Section 8-9
Section 8-9 Some Simplifying Generalizations
√
φ would then give us a “bond order” of 2/ 2 = 1.414. It is more convenient in practice to divide this number in half, because then the calculated π -bond order for ethylene turns out to be unity rather than two. Since ethylene has one π -bond, this can be seen to be a more sensible index.
As a result of these considerations, the π -bond order (sometimes called mobile bond
√
order) of the allyl radical is 1/ 2 = 0.707 in each bond. (Electrons in φ2 make no contribution to bond order since c2 vanishes. This is consistent with the nonbonding
label for φ2.)
Generalizing the argument gives, for pij , the π-bond order between nearest-neighbor atoms i and j : all MOs
pij =
nkcikcjk
(8-48)
k
where the symbols have the same meanings as in Eq. (8-46). In cases in which partially filled degenerate MOs are encountered, the averaging procedure described in connection with electron densities must be employed for bond orders as well.
EXAMPLE 8-4 Calculate p13 for the cyclopropenyl radical, using data in Fig. 8-7.
SOLUTION There are 2 electrons in φ1 and the coefficients on atoms 1 and 3 are 1 √ , so
3
this MO contributes 2 × ( 1 √ )2 = 2/3. We allocate 1 electron to φ2. Since c3 = 0 in this MO,
3
2
the contribution to p13 is zero. The remaining 1 electron goes to φ
2
3, yielding a contribution of 1 × 1 √ × −2 √ = −1 . So p − 1 = 1 .
2
13 = 2 6
6
6
3
6
2
8-9 Some Simplifying Generalizations
Thus far we have presented the bare bones of the HMO method using fairly small systems as examples. If we try to apply this method directly to larger molecules, it is very cumbersome. A ten-carbon-atom system leads to a 10 × 10 HMO determinant.
Expanding and solving this for roots and coefficients is tedious. However, there are some short cuts available for certain cases. In the event that the system is too complicated to yield to these, one can use computer programs which are readily available.
For straight chain and monocyclic planar, conjugated hydrocarbon systems, simple formulas exist for HMO energy roots and coefficients. These are derivable from the very simple forms of the HMO determinants for such systems.5 We state the results without proof.
For a straight chain of n unsaturated carbons numbered sequentially, x = −2 cos[kπ/(n + 1)], k = 1, 2, . . . , n
(8-49)
clk = [2/(n + 1)]1/2 sin[klπ/(n + 1)]
(8-50)
where l is the atom index and k the MO index.
5See Coulson [2].
Chapter 8 The Simple H ¨uckel Method and Applications
For a cyclic polyene of n carbons, x = −2 cos(2xk/n), k = 0, 1, . . . , n − 1
(8-51)
√
clk = n−1/2 exp[2πik(l − 1)/n], i = −1
(8-52)
The coefficients derived from Eq. (8-52) for monocyclic polyenes will be complex when the MO is one of a degenerate pair. In such cases one may take linear combinations of these degenerate MOs to produce MOs with real coefficients, if one desires.
There is also a diagrammatic way to find the energy levels for linear and monocyclic systems.6 Let us consider monocycles first. One begins by drawing a circle of radius 2 |β|. Into this circle inscribe the cycle, point down, as shown in Fig. 8-9 for benzene.
Project sideways the points where the polygon intersects the circle. The positions of these projections correspond to the HMO energy levels if the circle center is assumed to be at E = α (see Fig. 8-9). The number of intersections at a given energy is identical to the degeneracy. The numerical values for E are often obtainable from such a sketch by inspection or simple trigonometry.
For straight chains, a modified version of the above method may be used: For an n carbon chain, inscribe a cycle with 2n + 2 carbons into the circle as before. Projecting out all intersections except the highest and lowest, and ignoring degeneracies gives the proper roots. This is exemplified for the allyl system in Fig. 8-10.
Examination of the energy levels in Figs. 8-9 and 8-10 reveals that the orbital energies are symmetrically disposed about E = α. Why is this so? Consider the allyl system.
