Hybridization: Many of the organic elements (principally carbon, nitrogen, oxygen, etc.) use hybrids of their valence orbitals instead of the orbitals themselves in bonding. They do this because electrons repel each other, and hybrid orbitals allow the electrons to be as far apart as possible. On problem set 3, you will prove that the four sp3 hybrids form a tetrahedron; a tetrahedron is the farthest apart you can get 4 orbital lobes (try it!). By conservation of energy, the sp3 hybrid orbital has energy _ of the way up between the lower energy s orbital and the higher energy p orbitals.

Sometimes an atom can’t use all of its p orbitals to make the hybrid because it needs an untampered-with p orbital to form a pi bond. If a p orbital is left out of the hybridization, it is perpendicular to the remaining hybrid orbitals because the other hybrids were made from the other p’s, and the 3 p orbitals are perpendicular to each other. The perpendicularity of the 3 p orbitals is evident in derivatives of allene. Derivitives of allene form classic hybridization and symmetry problems that show up frequently on Chem 32 examinations.

The molecule allene is shown below. The central carbon is sp hybridized. It uses one of its p orbitals (say for example the one in the plane of the page) to make a double bond to the leftmost carbon. Since that carbon is using a p orbital in the page to bond, its remaining sp² hybrids must be perpendicular to that. Therefore the H’s on this carbon are coming out of and going into the paper, respectively. The central carbon uses the other of its p orbitals (in this example, the one perpendicular to the plane of the page) to make a double bond to the rightmost carbon. Therefore, the remaining sp² hybrids of the rightmost carbon are in the plane of the page. Note that this molecule has an S₄ improper rotation axis. If the H’s were some other group, however, the molecule would not possess this symmetry element.

Formal Charge: Each atom in a molecule has a "formal charge," which equals the number of valence electrons in the neutral atom minus the lone pair that the atom has in the molecule minus half the electrons in each bond. Note that it is possible to have an atom with "half a bond" via resonance (see below). The sum of the charges of the atoms in a molecule always equals the charge on the molecule.

Charged molecules or atoms are typically less stable than neutral molecules or atoms, and this is especially true when the positive or negative formal charge is on carbon. For this reason, charged derivatives of carbon are especially reactive (unstable, high energy) meaning that they want to get rid of their charge. In addition, atoms want to follow the octet rule, so radicals, or atoms possessing an odd number of total electrons, are also unstable and reactive.

Saturation: Saturated molecules are literally saturated with hydrogen atoms; because of this they have no double or triple bonds and they have two ends. Every ring and pi bond counts as one site of unsaturation. For pure hydrocarbons, the number of sites of unsaturation can be calculated mathematically using the formula sites of unsaturation = (2n+2-m)/2. For the purposes of this formula, it is necessary to subtract 1 from m for each nitrogen occuring in the molecule and to count halogens as a hydrogen atom. Oxygen and sulfur can be ignored. It is usually easier to ignore the formula and count sites of unsaturation directly from a picture of the molecule, but you should know the formula in case you’re not given the picture.

Resonance: The electrons in alternating pi bonds can delocalize over all the atoms involved in the pi bonded system. This stabilizes the system because electrons repel each other and want to be far apart. Anytime p orbitals are lined up next to each other this will occur, because electrons always jump at the chance to have more space. In a carbocation or carbonium ion (a carbon atom in a molecule that has a plus charge), the positive charge is localized in a p orbital. When this p orbital lines itself up parallel to a pi bond, the system is stabilized by resonance because the electrons in the pi bond bleed extensively over to the attractive positive charge next door. In such situations, the bond is really more like a fraction of a bond. This configuration is called an "allyl cation."

Another example is the carbonate anion, which has p orbitals on every atom. Can you draw the three resonance structures contributing to this overall picture?

Aromaticity: Long chains of alternating (conjugated) pi bonds have a special stability because there’s lots of opportunity for the electrons to move. Benzene has 6 p orbitals lined up parallel and in a ring: the electrons are in delocalization heaven. They whir around and around through the p orbital system like a snake chasing its tail. Anytime every atom in a ring has a parallel p orbital and there are 4n+2 pi electrons, the molecule will do the benzene thing and be: 1. flat as a pancake (this is necessary for the p orbitals to be able to line up parallel to each other), 2. exceptionally stable. Why exactly 4n + 2? This requires some rather complex MO theory. Basically, if you derive all of the MO’s, you’ll find there’s room for exactly 4n + 2 electrons in the "bonding" (i.e. below the energy of the atomic p orbitals) MO’s. This is Huckel’s Rule. Putting electrons in low energy orbitals makes a molecule stable. Molecules that have conjugated systems (i.e. alternating double bonds in a ring) but do not have exactly 4n + 2 electrons are anti-aromatic, meaning that they are: 1. buckled instead of flat, and 2. not especially stable.

The molecular orbital diagrams below illustrate why this should be the case for benzene and cyclo-octatetraene. Cyclo-octatetraene is buckled instead of flat because it wants to avoid having to put its electrons into the high energy (above zero) molecular orbitals; if it buckles its MO diagram changes so that its net energy is not so unfavorable. Note that these are simple cases of aromatic and anti-aromatic systems: the MO diagrams for larger and multi-cyclic molecules are far more complicated and their treatment is beyond the scope of this course.

Note that if a non-carbon atom has lone or lone pairs of electrons that could make the molecule aromatic, that atom will hybridize in such as way as to put those electron(s) in a p orbital so that that can happen. A couple of examples of this are shown below.