See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 — Identify the reactant structure: The starting material is (methylenecyclobutyl)acetylene or more likely methylenecyclobutane bearing a vinyl group. From the drawing, the reactant is 1-vinylmethylenecyclobutane, i.e., a cyclobutane ring with an exocyclic double bond (=CH2) and a vinyl group (–CH=CH2) both attached to the same carbon of the ring — effectively it is (1-vinylcyclobutylidene) or the alkene is between the ring carbon and the CH, making it an exocyclic alkene. The most consistent reading is: the compound has a cyclobutane ring, one carbon of which bears an exocyclic double bond to a CH2 group and also a –CH=CH2 group — i.e., it is a diene or a compound with one double bond. Given the single double bond symbol, the reactant is 1-(1-methylenecyclobutyl)ethene = (cyclobutylidene)acetaldehyde… Re-reading: the structure shows a cyclobutane ring with a carbon at top-right that has an exo double bond going up-right and a bond going down-right to a carbon that connects to H-I. The simplest reading: the molecule is methylenecyclobutane with a vinyl substituent, i.e., the alkene is exocyclic on cyclobutane AND there is a vinyl arm. This is 1-vinylmethylenecyclobutane. Step 2 — Addition of HI (Markovnikov): HI adds to the more substituted end of the double bond. Protonation occurs at the terminal carbon of the exocyclic double bond, generating a carbocation at the ring carbon (tertiary, on cyclobutane). This carbocation is secondary/tertiary at C1 of cyclobutane. Step 3 — Carbocation rearrangement: A tertiary carbocation on a cyclobutane ring is strained. Ring expansion via a 1,2-alkyl shift (ring-expansion of cyclobutane to cyclopentane) relieves ring strain and generates a more stable tertiary carbocation on a cyclopentane ring. This is a classic cyclobutylcarbinyl → cyclopentyl cation rearrangement. Step 4 — Identify the rearranged carbocation: After ring expansion, a tertiary carbocation on a cyclopentane ring forms. The vinyl group from the original molecule contributes to the ring expansion to give the 5-membered ring. Step 5 — Capture by iodide: Iodide ion attacks the tertiary carbocation on the cyclopentane ring, giving 1-iodo-1-(substituent)cyclopentane. This corresponds to option (b): a cyclopentane ring with iodine at a quaternary/tertiary carbon. Step 6 — Why other options fail: - (a) would be simple Markovnikov addition without rearrangement, retaining the strained cyclobutane ring — not favored. - (c) shows iodo on a side chain, not at the ring carbon — inconsistent with carbocation capture after ring expansion. - (d) shows iodo at a less substituted position — not the most stable carbocation product. Therefore, the correct answer is B.