Step 1: Understanding the Concept
Step 1 is electrophilic addition of HCl to the ring C=C. Protonation can leave the positive charge on either alkene carbon, and the more stable cation is preferred.
Step 2: Direction of addition
A cation next to the carbon bearing the nitro group is strongly destabilised by its powerful electron-withdrawing effect. Next to the carbon bearing the methoxy group the pull is weaker.
So H adds to the carbon near the nitro group and Cl bonds to the carbon adjacent to the methoxy-bearing carbon. A is 2-chloro-1-methoxy-4-nitrocyclohexane.
Step 3: Step 2
NaI in dry acetone is the Finkelstein halogen exchange. I replaces Cl on the same carbon, and NaCl precipitates, which drives the reaction.
Step 4: Conclusion
A has Cl and B has I on the carbon next to the methoxy-bearing carbon, with the other groups unchanged.
Final Answer:
A is the chloro compound and B is its iodo analogue, as drawn in option A.
\[ \boxed{\text{(A) }\text{the pair shown in option A}} \]
