See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Addition of HBr to Methylidene Cyclohexane **Step 1: Identify the substrate** The starting material is methylidenecyclohexane (exocyclic double bond: $C=CH_2$) **Step 2: Apply Markovnikov's Rule** When $HBr$ adds to an alkene, $H^+$ attaches to the carbon of the double bond with more hydrogen atoms, and $Br^-$ attaches to the carbon with fewer hydrogens. For the exocyclic $C=CH_2$: - One carbon (the $CH_2$) has 2 hydrogens → more substituted pathway - The other carbon (ring) has 0 hydrogens → less substituted **Step 3: Carbocation formation** $H^+$ adds to the $CH_2$ carbon, forming a secondary carbocation on the ring: $$\text{Ring-C}^+ \text{ (more stable secondary carbocation)}$$ This is more stable than the alternative (primary carbocation on the exocyclic carbon). **Step 4: Bromide attack** $Br^-$ attacks the secondary carbocation, placing $Br$ at the ring position adjacent to where the double bond was. **Result:** A cyclohexane ring with $CHBr$ (not $CH_2Br$ or $CH_3Br$) at one position. **Why option C is correct:** It shows $CHBr$ attached directly to the ring (Markovnikov addition). **Why others are wrong:** Options B and D have $Br$ on the exocyclic carbon (anti-Markovnikov); Option A has a methyl group instead of direct addition.