6.2: Bonding, Stereoisomerism, and Stability

So, why do alkenes exhibit stereoisomerism in the first place? Recall the molecular orbital description for a C=C double bond. The carbon atoms are sp² hybridized, which means that each carbon atom has 3 sp² orbitals and an atomic p orbital. In ethylene, the C=C double bond would look like this, in which there is net bonding between the carbon atoms ($σ_{C-C}$) AND above and below the plane of the molecule ($π_{C-C}$). There are four electrons total in the C=C bond, two in the $σ_{C-C}$ and two in the $π_{C-C}$. We say that the $π_{C-C}$ is the highest occupied molecular orbital (HOMO) and that the $π^{*}_{C-C}$ is the lowest unoccupied molecular orbital (LUMO). The HOMO and LUMO are known as the Frontier molecular orbitals.

The double bond increases the bond strength by 62 kcal/mol to 145 kcal/mol (a $σ_{C-C}$ bond has a bond strength of 83 kcal/mol). Since there is net bonding above and below the plane, there is no longer rotation about the central C=C bond. Thus, the creation of stereoisomers. Whether the two highest priority groups are on the same side or opposite sides will create different geometric isomers.

One other important feature of alkenes is their relative stability. Alkenes that are more highly substituted are more stable than less highly substituted alkenes. Monosubstituted alkenes are higher in energy than disubstituted alkenes, which are higher in energy than trisubstituted, and so on.

The reason for this is $σ$-donation. Let’s consider 1-propene and 2-methyl-1-propene. We say that greater substitution stabilizes the molecule, which it does this through a-donation. Stability can be calculated by summing all of the bond enthalpies in the molecule. When $σ_{C-H}$ and $π^{*}_{C-C}$ orbitals mix, a new orbital is created which lowers the energy of the $σ_{C-H}$, making the alkene more stable overall. How does this happen? Well, remember that $σ_{C-H}$ bonds are electron-donating. Since $σ$-donation occurs many more times in a disubstituted alkene than a monosubstituted alkene, there is greater stabilization of the $σ_{C-H}$ bonds. There are six $σ_{C-H}$ bonds that can donate in a disubstituted system, but only three $σ_{C-H}$ bonds that can do so in a monosubstituted alkene.

Besides steric strain and $σ$-donation, alkene stability can also be effected by ring strain. Placing a double bond in a ring alters the conformation and contributes to angle strain. Alkene carbon atoms that are sp² hybridized want to have bond angles of 120°, but when they are placed in a ring, sometimes they are forced to be much smaller. This increases the energy of the cycloalkene. Larger rings (7-, 8-, 9-, etc.) do not suffer from angle strain in the opposite sense (>120° bond angles) because the ring can distort its conformation to keep the alkene carbons at 120°.

All of the molecules above are cis-cycloalkenes. In trans-cycloalkenes, the strain is even greater. Because trans-substituted groups are much farther apart, the smallest trans-cycloalkene possible is cyclooctene. Anything smaller than that is too strained.

Finally, we must consider alkenes in bicyclic systems. Bredt’s rule states that you cannot have a double bond at a bridgehead carbon. The reason for this is that there is no overlap of atomic p orbitals to create a $π_{C-C}$ bond. The atomic p orbital on the bridgehead is orthogonal to the atomic p orbital on the adjacent carbon, therefore there is no net bonding – the orbitals cannot mix because they are perpendicular to one another.