Find the major product for the reaction shown | vedclass

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(A) The reaction is an iodolactonization followed by dehydroiodination. $1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to

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(A) The reaction is an iodolactonization followed by dehydroiodination. $1$. Treatment of the unsaturated carboxylic acid with $I_2/NaHCO_3$ leads to the formation of an iodolactone intermediate. The carboxylate oxygen attacks the iodonium ion formed across the double bond,resulting in a bicyclic lactone with an iodine atom at the bridgehead position. $2$. Subsequent treatment with pyridine and heat $(\Delta)$ promotes an $E2$ elimination reaction,removing the iodine atom and a hydrogen atom from the adjacent carbon to form a double bond. $3$. The final product is a bicyclic lactone containing a double bond in the six-membered ring,which corresponds to option $A$.

Find the major product for the reaction shown below.

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