To solve this problem, we need to understand the reaction conditions.$\\$The reaction involves an alkene, HBr, and a peroxide $((C_6H_5CO)_2O_2). \\$In the presence of peroxides, the addition of HBr to an alkene follows the anti-Markovnikov rule.$\\$This is known as the Kharasch effect or peroxide effect.$\\$Step-by-step solution:$\\$1. Identify the alkene:$\\$ The given alkene is 2-methylpropene: $\\$ $(CH_3)_2C=CH_2 \\$2. Apply the anti-Markovnikov rule:$\\$ In the presence of peroxides, the bromine radical adds to the less substituted carbon.$\\$3. Radical mechanism:$\\$ • Initiation: Peroxide decomposes to form radicals.$\\$ • Propagation: Bromine radical adds to the less substituted carbon of the alkene, forming a more stable radical.$\\$ • Termination: The radical reacts with HBr to form the product.$\\$4. Product formation:$\\$ The bromine adds to the terminal carbon, resulting in:$\\$ $CH_3CH_2CH_2Br \\$Therefore, the major product is:$\\$Option 1: $CH_3CH_2CH_2Br \\$This corresponds to the structure shown in Option 1.