The other six filled orbitals describe C-H bond- ing, and the six empty orbitals are C-H antibonding. (a) Notice on page 147 that all of the nonbonding orbitals are either p orbitals, or hybrids which are directed away from the AH bonds. (b) Thus, when AHn units are interacted, it is the nonbonding fragment orbitals which create the largest overlap.  8.27b 151 (c) We can bring A-H bonding and antibonding orbitals into the picture as they are needed. c. Second example, hyperconjugation in the ethyl cation. i) To show this effect, we shall need one C-H σ bonding combination, in addition to the nonbonding orbitals. ii) This picture can be compared with our earlier V.B. description of the ethyl cation (a) Hyperconjugation was represented by two resonance states: (b) The combination of these resonance states corresponds to the π overlap of the C-H σ bond and the p orbital shown in the fragment diagram. iii) Notice that we can stabilize the C-H σ bonding MO even more by moving the hydrogen in a bridg- ing situation: 152 (a) This should mean that the bridged form is more stable than the classical structure. (i) The best theoretical estimate does indeed show this— the bridged isomer is 7 kcal/mol more stable than the classical structure. (ii) The calculations are high enough in quality so there is no doubt but that this is experimentally true in the gas phase, but this is not so clear in solution. (iii) A computation (at a lower level of theory) with four HC1 molecules to represent solvent mol- ecules has shown that the classical structure is more strongly solvated in this system, and becomes 15 kcal/mol more stable than the bridged isomer. C C H H H H H Cl H Cl H Cl H Cl H C C H H H H H Cl H Cl H Cl H Cl H 8.31 2. The simplification involving replacement of H’s by other atoms or groups is quite straightforward. a. The essential shape of the fragment orbitals does not change upon substitution of another atom or group for H, e.g. for CH3OCH3, there still are two bonding and two nonbonding MO’s.    
8.32 There are still two lone-pair MOs which have the same form as before 155 i) At the presumed T.S. there should be greater overlap for the triangular form than there is in the linear geometry with the provision that r1' is the same in both cases. ii) Putting this all together in one diagram: D1 H2 H3 r1’ r2’ D1 H2 H3 r1’ r2’ S12 S13 ~ 0 S12 = S13 8.37
   D H H r1’ r2’ |∆E1| < |∆E1’| ∆E2’ ∆E1’ D H H r1’ r2’ ∆E 2 ∆E 1 8.38 iii) On your own: take the linear and triangular endpoints of ψ1, ψ2 & ψ3 and make a Walsh diagram of what happens to them as the H-H-H angle, α, varies from 180° to 60°, keeping rl = r2 . Give a reason why each orbital goes up or down in energy. D1 H2 H3 D1 H2 H3 8.39 3. With these orbitals in hand, we can now evaluate the three reactions: a. Case I: D+ + H-H → D-H + H+ i) There are 2 e¯s, so ψ1 (only) is filled. (a)The triangular pathway is favored over a linear path, because ∆E1' > ∆E1. (b) Note that 2 e¯ are delocalized over 3 centers, i.e. 3 bonds, we’ll comeback to this point later. ii) Looking more closely at this: 156 (a) ψI was originally σ of H2. It goes down in energy as D+ approaches. (b) The H-H distance, rl, becomes a little larger, but provided that rl (= r2) does not happen to be extraordinarily large (and there is no reason to suspect that it is), the transition state that we have proposed should be at a lower energy than our reactants! (i) This is indeed the case: (ii) The proposed transition state should be consid- ered a product. D+ H H + D H H D H H+ 8.40 Reaction Path ∆E b Case II: D¯ + H2 → D-H + H¯ i) We now have 4 e¯s. (a) In the triangular approach ψ1 goes down in energy, but ψ2 goes up very high. (b) In the linear approach, on the other hand, ψ1 goes down slightly and ψ2 stays at a relatively constant energy. (c) This latter path should be favored. (i) Note that now 4 e¯ are distributed over 2 bonds (3 centers). Is the Lewis octet rule violated for the central H ? (ii) Clearly there are 4e¯s, rather than 2e¯s “around” it. We will come back to this question later. ii) We might expect that, as in the previous case, that our proposed transition state may really be an energy minimium. (a) Although the H-H distance, rl, must increase, bonding with the s orbital of D more than compen- sates. ψ1 is stabilized. 157 (b) Again as long as the rl (= r2) distance is not too long. ψ2 stays at about the same energy as D itself. (c) Then the linear DH2- system would be pre- dicted as a stable point. It is not. (i) What is not accounted for in our orbital energy variation over the reaction path (Walsh diagram) is electron-electron repulsion. (ii) In the H3+ system, there are 2 e¯s spread out over 3 nuclei. (iii) In the H3¯ case there are twice as many elec- trons and the same number of nuclei. Clearly there is more e¯ - e¯ repulsion. (iv) Thus, it is a real transition state. 