VIII. Perturbation Theory and Symmetry

A. Interaction between orbitals

1. Degenerate case (both orbitals are at the same energy)

a. Molecular orbitals:

i) ^y₁ = c_lf_a + c_lf_b c_lc₂ (c_l < c₂ )

ii) ^y₂ = c₂f_a c₂ f_b

b. Energy: IDE₁I < IDE₂1

i) \ 2e^- Æ net bonding ( attraction )

ii) 4e^- Æ net antibonding (destabilization)

iii) DE₁ and DE₂ µ S_ab (overlap between f_a and f_b)

c. Overlap:

i) This depends on both orbita1 type and geometry.

In general: s > p > d

ii) Some common cases for the dependence of S_ab on geometry changes:

note at q=90° bonding and antibonding cancel, S_ab = 0

note that at f=45° only 30% of the overlap is lost

2. Orbitals at different energies

a. Molecular orbitals:

i) ^y₁ = c_lf_a + c₂f_b c_l > c₂

ii) ^y₂ = c₃f_a c₄f_b c₃< c₄

iii) In other words, the resultant MO most strongly resembles that starting orbital closest to it in energy.

b. Again, for the energies:

i) IDE₁I < IDE₂1 and

ii) DE₁, DE₂ µ S_ab / (e_a e_b) note: inverse depen- dence on energy difference

3. Interacting 3 orbitals

a. In a linear fashion (r_AB r_BC and ABC = 180°):

Note: the phases on the orbitals are set so that overlap between f_a & f_c and f_b & f_c are both positive.

i) Molecular orbitals:

(a) ^y₁ = c_lf_a + c₂f_c+ c₃f_b c_l > c₂ >> c₃

(b) ^y₂ = c₄f_c - c₅f_a+ c₆f_b c₄ > c₅ c₆

(c) ^y₃ = c₇f_b + c₈f_c+ c₉f_a c₇ > c₈ >>c₉

b. Triangular interaction:

(for the special case when r_AB = r_BC = r_AC, and atom A = B = C )

For this special case, ^y₂ & ^y₃ are the same energy; they are degenerate. Also note that there is no overlap between f_c and f_b, i.e.

B. Symmetry Correct Orbitals

1. Molecules have elements of symmetry.

2. Molecular orbitals must be "symmetry correct".

a. A molecular orbital must be either symmetric (S) or antisymmetric (A) with respect to all of the symmetry elements of the molecule if this MO is to bear any relationship to the physical properties of the molecule.

i) In other words, an MO must be symmetry cor- rect.

ii) MO's from VB calculations do not necessarily have this property.

b. We shall discuss two methods to generate symmetry correct orbitals for simple molecules.

i) One starts from a sort of VB idea of their form.

ii) The other uses an LCAO, linear combination of atomic orbitals, approach.

(a) First we shall use the latter method for AH₂ and AH₃ where A is an arbitary atom.

(b) We will use only the s and p atomic orbitals

3. Application of the LCAO approach to AH_n molecules.

a. There are two steps to this approach:

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:

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 ^y₂, ^y₄ and ^y₅ are easy to see:

ii) ^y₁, ^y₃ and ^y₆ show pattern typical of mixing three orbitals:

iii) Notice that we started with 6 atomic orbitals

(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 nonbonding 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.

(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 s 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, hn, ionizes electrons ejecting them from the MOs with a kinetic energy, KE. Then

IP = hn - KE and

-e_i = IP (Koopmanns' theorem).

(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 ^y₃ and ^y₄ 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:

(ii) The form of these orbitals are:

(iii) For BH₃, there are six valence electrons so ^y₁, ^y₂, ^y₃ are filled and ^y₄ is empty.

d. Consider a distortion of the AH₃ system to give a pyramidal molecule:

(i) We now do not have a mirror plane of symmetry containing all four atoms.

(ii) The forms of ^y₂, ^y₃, and ^y₅, ^y₆ (formerly ^y₆, ^y₇) are very much like those we had before, except that the interacting orbitals are not now in the same plane.

(iii) What has changed substantially is that one p orbital on A, which is nonbonding in the planar system, can interact (has the same symmetry)

(a) with the s orbital on A

(b) and with the totally symmetric H₃ combination.

