### Core Logic Let's analyze the steps of the reaction sequence: 1. **Step 1 (Br_2 / hnu$Br_2 / h\nu$):** Light-induced free radical substitution targeted at the most stable tertiary position, producing 1-bromo-1-methylcyclohexane. 2. **Step 2 (Alcoholic KOH, Delta$KOH, \Delta$):** Dehydrohalogenation occurs via an E2 mechanism. Following Saytzeff's rule, elimination favors the formation of the more highly substituted, stable alkene: 1-methylcyclohexene. 3. **Step 3 (HBr / R-O-O-R, hnu$HBr / R-O-O-R, h\nu$):** Radical hydrobromination across the unsymmetrical alkene. The presence of peroxide shifts addition toward the **Anti-Markovnikov** path, placing the bromine atom cleanly at the less-substituted secondary carbon to yield **1-bromo-2-methylcyclohexane**. ### Pattern Recognition Combining Saytzeff elimination with a peroxide-promoted HBr$HBr$ addition allows you to reposition functional groups from highly substituted tertiary carbons to adjacent secondary positions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes