Elimination Reactions (2): Zaitsev’s Rule

by James

in Alkenes, Alkyl Halides, Chemical Bonds, Organic Chemistry 1, Organic Reactions

So far, we’ve only looked at some very simple elimination reactions. 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 treat the alcohol below with a strong acid (like sulfuric acid, H2SO4) and heat, you obtain one major product (an alkene) and a minor product (also an alkene). As we talked about last time, the fact that we’re forming new C-C π bonds here is a sign that these are elimination reactions.

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

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”:

So what’s going on here? Note that in both of the elimination reactions we’re seeing that the major product is the one where the more substituted alkene is being formed (that is, the alkene attached to the most carbons)? Why might this be?

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 – click here for data]
This agrees with the trend that’s observed for elimination reactions.  The major product of an elimination reaction is 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

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{ 10 comments… read them below or add one }

Emily G.

Very helpful content review for MCAT studying! Thanks!



Glad it helped!


Jolene Wang

It is very helpful for Ochem study!! Thanks a lot!


John Stuart

Thank you so much! This is a lot clearer than my textbook. I’m finally starting to understand this stuff.



plz post on formation of alkanes alkynes alkenes wurtz reaction



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)?



Why is the major product in the first reaction “tetrasubstituted”? Isn’t it only bound to 3 carbons?



The major product in the first reaction is attached to four individual CH3 groups, hence, “tetrasubstituted”.



should we note here that double bond stability is not just about substitution but also about conjugation?



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.


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