Alkyl Halide Reactions: Alkene formation using Strong Bases (E2 Zaitsev Product)

Alkyl halides can sometimes undergo elimination reactions when exposed to strong bases (NaOH, NaOMe, etc…), and elimination reactions almost exclusively occur when the reaction is run with heat (Δ):

This reaction proceeds through an E2 elimination reaction, where the halide is removed and a double bond is formed with the most substituted carbon atom (Zaitsev rule). An elimination is more likely to occur with secondary and tertiary alkyl halides as opposed to primary alkyl halides, where an SN2 substitution will occur. Heat can be applied to cause the SN2 substitution to become an E2 elimination:

SN2 vs E2 Reaction Conditions

The reaction mechanism is depicted below using sodium methoxide (NaOMe) however, it would proceed the same way using the other sterically free strong bases:

One of the hallmarks of SN2 and E2 reactions is that they occur in 1 step! The free electrons from the negatively charged oxygen atom on the NaOMe molecule pull a proton off a carbon atom (the most substituted carbon is the most likely to “give up” a proton). The free electrons then form a bond between this carbon and the carbon bound to the halide (Cl in this example). This pushes the chloride ion off, resulting in the formation of an alkene.

To learn more about SN2 and E2 reactions and how to distinguish them, be sure to check out our Lessons!

Practice this reaction using our Reaction Solver!