The major product formed from the reaction between 1-bromo-3-methylcyclohexane and sodium ethoxide in ethanol solvent with a heat source is 1-ethoxy-3-methylcyclohexane. This reaction proceeds via an E2 bimolecular elimination mechanism.Mechanism of the reaction:1. Sodium ethoxide NaOEt acts as a strong base in this reaction. The ethoxide ion OEt- abstracts a proton H+ from the -carbon carbon adjacent to the carbon bearing the bromine atom of 1-bromo-3-methylcyclohexane. In this case, the -carbon is the one in the axial position, as it is anti-periplanar to the bromine atom, which is a requirement for the E2 mechanism.2. As the ethoxide ion abstracts the proton, the electrons from the C-H bond move to form a double bond between the -carbon carbon bearing the bromine atom and the -carbon.3. Simultaneously, the bromine atom leaves as a bromide ion Br- , taking its bonding electrons with it. This results in the formation of 1-ethoxy-3-methylcyclohexane as the major product.4. The byproduct of this reaction is sodium bromide NaBr , which is formed when the bromide ion associates with the sodium ion Na+ from sodium ethoxide.Overall, the reaction is an E2 elimination reaction, resulting in the formation of 1-ethoxy-3-methylcyclohexane as the major product.