Describe the mechanism of elimination reactions (E1 and E2) in haloalkanes with examples.

Elimination reactions in haloalkanes involve the removal of a halogen atom and a hydrogen atom to form an alkene. There are two types of elimination reactions: E1 and E2.

1. E1 (First Order Elimination) Reaction: In an E1 reaction, the reaction rate is proportional to the concentration of the haloalkane only, hence the name first order elimination.

Step 1: Ionization The reaction begins with the ionization of the haloalkane in the presence of a polar solvent, leading to the formation of a carbocation and a halide ion. This is the slow, rate-determining step.

R-X → R+ + X-

Step 2: Deprotonation The carbocation then loses a proton to a base, forming an alkene.

R+ + :B → R=B + H-B

For example, in the dehydration of 2-bromo-2-methylpropane, the bromine leaves first to form a tertiary carbocation. Then a base (water in this case) removes a proton from an adjacent carbon, forming an alkene (2-methylpropene).

2. E2 (Second Order Elimination) Reaction: In an E2 reaction, the reaction rate is proportional to the concentrations of both the haloalkane and the base, hence the name second order elimination.

Step 1: Single concerted step In this mechanism, the base abstracts a proton from the haloalkane while the halogen leaves, forming a pi bond. This happens in one concerted step, meaning it happens simultaneously.

R-X + :B → R=B + H-B + X-

For example, in the reaction of 2-bromo-2-methylpropane with a strong base like hydroxide, the base abstracts a proton from an adjacent carbon while the bromine leaves, forming 2-methylpropene.

In both E1 and E2 reactions, the product is an alkene. The major product can be predicted using Zaitsev's rule, which states that the most substituted alkene (the one with the most alkyl groups attached to the carbon atoms of the double bond) will be the major product.