See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When HBr adds to an alkene in the presence of peroxide (peroxide effect / anti-Markovnikov addition), the reaction proceeds via a free-radical mechanism. In this mechanism, the bromine radical adds to the less substituted (terminal) carbon of the double bond, placing the bromine at the terminal carbon and the hydrogen at the internal carbon. This is the opposite of normal Markovnikov addition. Step 1: Identify the reactive site. The molecule CH2=CH(CH2)8COOH has a terminal alkene (CH2=CH-) at one end and a carboxylic acid group at the other end. The carboxylic acid group does not react with HBr under these conditions. Step 2: Apply anti-Markovnikov (peroxide) addition. In free-radical addition: - A bromine radical adds to the terminal CH2= carbon (less substituted end) because this gives the more stable secondary radical intermediate at the internal carbon. - The hydrogen then adds to the internal =CH- carbon. Step 3: Determine the product. Bromine adds to the terminal carbon (CH2=), giving CH2Br-CH2-(CH2)8COOH, i.e., CH2BrCH2(CH2)8COOH. Why other options fail: - Option (a) CH3-CHBr(CH2)8COOH: This is the Markovnikov product (Br on internal carbon), which would form without peroxide. - Option (b) CH2=CH(CH2)8COBr: This would imply reaction at the carboxylic acid group, which does not occur under these conditions. - Option (d) CH2=CH(CH2)7CHBrCOOH: This implies addition of HBr adjacent to the COOH group with retention of the double bond, which is not consistent with the reaction mechanism. Therefore, the correct answer is C.