The electrophilic addition of HBr to propene in the presence of a peroxide catalyst follows a different mechanism than the standard electrophilic addition. This reaction is known as the free radical addition or anti-Markovnikov addition of HBr to propene. The presence of a peroxide catalyst leads to the formation of a bromine radical, which then reacts with propene to form the final product. The mechanism can be broken down into the following steps:1. Initiation:The peroxide catalyst ROOR' undergoes homolytic cleavage, usually induced by heat or light, to form two alkoxy radicals RO and R'O . These radicals are highly reactive and can initiate the reaction.ROOR' RO + R'OOne of the alkoxy radicals reacts with HBr, breaking the H-Br bond homolytically and generating a bromine radical Br and an alcohol ROH .RO + HBr ROH + Br2. Propagation:The bromine radical Br reacts with propene CH2=CH-CH3 in a free radical addition, forming a secondary free radical at the allylic position CH2-CH-CH3 .Br + CH2=CH-CH3 CH2-CH Br -CH3The secondary free radical reacts with another molecule of HBr, breaking the H-Br bond homolytically and generating a new bromine radical Br and the final product, 1-bromopropane CH3-CH Br -CH3 .CH2-CH-CH3 + HBr CH3-CH Br -CH3 + Br3. Termination:The reaction can be terminated by the combination of two free radicals, such as two bromine radicals or a bromine radical and an alkoxy radical, to form a stable molecule.Br + Br Br2Br + RO R-O-BrThe overall reaction is as follows:CH2=CH-CH3 + HBr CH3-CH Br -CH3In the presence of a peroxide catalyst, the electrophilic addition of HBr to propene follows the anti-Markovnikov rule, where the bromine atom is added to the less substituted carbon of the double bond, resulting in the formation of 1-bromopropane.