The lowest-energy MO has two bonding interactions. The highest-energy MO differs only in that these interactions are now antibonding. [See Fig. 8-5 and note that the coefficients in φ1 and φ3 are identical except for sign in Eqs. (8-27) and (8-29).] The role of the β terms is thus reversed and so they act to raise the orbital energy for φ3 just as much as they lower it for φ1. A similar situation holds for benzene. As we will see shortly, the lowest energy corresponds to an MO without nodes between atoms, so this is a totally bonding MO. The highest-energy MO has nodal planes between all neighbor carbons, and so every interaction is antibonding. An analogous argument holds for the degenerate pairs of benzene MOs. These observations suggest that the energy of an Figure 8-9 HMO energy levels for benzene produced by projecting intersections of a hexagon with a circle of radius 2 |β|.
6See Frost and Musulin [3].
Section 8-9 Some Simplifying Generalizations
Figure 8-10 HMO energy levels for the allyl system (n = 3) produced by projecting the inter sections of an octagon (n = 2 × 3 + 2) with a circle of radius 2 |β|.
MO should be expressible as a function of the net bond order associated with it, and this is indeed the case. The energy of the ith MO is given by the expression
E
ˆ
ˆ
i = φiHπ φi dv =
ckiχkHπ
cliχl dv
(8-53)
k
l
=
c
ˆ
ki cli χkHπ χl dv
(8-54)
k
l
When the atom indices k and l are identical, the integral is equal to α; when k and l are neighbors, it equals β. Otherwise it vanishes. Hence, we may write neighbors
Ei = c2kiα +
ckicliβ
(8-55)
k
k,l
However, c2 is q
ki
k,i , the electron density at atom k due to one electron in MO φi , and ckicli is pkl,i, the bond order between atoms k and l due to an electron in φi.
Therefore,
neighbors
Ei = qk,iα + 2 pkl,iβ
(8-56)
k
k
We have seen that the sum of electron densities must equal the total number of electrons present. For one electron in φi, this gives additional simplification.
bonds
Ei = α + 2β pkl,i
(8-57)
ktotal π -electron energy is the sum of one-electron energies. For nπ electrons bonds
Eπ = nα + 2β
pkl
(8-58)
k
Chapter 8 The Simple H ¨uckel Method and ApplicationsEXAMPLE 8-5 The total π energy of cycloheptatrienyl radical (C7H7) is 7α + 8.5429β. What is the bond order for any bond in this molecule, assuming it to be heptagonal?
SOLUTION The total π-bond order must be 8.5429/2 or 4.2714. This results from seven identical bonds, so each bond order is 4.2714/7 = 0.6102.
Does this pairing of energy levels observed for allyl and benzene always occur? It is easy to show that it cannot in rings with an odd number of carbon centers. Consider the cyclopropenyl system. The lowest-energy MO is nodeless, totally bonding and has an energy of 2α + 2β. [Note from Eq. (8-51) and also from the diagram method that every monocyclic system has a totally bonding MO at this energy.] To transform these three bonding interactions into antibonding interactions of equal magnitude requires that we cause a sign reversal across every bond. This is impossible, for, if c1 disagrees in sign with c2 and c3, then c2 and c3 must agree in sign and cannot yield an antibonding interaction.
Not surprisingly, this has all been considered in a rigorous mathematical fashion.
Systems containing a ring with an odd number of atoms are “nonalternant” systems. All other homonuclear unsaturated systems are “alternant” systems. An alternant system can always have asterisks placed on some of the centers so that no two neighbors are both asterisked or unasterisked. For nonalternants, this is not possible (see Fig. 8-11).
It is convenient to subdivide alternant systems into even alternants or odd alternants according to whether the number of centers is even or odd. With this terminology defined, we can now state the pairing theorem and some of its immediate consequences.
The theorem states that, for alternant systems, (1) energy levels are paired such that, for each level at E = α + kβ there is a level at E = α − kβ; (2) MOs that are paired in energy differ only in the signs of the coefficients for one of the sets (asterisked or unasterisked) of AOs.
It is easy to see that an immediate result of this theorem is that an odd-alternant system, which must have an odd number of MOs, must have a nonbonding (E = α) MO that is not paired with another MO. It is also possible to show that the electron density is unity at every carbon for the neutral ground state of an alternant system. The proofs of the pairing theorem and some of its consequences are given in Appendix 5.
Another useful short cut exists that enables one to sketch qualitatively the MOs for any linear polyene. The HMOs for the allyl and butadiene systems are given in Fig. 8-12. Notice that the envelopes of positive (or negative) phase in these MOs are similar in appearance to the particle in a one-dimensional “box” solutions described
Figure 8-11 (a) Even and (b) odd alternants have no two neighbors identical in terms of an asterisk label. (c) Nonalternants have neighbors that are identical.
