The Chemical Bond
Reactions of Alkyl Halides
|Picture the mechanism for E1 Elimination. Imagine that the halide has left, leaving the carbocation. In some molecules, there may be a choice of protons from either side of the carbocation for the deprotonation step, which follows, forming the double bond. Which one will be removed? Zaitsev's rule states that E1 Elimination favors the product with the double bond oriented towards the more highly substituted carbon. This is because the transition state in the product determining step has some double bond character, and more highly substituted double bonds are lower energy.|
An interesting complexity is that the more basic the counterion that pulls off the proton, the more the transition state will look like the carbocation, and the less double bond character it will have. A strong base is not required for E1 elimination because deprotonation is not the rate determining step, however the basicity does determine the regioselectivity.
Notice that there is a carbocation intermediate, so product can form with an altered carbon skeleton due to carbocation rearrangement.