See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is isobutane (2-methylpropane), CH3-CH(CH3) with an additional CH3, making it actually isobutylene precursor. Wait - the structure shown is CH3-CH with a CH3 branch, i.e., isobutane (2-methylpropane): (CH3)2CH-... but looking again, the structure is CH3-CH with CH3 below, which is propane with a methyl branch at C2... Actually the starting material is isobutane: (CH3)2CH2 is not right either. The structure is CH3-CH(-CH3) which with the implied H is isobutane (2-methylpropane), (CH3)3CH. Step 2 - Reaction with KMnO4: KMnO4 oxidizes isobutane. For a tertiary C-H, oxidation with KMnO4 gives the tertiary alcohol: (CH3)3COH (tert-butanol). So (A) = (CH3)3C-OH (2-methyl-2-propanol / tert-butanol). Step 3 - Reaction A with H+/Delta: Acid-catalyzed dehydration of tert-butanol gives isobutylene (2-methylpropene): (CH3)2C=CH2. So (B) = (CH3)2C=CH2. Step 4 - Reaction B with HBr/ROOR: ROOR is a peroxide initiator, which causes anti-Markovnikov (free radical) addition of HBr to the alkene. For (CH3)2C=CH2 (isobutylene), Markovnikov addition gives (CH3)3CBr (tert-butyl bromide), while anti-Markovnikov (radical) addition places Br at the less substituted carbon (the terminal CH2), giving (CH3)2CH-CH2Br (1-bromo-2-methylpropane). Step 5 - Identify product C: Anti-Markovnikov addition of HBr to (CH3)2C=CH2 gives (CH3)2CH-CH2Br, which is 1-bromo-2-methylpropane. This corresponds to option (d): CH3-CH(CH3)-CH2-Br. Why other options fail: - (a) 2-bromopropane would require a 3-carbon chain without branching. - (b) 2-bromo-2-methylpropane (tert-butyl bromide) would be the Markovnikov product, not anti-Markovnikov. - (c) does not correspond to a logical product of this reaction sequence. Therefore, the correct answer is D.