Final answer:
The final product of the reaction sequence involving trans-3-Methylcyclopentanol treated with CH₃SO₂Cl and then heated with KI in methanol is 3-methylcyclopentene, due to the stability preference in elimination reactions.
Step-by-step explanation:
When trans-3-Methylcyclopentanol is treated with CH₃SO₂Cl in the presence of a base, the hydroxyl group is substituted by a methanesulfonyl group, forming a mesylate ester. This ester is then heated with KI in methanol. During this process, the mesylate group is replaced by iodide due to a nucleophilic substitution reaction, yielding 3-methylcyclopentyl iodide.
Next, when the 3-methylcyclopentyl iodide is heated in methanol, it undergoes an elimination reaction, where the iodide leaves, and a double bond forms within the cyclopentane ring. Given the nature of elimination reactions to favor the formation of more stable, highly substituted alkenes, the reaction most likely leads to the formation of 3-methylcyclopentene as a final product due to its greater alkene stability over the alternative, which is 1-methylcyclopentene.