The major product formed when 1-bromo-2-methylcyclohexane undergoes an E2 elimination reaction in the presence of a strong base is 1-methylcyclohexene.Mechanism of the reaction:1. The strong base e.g., ethoxide ion, -OEt abstracts a proton from one of the -carbon atoms adjacent to the carbon bearing the bromine atom. In this case, the -carbon is the 2-position on the cyclohexane ring.2. As the base abstracts the proton, the electrons from the C-H bond move to form a double bond between the - and -carbon atoms.3. Simultaneously, the C-Br bond breaks, and the bromide ion leaves as a leaving group.The stereochemistry of the product is determined by the fact that E2 elimination reactions require an antiperiplanar arrangement of the hydrogen atom being abstracted and the leaving group. This means that the hydrogen atom and the bromine atom must be in opposite planes of the cyclohexane ring.In the case of 1-bromo-2-methylcyclohexane, the most stable conformation is with the bromine atom in the axial position and the methyl group in the equatorial position. This arrangement allows for an antiperiplanar hydrogen atom to be abstracted by the strong base, leading to the formation of 1-methylcyclohexene as the major product.The stereochemistry of the product is determined by the fact that E2 elimination reactions require an antiperiplanar arrangement of the hydrogen atom being abstracted and the leaving group. This means that the hydrogen atom and the bromine atom must be in opposite planes of the cyclohexane ring.In the case of 1-bromo-2-methylcyclohexane, the most stable conformation is with the bromine atom in the axial position and the methyl group in the equatorial position. This arrangement allows for an antiperiplanar hydrogen atom to be abstracted by the strong base, leading to the formation of 1-methylcyclohexene as the major product.