If you’re asked to calculate the number of stereoisomers a compound forms in the exam, you’ve got 2 choices: either do it the long way round and draw them all out until you’re convinced you’ve got them all, or you can follow this quick tip that will allow you to calculate it in 2 seconds flat.
Before I teach you the tip, we need to revise stereoisomers.
What are stereoisomers?
Stereoisomers are compounds that have the same structural formula but a different arrangement in space. In order to form stereoisomers, a compound needs to have a carbon-carbon double bond (and each carbon needs to have 2 different groups), a chiral centre (a carbon with 4 different groups), or both of these structures. Stereoisomers formed by compounds with a c=c are referred to as geometrical isomers (or E/Z isomers, or cis-trans isomers) and stereoisomers formed by compounds with a chiral carbon are optical isomers.
Calculate the number of stereosiomers without drawing them all out
To quickly calculate the number of stereoisomers, you can use this formula:
The number of stereoisomers = 2ⁿ, where n is the number of c=c &/or chiral carbons.
For example, if you have a compound that has 3 c=c (where all of the double-bonded carbons 2 different groups each) n = 3. This means the compound will have 3 stereoisomers (2³ =8).
And if your compound has 1 chiral carbon and 1 c=c (each carbon has 2 different groups) n is 2, so the compound forms 4 stereoisomers.
This tip couldn’t be simpler, but as you can see, you need to make sure you’ve correctly worked out how many chiral carbons and relevant c=c you have. You’re likely to find it easier if you have the displayed formula, so if they’ve given you anything else (structural formula, skeletal formula 😱 etc.) it’s worth taking the time to draw our the displayed formula.