(c) This gives rise to orbitals ^y₁, ^y₄, and ^y₇.

e. We can do exactly the same thing for the H-A mol- ecule:

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 ^c_a and ^c_b (two equivalent VB orbitals)

^y₊ = ^c_a + ^c_b

^y_- =^c_a - ^c_b

(b) If there are three equivalent VB orbitals and the molecule has a C₃ axis, then for ^c_a, ^c_b. ^c_c

^y₁ = ^c_a + ^c_b + ^c_c

^y₂ = 2^c_a -^c_b -^c_c

^y₃ = ^c_b - ^c_c

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

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 ^y₃. ^y₁ and ^y₅ in the B.O. approach also mix into ^y₃, however this is a sec- ondary effect, and is not so important here for what we want to do with the MOs

here.

c. Example two: BH₃

Notice that the form of ^y₁ to ^y₇ on the previous page are exactly those that we derived from the LCAO model

d.The MO shapes of AH₄, AH₃, AH₂, and AH are listed below. Study these fragment orbitals carefully.

(a) Learn their shapes and relative energies.

(b) Work out the forms of the MO's not covered previously by the LCAO and BO methods.

ii) There are several points of interest in this series:

(a) First of all notice that there are as many bond- ing and antibonding AH orbitals as there are H's; i.e. AH₄: 4; AH₃: 3; AH₂: 2; AH: 1.

(b) Notice also that there are nA A-H bonding and 4-n (n = number of hydrogens) nonbonding levels, i.e. AH₄: 0, AH₃: 1, AH₂: 2, AH: 3.

(c) A stable molecule will be one with 2n + 2 (4 - n) electrons (= 8 electrons).

iii) Finally, notice that in AH₄ there is one triply degenerate orbital and one nondegenerate one. Thus, our picture of the photoionization spectrum of CH₄ (page 15) is consistent with this delocalized, symmetry correct picture.

C. Interaction of fragment orbitals

1. We are now going to use the AH_n orbitals as building blocks to construct the MO's of more complicated mol- ecules, applying the following simplifications:

a. Concentrate only on those orbitals which are the most important ones.

b. Replacing a H atom with a CH₃, F. etc. group does not change the basic form of these orbitals.

2. The first simplification above is problematic: which are the "most important ones"?

a. As general rules:

i) The nonbonding orbitals for AH_n are important.

ii) The s and s* of AH_n orbitals will only be used when we need them.

b. First example, the orbitals of ethane derived from two methyl radicals:

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 s and s* orbital as shown below

The essential fragment bonding interaction

ii) The other six filled orbitals describe CH bond- ing, and the six empty orbitals are CH 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 AH_n units are interacted, it is the nonbonding fragment orbitals which create the largest overlap.

(c) We can bring AH 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 CH s 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 p overlap of the CH s bond and the p orbital shown in the fragment diagram.

iii) Notice that we can stabilize the CH s bonding MO even more by moving the hydrogen in a bridg- ing situation:

(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.

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.

There are still two lone-pair MOs which have the same form as before

b. There is one modification which is important:

i) If a group or an atom which is much more elec- tronegative than H is substituted, then the bonding MO is stabilized and becomes more localized on the electronegative substituent.

ii) At the same time, the antibonding MO is also stabilized and becomes much more localized on the central atom.

e.g. Consider the relationships in CH₃Li, CH₄ and CH₃F:

more concentrated on Li

more concentrated on C

more concentrated on F

more concentrated on C

iii) In general, the bonding orbital is more concen- trated on the more electronegative atom and the antibonding MO is more concentrated on the less electronegative atom.

D. Interactions Between Molecules: Reactions

1. LCAOMO representations of these are handled in exactly the same way as before, except that they are not static, so:

a. One must decide on a reaction path, and then

b. evaluate the orbital energies as a function of the

reaction path.

