Zaitsev's Rule: Definition & Examples

Instructor: Amanda Robb

Amanda holds a Masters in Science from Tufts Medical School in Cellular and Molecular Physiology. She has taught high school Biology and Physics for 8 years.

In this lesson we'll be learning about elimination reactions in alkenes. We'll discuss Zaitsev's rule and what conditions favor Zaitsev products versus Hoffman products.

What Are Elimination Reactions?

On a last minute trip to the beach, you pick up a cooler at the grocery store. Made of fluffy, white Styrofoam, it's sure to keep your drinks cool for the day. You bag your groceries in convenient plastic bags and head out. Have you ever wondered how these modern miracles are made? Styrofoam, plastic, and many other common materials are made from alkenes.

Alkenes are hydrocarbon chains with at least one double bond. Chemists synthesize all sorts of different alkenes for different purposes, such as insulation, packaging, manufacturing and fuel. To do this, they use a chemical reaction called an elimination reaction. In this reaction, a hydrocarbon chain has two substituents leave the molecule and the electrons are converted into a double bond between carbon atoms.

Alkenes are synthesized through elimination reactions
elimination reaction

Zaitsev's Rule

During elimination reactions, often there is more than one place that a double bond can be formed around the leaving group. Let's look at an example. Say you treat 2,2-bromo-methyl-pentane with a strong acid, which will result in an elimination reaction. Clearly the bromide ion will leave, but will the double bond form between the methyl group and the main chain carbon, or within the main chain carbons themselves?

2,2-bromo-methyl-pentane is reacted with a strong base during an elimination reaction to produce 2,2 methyl-pentene
elimination reaction

When this reaction is carried out, the overwhelming product is 2,2 methyl-pentene. The double bond forms between the carbons in the main chain, not the methyl groups. Other reactions produce similar results.

Let's say you combined 1,1 chloro-methyl-benzene with a strong base and heat to carry out an elimination reaction. Will the double bond form within the ring or with the methyl group? Again, most of the product will have a double bond within the ring.

As it turns out, elimination reactions tend to produce products where the double bond creates the most highly substituted alkene. But, like all things in chemistry, this isn't just for looks. There's a good reason why alkenes with more substituents are the greater product.

More highly substituted alkenes tend to be more thermodynamically stable compared to less substituted alkenes. This process was observed and reported by a Russian scientist, Alexander Zaitsev, in the 19th century. Today, we refer to this as Zaitsev's rule, which states that the more highly substituted alkene is the more likely product of an elimination reaction. Thus tetra substituted alkenes are more stable compared to tri-substituted, compared to di-substituted, compared to mono-substituted alkenes.

But what makes more substituted alkenes more stable? The answer is a force called hyperconjugation. Here, the electrons are delocalized over adjacent pi orbitals of neighboring carbon atoms around the double bond. Thus, the more substituents an alkene has around the double bond, the more hyperconjugation that can occur and the more stable the molecule will be.

Reactions that Favor Hoffman Products

It seems like based on thermodynamics, we should always get the Zaitsev product. But remember, most elimination reactions produce a mixture of products, the Zaitsev product and the lesser substituted alkene, called the Hoffman product. Why do Hoffman products form at all if they are less stable?

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