We’ve recently talked about Zaitsev’s rule in elimination reactions, and how the transition state leading to the more substituted alkene is lower in energy. The bottom line in elimination reactions is that you should expect to form the “more substituted” alkene if at all possible. Like in the elimination reaction below, for instance, we get 80% of the tetrasubstituted alkene [“Zaitsev” – more substituted because there are 4 carbons attached to the alkene] and 20% of the disubstituted “non-Zaitsev” product.
However, today we’ll talk about one interesting exception to this “rule” and how under certain conditions we actually end up with the “non-Zaitsev” alkene product instead.
For instance, instead of using sodium methoxide, (NaOMe) if you use the base NaOC(CH3)3 [or KOC(CH3)3, changing sodium for potassium doesn’t really matter here] you end up with an interesting reversal of products!
So what’s going on here? Why might we get less of the Zaitsev product here and more of the “non-Zaitsev” product?
Well, if we draw out what the structure of the reactants might look like in their transition state, we can start to see why. [Note: this is not technically a transition state since we’re not drawing partial bonds, but you can at least see how the reactants are assembled]. The base in this instance – potassium t-butoxide – is an extremely bulky base, and the proton we remove to form the Zaitsev product is on a tertiary carbon. As the oxygen from the base draws nearer to this proton, a steric clash occurs. In essence the electron clouds around the methyl groups are interacting with each other, and the repulsion between these clouds will raise the energy of the transition state [remember – opposite charges attract, like charges repel]. This will slow down the reaction.
Looking at the reactant assembly that produces the non-Zaitsev product, the bulky base is removing a proton from a primary carbon. Steric clash is considerably reduced in comparison to that for the Zaitsev product. Elimination is faster, and we therefore end up with the less substituted alkene as our major product. Note: for an excellent set of 3D models that show the differences in excellent detail, visit this page on Reusch’s site.
This is one example of a reaction where the more thermodynamically stable product is not formed [recall that alkene stability increases with the number of carbons directly attached to the alkene].
So the bottom line for this post is that when performing an E2 reaction, using a bulky base will produce a greater proportion of non-Zaitsev alkene products relative to a less bulky base.
As far as we’ll see, the most common “bulky base” we need to consider is the t-butoxide ion, which can be drawn in many forms [see diagram]; occasionally you might see lithium di-isopropyl amide (LDA) used as well. For our purposes this completes the roster of bulky bases.
In the next post we’ll talk about an interesting observation we can make during certain E1 reactions.