## Step 1: Analyze the Reactants and Conditions - **Substrate:** The substrate is a secondary alkyl bromide (bromocyclopentane). - **Nucleophile/Base:** attacking species is ethoxide (NaOEt, a strong base/nucleophile) in ethanol (EtOH, a polar protic solvent). ### Explanation: - **Secondary alkyl halide** can undergo \( S_N1 \), \( S_N2 \), \( E1 \), or \( E2 \) mechanisms. - **NaOEt** is a strong nucleophile and strong base, which favors \( S_N2 \) and \( E2 \). - **EtOH** (polar protic solvent) slightly favors \( S_N1 \) and \( E1 \), but the strength of the base/nucleophile is more crucial here. **So, the likely mechanisms are \( S_N2 \) or \( E2 \).** --- ## Step 2: Decide Between \( S_N2 \) and \( E2 \) - **Strong, bulky base (NaOEt) and secondary substrate** favor **E2** over \( S_N2 \), especially with a strong base present. - \( S_N2 \) is less favored due to steric hindrance at the secondary carbon. - \( S_N1 \) and \( E1 \) are less likely because those are favored by weak nucleophiles/bases and polar protic solvents without a strong base. ### Explanation: - **E2 eliminates a proton and the leaving group in a single concerted step,** typical under these conditions (secondary halide + strong base). --- ## Final Answer **The most likely mechanism is \( \boxed{E2} \).** --- ### Mechanism (Summary): 1. **Ethoxide (EtO-) abstracts a \(\beta\)-hydrogen** from the cyclopentyl ring. 2. **Simultaneously, the bromide (Br-) leaves**, forming a double bond (alkene) in the cyclopentane ring. **Major product:** Cyclopentene --- #### **Circled Product Structure:** - Draw a cyclopentene (a five-membered ring with one double bond). --- **Summary Table:** | Step | Explanation | Conclusion | |--------------|------------------------------------------------|-----------------| | Step 1 | Secondary halide + strong base/nucleophile | E2 or SN2 | | Step 2 | Strong base + secondary substrate = E2 favored | E2 mechanism | **Final product:** Cyclopentene (via E2 elimination)

## Step 1: Analyze the Reactants and Conditions - **Substrate:** The substrate is a secondary alkyl bromide (bromocyclopentane). - **Nucleophile/Base:** The attacking species is ethoxide (NaOEt), a strong base/nucleophile, in ethanol (EtOH), a polar protic solvent. ### Explanation: - A secondary alkyl halide can undergo various mechanisms: \( S_N1 \), \( S_N2 \), \( E1 \), or \( E2 \). - **NaOEt** is a strong nucleophile and base, favoring \( S_N2 \) and \( E2 \). - **EtOH** (polar protic solvent) favors \( S_N1 \) and \( E1 \) slightly, but the presence of a strong base is more influential. **Conclusion:** The mechanisms under consideration are \( S_N2 \) and \( E2 \). --- ## Step 2: Decide Between \( S_N2 \) and \( E2 \) - The strong base (NaOEt) and the secondary substrate favor **E2** over \( S_N2 \). - **Steric hindrance** at the secondary carbon makes \( S_N2 \) less favorable. - \( S_N1 \) and \( E1 \) mechanisms are unlikely due to the requirement for weak nucleophiles/bases. ### Explanation: - The **E2 mechanism** involves a concerted reaction where the base abstracts a \(\beta\)-hydrogen while the leaving group departs, resulting in the formation of a double bond. --- ## Final Answer **The most likely mechanism is \( \boxed{E2} \).** --- ### Mechanism (Summary): 1. **Ethoxide (EtO-) abstracts a \(\beta\)-hydrogen** from the cyclopentyl ring. 2. **Simultaneously, the bromide (Br-) leaves**, forming a double bond (alkene) in the cyclopentane ring. **Major product:** Cyclopentene --- #### **Circled Product Structure:** - Draw a cyclopentene structure (a five-membered ring with one double bond). --- **Summary Table:** | Step | Explanation | Conclusion | |--------------|------------------------------------------------|-----------------| | Step 1 | Secondary halide + strong base/nucleophile | E2 or SN2 | | Step 2 | Strong base + secondary substrate = E2 favored | E2 mechanism | **Final product:** Cyclopentene (via E2 elimination)