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 a review 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 nomenclature 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-Prelog 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: List the attached atoms looking for the first point of difference. Here we have C and H atoms attached.	Step 3: Assign the relative priorities. Since the atomic numbers C > H then the -CH₃ group is higher priority.	Step 4: Look at the relative positions of the higher priority groups: same side = Z, hence (Z)-but-2-ene


Step 1: Split the alkene	Step 2: List the attached atoms looking for the first point of difference. Here we have C and H atoms attached.	Step 3: Assign the relative priorities. Since the atomic numbers C > H then the -CH₃ group is higher priority.	Step 4: Look at the relative positions of the higher priority groups: opposite side = E, hence (E)-but-2-ene

Example: 3-methylpent-2-ene


Step 1: Split the alkene	Step 2: Assign the relative priorities. The two atoms attached to the left end C are C and H, so since the atomic numbers C > H then the -CH₃ group is higher priority.	Step 3: The two atoms attached to the right end C are both C. So we need to look at the atoms these C are attached to. List them out...we have C (H,H,H) and C (C,H,H) systems. This means that at the first point of difference (see the arrow), a C > H meaning that the ethyl group is higher priority.	Step 4: Look at the relative positions of the higher priority groups : opposite side = E, hence (E)-3-methylpent-2-ene.