√
φ would then give us a “bond order” of 2/ 2 = 1.414. It is more convenient in practice to divide this number in half, because then the calculated π -bond order for ethylene turns out to be unity rather than two. Since ethylene has one π -bond, this can be seen to be a more sensible index.
As a result of these considerations, the π -bond order (sometimes called mobile bond
√
order) of the allyl radical is 1/ 2 = 0.707 in each bond. (Electrons in φ2 make no contribution to bond order since c2 vanishes. This is consistent with the nonbonding
label for φ2.)
Generalizing the argument gives, for pij , the π-bond order between nearest-neighbor atoms i and j : all MOs
pij =
nkcikcjk
(8-48)
k
where the symbols have the same meanings as in Eq. (8-46). In cases in which partially filled degenerate MOs are encountered, the averaging procedure described in connection with electron densities must be employed for bond orders as well.
EXAMPLE 8-4 Calculate p13 for the cyclopropenyl radical, using data in Fig. 8-7.
SOLUTION There are 2 electrons in φ1 and the coefficients on atoms 1 and 3 are 1 √ , so
3
this MO contributes 2 × ( 1 √ )2 = 2/3. We allocate 1 electron to φ2. Since c3 = 0 in this MO,
3
2
the contribution to p13 is zero. The remaining 1 electron goes to φ
2
3, yielding a contribution of 1 × 1 √ × −2 √ = −1 . So p − 1 = 1 .
2
13 = 2 6
6
6
3
6
2
8-9 Some Simplifying Generalizations
Thus far we have presented the bare bones of the HMO method using fairly small systems as examples. If we try to apply this method directly to larger molecules, it is very cumbersome. A ten-carbon-atom system leads to a 10 × 10 HMO determinant.
Expanding and solving this for roots and coefficients is tedious. However, there are some short cuts available for certain cases. In the event that the system is too complicated to yield to these, one can use computer programs which are readily available.
For straight chain and monocyclic planar, conjugated hydrocarbon systems, simple formulas exist for HMO energy roots and coefficients. These are derivable from the very simple forms of the HMO determinants for such systems.5 We state the results without proof.
For a straight chain of n unsaturated carbons numbered sequentially, x = −2 cos[kπ/(n + 1)], k = 1, 2, . . . , n
(8-49)
clk = [2/(n + 1)]1/2 sin[klπ/(n + 1)]
(8-50)
where l is the atom index and k the MO index.
5See Coulson [2].
Chapter 8 The Simple H ¨uckel Method and Applications
For a cyclic polyene of n carbons, x = −2 cos(2xk/n), k = 0, 1, . . . , n − 1
(8-51)
√
clk = n−1/2 exp[2πik(l − 1)/n], i = −1
(8-52)
The coefficients derived from Eq. (8-52) for monocyclic polyenes will be complex when the MO is one of a degenerate pair. In such cases one may take linear combinations of these degenerate MOs to produce MOs with real coefficients, if one desires.
There is also a diagrammatic way to find the energy levels for linear and monocyclic systems.6 Let us consider monocycles first. One begins by drawing a circle of radius 2 |β|. Into this circle inscribe the cycle, point down, as shown in Fig. 8-9 for benzene.
Project sideways the points where the polygon intersects the circle. The positions of these projections correspond to the HMO energy levels if the circle center is assumed to be at E = α (see Fig. 8-9). The number of intersections at a given energy is identical to the degeneracy. The numerical values for E are often obtainable from such a sketch by inspection or simple trigonometry.
For straight chains, a modified version of the above method may be used: For an n carbon chain, inscribe a cycle with 2n + 2 carbons into the circle as before. Projecting out all intersections except the highest and lowest, and ignoring degeneracies gives the proper roots. This is exemplified for the allyl system in Fig. 8-10.
Examination of the energy levels in Figs. 8-9 and 8-10 reveals that the orbital energies are symmetrically disposed about E = α. Why is this so? Consider the allyl system.
The lowest-energy MO has two bonding interactions. The highest-energy MO differs only in that these interactions are now antibonding. [See Fig. 8-5 and note that the coefficients in φ1 and φ3 are identical except for sign in Eqs. (8-27) and (8-29).] The role of the β terms is thus reversed and so they act to raise the orbital energy for φ3 just as much as they lower it for φ1. A similar situation holds for benzene. As we will see shortly, the lowest energy corresponds to an MO without nodes between atoms, so this is a totally bonding MO. The highest-energy MO has nodal planes between all neighbor carbons, and so every interaction is antibonding. An analogous argument holds for the degenerate pairs of benzene MOs. These observations suggest that the energy of an Figure 8-9 HMO energy levels for benzene produced by projecting intersections of a hexagon with a circle of radius 2 |β|.
