See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material. The starting material is methylcyclopentane (a cyclopentane ring with a methyl substituent). Step 2: Reaction 1 - Br2/hv (free radical bromination). Free radical bromination is highly selective and preferentially abstracts the most stable (most substituted/weakest) C-H bond. In methylcyclopentane, the tertiary C-H at the ring carbon bearing the methyl group is the weakest and most reactive. Therefore, bromine radical abstracts the tertiary hydrogen, and Br adds to the tertiary carbon. Major product X = 1-bromo-1-methylcyclopentane (tertiary bromide). Step 3: Reaction 2 - Alcoholic KOH/heat (E2 elimination). Alcoholic KOH causes dehydrohalogenation (E2). From 1-bromo-1-methylcyclopentane, elimination removes HBr. The Br is on a quaternary-like tertiary carbon; elimination occurs to give the more substituted alkene by Zaitsev's rule. The product is 1-methylcyclopentene (methyl group and double bond on the ring, endocyclic alkene). Major product Y = 1-methylcyclopentene. Step 4: Reaction 3 - HBr/Peroxide (anti-Markovnikov addition). In the presence of peroxide, HBr adds via a free radical mechanism following anti-Markovnikov regioselectivity. The bromine radical adds to the less substituted carbon of the double bond (the terminal/less substituted end). In 1-methylcyclopentene, the double bond is between C1 (bearing methyl) and C2. Anti-Markovnikov addition means Br goes to C2 (the less substituted carbon of the double bond, i.e., the carbon without the methyl group), and H goes to C1 (the more substituted carbon bearing the methyl group). This gives 1-methyl-2-bromocyclopentane. Major product Z = 1-methyl-2-bromocyclopentane. Step 5: Stereochemistry note. Radical addition gives predominantly the more stable product; the major stereoisomer would be the trans (1-methyl-2-bromocyclopentane, trans arrangement), consistent with option (c). Why other options fail: - (a) bromocyclohexane: no ring expansion occurs under these conditions. - (b) 1,3-dibromocyclopentane: would require two bromination steps, not consistent with the reaction sequence. - (d) cyclopentylmethyl bromide (bromomethyl cyclopentane): this would require Br to end up on the exocyclic methyl carbon, which is not consistent with anti-Markovnikov addition to 1-methylcyclopentene (the methyl is not part of the double bond in the endocyclic alkene). Therefore, the correct answer is C.