The product formed in the following reaction is:

Step-by-step solution

The given reaction is the addition of HBr to an alkene in the presence of benzoyl peroxide. This proceeds via a free-radical mechanism (Kharasch effect or anti-Markovnikov addition). First, the peroxide initiates the reaction to generate a bromine radical ( Br^ ). The Br^ radical attacks the double bond to form the most stable carbon radical intermediate. There are two possible sites for the attack on the HO-C₆H₄-CH=CH-CH₃ molecule: 1. Attack at the -carbon gives a secondary alkyl radical: HO-C₆H₄-CH(Br)- C H-CH₃ 2