Introduction
In comparison to the pi electrons of simple alkenes, the pi electrons of benzene are extremely non-reactive towards electrophiles. Such low reactivity suggests that the pi electrons of benzene are unusually stable. Valence bond theory attempts to rationalize this unusual stability by invoking a resonance argument. According to this line of thinking neither valence bond notation 1 nor 2, Figure 1, can adequatley represent the true structure of benzene. Rather, the true structure is closer to 3, which is described as a hybrid of structures 1 and 2.
Figure 1
Rationalizing Benzene
The implication of valence bond structure 1 is that each pi bond is localized between two carbon atoms, which means that the pi electron pair in each of these bonds experiences the nuclear attraction of two carbon atoms. The same goes for the pi electrons in structure 2. In structure 3 however, the pi electrons are no longer confined to regions between two specific carbon atoms. Rather they are free to roam over the entire 6-carbon atom framework. In essence, each electron experiences the nuclear attraction of 6 nuclei. Valence bond theory ascribes the added stability of the pi electrons in benzene to this increased nuclear attraction. Let's be clear- This is nonsense. Valence bond theory fails to adequately describe the physical and chemical properties of benzene. Modifying valence bond theory by incorporating the concept of resonance brings us closer to reality, but it's still a sham. We have seen that molecular orbital theory (MO theory) offers an alternative description of the bonding in molecules where a single Lewis structure does not adequately describe the structure of a compound. We are now going to take another look at MO theory.
MO Theory Revisited
In our introduction to molecular orbital theory we saw the molecular orbitals are formed by taking linear combinations of atomic orbitals. To describe the pi system in benzene we will combine one AO from each of the six carbons to form six MOs. Restricting our attention to those AOs that are involved in the formation of the pi system of benzene is justified because the pi system is orthogonal (perpendicular) to the sigma bonded framework of the molecule.
Figure 2 dissects the bonding in a molecule of benzene from a molecular orbital point of view.
Figure 2
Caution! Deconstruction Ahead
Note the following points:
- This model assumes that each carbon is sp2 hybridized.
- The s-bonded system is treated independently of the p-bonded system. This is justified by the assumption that the sp2 hybridized orbitals are orthogonal to the p orbital on each carbon.
- Electrons in the s-bonded orbitals are much lower in energy than those in the p-bonded system. Head-to-head overlap of orbitals provides greater nuclear attraction for the electrons than does side-to-side overlap.
- There are three types of bonds in benzene
- 6 C-C s-bonds formed by sp2- sp2 overlap
- 6 C-H s-bond formed by sp2-s overlap
- 3 C-C p-bonds formed by p-p overlap
While MO theory deals with sigma bonds as well as pi bonds, we will restrict our attention to the pi-bonded system since electrophilic aromatic substitution reactions involve the pi system of the aromatic ring. The MO diagram of the pi system of benzene is shown in Figures 3 and 4. This treatment assumes a pre-constructed sigma-bonded framework upon which we build the system of pi orbitals. The view in Figure 3 is looking down on the top of the orbitals, while that in Figure 4 is edge-on. The C and H nuclei appear as black dots. The dashed red line indicates the energy level of an isolated p orbital.
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| Students often ask the question "How do the electrons get from the red zone to the blue zone in y2 if the probability of finding an electron between these two zones is zero?" This question stems from the fact that these students are thinking about electrons as discrete particles, like tiny BBs. MO theory does not regard electrons as particles. Rather, it describes their behavior as waves. Einstein pointed out that at the molecular and sub-molecular level matter exhibits behaviors characteristic of both particles and waves; the so-called "wave-particle duality". In MO theory orbitals are pictures of three-dimensional standing waves, similar to a two-dimensional standing wave such as a sine wave. In a sine wave, e.g. y = sin x, a node simply represents that point at which the value of y changes from positive to negative. It has similar meaning in MO theory.  |
As the number of nodes in an orbital increases, so does the energy of the electrons that occupy that orbital. Since y2, and y3 each have two nodes, they have the same energy. They are said to be degenerate.
Exercise 1 How many nodes do y4 and y5 have? How many nodes does y6 have?
