See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When an unsaturated carboxylic acid is treated with I2 in the presence of NaHCO3 (a mild base that deprotonates the carboxylic acid to form the carboxylate anion), an iodolactonization reaction occurs. Step 1: Identify the substrate. 4-Pentenoic acid is CH2=CH-CH2-CH2-COOH (the double bond is between C4 and C5, i.e., a terminal alkene, with the carboxylic acid at C1). Step 2: NaHCO3 deprotonates the carboxylic acid to generate the carboxylate anion (COO-), which acts as an internal nucleophile. Step 3: I2 acts as an electrophile and activates the double bond by forming an iodonium ion intermediate across the C4=C5 double bond. Step 4: The carboxylate oxygen performs intramolecular nucleophilic attack on the more substituted or accessible carbon of the iodonium ion (C4, which is the internal carbon bearing the chain) via an anti addition, following Baldwin's rules. A 5-membered ring transition state (5-exo-tet) is favored over a 6-membered ring. Step 5: The 5-exo-tet cyclization places the oxygen from the carboxylate at C4, forming a 5-membered lactone ring (gamma-lactone = dihydrofuran-2-one), with the iodine ending up on the exocyclic CH2 group (C5, the terminal carbon), giving 5-(iodomethyl)dihydrofuran-2-one. Step 6: Evaluate options: - (a) 4,5-diiodopentanoic acid: This would result from simple addition of I2 to the double bond without lactonization — incorrect. - (b) 5-iodomethyl-dihydrofuran-2-one: This is the product of 5-exo-tet iodolactonization forming a 5-membered lactone with -CH2I substituent — correct. - (c) 5-iodo-tetrahydropyran-2-one: This would require a 6-membered lactone via 6-endo-tet cyclization, which is disfavored by Baldwin's rules compared to 5-exo-tet — incorrect. - (d) 4-pentenolyiodide: This implies simple acid iodide formation, not relevant here — incorrect. Therefore, the correct answer is B.