2. Let's examine two simple examples.

a. A comparison of the pathways:

b. A comparison of the same two possible reaction geometries for a system with two additional elec- trons:

D + H₂ Æ DH + H

c. For the collinear approach we have:

i) Notice that the energy of ^y₂ stays relatively constant, it is "caught" between H₂ s and H₂ s*.

ii) Once again, the fact that DE₁ is smaller than DE₂ is a simple consequence of the fact that antibonding combinations are destabilized more than the bond- ing ones are stabilized.

d. Now let us evaluate the triangular approach:

This orbital actually drops a bit in energy since it is it is expected that r_2' > r₂ .

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:

iii) On your own: take the linear and triangular endpoints of y₁, y₂ & y₃ and make a Walsh diagram of what happens to them as the HHH angle, a, varies from 180° to 60°, keeping r_l = r₂ . Give a reason why each orbital goes up or down in energy.

3. With these orbitals in hand, we can now evaluate the three reactions:

a. Case I: D⁺ + H-H Æ D-H + H⁺

i) There are 2 es, so ^y₁ (only) is filled.

(a)The triangular pathway is favored over a linear path, because DE₁' > DE₁.

(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:

(a) ^y_I was originally s of H₂. It goes down in energy as D⁺ approaches.

(b) The HH distance, r_l, becomes a little larger, but provided that r_l (= 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.

b Case II: D + H₂ Æ D-H + H

i) We now have 4 es.

(a) In the triangular approach ^y₁ goes down in energy, but ^y₂ goes up very high.

(b) In the linear approach, on the other hand, ^y₁ goes down slightly and ^y₂ 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 4es, rather than 2es "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 HH distance, r_l, must increase, bonding with the s orbital of D more than compen- sates. ^y₁ is stabilized.

(b) Again as long as the r_l (= r₂) distance is not too long. ^y₂ 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 electronelectron repulsion.

(ii) In the H3⁺ system, there are 2 es 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.

c. Case III: D + H₂ Æ DH + H

i) We now have 3 es the choices before were very clear now we have to evaluate the differing behavior of ^y₁ and ^y₂.

(a) Clearly the difference between ^y₁ in both paths is much smaller than that for ^y₂

(b) However, there are 2 electrons in ^y₁ and only one in ^y₂. 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 HHH angle at the T.S. from 180° 120°

d. But is it all relevent?

i) The H₃⁺ and H₃ systems actually relate to much

chemistry that all of us do.

ii) We will extend our treatment of H₃and H₃⁺ in two directions:

(a) nucleophilic and electrophilic substitution reactions in organic chemistry,

(b) and the basis for the Lewis octet or electron counting rules.

3. Nucleophilic substitution

a. H₃ and S_N2

i) We have shown in detail that the T.S. for H₃ is linear rather than triangular. So too is the S_N2 reaction. i e.

as opposed to:

ii) Obviously we don't want to cover explicitly all of the interactions from all of the obitals of the nu- cleophile and CH₃X. In general, we will only want to concentrate those interactions which will create the most stabilization (or destabilization) between the two reactants in a reaction.

b. HOMO and LUMO

i) When we interact two orbitals, f_i and f_j, the stabilization energy, DE, depends on two major factors:

overlap; used to optimize TS geometry

energy difference; used to single out important interactions

ii) For stabilization to occur one orbital must be

filled and the other empty.

(a) Therefore the smallest energy difference be- tween filled and empty orbitals must come from the highest occupied molecular orbital (HOMO) of one molecule and the lowest unoccupied molecular orbital (LUMO) of the other.

(b) For a general system we normally have the following situation:

low-lying LUMO, Lewis acid, electrophile

high-lying HOMO, Lewis base, nucleophile

A less frequent occurrence is a molecule with both a low-lying LUMO and a high-lying HOMO - nor- mally one interaction dominates.

iii) There will be interactions between the filled stack of A and the filled stack of B. But the overlap is normally small and one will want to choose a reaction path that minimizes them.

(a) Recall our examples using AH_n fragments to build the MO's of larger molecules we concen- trated on the nonbonding orbitals, which are at moderate energies, neither too high nor too low.

(b)They could form HOMOLUMO interactions and being hybridized out away from the other AH bonds, were capable of the greatest overlap.

(c) Here in a reaction, where we are building a "supermolecule" (TS) from the MO's of two mol- ecules, the same principles apply.

iv) The value, in numerical terms, of the HOMO and LUMO in any molecule is set before chemical reaction, ie. it is independent of the reaction path.

