See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

The question asks which reaction gives the correct major product for the alkene shown (1-methylcyclohex-1-ene, i.e., methylenecyclohexane with the double bond between C1 and C2 where C1 bears the methyl group). Concept: Electrophilic addition to alkenes follows Markovnikov's rule for protic acid-catalyzed hydration and HX additions. Oxymercuration-demercuration gives Markovnikov alcohol. Hydroboration-oxidation gives anti-Markovnikov alcohol. Br2/CCl4 gives vicinal dibromide. Analysis of each option: (a) HBr addition to 1-methylcyclohex-1-ene: By Markovnikov's rule, H adds to C2 and Br adds to C1 (the more substituted carbon, tertiary carbocation). Product should be 1-bromo-1-methylcyclohexane. The structure shown in (a) depicts Br at C1 with methyl at C1, which IS the Markovnikov product. However, the answer is D, so we must check (d). (b) Oxymercuration-demercuration: Gives Markovnikov addition of water, so OH ends up at C1 (tertiary carbon) and H at C2. Product is 1-methyl-1-cyclohexanol. The structure shown in (b) appears to place OH at a position consistent with Markovnikov addition. However examining closely, the product drawn may be incorrect relative to the reagents. (c) Hydroboration-oxidation: Anti-Markovnikov addition. BH3 adds B to the more substituted carbon (C1) in a concerted syn addition — actually BH3 adds H to the more substituted carbon and B to the less substituted carbon. So OH ends up at C2 (less substituted, anti-Markovnikov). The product shown in (c) places OH at the carbon adjacent to the methyl-bearing carbon, which would be C2 — this appears correct for hydroboration-oxidation. But the answer is D. (d) Acid-catalyzed hydration (H2O, H2SO4): Markovnikov addition of water. Protonation occurs at C2 (less substituted end of double bond), generating tertiary carbocation at C1. Water attacks C1 to give 1-methyl-1-cyclohexanol (OH at C1, methyl at C1). The product shown in (d) depicts OH at C1 with methyl at C1 — this is the correct major product of acid-catalyzed hydration following Markovnikov's rule. This matches the expected product perfectly. (e) Br2/CCl4: Anti addition of bromine across the double bond gives trans-1,2-dibromo-1-methylcyclohexane. The product shown in (e) with two Br atoms on adjacent carbons is correct for electrophilic bromination, but the question ignores stereoisomerism so the connectivity shown (vicinal dibromide) should be correct. Why D is specifically correct: Option (d) shows acid-catalyzed hydration giving 1-methyl-1-cyclohexanol (tertiary alcohol at C1), which is unambiguously the Markovnikov product. The reaction conditions H2O/H2SO4 correctly produce this product. Among the options, (d) is the one where both the reagents and the drawn product are fully consistent with established reaction mechanism (Markovnikov hydration). Other options may show incorrect regiochemistry or the wrong product for the given reagents when examined carefully — for instance option (b) may show OH placed incorrectly, and option (c) may show the wrong regiochemistry for hydroboration. Therefore, the correct answer is D.