Animated Solution for Chemistry - Organic Chemistry: Identify $A$ and $B$ in the chemical reaction.

Welcome to a beautiful problem that perfectly marries electrophilic addition with nucleophilic substitution!

Let's start by looking closely at our reactant. We have a cyclohexene ring adorned with two very different substituents: a methoxy group ($-OC{H}_3$) at the top and a nitro group ($-N{O}_2$) at the bottom.

Our first reagent is $HCl$. This immediately tells us that we are going to see an electrophilic addition across that double bond. But the question is, which way will it add?

Markownikoff's Rule

According to Markownikoff's rule, the hydrogen ion ($H^{+}$) will attach to the double bond in a way that forms the most stable carbocation intermediate.

Let's visualize the two possibilities. If the $H^{+}$ attaches to the bottom-right carbon, the positive charge will appear on the top-right carbon. Conversely, if it attaches to the top-right carbon, the positive charge appears on the bottom-right carbon.

Here is where we must evaluate the electronic effects of our substituents. The nitro group is a powerful electron-withdrawing group. Through its strong $-I$ and $-M$ effects, it highly destabilizes any nearby positive charge.

Therefore, the carbocation that forms further away from the nitro group—at the top-right carbon—is significantly more stable.

The Finkelstein Exchange

Since the top-right carbocation is more stable, the chloride ion ($C{l}^{-}$) will attack right there. This gives us our major Product A, where the chlorine atom is attached to the carbon adjacent to the methoxy group.

Now, we move to the second step. Product A is treated with sodium iodide ($NaI$) in dry acetone. This is a classic named reaction: the Finkelstein reaction!

The Finkelstein reaction is a textbook $S_{N}2$ halogen exchange. The iodide ion acts as a nucleophile and cleanly displaces the chloride ion.

So, the chlorine at the top-right position is simply replaced by an iodine atom, giving us our final major Product B. Comparing this with our options, we find a perfect match with Option (b).