## What is Zaitsev's Rule?

Alexander Zaitsev (also pronounced as Saytzeff), in 1875, prepared a rule to help predict the result of elimination reactions which stated, "The favored product in dehydrohalogenation reactions is that alkene that has the majority of alkyl groups attached to the double-bonded carbon atoms."

## What are Haloalkanes?

Haloalkane or alkyl halide is formed after replacing hydrogen atom in the respective aliphatic (compounds in which C and H are attached in straight chains or branched chains and are non-aromatic in nature) or aromatic (having cyclic rings in structure) hydrocarbon with a halogen.

## Reactions of Haloalkanes

### Elimination reaction

Elimination reaction as the name suggests is the process of eliminating one or more atoms either in pairs or in groups. It is usually carried out in the presence of acids or bases.

• Alpha C and alpha H:

Alpha carbon is the first C attached to a functional group, say carbonyl, in a hydrocarbon. Alpha H is the hydrogen attached to alpha carbon.

• Beta C and beta hydrogen:

ẞ hydrogen atom is the H attached to a ß carbon. ẞ C atom is the second C atom attached to a functional group in a hydrocarbon. This naming system is derived from Greek literature.

Elimination reaction of haloalkanes, also known as dehydrohalogenation is when haloalkane along with ẞ hydrogen atom is heated along with alcoholic KOH (potassium hydroxide). Here elimination of a hydrogen attached to ẞ C (ẞ H) and a halogen (X) atom from alpha C atom takes place. This reaction results in the formation of an alkene. Due to the removal of beta H atom, this reaction is also sometimes referred to as ẞ elimination reaction.

In the elimination reaction of haloalkanes, two different processes can be defined for the formation of products. The result of this reaction is the formation of alkenes.

Zaitsev/Saytzeff stated alkenes with less hydrogen on the C atom of the double bond would be observed as the preferred product.

### Elimination reaction when halogen is present on the terminal carbon

If there's presence of X in the terminating end of the haloalkane chain, only one possibility for elimination reaction exists, that is the one which results in terminal alkene as the product.

For example: 1-chlorobutane on elimination reaction forms but-1-ene.

$C{H}_{3}C{H}_{2}-\underset{\begin{array}{l}|\\ H\end{array}}{\stackrel{\beta }{C}}H-{\underset{\begin{array}{l}|\\ Cl\end{array}}{\stackrel{\alpha }{C}H}}_{2}+KOH\left(alc.\right)\stackrel{heat}{\to }C{H}_{3}C{H}_{2}CH=C{H}_{2}+KCl+{H}_{2}O$

Steps to proceed with such a reaction:

1. Identify alpha X and beta H.

2. Eliminate them from the compound ( the metal of base might react with halogen to form salt and hydroxide , OH- of base might react with H+ to form water).

3. Saturate the bond between alpha and beta C to give an alkene.

Contrary to this if the X atom is present on any C atom, not just the terminal atom, the haloalkane can undergo elimination reaction in various directions (two or more) relying on number of different types of beta H available. As Zaitsev claims, out of all these, alkene which is more highly substituted would be considered as the major product of this reaction. Zaitsev product is the major product, typically favored over the least substituted alkene, known as the Hofmann product. Zaitsev product is considered more stable as thermodynamically more substituted alkene gives out less heat of hydrogenation.

For example, dehydrogenation of 2- bromobutane can give either but-1-ene or but-2-ene, out of which but-1-ene has a double bond between CH and CH2 and but-2-ene has a double bond between CH and CH.

$C{H}_{3}-C{H}_{2}-CH=C{H}_{2}\underset{-KBr}{\overset{heat,KOH\left(alc.\right)}{←}}C{H}_{3}-\underset{\begin{array}{l}|\\ H\end{array}}{\stackrel{\beta }{C}}H-\underset{}{\underset{\begin{array}{l}|\\ Br\end{array}}{\stackrel{\alpha }{C}}H-\stackrel{\beta }{C}{H}_{3}}\underset{-KBr}{\overset{heat,KOH\left(alc.\right)}{\to }}C{H}_{3}-CH=CH-C{H}_{3}$

So but-2-ene being more highly substituted alkene is the major product (81%, the Saytzeff product) in comparison to but-1-ene being less highly substituted minor product (19%, the Hofmann product).

If the haloalkane consist of multiple beta H atoms, such as in the case of secondary and tertiary C attached to X in haloalkanes, there is no formation of multiple alkenes, instead one alkene is preferred and formed as the major product. This is where Zaitsev's rule comes into action.

### Elimination reaction when the halogen is not on the terminal carbon

1. Identify alpha X and beta H atoms.

2.Eliminate them from the compound one at a time.

3. Saturate the bond between alpha and beta C to give alkenes of each case.

4. Identify the most highly substituted alkene.

5. Categorize it as the major product known as Saytzeff product and the minor product as the Hofmann product.

If the end products formed during Zaitsev elimination show cis - trans isomerism the trans alkene being more stable will always be made as the major product of the reaction.

For example, 2-bromobutane gives trans-2-butene as major product after elimination.

$C{H}_{3}-\underset{\begin{array}{l}|\\ H\end{array}}{\stackrel{\beta }{C}}H-\underset{\begin{array}{l}|\\ Br\end{array}}{\stackrel{\alpha }{C}H-\stackrel{\beta }{C}{H}_{3}}\underset{-KBr}{\overset{heat,KOH\left(alc.\right)}{\to }}\underset{\begin{array}{l}|\\ H\end{array}}{\overset{\begin{array}{l}C{H}_{3}\\ |\end{array}}{C}}=\underset{\begin{array}{l}|\\ C{H}_{3}\end{array}}{\stackrel{\begin{array}{l}H\\ |\end{array}}{C}+}\underset{\begin{array}{l}|\\ H\end{array}}{\overset{\begin{array}{l}C{H}_{3}\\ |\end{array}}{C}}=\underset{\begin{array}{l}|\\ H\end{array}}{\overset{\begin{array}{l}C{H}_{3}\\ |\end{array}}{C}}$

2-bromobutane                    trans-2-butene(71%)  cis-2-butene(10%)

## Ease of Dehydrohalogenation

As Saytzeffs rule suggests any haloalkane which forms more stable and more highly substituted alkene has to undergo elimination reaction at a quicker rate than the alkyl halide which gives lesser stable alkene. Say we have tertiary, secondary and primary alkyl halides chloroethanes, 2-chloropropane and 2-chloro-2-methylpropane.

The ease with which elimination reaction of these 3 alkyl halides is carried out, increases as number of beta H in the compound increases.

In general terms the dehydrogenation of different haloalkanes having the same X atom decrements in the form: tertiary>secondary>primary.

And for the same haloalkane having X atoms it decrements in the form:

R-I > R-Br > R-Cl > R-F.

## Hofmann Elimination

August Wilhelm von Hofmann discovered that the elimination reaction of amines forms the least stable and least substituted alkene as the product, also referred to as the Hofmann product.

The other name of this reaction is the Hofmann alkene synthesis rule and this is an exception of prevalent elimination reactions in which the Zaitsev rule predicts the formation of the most predominant alkene.

## Context and Applications

• Bachelors and Masters in Chemistry.
• Analytical Chemistry Research.
• Chemical Engineering
• Research in Biochemistry and polymers.
• Alcohols, carboxylic acids, acetic acid, acyl group, aryl group, and esters.
• Biomolecules.
• Nitration.
• Phenols, haloalkanes, halo arenes.

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