In an elimination reaction involving 2-methylcyclohexanol and a strong base like sodium ethoxide, the major product formed is 1-methylcyclohexene. The mechanism for this reaction is as follows:1. Deprotonation: The strong base sodium ethoxide abstracts a proton from the -carbon carbon adjacent to the alcohol group of 2-methylcyclohexanol. This forms an alkoxide intermediate and a sodium ion.2. Formation of the double bond: The alkoxide intermediate undergoes an intramolecular reaction, with the lone pair of electrons on the oxygen atom forming a double bond with the -carbon. Simultaneously, the carbon-oxygen bond breaks, and the electrons are transferred to the oxygen atom, resulting in the formation of 1-methylcyclohexene and a hydroxide ion.The major product, 1-methylcyclohexene, is formed due to Zaitsev's rule, which states that the more substituted alkene the one with more alkyl groups attached to the double bond will be the major product in an elimination reaction. In this case, 1-methylcyclohexene is more substituted than the alternative product, cyclohexene, and therefore is the major product.