1. For cis-trans isomers to exist
(stereogenic atoms)
Alkenes exist as cis-trans isomers when the two groups on each end of the C=C are different.
If A is not the same as B, and D is not the same as E, then cis-trans isomers exist.
In this case, the two C’s of the C=C are called stereocenters or stereogenic atoms.
stereocenter: atom at which interchange
of two groups produces a stereoisomer

2. Designating the Configuration of Cis-Trans Isomers
configuration: the three-dimensional arrangement of groups
about a stereocenter
Cis and trans are one way to designate the configuration of an alkene. However, these terms can be ambiguous.
Is this the cis isomer or the trans isomer? We can say the methyl group is cis to the Cl or trans to the F. We need a general way to choose a group on each end of a C=C to compare to determine whether the compound is the cis isomer or the trans isomer. A method has been developed to determine which groups we should compare. To avoid confusion, the newer method does not use cis and trans. Z
(zusammen – together) E
(entgegen – opposite)

4. Cahn-Ingold-Prelog Sequence Rules
Rule 1: Of the two atoms attached to one C of the C=C, the one
with the higher atomic number has the higher priority.
C has higher
priority than H
Cl has higher
priority than F
(Z)-1-chloro-1-fluoro-1-propene
The higher priority group one each end of the C=C is chosen using the Cahn-Ingold-Prelog sequence rules, which are based on atomic number.
Rule 1 is enough to assign the configuration of the molecule we examined before.
So this is the Z stereoisomer.

5. Rule 2: If the two atoms attached to the carbon are the same,
compare the atoms attached to them in order of
decreasing priority. The decision is made at the first
point of difference.
C bonded to O and 2 H’s
C bonded to 3 H’s
C bonded to 2 C’s
and one H
It may be necessary to continue farther out a chain of atoms until the first point of difference is reached. Let’s look at an example.
The higher priority groups are trans, so this is the E stereoisomer.
C bonded to C and 2 H’s

6. Rule 3: Double bonds and triple bonds in the groups attached
to the C=C are treated as though they are constructed
from two or three single bonds, respectively.
An easy way to visualize this is to replace the pi bond with a single bond to the same kind of atom.

7. This is the Z-stereoisomer.
So let’s try an example.

8. FIGURE 6.5: PLOT OF ENERGY VERSUS DIHEDRAL ANGLE FOR CONFORMATIONS OF ETHANE.
Fig. 6-5, p. 188

10. the various shapes that a molecule can assume
Conformations:
the various shapes that a molecule can assume
by rotations about single bonds
Newman projections
Now we’re going to look at another type of 3D shapes - conformations.
Recall that we said that there was free rotation about single bonds. Let’s examine the shapes that are produced by these bond rotations.
Consider ethane. There are two extreme conformations.
Remember the meaning of the wedged and dashed bonds.
Easier to see in an Newman projections. Looks down the C-C bond.
Note the dihedral angle.
Experiments have shown that the eclipsed conformation is 2.9 kcal/mol less stable than the staggered. The MO’s interact more favorably in staggered than eclipsed. We say that the eclipsed has 2.9 kcal/mol of torsional strain.
less stable
more stable by 2.9 kcal/mol
torsional strain: the destabilization caused by eclipsed bonds

11. ACTIVE FIGURE 6.7: PLOT OF ENERGY VERSUS DIHEDRAL ANGLE FOR CONFORMATIONS OF BUTANE.
Fig. 6-7, p. 191

12. FIGURE 6.8: CYCLOPROPANE. (A) BOND ANGLES, (B) ANGLE STRAIN, AND (C) TORSIONAL STRAIN.
Fig. 6-8a, p. 195

13. Table 6-1, p. 194

14. less angle strain than cyclopropane
Cyclobutane
less angle strain than cyclopropane
lots of torsional strain (if planar)
lowest energy conformation is nonplanar
Lowest energy conformation is actually nonplanar, decreases torsional, increases angle.
Cyclobutane rings are less common because they are more difficult to prepare.
angle strain + torsional strain = 26.0 kcal/mol
Cyclobutane rings are less common.

16. equatorial hydrogens axial hydrogens
There are two types of hydrogens in the chair conformation.
Axial H’s are parallel to axis of ring.
Equatorial H’s extend outward from the equator of the ring.

18. axial methyl is destabilized by 1.7 kcal/mol (axial strain energy)
Methylcyclohexane
axial methyl is destabilized by
1.7 kcal/mol (axial strain energy)
K = 18 5%
95%
Next let’s look at a compound that has a substituent on the ring of cyclohexane. The methyl group is equatorial in one chair conformation and axial in the conformation produced by ring-flipping. Let’s see how to draw and analyze.
We often begin with a structure like this with a flat ring (even though we know it is a chair). The one bond on each C is in front of the plane of the ring and one is behind. Turn it and convert to chair.
Axial methyl is destabilized. 1,3-diaxial interactions = 1.7 (in this case same as two gauche = 1.6)
So equilibrium favors equatorial. Small energy difference still gives a large K because of log factor.
1.7 gives K = 18.
CH3 is equatorial
CH3 is axial