This is an elimination reaction (E2 mechanism) where the hydroxide ion acts as a strong base. The steps of the reaction are as follows: Step 1: Identify the base and leaving group In the presence of \(\text{OH}^-\) and EtOH, the bromine atom (\(\text{Br}\)) acts as the leaving group. The hydroxide ion abstracts a proton (\(\text{H}^+\)) from the adjacent carbon atom, resulting in the formation of a double bond. Step 2: Formation of the double bond Since the reaction proceeds via the E2 mechanism, the elimination occurs in a single step. The base removes a \(\beta\)-hydrogen atom (attached to the carbon next to the carbon with \(\text{Br}\)), and the \(\text{Br}^-\) leaves, resulting in the formation of a double bond between the \(\alpha\)- and \(\beta\)-carbons. Step 3: Determine the major product The double bond forms between the \(\alpha\)-carbon (the one originally bonded to the bromine atom) and the \(\beta\)-carbon. Due to Zaitsev's rule, the most substituted alkene is the major product. Hence, the major product is cyclopentene with a methyl (\(\text{CH}_3\)) substituent. Final Answer: (3) Cyclopentene with a \(\text{CH}_3\) substituent.
