Find out the major product for the above reaction.

Explanation: The reaction begins with iodolactonization.

The carboxylic acid is converted into a carboxylate ion by $$NaHCO_3$$.

The alkene reacts with $$I_2$$ to form an iodonium ion intermediate.

The carboxylate ion attacks intramolecularly to form a stable five-membered lactone ring, placing iodine on the more substituted carbon.

In the next step, pyridine and heat promote dehydrohalogenation $$-HI$$ is eliminated to form a double bond.

Elimination occurs preferentially at the $$\alpha$$-carbon adjacent to the carbonyl group because it produces an $$\alpha,\beta$$-unsaturated lactone. The resulting conjugated system is thermodynamically more stable than non-conjugated alternatives.

Therefore, the major product is the conjugated $$\alpha,\beta$$-unsaturated lactone corresponding to Option C.