See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is a cyclohexene derivative with a trisubstituted double bond. Specifically, it is 1-methyl-2-methylcyclohex-2-ene (or equivalently, a methylcyclohexene with a methyl on the double bond carbon), where C1 bears a methyl group on a wedge bond, indicating a defined stereocenter. The double bond is between C1 and C2, with C1 being sp2 (loses its stereocenter upon reaction) and C2 also sp2. Step 2 - Identify the reaction: Addition of HBr to the alkene follows Markovnikov's rule. The proton adds to the carbon that gives the more stable carbocation (Markovnikov addition). Step 3 - Determine the carbocation intermediate: Upon protonation, the more stable (more substituted) carbocation forms. For a 1-methylcyclohexene with a methyl at C1 and C2 being part of the ring double bond, protonation of the less substituted carbon gives a tertiary carbocation at the more substituted carbon. This carbocation intermediate is planar (sp2, trigonal), meaning it has no facial selectivity — Br⁻ can attack from either face with equal probability. Step 4 - Analyze the stereochemical outcome: The starting material has one defined stereocenter (the wedge methyl at C1), but upon forming the alkene and then the carbocation, C1 becomes sp2 and planar. After HBr addition, two new stereocenters are generated (C1 and C2, or whichever two carbons are involved). Since the carbocation is planar, attack of Br⁻ from both faces is equally likely, generating two products. Step 5 - Determine the relationship between the two products: Because the starting alkene is already chiral (has a defined configuration), the two products formed by attack from the two faces of the carbocation are NOT mirror images of each other — they are diastereomers. They have the same connectivity but differ in the relative configuration of the stereocenters produced. One face attack gives one diastereomer; the other face attack gives the other diastereomer. Step 6 - Why not the other options: - (a) Racemic mixture: A racemic mixture consists of equal amounts of two enantiomers. The two products here are diastereomers, not enantiomers, because the starting material was already chiral. - (b) Single enantiomer: Not possible since the planar carbocation allows attack from both faces. - (d) Achiral molecule: The products contain multiple stereocenters and are not achiral (unless meso, which is not the case here). Therefore, the correct answer is C.