8.41 Reaction Path ∆E D + H H H HD H+D H- - - 15 kcal/mol c. Case III: D + H2 → DH + H i) We now have 3 e¯s - the choices before were very clear - now we have to evaluate the differing behavior of ψ1 and ψ2. (a) Clearly the difference between ψ1 in both paths is much smaller than that for ψ2 (b) However, there are 2 electrons in ψ1 and only one in ψ2. So it is extremely difficult to predict what happens. (c) This in general is true - radical reactions are very difficult to make predictions about. ii) Actually, as we saw before in the discussion of energy surfaces, a linear T.S. is prefered. It does not cost much energy, however, to vary the H-H-H angle at the T.S. from 180° - 120° d. But is it all relevent? i) The H3+ and H3¯ systems actually relate to much 160 v) How a reaction will proceed - its reaction path - is (or can be) determined by the overlap between the LUMO on A and HOMO on B, etc.— the Sij term in the perturbation expression. (a) Thus, first we determine what the critical interaction(s) between the HOMO and LUMO in a chemical reaction will be. (b) Then we evaluate details of the reaction path by evaluating changes in the overlap for these critical interactions. c. The SN2 reaction revisited. i)For the nucleophile the HOMO is obvious. It must be a simple or hybrid AO containing a lone pair: or or 8.45It does not matter which we choose, since all haveone shaded lobe of spherical symmetry pointedtowards the carbon.ii) For the electrophile, as we saw before, if X is an electronegative element or group, one bonding and one antibonding orbital are stabilized. 
C-H σ C-X σ C-X σ* C-H σ* 8.46 more concentrated on C more concentrated on X iii) The critical interaction then is between the highest occupied orbital of the nucleophile (the base) and the lowest unoccupied orbital of the electrophile (the acid). 161  
C-H σ C-X σ C-X σ* C-H σ* 8.47 HOMO LUMO C XNuc- iv) Backside versus frontside attack. (a) Overlap is better in the case of backside attack:  8.48 C XNuc - versus C X Nuc (b) Just as it is in the case of D¯ + H2 the overlap is maximized if the bond being formed and bond being broken form a 180° angle. v) We can also make a number of other predictions about the reactivity order (in the gas phase!) using this model: (a) reactivity should increase in a series of nucleo- philes as the HOMO goes up in energy, i.e. ¯:CH3 > ¯:NH2 > ¯:OH > ¯:F (increasing electronegativity on the central atom) (CH3)3N > (CH3)2NH > CH3NH2 > NH3 (CH3 groups donate e _ density - HOMO goes to higher energy, i.e.)  8.49 162 (b) reactivity should increase as the LUMO in CH3X goes down in energy, i.e. with an increased electronegativity difference between C and X: CH3F > CH3CI > CH3Br > CH3I (in gas phase!) CH3OH2+ > CH3OH (c) As a corollary - the propensity to undergo frontside displacement should increase as the coefficient on the leaving group decreases and the carbon orbital becomes more delocalized and diffuse. (i) For example SiH3X vs. CH3X. (ii) Si is more electropositive and the 3s/3p orbital is more diffuse than the corresponding 2s/2p orbital of C. (iii) This means that there is more possibility for frontside displacement of the leaving group from Si. d. Let us now go back to a previous question concern- ing e¯ counting: H- + H H H H H H- + H C H H H H C H HH H - 8.50 Are there 4 e- around the central H and are there 10 e¯ round the carbon? The answer is no in both cases.  no electron density on the central hydrogen H H H 8.51 This is the only filled MO with a coefficient on the central H. Therefore, only 2e¯s are shared between the central hydrogen and its neighbors. ii) To build the orbitals of our CH3¯ transition state let us consider the following method: 165 iii) Other examples - consider the dimerization of BH3 to give diborane: Here two sets of HOMOs and LUMOs are used. Each B-H bond is a two center - two electron one. The p AO on each boron is empty so when the dimer forms, we now have two 3 center - 2 elec- tron bonds formed for the bridging hydrogens. f. If, normally because of some symmetry constraint, one HOMO interacts with another HOMO, then this cannot be a favorable reaction path. i) The antibonding, filled combination is destabilized. ii) This destabilization is greater than the stabiliza- tion from the bonding combination (a) Example: consider the least-motion path for the dimerization of singlet :CH2. It will look like: C H H C H H + 8.59C H H C H H These are all but the highest MO of CH2 along with 3-D representations of the orbitals. Singlet methylene has two electrons in 2a1 and none in b1. The b1 and 2a1 orbitals are the nonbonding ones for AH2 166 Therefore, a Walsh diagram for the least motion dimerization process will look like:
