REACTION OF 2-CHLORO-2-METHYLPENTANE WITH SODIUM METHOXIDE IN METHANOL

Introduction:
The reaction of 2-chloro-2-methylpentane with sodium methoxide in methanol is an example of an elimination reaction. The reaction takes place via an E2 mechanism and results in the formation of an alkene.

Reaction Mechanism:
The reaction proceeds through a concerted mechanism in which the leaving group (chlorine) and a proton are eliminated from the substrate simultaneously. This results in the formation of a double bond between the two adjacent carbon atoms, resulting in the formation of an alkene. The reaction mechanism is as follows:

Substrate + Base → Elimination → Alkene + Base

Products:
The products of the reaction are an alkene and the conjugate base of the base used in the reaction. In this case, the products are 2-methyl-2-pentene and sodium chloride.

Overall Reaction:
The overall reaction can be represented as follows:

2-Chloro-2-methylpentane + NaOCH3 → 2-Methyl-2-pentene + NaCl + CH3OH

Conclusion:
The reaction of 2-chloro-2-methylpentane with sodium methoxide in methanol results in the formation of 2-methyl-2-pentene, sodium chloride, and methanol. The reaction proceeds via an E2 mechanism and is an example of an elimination reaction.

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields 2-methoxy-2-methylpentane (substitution product) and 2-methylpent-2-ene (elimination product). B will be the minor product. Therefore, all the products will be formed.