Elimination Reaction – Core Concepts
In this tutorial, you will be introduced to a type of reaction in organic chemistry, the elimination reaction. More specifically, you will learn about one type of elimination reaction, E1, by walking through the mechanism and an example problem.
Topics Covered in Other Articles
- Alkene– a double (pi) bond
- Carbocation– molecule containing positively charged carbon ion
- Deprotonation– the act of removing a proton (H+ ion) from a molecule
- Intermediate– a compound that is formed then used during a reaction
- Ionization– a process where a molecule obtains a positive or negative charge from the removal or addition of electrons
- Leaving Group– an atom or group that can break away from a molecule
- Mechanism– visual process of reactions to show the movement of electrons and molecules through arrows
What are Elimination Reactions?
An elimination reaction consists of a formation of a C-C double bond by the removal of 2 single bonds from adjacent carbons; a leaving group attached to a carbon and a hydrogen attached to an adjacent carbon will usually be the two bonds broken in these types of reactions.
There are two types of elimination reactions, E1 and E2, which consist of different amount of steps. We will walk through E1 elimination below.
Although it may be confusing at first, keep in mind that the number associated with the reaction does not correspond to the number of steps, but rather the rate of reaction! E1 is a first-order unimolecular reaction, meaning the rate depends only on the concentration of the molecule.
The E1 elimination reaction has a two step mechanism that involves a carbocation intermediate:
The first step is an ionization step, in which the leaving group breaks away from the molecule, leaving a positive carbocation intermediate. The leaving group takes a full octet of electrons with it, including the electron it was sharing with the carbon to form the single bond. This electron movement is indicated by the arrowhead pointing from the bond to the X.
The second step is deprotonation of an adjacent carbon by a nucleophile, or Lewis Base. The base uses its valence electrons to grab onto the hydrogen, shown by the arrow going from the B to the hydrogen. The arrow pointing from the C-H bond to the space between carbons indicates the extra electrons this carbon now has will be shared with the positively charged carbon next to it, creating a neutral alkene product.
In the second step of the mechanism, a hydrogen is taken from an adjacent carbon in order for the double bond to form. But what if there are multiple adjacent carbons with hydrogens? Which do you choose?
E1 elimination favors the most substituted alkene product. There are four bonds branching from a double bond; if they contain any molecule apart from hydrogen, they play a part in the degree of substitution. The most stable configuration of an alkene is one that is the most substituted.
For example, here are two possible products of an E1 elimination reaction. The tetra-substituted alkene is more likely to be the major product than the disubstituted option.
Example of E1 Elimination
This is an example of a hydrocarbon chain undergoing E1 elimination. The first step, ionization, occurs when the bromine leaving group breaks away from the molecule, leaving a positive charge on the carbon. Then a nucleophile, in this case, water, grabs hydrogen from an adjacent carbon, creating a temporary negative charge. There were multiple hydrogens that the water could grab, but keep in mind that the most substituted alkene product is always favored. The electrons are then shared between carbons to create the neutral final product.