6See Frost and Musulin [3].
Section 8-9 Some Simplifying Generalizations
Figure 8-10 HMO energy levels for the allyl system (n = 3) produced by projecting the inter sections of an octagon (n = 2 × 3 + 2) with a circle of radius 2 |β|.
MO should be expressible as a function of the net bond order associated with it, and this is indeed the case. The energy of the ith MO is given by the expression
E
ˆ
ˆ
i = φiHπ φi dv =
ckiχkHπ
cliχl dv
(8-53)
k
l
=
c
ˆ
ki cli χkHπ χl dv
(8-54)
k
l
When the atom indices k and l are identical, the integral is equal to α; when k and l are neighbors, it equals β. Otherwise it vanishes. Hence, we may write neighbors
Ei = c2kiα +
ckicliβ
(8-55)
k
k,l
However, c2 is q
ki
k,i , the electron density at atom k due to one electron in MO φi , and ckicli is pkl,i, the bond order between atoms k and l due to an electron in φi.
Therefore,
neighbors
Ei = qk,iα + 2 pkl,iβ
(8-56)
k
k
We have seen that the sum of electron densities must equal the total number of electrons present. For one electron in φi, this gives additional simplification.
bonds
Ei = α + 2β pkl,i
(8-57)
ktotal π -electron energy is the sum of one-electron energies. For nπ electrons bonds
Eπ = nα + 2β
pkl
(8-58)
k
Chapter 8 The Simple H ¨uckel Method and ApplicationsEXAMPLE 8-5 The total π energy of cycloheptatrienyl radical (C7H7) is 7α + 8.5429β. What is the bond order for any bond in this molecule, assuming it to be heptagonal?
SOLUTION The total π-bond order must be 8.5429/2 or 4.2714. This results from seven identical bonds, so each bond order is 4.2714/7 = 0.6102.
Does this pairing of energy levels observed for allyl and benzene always occur? It is easy to show that it cannot in rings with an odd number of carbon centers. Consider the cyclopropenyl system. The lowest-energy MO is nodeless, totally bonding and has an energy of 2α + 2β. [Note from Eq. (8-51) and also from the diagram method that every monocyclic system has a totally bonding MO at this energy.] To transform these three bonding interactions into antibonding interactions of equal magnitude requires that we cause a sign reversal across every bond. This is impossible, for, if c1 disagrees in sign with c2 and c3, then c2 and c3 must agree in sign and cannot yield an antibonding interaction.
Not surprisingly, this has all been considered in a rigorous mathematical fashion.
Systems containing a ring with an odd number of atoms are “nonalternant” systems. All other homonuclear unsaturated systems are “alternant” systems. An alternant system can always have asterisks placed on some of the centers so that no two neighbors are both asterisked or unasterisked. For nonalternants, this is not possible (see Fig. 8-11).
It is convenient to subdivide alternant systems into even alternants or odd alternants according to whether the number of centers is even or odd. With this terminology defined, we can now state the pairing theorem and some of its immediate consequences.
The theorem states that, for alternant systems, (1) energy levels are paired such that, for each level at E = α + kβ there is a level at E = α − kβ; (2) MOs that are paired in energy differ only in the signs of the coefficients for one of the sets (asterisked or unasterisked) of AOs.
It is easy to see that an immediate result of this theorem is that an odd-alternant system, which must have an odd number of MOs, must have a nonbonding (E = α) MO that is not paired with another MO. It is also possible to show that the electron density is unity at every carbon for the neutral ground state of an alternant system. The proofs of the pairing theorem and some of its consequences are given in Appendix 5.
Another useful short cut exists that enables one to sketch qualitatively the MOs for any linear polyene. The HMOs for the allyl and butadiene systems are given in Fig. 8-12. Notice that the envelopes of positive (or negative) phase in these MOs are similar in appearance to the particle in a one-dimensional “box” solutions described
Figure 8-11 (a) Even and (b) odd alternants have no two neighbors identical in terms of an asterisk label. (c) Nonalternants have neighbors that are identical.