Bulky Bases in Elimination Reactions

by James

in Alkenes, Alkyl Halides, Organic Chemistry 1

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)[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.

Next Post: Comparing the E1 and SN1 Reactions

Related Posts:

{ 6 comments… read them below or add one }

azmanam

Do you have a primary literature reference for this? I sometimes clash (see what I did there?) with my colleagues who don’t teach the t-butoxide non-Zaitsev elimination. I’d love to prove them wrong once and for all :)

Reply

james

I am stealing my data from Reusch’s site. My version of March talks about leaving groups and base strengths re: Zaitsev/ non-Zaitsev, but does not mention t-butoxide. It’s unevenly taught across the country, I can tell you that.
Someday when I have my own chemistry lab I am going to go through all these substitution and elimination reactions and churn out real data so I don’t have to feel like I’m talking ex recto all the time.

Reply

azmanam

Gotcha. That’s where I always wind up, too. It’s frustrating for me, because students change professors at semester break, so I can’t very well hold new students responsible for a reaction their fall semester prof didn’t teach them… so I can’t really hold anyone in the class responsible for it 2nd semester… so why teach it at all in 1st semester?

Reply

Amir

Dear James:

I am studying for MCAT. and things are very confusing to me regarding this subject. Organic Chemistry by Smith does not recognize hofmann product at all and gives zaitsev product with tertiary butoxide even with tertiary. Organic Chemistry by Klein and Solomon give hofmann product with both tertiary and secondary alkyl halides (but no example of what happens with primary), and Organic chemistry by Bruice explicitly says tertiary butoxide with only tertiary gives hofmann but says secondary gives still zatisev. what am I supposed to do on a national exam when I don’t know which reference is used? what is the product with primary and secondary alkyl halide with tertiary butoxide?

Thanks

Reply

James

I don’t know what to tell you other than this is another issue which is a rat’s nest. It’s not taught consistently.

Reply

Alan

That’s an easy question to answer.

#1 rule is that the MCAT writers are ALWAYS RIGHT. So on a passage they will tell you what you need to know for the ochem passages and you use that info along with your background knowledge. The MCAT would never in a million years force you to rely on varying types of information for controversial topics. They will tell you what you think on the test for those types of questions.

Reply

Leave a Comment