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 S_ij 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 S_N2 reaction revisited.

i)For the nucleophile the HOMO is obvious. It must be a simple or hybrid AO containing a lone pair:

It does not matter which we choose, since all have one shaded lobe of spherical symmetry pointed towards 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.

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).

iv) Backside versus frontside attack.

(a) Overlap is better in the case of backside attack:

(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.)

(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:

Are there 4 e^- around the central H and are there 10 e round the carbon? The answer is no in both cases.

This is the only filled MO with a coefficient on the central H.

Therefore, only 2es 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:

(a) Only the electrons in y₁ through y₄ occupy orbitals on the central carbon; total 8 e, all are bonding: y₅ is an orbital that is nonbonding be- tween the carbon and surrounding hydrogen atoms. Again there is no electron density in it on carbon.

(b) Notice that we have not had to use d orbitals to describe the bonding.

(i) Pentavalent carbon compounds do exist in the ground state.

(ii) So do isoelectronic SiR₅ and PR₅ molecules (where d orbitals again probably are not very important).

(iii) In fact there are large classes of compounds which one could count 10 e or 12 e around the central atom. They are stable because there are I or 2 nonbonding MO's that contain a node at the central atom.

e. Electrophilic attack at a tetrahedral carbon

i) Once again there are two different sides for approach:

ii) Orbital considerations

(a) For X⁺ to be a leaving group here, it must be pretty electropositive.

(b) Thus our primary interactions at the T.S. are an empty orbital (LUMO) on E⁺ and the HOMO on CH₃X.

(c) Since X is electropositive, the form of the HOMO must be:

Therefore of the two transition states:

There is increased bonding (overlap) for frontside attack. So this gives the more stable T. S. In fact, very good calculations have shown that the structure on the left below is 11 kcal/mol less stable that on the right side. (Note that the trigonal bipyramid is geometrically identical to the proposed backside attack while the square pyramid is like the transi- tion state proposed for frontside attack)

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 leastmotion path for the dimerization of singlet :CH₂. It will look like:

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

Therefore, a Walsh diagram for the least motion dimerization process will look like:

(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:

(c) A more favorable path would match the HOMO in one :CH₂ with the LUMO in another :CH₂ and vice versa i.e.:

(i) This will give two stabilizing interactions.

(ii) The reaction path then looks like:

or

(iii) In fact there are a series of xray 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 xray 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:

(i) We again have a very destabilizing HOMOHOMO interaction and a nonproductive LUMOLUMO inter- action.

(ii) If we drew out a Walsh diagram we would again see a HOMOLUMO crossing. (Try it).

(b) An alternative nonleast motion path which shows stabilization would be:

which then leads to the reaction path shown below

Refer back to Chpt. 6G for a calculation.

g. Let us consider what seems to be on the surface an exttrodinarily simple reaction - the breaking of a C-C bond to form two radicals and we will tie them to- gether by a methylene chain.

i) Here in the kinetic scheme

k₁ = one bond rotation of -CHD when C1-C2 is cleaved

k₁' = one bond rotation of -CHD when C1-C3 is cleaved

k₁₃ = two bond rotation when C1-C3 is cleaved

k₁₂ = two bond rotation when C1-C2 is cleaved

where k = k₁ + k₁' + k₁₃

k_i = the observed rate constant for the cis Æ trans rearrangement

k_a = the observed rate constant for the loss of optical activity

ii) One can show(do this) that k_i = 4k

and k_a = 2k + 2k₁₂.

iii) This reaction has been studied many times by many people. Most recently it was found by Berson at 695K that k_i/k_a = 1.07 ±0.04 and this lead them to conclude that k₁₂/k₁ = 5 - 42. In other words, double rotations of the CHD groups are much more common than single rotations. On the other hand, Baldwin found at 680K that k_i/k_a = 1.48 ±0.04 and this lead them to conclude that k₁₂/k₁ = 1.0. In other words, CHD rotation is random, neither single nor double rotations are preferred.

iv) Let us see what some of the difficulties are in this "simple" reaction. The minima for the diradical intermediates are:

Since the two CHD units can never get too far apart, at least one CHD group must rotate by 90° and this will provide the barrier for a very exother- mic ring closure.

v) Let us consider stereochemically how the two CHD goups could break apart and undergo double rotation. They are called conrotatory (rotation in the same direction) and disrotatory (rotation in opposite directions). As shown on the right side, the C-C s bond becomes a different linear combination of p AOs at the 0°, 0° minima.

