See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction conditions: I2/NaHCO3 is the classic iodolactonization condition. NaHCO3 is a mild base that deprotonates the carboxylic acid to form the carboxylate nucleophile in situ, while I2 acts as the electrophilic iodine source. Step 2 - Identify the substrate: The starting material is a cyclohex-2-ene bearing a -CH2COOH (acetic acid) side chain at C1, with specific stereochemistry (H and the side chain both on wedge bonds at C1, indicating a defined stereocenter). The double bond is between C2 and C3 of the cyclohexene ring. Step 3 - Mechanism of iodolactonization: I2 activates the double bond by forming an iodonium ion (cyclic, bridging C2-C3). The carboxylate oxygen, acting as the intramolecular nucleophile, attacks the iodonium ion from the back side (anti addition, SN2-type). The carboxylate attacks the more electrophilic carbon of the iodonium (the more substituted or geometrically accessible carbon), forming a five-membered (gamma) or six-membered (delta) lactone ring. With an acetic acid side chain (-CH2COO-) at C1, the oxygen is two carbons away from C1, and C2 is adjacent to C1, so attack on C2 gives a five-membered lactone (gamma-lactone, butyrolactone ring fused to cyclohexane). Step 4 - Stereochemical outcome: The iodonium ion forms on one face of the double bond. Anti addition means I ends up trans to the lactone oxygen. The product is a bicyclic compound: a cyclohexane ring fused with a five-membered lactone ring, with I on the exo/axial position and the ring junction stereochemistry as shown in option (b). NaHCO3 does not convert the lactone to a sodium salt under these mild conditions; the lactone ring closes spontaneously. Step 5 - Why other options fail: - Option (a): Shows a diiodide with sodium carboxylate and intact double bond - incorrect; iodolactonization gives one I and a lactone, not two iodines. - Option (c): Shows an alpha-iodo acid with intact double bond and no ring closure - incorrect; iodolactonization proceeds with ring closure under these conditions. - Option (d): Shows a sodium carboxylate with iodine and an intact double bond but no lactone - incorrect; the carboxylate nucleophile closes the ring to form the lactone. Step 6 - Option (b) correctly depicts the bicyclic iodolactone product with anti stereochemistry (I and the lactone oxygen on opposite faces), consistent with the iodonium ion mechanism. Therefore, the correct answer is B.