Hydrogen chloride adds across the ring double bond by Markovnikov's rule, placing the chlorine on the carbon next to the one bearing the methoxy group, because the carbocation forms away from the strongly electron-withdrawing nitro group. The second step, sodium iodide in dry acetone, is the Finkelstein reaction, which exchanges that C−Cl for C−I; it acts only on the alkyl halide and leaves the methoxy ether unchanged. The final product therefore retains the −OCH3 group and carries iodine where the chlorine had been.
JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option