See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The structure shown is 3-bromocyclohex-1-ene (a cyclohexene with a Br substituent on a wedge bond at C3, allylic position relative to the double bond at C1-C2). Step 2 - Understand the reaction pathway: The problem states the reaction proceeds through a classical carbocation (not a bridged or nonclassical ion). HBr adds to the double bond. Protonation of the double bond occurs first (Markovnikov addition), placing H at C1 and generating a carbocation at C2. However, since the Br is at C3, and the carbocation at C2 is adjacent to C3, allylic stabilization or direct consideration is needed. Step 3 - Protonation of the double bond: H+ adds to C1 (less substituted end, or following Markovnikov), generating a carbocation at C2. Alternatively, H+ adds to C2, giving carbocation at C1. Since C1 is allylic to the existing system and we consider Markovnikov, the carbocation forms at C1 (secondary, allylic with respect to C3-Br). Actually, protonation at C2 gives a carbocation at C1, which is adjacent to C2 and beta to C3-Br. Step 4 - Classical carbocation at C1: With a carbocation at C1 and Br already on C3 (wedge = up/beta face), the bromide ion (Br-) attacks the carbocation at C1. Since this is a classical carbocation (planar, sp2), Br- can attack from either face. However, the existing Br at C3 influences the preferred approach. Step 5 - Stereochemical outcome: The carbocation at C1 is classical (planar). The existing Br at C3 is on the wedge (beta/up face). Br- attacks C1 from the bottom face (alpha face) preferentially due to steric interactions with the C3-Br on the top face, giving Br at C1 on the alpha (dash) face. With Br at C3 on the wedge (up) and Br at C1 on the dash (down), the two Br groups are on opposite faces = trans relationship. This gives trans-1,3-dibromocyclohexane. Step 6 - Why other options fail: - (b) cis-1,3: Would require both Br groups on the same face, but attack from the same face as C3-Br is sterically disfavored. - (c) trans-1,2 and (d) cis-1,2: These would result from 1,2-addition where Br- attacks C2 (the carbon adjacent to C1), but the carbocation is at C1, not C2, and allylic rearrangement leads to C1/C3 products. Therefore, the correct answer is A.