In this tutorial, you will be introduced to another type of elimination reaction found in organic chemistry, E2. To further understand this reaction, you will go through the mechanism and walk through an example.
First, be sure to read the tutorial on E1 Elimination Reactions!
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
- 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
- Stereochemistry– physical orientation of the components in a molecule
- Stereospecificity– stereochemistry of the beginning reactant plays a role in how the molecule can react and determining stereochemistry of the final product
What are Elimination Reactions?
An elimination reaction consists of the 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.
E2 is a second-order bimolecular reaction, meaning the rate depends both on the concentration of the molecule and of the assisting base. Let’s walk through E2 elimination below!
E2 is a one step reaction mechanism with no intermediate:
The arrow pointing from the B to the H indicates the nucleophile, or Lewis base, grabbing onto the hydrogen, which deprotonates the carbon. At the same time, the leaving group breaks away from the molecule and takes a full octet of electrons with it, ionizing that carbon; this is indicated by the arrow pointing from the C-X bond to the X. The deprotonated carbon then shares its temporary negative charge with the adjacent carbocation; this is shown from the arrow going from the C-H bond to the C-C bond. This creates the double bond in the neutral alkene product.
This entire process takes place in one step, unlike the E1 reaction, which takes place in two.
Stereospecificity of E2 reactions
Again, there is the problem of deciding which hydrogen the base will grab onto. In E2 reactions, the product is dependent on the stereochemistry of the molecule. The leaving group and hydrogen must be “anti” to each other, meaning they must rest on opposite sides of the molecule (180 degrees apart). Since the deprotonation and the ionization are happening at the same time in this reaction, for there to be enough space, the two parts of the molecule must be as far away from each other as possible.
This concept might be tricky to visualize in your head, so here is an example of the correct versus incorrect stereochemistry in an E2 reaction. In the correct molecule, the leaving group is on a wedge and the hydrogen is on a dash, showing that they are ‘sticking out’ in opposite directions. This allows for the nucleophile to deprotonate the molecule at the same time that the leaving group breaks away. In the incorrect molecule, both the leaving group and hydrogen are on wedges; this does not give enough space for the reaction to occur.
Example of E2 Reaction
This is an example of a hydrocarbon chain undergoing E2 elimination. The iodine leaving group breaks away, shown by the arrow pointing from the C-I bond to the I. At the same time, the base, tert-butyl oxide, grabs onto the hydrogen that is “anti” to the leaving group; this is displayed through the arrow pointing from the oxygen of the base to the hydrogen. Then, the temporary negatively charged carbon and adjacent carbocation form another bond to create the neutral alkene product. This is shown by the arrow pointing from the C-H bond to the C-C bond.
- E1 vs. E2