The major product of the reaction is:

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(B) The reaction is an iodolactonization. $1$. The $I_2$ reacts with the alkene to form a cyclic iodonium ion intermediate. $2$. The $NaHCO_3$ acts as

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(B) The reaction is an iodolactonization. $1$. The $I_2$ reacts with the alkene to form a cyclic iodonium ion intermediate. $2$. The $NaHCO_3$ acts as a base to deprotonate the carboxylic acid group,forming a carboxylate ion. $3$. The carboxylate oxygen performs an intramolecular nucleophilic attack on the more substituted carbon of the iodonium ion,leading to the formation of a five-membered lactone ring. $4$. The final product is an iodolactone,where the iodine atom and the oxygen atom are added across the double bond in an anti-fashion. Based on the structure,the correct product is the one where the iodine is at the bridgehead position and the lactone ring is fused to the cyclohexane ring.

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