The first thing that would happen is the loan pair on the oxygen (the alcohol) will act as a nucleophile and attack the sulfur on the other molecule, which will kick of the chlorine. So X should be those two molecules joined together from the nucleophilic attack. If you then reacted that with LiBr, the bromide ion will act as an excellent nucleophile and attack the carbon that is attached to the oxygen, which is an excellent leaving group because the negative charge is stabilized through resonance across THREE oxygen atoms. (sn2 type reaction). So structure Z should be two molecules, the non aromatic ring with a t-butly and a bromine (remember sn2 inverts the stereochemistry). The second molecule should be the aromatic ring with the NO2 and SO3- para to eachother.
The answer is:
1. Et-Br, Ether
------------------->
2. KOC(CH3)3
I drew out what you said, and I agree. But the question doesn't seem to want any of that. It's asking for reagents, but they've already given us some. Pyridine and acetone are solvents, but I would think LiBr would be a Nu: like you just said. So I'm with you, maybe the question was worded wrong or they made a mistake.
Yeah, the answer they have is a mistake; it does not go with this question. As long a you fully understand where the two products are coming from you are good to go!
Pyridine is just there to deprotonate after nucleophilic attack. Then in the second step acetone is an excellent solvent for the sn2 reaction. Remember, polar aprotic solvents (like acetone, DMSO ect) speed up sn2 reactions.
