The acid-catalyzed dehydration of 1-methoxy-2-methylcyclohexane involves the following steps:1. Protonation of the methoxy group by a strong acid e.g., H2SO4 or H3PO4 to form a good leaving group methanol .2. Formation of a carbocation intermediate by the departure of the methanol group.3. Rearrangement of the carbocation if necessary to form a more stable carbocation.4. Elimination of a proton from a neighboring carbon to form a double bond alkene .There are two possible alkenes that can be formed from the dehydration of 1-methoxy-2-methylcyclohexane:A 1-methylcyclohexeneB 3-methylcyclohexeneHere's the step-by-step mechanism:1. Protonation of the methoxy group by a strong acid to form a good leaving group methanol . O ||CH3-O-CH CH3 C6H11 + H+ --> CH3-OH + CH CH3 C6H11+2. Formation of a carbocation intermediate by the departure of the methanol group. CH CH3 C6H11+ 3. Rearrangement of the carbocation if necessary to form a more stable carbocation. In this case, no rearrangement is needed because the carbocation is already at the more substituted carbon 2 carbocation .4. Elimination of a proton from a neighboring carbon to form a double bond alkene . There are two possible eliminations:A Elimination from the carbon next to the methyl group: CH2=C CH3 C6H11 1-methylcyclohexene B Elimination from the carbon opposite to the methyl group: CH2C CH3 =C6H11 3-methylcyclohexene The major product will be 1-methylcyclohexene because it is formed via the Zaitsev's rule, which states that the more substituted alkene is the major product in an elimination reaction. In this case, 1-methylcyclohexene has the double bond between two secondary carbons, while 3-methylcyclohexene has the double bond between a primary and a secondary carbon. Therefore, 1-methylcyclohexene is more substituted and will be the major product.