    C H H C H H M1 C H H C H H AA SA AS SS M2 M2 M1 SS SA AA AS M1M2 M1M2 8.61 (b) The reaction is said to be symmetry forbidden. (i) When the mirror plane of symmetry is con- served, a filled AS HOMO crosses the SA LUMO. (ii) Since the AS HOMO must go up far in energy there will also be an appreciable barrier associated with the reaction. (iii) Another way of looking at this path is: 8.62 repulsion no consequence (c) A more favorable path would match the HOMO in one :CH2 with the LUMO in another :CH2 and vice versa i.e.: 8.63 HOMO LUMO LUMO HOMO 167 (i) This will give two stabilizing interactions. (ii) The reaction path then looks like: H C H H C H 8.64 H C H H C H H C H H C H or (iii) In fact there are a series of x-ray structures for R2AAR2 . where R = bulky group, e.g. CH2C(CH3)3 and A= C,Si,Ge,Sn. [a] When A = Ge or Sn, the x-ray structure is like that proposed for the T.S. [b] When A = C, it is at the product. [c] When A = Si, it lies between the T.S. and prod- uct. (iv) Example two; an analogous reaction, the addi- tion of singlet :CH2 to ethylene. (a) First, let us examine the least motion path: H C H 8.66a H C H H C H H C H H C H H C H H C H H C H H C H (i) We again have a very destabilizing HOMO-HOMO interaction and a nonproductive LUMO-LUMO inter- action. (ii) If we drew out a Walsh diagram we would again see a HOMO-LUMO crossing. (Try it). (b) An alternative non-least motion path which shows stabilization would be: 170

 
 
 H D D H 8.71 H H H H H D D H conrotatory A A A H H H D D H H D H D S S S disrotatory H D D H H D H D p σp* σ As shown by the interaction diagram above, this is not so simple of a matter. As long as the CH2 group is a good stong σ - donor, then S will lie above A and the conrotatory openning is to be preferred over the disrotatory route. vi) A computed potential energy surface for the reaction is illustrated below: 171 E. Hard-Soft Acid-Base (HSAB) Theory 1. In the mid to late 1960’s, Ralph Pearson and others recast these frontier orbital ideas of maximizing HOMO-LUMO interactions in terms of the following equations: The energies here are in kcal/mol relative to the 0°, 0° minimum. Conrotatory and disrotatory double rotations follow the dotted diagonal lines. The shaded areas correspond to the very deep minima corresponding to the cyclopropanes. vii) Trajectory studies were done using this poten- tial energy surface [J. Am. Chem. Soc., 102, 3648 (1998)]. These guys found that k12/k1 = 2.3 - 3.5!! This is right in between the two experimental determinations - a safe place to be for a theoreti- cian! What they find is a very complicated situation. Most (~60%) of the trajectories are double rota- tion, conrotatory ones. They occur over very short time periods (τ≈130 fs). However, a significant number (~25%) have very long lifetimes - this is a twixtyl surface - with τ>400 fs. This gives k12/k1 = 1.4. The actual results are then likely to be very dependant on a number of variables. viii) But the situation is even more complicated! Very good calculations predict that the triplet state is 0.7 kcal/mol more stable than the singlet. There may very well be a singlet-triplet interconversion process. In the examples below, X = SiH3 is a very good donor substituent so S in the interaction diagram on the previous page lies much higher than A. When X = F, the σ* orbital is very low and the S MO is stablized well below A. A disrotatory double rotation is now preferred. X X H H H H X singlet - triplet SiH3 11.1 kcal/mol F 3.8 H -0.7 8.73 172 a. For the reaction: A + :B → A - B Lewis acid Lewis base (electrophile) (nucleophile) b. we get the following energy relationship: ∆E ∝ qAqB rABε + 2 2 cimcjnβij( ) En −Emn ∑ m ∑ where βij = K ei + ej( )Sij 2 c. Definitions: (m are unoccupied MOs on A and n are occupied MOs on B) i) EAB = interaction energy between A and B ii) qA, qB = charges on A and B iii) rAB = distance between A and B iv) ε = dielectric constant of the solvent v) Smn = overlap between an unoccupied orbital on A (m) and an occupied orbital on B (n) vi) Em = orbital energy of an unoccupied orbital (m) on A (related to the electron affinity of A) vii) En = orbital energy of an occupied orbital (n) on B (related to the negative of the ionization potential on B) viii) cim= mixing coefficient for AO i in MO m and cjn = mixing coefficient for AO j in MO n. ix) βij=a resonance integral between AO i and AO j x) ei and ej = energies of AOs i and j, respectively xi) Sij = overlap integral between AOs i and j xii) K = a constant (normally ~ 2.7) d. Interpretation i) The charge term represents the electrostatic attraction term between A and B ii) The dominant part of the overlap term will be the HOMO-LUMO interaction. iii) Compare this to the equation in Lowry & Richardson (p.322). 2. Classfication of acids and bases