The reaction between 2-bromo-2-methylpentane and sodium ethoxide to form 2-methyl-2-pentene is an E2 elimination bimolecular reaction. In this reaction, the bromine atom is eliminated along with a hydrogen atom from the neighboring carbon, resulting in the formation of a double bond. Here's the detailed step-by-step process of the reaction:1. Formation of ethoxide ion: Sodium ethoxide NaOEt is a strong base, and it exists in an equilibrium with its conjugate acid, ethanol EtOH . In this step, sodium ethoxide dissociates into sodium ion Na+ and ethoxide ion EtO- .NaOEt Na+ + EtO-2. Deprotonation: The ethoxide ion EtO- acts as a strong base and abstracts a proton H+ from the -carbon carbon next to the carbon attached to the bromine atom of 2-bromo-2-methylpentane. This step generates an alkoxide intermediate.EtO- + CH3 2CBrCH2CH3 EtOH + [ CH3 2CBrCHCH3]-3. Formation of the double bond: The alkoxide intermediate formed in the previous step undergoes a concerted process where the electrons from the carbon-hydrogen bond move to form a carbon-carbon double bond C=C between the - and -carbons. Simultaneously, the carbon-bromine bond breaks, and the bromide ion Br- is eliminated.[ CH3 2CBrCHCH3]- CH3 2C=CHCH3 + Br-4. Final product: The resulting product is 2-methyl-2-pentene, which is formed along with the byproducts ethanol and bromide ion.Overall reaction: CH3 2CBrCH2CH3 + NaOEt CH3 2C=CHCH3 + EtOH + NaBr