Alkene stereochemistry

Basic IUPAC Organic Nomenclature

Stereoisomers of Alkenes

As we have already described, alkenes with two different substituents at each end of the C=C can exist as a pair of stereoisomers.

The alkene can only exist as stereoisomers if R1 is not equal to R2 AND R3 is not equal to R4.

As we have already seen there are two ways to name these types of isomers: (for more details follow the links)

The E- and Z- style is more reliable (i.e. potentially less ambiguous) and particularly suited to highly substituted alkenes, especially when the substituents are not alkyl groups.

The E- and Z-alkene nomenclautre system is based on the Cahn-Ingold-Prelog priority rules. They can also be used for naming chirality center stereoisomers (see later).
In order to apply the Cahn-Ingold priority rules to alkenes:

Imagine each alkene as two pieces, each piece containing one of the sp² C atoms
Assign the priority (high = 1, low = 2) to each atom attached to each sp² C based on atomic number

Subrules:
- Isotopes: H vs D ? Since isotopes have identical atomic numbers, the mass number is used to discriminate them so D > H
- Same atom attached ? By moving out away from the C=C one atom at a time, locate the first point of difference and apply the priority rule there.
Determine the relative position of the two higher priority groups
- If they are on the same side then it is a (Z)-alkene (German; zusammen = together)
- If they are on opposite sides then it is an (E)-alkene (German; entgegen = opposite)
If there is more than one C=C that can be E/Z, then the locant and the stereochemistry of each alkene needs to be included e.g. (2E,4Z)-

Example: but-2-ene


Step 1 : split the alkene	Step 2 : assign the relative priorities. The two attached atoms are C and H, so since the atomic numbers C > H then the -CH₃ group is higher priority.	Step 3 : look at the relative positions of the higher priority groups : same side = Z, hence (Z)-but-2-ene.


		Step 3 : look at the relative positions of the two high priority groups are on the opposite side, then this is (E)-but-2-ene

Example: 3-methylhex-2-ene


Step 1 : split the alkene	Step 2a : assign the relative priorities. The two attached atoms on the left end are C and H, so since the atomic numbers C > H then the -CH₃ group is higher priority.	Step 2b : The two attached atoms on the right end are both C. So we need to look at the atoms these C are attached to. We have C (H,H,H) and C (C,H,H) systems. This means that at the first point of difference, a C > H meaning that the ethyl group is higher priority.	Step 3 : look at the relative positions of the higher priority groups : opposite side = E, hence (E)-3-methylpent-2-ene.