Exercise 2 When benzene reacts with an electrophile, which MO is involved? y2 y3 y2 or y3 y4 or y5
The most important feature of Figures 3 and 4 is the arrangement of the orbitals. Examine the pattern of MOs shown in Figure 5 for compounds containing 4, 6, 8, and 10-membered rings.
Figure 5
Stackin' 'em Up
The orbitals are arrayed symmetrically about a center line (------) which represents the energy level of an isolated p orbital. Orbitals that are lower in energy than an isolated p orbital are classified as bonding orbitals. Those that are higher are called antibonding orbitals. Those at the same level as an isolated p orbital are called non-bonding orbitals.
Aromaticity, Stability, Filled Shell Rules, and Hückel's Rule
There is a parallel here with the filled shell rules in that the most stable electronic arrangement occurs when all of the bonding molecular orbitals are filled such as in benzene and naphthalene. These systems are considered to be aromatic. The connotation of the word is that the pi electrons of aromatic molecules possess unusual stability in comparison to pi electrons in simple alkenes.
Exercise 3 How many pi electrons are there in anthracene,
Exercise 4 How many pi electrons are there in 
If you consider the MO diagrams of benzene, naphthalene, and anthracene, you will notice that all contain a single low-energy orbital, y1, which holds 2 electrons. Benzene also contains 1 pair of degenerate bonding orbitals, y2 and y3 which hold 4 electrons. Naphthalene contains 2 pairs of degenerate orbitals, y2 and y3 and y4 and y5 which hold 4 electrons each. Anthracene contains 3 pairs of degenerate orbitals, y2 and y3, y4 and y5, and y6 and y7 each of which holds 4 electrons. It is possible to generalize this idea in terms of a simple formula that indicates the number of pi electrons required for a cyclic system to be considered aromatic. This formula was devised by a physicist named Erich Hückel, and it is commonly referred to as Hückel's 4n+2 rule: cyclic planar molecules in which each atom has a p orbital are aromatic if they contain 4n + 2 pi electrons. Here n is an integer, 0,1,2,3,...., that represents the number of sets of degenerate bonding orbitals in the MO diagram of the molecule. It takes 4 electrons to fill each set. The number 2 refers to the two electrons in y1. In other words, pi systems containing 2, 6, 10, 14,...... electrons display unusual stability.
Anti-aromaticity
Molecules in which the highest occupied molecular orbitals are not completely filled, like those shown for cyclobutadiene and cyclooctatetraene are regarded as anti-aromatic. Here the connotation is that the pi electrons in anti-aromatic molecules are less stable than their counterparts in simple alkenes. In the most common situations anti-aromatic molecules contain 4n pi electrons, i.e. 0, 4, 8, 12, ..... electrons.
Aromatic Ions
There are many examples of organic cations and anions that display greater stability than you might expect at first glance. Several of these are shown in Figure 6.
Figure 6
I Get A Charge Out of You
The MO diagrams of these four ions are shown in Figure 7. You should compare the patterns of orbitals in this figure with the patterns shown in Figure 5.
Figure 7
Stackin' 'em Up: The Sequel
Exercise 5 How many pi electrons are there in the cyclopropenium ion? the cyclopentadienyl ion? the cycloheptatrienyl ion? the cylcooctadienyl dianion? Draw the MO diagrams of these compounds.
Exercise 6 Valence bond theory uses resonance to rationalize the stability of the ions in Figure 6. Draw structures to indicate how the charge is delocalized by resonance in the cyclopropenium ion, the cyclopentadienyl ion, and the cycloheptatrienyl ion. You should show 3, 5, and 7 resonance structures for these ions , respectively. Use curved arrows to show the movement of electrons.
While each of the ions in Figure 6 makes for an interesting case study, we will consider only one, namely the cyclopentadienyl anion. Compare the acid-base reactions shown in Equations 1 and 2.
The pKa of cyclopentadiene is 16!!! The equilibrium constant for reaction 1 is approximately 1. The pKa of 1,4-pentadiene is approximately 45. Why is cyclopentadiene so much more acidic than 1,4-pentadiene? More to the point, why is the conjugate base of cyclopentadiene so much more stable than the conjugate base of 1,4-pentadiene? According to MO theory the pi system of the cyclopentadienyl anion has a filled shell electron configuration. The unusual stability associated with this arrangement means is responsible for the ease of deprotonation of cyclopentadiene.
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