Union of even AHs

From the pairing theorem, it follows that all the bonding MOs in R and S are filled with pairs of electrons and all the antibonding MOs are empty (Fig. 3.7a). If an MO of R has the same energy as one of S (i.e., if the two MOs are degenerate), it follows that either both are filled with pairs of electrons or both are empty. The first-order interaction between the MOs therefore leads to no change in the total energy of the system (see Section 2.5 and Fig. 

Any change in total energy when R and S unite is therefore due to second-order perturbations. 

Since the second-order interaction between two MOs raises one up in energy by the same amount that the other is reduced, the sum of the energies of the two orbitals remains unchanged. The second-order interactions between filled MOs of R and S therefore also fail to alter the total energy (Fig. 3.7c). 

It can be shown (see the appendix at the end of this chapter) that this overall change in n energy, i.e., the n energy of union, is not only small but also has approximately the same value for union of any two even AHs, 

Since the bonds in ethylene are localized (Section 1.13), its heat of atomization El can be written as minus a sum of the corresponding bond energies, 

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Since the union of even AHs involves a second-order perturbation, and since second-order perturbations are not additive (Section 2.6), the additivity of bond energies does not extend to cyclic polyenes. 

Second-order perturbations are not additive. In double union of even AHs, the second-order terms are large and cannot be ignored. 

Ethylene can be regarded as the simplest possible even AH, with one starred atom and one unstarred atom ((5H2=CH2) and butadiene (CH2=CH—CH=CH2) can be formed by union of two molecules of ethylene. Now equation (3.14) applies to the union of any pair of even AHs the energy of union 3E of ethylene to butadiene is therefore given by... 

There is one type of even AH in which the numbers of starred and unstarred atoms are the same but for which one can nevertheless write no classical structure. These are AHs derived by union of two odd AHs through an inactive position in each 3,3 -bibenzyl (Fig. 3.17a) is a simple example, containing equal numbers of starred and unstarred positions (Fig. 3.17b). The reason for the lack of a classical structure for these molecules is easily seen. Any classical structure for an odd AH radical necessarily has the odd electron at a starred position. In order to get a classical, even AH from it, we must unite it with another odd AH through one of its starred atoms. The... 

In our PMO treatment of even AHs by union of pairs of odd AHs (Section 3.8), it is clear that the perturbed NBMOs represent approximations to the HOMO and LUMO of the resulting even AH (see Fig. 3.9). The separation between them is thus given [see equation (3.23)] by... 

The relationship between lOAHs and ROAHs is well-defined. ROAHs are formed by union (<-u->) between even AHs and inactive positions of lOAHs (e.g., 20 -> 21). ROAH anions, therefore, can be conveniently regarded as derivatives of lOAH anions. The even AH substituents, which have little or no influence on the nature of the anion, can be treated independently. These even and odd fragments are described as cross-conjugated. °... 

This treatment can be extended at once to the larger even cyclic polyenes. These can be formed by union of a linear, odd AH with methyl,... 

Bicyclic polyenes, even nonalternant ones, can be treated in a similar manner since they can be derived by union of an open-chain, odd AH and methyl. Figure 3.10 illustrates the treatment of the Cio series by union of methyl and nonatetraenyl. 

The procedure developed above can be used quite generally to study the stability of cyclic even AHs, and also of nonalternant even hydrocarbons in which there are two adjacent odd-numbered rings. It cannot, however, be extended to other odd systems since an odd conjugated system cannot be derived by union of two odd AHs. Here we need the third type of perturbation considered in Section 2.4, i.e., intramolecular union between two positions in an AH. 

FIGURE 3.15. Union of an odd AH (b) with methyl to form a nonclassical, even AH (a) involves no first-order change in n energy. The nonclassical AH is therefore less stable than classical isomers (c, d). 

MO vanish if the coefficient at the point of union vanishes. The nonclassical, even AH therefore also possesses an NBMO. But from the pairing theorem, all the MOs in an even AH are paired. NBMOs in such a system must occur in pairs. The nonclassical, even AH therefore possesses two NBMOs. These, however, have to accommodate only two electrons between them. From Hund s rule, the stable state of the system will then be a triplet in which the electrons occupy different MOs with parallel spins. Such a structure, with two unpaired electrons, should behave as a biradical. 

One type of intermolecular union which we have not yet considered is that of an even AH to an odd AH. The discussion will be given only for the simplest case, that of union of an even AH with methyl, since the extension to other odd AHs will be self-evident. 

FIGURE 3.18. (a) Interaction of orbitals in union of an even AH R with methyl (b) calculation of the n energy of union. 

The even AH R can be constructed by union of methyl with the odd AH R derived from R by removal of atom k (Fig. 3.23a). To a first approximation, this union involves only an interaction between the NBMO Oq of R and the 2p AO ij/ of methyl (Fig. 32.3b). Since these orbitals are degenerate, the interaction between them is analogous to that between two hydrogen atoms when they combine to H2 (Section 1.5 p. 13). The resulting bonding MO S in Fig. 3.23(b) will then be composed equally of the two interacting orbitals. 

FIGURE 3.23. (a) Union of methyl with odd AH R to regenerate the even AH R (b) first-order perturbation of orbitals during union (c) corresponding perturbation when atom k is replaced by a heteroatom electronegative relative to carbon. 

In this appendix, we will complete the proof that the union of two even AHs results in a small and constant change in n energy. Consider the union of two even AHs R and S through atom r in R to atom s in S (see Fig. 2.2). 

We use the notation of the previous chapter (Section 2.5). Since the bonding MOs in an even AH are all filled and the antibonding ones are empty, any degeneracy between R and S involves pairs of filled MOs or pairs of empty MOs. The n energy of union A rs is then a sum of second-order perturbations [equation (2.19)], involving interactions between filled MOs of R and empty MOs of S, and between filled MOs of S and empty MOs of R i.e.. 

Thus to this approximation, the n energy of union has a constant value, independent of the even AHs. 

Let us next consider the effect of introducing heteroatoms. To get a neutral substituent rather than an ionic one, it is necessary that one of the active atoms in the carbanion HR should be replaced by a heteroatom. The 7r-electron density at that position will then be 1 + with or the corresponding NBMO coefficient. Consider now introduction of the substituent into an odd AH cation HS". The resulting derivative RS will be an even AH (identical with that produced by union of the odd AH radicals HR- and HS Fig. 4.15a) the 7r-electron density of the heteroatom will now be unity. The energy of union dEj, of the heteroatomic substituent (HR )" with HS" will therefore be less than that (5Ej,) of the carbanion HR" with HS", the difference between the two energies being given by... 

If we introduce such a substituent into an even AH, HS, to form an odd AH anion, RS", the group S will form an inactive segment. The 7r-electron densities of the atoms in S will therefore still be unity. The substituent R therefore behaves just like a + substituent, and indeed it can be shown that the second-order n energy of union is the same as that between two even AHs. A substituent derived from an odd AH anion by loss of a hydrogen atom from an inactive position therefore behaves as a substituent. 

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