See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When HBr is added to an alkene in the presence of peroxide (radical initiator), the reaction proceeds via a free radical mechanism (anti-Markovnikov addition) rather than the ionic Markovnikov mechanism. Step 1: Identify the reaction conditions. The reagent is HBr with (C6H5CO)2O2 (benzoyl peroxide), which is a radical initiator. This triggers free radical addition of HBr. Step 2: Recall the mechanism of free radical addition of HBr. Benzoyl peroxide generates radicals that abstract H from HBr, producing a bromine radical (Br•). Step 3: The bromine radical adds to the less substituted carbon (terminal carbon) of the double bond in propene (CH3—CH=CH2), generating the more stable secondary carbon radical at C2. This is because Br• adds to the terminal CH2 to give the secondary radical at the internal carbon. Step 4: The secondary carbon radical then abstracts H from HBr, giving CH3—CH2—CH2—Br (1-bromopropane) as the product. Step 5: This is anti-Markovnikov addition — Br ends up on the terminal (less substituted) carbon, which is option (a): CH3—CH2—CH2—Br. Why other options fail: - Option (b) CH3CH(Br)—CH3 (2-bromopropane) is the Markovnikov product, formed under ionic conditions (no peroxide). - Option (c) BrCH2—CH=CH2 would require allylic substitution, not addition. - Option (d) a bromocyclopropane requires a different type of reaction entirely. Therefore, the correct answer is A.