See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is 1-methyl-1-vinylcyclopentane, which has a cyclopentane ring with both a methyl group and a vinyl group (CH2=CH-) attached to the same carbon (C1). Step 2 - Mechanism of HBr addition: In CCl4 (non-polar solvent, no peroxides), HBr adds via electrophilic addition (Markovnikov addition). The proton adds to the terminal carbon of the vinyl group (CH2=), generating the more stable carbocation at the internal carbon directly attached to the ring. Step 3 - Carbocation formed: Protonation of CH2=CH- at the terminal =CH2 gives a secondary carbocation at the carbon adjacent to the ring: cyclopentane-C1(CH3)-+CH-CH3. Wait - more precisely, the vinyl group is CH2=CH2 terminal, so H+ adds to =CH2 giving +CH2- attached to C1? No: the vinyl group is -CH=CH2. H+ adds to the terminal CH2 (Markovnikov), giving carbocation at the internal carbon: C1(ring)(CH3)-+CH2... actually Markovnikov means H adds to carbon with more H's (=CH2), carbocation forms at =CH- carbon which is directly bonded to C1 of cyclopentane. Step 4 - Ring expansion via 1,2-shift: The carbocation at the carbon adjacent to C1 of cyclopentane (which bears a methyl group) is secondary. A 1,2-alkyl shift (ring expansion) can occur: one of the C-C bonds of the cyclopentane ring migrates to the adjacent carbocation, expanding the cyclopentane to a cyclohexane ring. This generates a tertiary carbocation at C1 of the new cyclohexane ring (since C1 had methyl + ring bonds). Step 5 - Product after ring expansion: The ring expansion converts cyclopentane + pendant carbocation into a cyclohexyl tertiary carbocation. The tertiary carbocation is at a ring carbon that also bears a methyl group. Bromide then attacks this tertiary carbocation. Step 6 - Structure of product: The result is 1-bromo-1-methylcyclohexane... but we need to account for the methyl. After ring expansion of 1-methylcyclopentane with an adjacent carbocation, we get a cyclohexane ring where the carbon bearing the positive charge has one methyl group. The Br attacks to give 1-bromo-1-methylcyclohexane. However, looking at option (c): it shows a cyclohexane ring with one carbon bearing two substituents (gem arrangement) and Br on an adjacent carbon. This corresponds to the ring-expanded product where the tertiary carbocation (bearing methyl) is trapped by Br-, giving a cyclohexane with a quaternary-like carbon (two methyls from the original methyl + ring carbon) and Br. Specifically, option (c) shows 1,1-dimethyl-2-bromocyclohexane or 1-bromo-1-methyl-2-methylcyclohexane, consistent with ring expansion of 1-methyl-1-vinylcyclopentane via Markovnikov addition followed by 1,2-carbon shift. Step 7 - Why other options fail: - Option (a): Shows an aromatic/cyclohexene product which requires elimination, not addition under these conditions. - Option (b): Shows a cyclopentene with Br and no ring expansion, which would be a minor or incorrect regiochemistry product. - Option (d): Shows a cyclopentane product with no ring expansion, representing simple Markovnikov addition without rearrangement, which is not the major product when a more stable tertiary carbocation is accessible via ring expansion. Therefore, the correct answer is C.