Mechanism of Elimination Reactions

As noted earlier, the halogen-carbon bond in an alkyl halide is polarized due to the electronegativity difference between the atoms. This polarization can lead to the formation of a partial or fully positive charge on the carbon atom

The full or partial positive charge on the carbon atom is delocalized (dispersed) down the carbon chain. This, in turn, makes the hydrogen atoms attached to these carbons very slightly positive and thus very weakly acidic. Therefore, a very strong base can now remove slightly positive hydrogen with the resulting release of electrons down the chain, forming a π bond between the carbon atoms. The actual mechanism can be one of two types, E1 or E2, depending upon the structure of the activated complex.
E1 mechanism
An atom that bears a pair of unshared electrons takes on one of two roles. The atom may share these electrons with a carbon atom that bears a leaving group, or it may share these electrons with a hydrogen atom. In the former case, the atom acts as a nucleophile, while in the latter case it acts as a base. Therefore, depending on reaction conditions, the atom may be involved in a substitution reaction or an elimination reaction.
The reaction of an OH− ion with tertiary butyl bromide leads to little or no substitution product because steric hindrance blocks the rear lobe of the carbon atom to which the bromine atom is bonded. With the aid of a polar solvent, the bromine-carbon bond ionizes to form a tertiary carbocation and a bromide ion. The hydrogen atoms on the carbons adjacent to the carbocation carbon acquire a slight positive charge, allowing the OH− ion to employ its basic characteristics. Thus, the OH− ion abstracts a hydrogen atom, and the electrons migrate down the chain, forming a double bond. 

The activated complex for this reaction contains only the alkyl halide and is, therefore, unimolecular. The reaction follows an E1 mechanism

Elimination Reactions

During an elimination reaction, a bond forms by the removal of two atoms or groups from the original molecule. In most instances, the bond that forms is a π bond. Elimination reactions compete with substitution reactions when alkyl halides react with a nucleophile. 

The elimination of hydrogen halide (a halogen acid) from an alkyl halide requires a strong base such as the alkoxide ion, RO−. Weaker bases such as the OH− ion give poor yields of elimination product. 
If an alkyl halide contains more than two carbons in its chain, and the carbon atoms adjacent to the carbon atom bonded to the halogen each have hydrogen atoms bonded to them, two products will form. The major product is predicted by Zaitsev's Rule, which states that the more highly branched alkene will be the major product. For example, in the dehydrohalogenation reaction between 2-chlorobutane and sodium methoxide, the major product is 2-butene. 

SN1 mechanism

The second major type of nucleophilic substitution mechanism is the SN1 mechanism. This mechanism proceeds via two steps. The first step (the slow step) involves the breakdown of the alkyl halide into an alkyl carbocation and a leaving group anion. The second step (the fast step) involves the formation of a bond between the nucleophile and the alkyl carbocation

Because the activated complex contains only one species—the alkyl carbocation—the substitution is considered unimolecular.
Carbocations contain sp2 hybridized orbitals and thus have planar structures. SN1 mechanisms proceed via a carbocation intermediate, so a nucleophile attack is equally possible from either side of the plane. Therefore, a pure, optically active alkyl halide undergoing an SN1 substitution reaction will generate a racemic mixture as a product, as shown in Figure .