Here’s an important little fact about alkenes.
The more carbons an alkene is attached to, the more stable it is. Like this:
A note on lingo: as we replace hydrogens with carbons, we usually say that the alkene becomes “more substituted”.
So alkene stability increases with substitution.
This also has an impact on elimination reactions.
Remember the pattern for elimination reactions:
- We’re breaking the C-(leaving group) bond
- We’re breaking a C-H bond that’s adjacent to the C-(leaving group) bond
- A new C-C Pi bond is formed between these two carbons
This brings up an important question, though. When there’s multiple C-H bonds to choose from, how do you know which one to pick?
Here’s the rule: break the C-H bond that will lead to the more substituted alkene.
Look at this reaction:
Here, see how there are two different alkenes that could be formed, but only one is the major product.
Note how the more substituted alkene is the major product.
We broke the C-H bond from the CH2, and not the CH3. In other words, we’re removing a hydrogen from the carbon attached to the fewest hydrogens.
This is called “Zaitsev’s rule”.
So when you form an alkene in an elimination reaction, make sure you form the most substituted alkene.
Thanks for reading! James
P.S. Another way of stating Zaitsev’s rule is that “the poor get poorer”. In other words, the carbon with the fewest hydrogens loses the hydrogen.
P.P.S. If you have the potential to form cis and trans alkenes, the trans alkene will be favored.
P.P.P.S There’s one prominent exception, using “bulky bases” – I’ll talk about it later this week.