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 s - 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:

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 (t130 fs). However, a significant number (~25%) have very long lifetimes - this is a twixtyl surface - with t>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 s* orbital is very low and the S MO is stablized well below A. A disrotatory double rotation is now preferred.

E. HardSoft AcidBase (HSAB) Theory

1. In the mid to late 1960's, Ralph Pearson and others recast these frontier orbital ideas of maximizing HOMOLUMO interactions in terms of the following equations:

a. For the reaction: A + :B Æ A B

Lewis acid Lewis base

(electrophile) (nucleophile)

b. we get the following energy relationship:

c. Definitions: (m are unoccupied MOs on A and n are occupied MOs on B)

i) E_AB = interaction energy between A and B

ii) q_A, q_B = charges on A and B

iii) r_AB = distance between A and B

iv) e = dielectric constant of the solvent

v) S_mn = overlap between an unoccupied orbital on A (m) and an occupied orbital on B (n)

vi) E_m = orbital energy of an unoccupied orbital (m) on A (related to the electron affinity of A)

vii) E_n = orbital energy of an occupied orbital (n) on B (related to the negative of the ionization potential on B)

viii) c_im= mixing coefficient for AO i in MO m and c_jn = mixing coefficient for AO j in MO n.

ix) b_ij=a resonance integral between AO i and AO j

x) e_i and e_j = energies of AOs i and j, respectively

xi) S_ij = 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 HOMOLUMO interaction.

iii) Compare this to the equation in Lowry & Richardson (p.322).

2. Classfication of acids and bases

a. Soft Base The lone pair is highly polarizable (dif- fuse), easily oxidized (low IP), and on an atom of low electronegativity.

b. Hard Base The lone pair is completely opposite to the above, and e density is concentrated on this atom.

c. Soft Acid The unoccupied orbital is highly polariz- able (diffuse), easily reduced (high electron affintiy), and on an atom of high electronegativity.

d. Hard Acid The unoccupied orbital is completely opposite to the above, and positive charge is concen trated on this atom.

*** Your book lists an extensive table of acids and bases by these classifications***

e. Parr and Pearson have proposed a hardness param- eter, h, defined by

h = (I - A)/2

where I = ionization potential of the (HOMO) base

A = electron affinity (LUMO) of the acid

therefore, I - A = the HOMO - LUMO energy gap.

h (eV) h(eV)

Li+ 35.1 H- 6.8

Na+ 21.1 F- 7.0

Mg(2+) 32.5 Cl- 4.7

Cu+ 6.9

Ag+ 6.9

Au+ 6.9

Tl+ 4.2

Hg+ 4.2

For atoms in neutral molecules:

h (eV) h (eV)

O: R₂O 16.2 F: RF 45.3

R₂CO 18.8 Cl: RCl 10.9

S: R₂S 9.9 Br: RBr 9.5

N: RNH₂ 14.7 I: RI 8.1

ArNH₂ 12.1

P: R₃P 9.4

3. HSAB interactions

a. Two types of interaction are favored:

HARD HARD

Most of E_ab comes

from the charge term.

SOFT SOFT

Most of E_ab comes

from the overlap term.

b. The other two cases are less favorable.

i) Hard acid with soft base or soft acid with hard base have small charge terms.

ii) These combinations also have small overlap terms.

iii) Thus they are not prefered to hardhard or softsoft interactions, in which either the charge or the overlap interaction is enhanced.

c. Your book covers some ways that HSAB theory can be used

i) The equation used in an exact quantitative man- ner does not work very well. It does show us semiquantitatively (or qualitatively) why hard bases prefer to react with hard acids and soft bases with soft acids.

ii) This concept can then be used to explain why compounds react at different rates with a particu- lar reagent, or why reactions are favored at one over another position on a molecule.