Ace Your Next Organic Chemistry
Exam.

With these Downloadable PDF Study Guides

Our Study Guides

Elimination Reactions

By James Ashenhurst

Elimination Reactions (2): Zaitsev’s Rule

Last updated: March 26th, 2019 |

So far, we’ve only looked at some simple elimination reactions where only one product is possible.  In this post we’ll look at some examples where we start to see some of the extra “wrinkles” that can be present in elimination reactions.

For example, if you heat the alcohol below with a strong acid (like sulfuric acid, H2SO4)  you obtain one major product (an alkene) and a minor product (also an alkene).

What’s interesting about this? Well, if you look closely you should see that actually two elimination products are possible here, but only one is formed as the major product.

The alkene which is “tetrasubstituted” – that is, attached to four carbon atoms – is the major product, and not the “disubstituted” alkene, which is attached to two carbon atoms and two hydrogen atoms.

(The fact that we’re forming a new C-C π bond at the expense of sigma bonds on adjacent carbons is characteristic of elimination reactions.)

Similarly, look at the product of this next reaction. Taking an alkyl bromide and adding a strong base, we again get a “major” product and a “minor product”.

Again, the major product is “more substituted” than the minor product. Of the 4 atoms directly attached to the alkene in the major product, 3 of them are carbon and 1 is hydrogen.  In the minor product,   2 carbon atoms and 2 hydrogen atoms are directly attached to the alkene.

So what’s responsible for this preference for the “more substituted” alkene in elimination reactions?

Well, this correlates nicely with an observation that’s been made regarding the heats of formation of various alkenes. As an alkene becomes more substituted (i.e. more carbons attached, fewer hydrogens attached) it becomes more thermodynamically stable. [This observation comes from measuring the enthalpy of hydrogenation for various alkenes –

This agrees nicely with the trend that’s observed for elimination reactions.  The major product of an elimination reaction tends to be the more substituted alkene. This is because the transition state leading to the more substituted alkene is lower in energy and therefore will proceed at a higher rate.

It was a Russian chemist named Alexander Zaitsev who published a paper making this observation back in the late 19th century, and therefore this observation has become known as Zaitsev’s Rule. Formally, the rule is that an elimination will occur such that a hydrogen is removed from the “β-carbon” with the fewest hydrogens. [Organic chemists and their terms: the “α-carbon” is the carbon attached to the leaving group, while “β-carbons” are all carbons attached to the alpha carbon.]

In the next post, we’ll go into a little more detail as to why so many elimination reactions seem to have the word “heat” written under them.

Next Post: Elimination Reactions Are Favored By Heat

Related Posts:

Comments

Comment section

10 thoughts on “Elimination Reactions (2): Zaitsev’s Rule

  1. Would you be able to include examples of when an elimination reaction is Anti-zaitsev’s? Is it only for bases that are strong and bulky (i.e. KOtBu, DBN or DBU)?

    1. Conjugation indeed plays a role in alkene stability, but this post focuses on the relationship between carbon substitution and alkene stability. It can be hard to predict situations where multiple variables are in play and conflict, such as deciding whether a tetrasubstituted non-conjugated alkene will form as opposed to a disubstituted conjugated alkene. Hence this post tries to keep things simple.

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.