The major end product Z in the given sequence of reactions is:
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Phenol shows ortho/para directing behavior, but in mild non-polar conditions para product dominates due to lower steric hindrance compared to ortho substitution.
Concept:
Phenol undergoes a multi-step transformation involving reduction to benzene, Friedel–Crafts alkylation, cumene process, and finally electrophilic substitution. The final bromination depends strongly on the activating nature of the hydroxyl group and reaction conditions.
Step 1: Conversion of phenol to A (Zn dust distillation).
When phenol is heated with zinc dust, it is reduced to benzene:
\[
\text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO}
\]
Thus, A = benzene.
Step 2: Friedel–Crafts alkylation to form B.
Benzene reacts with isopropyl chloride in presence of anhydrous AlCl$_3$ to form cumene (isopropylbenzene):
\[
\text{C}_6\text{H}_6 \xrightarrow{(\text{CH}_3)_2\text{CHCl}, \, \text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2
\]
Thus, B = cumene.
Step 3: Cumene oxidation (cumene process).
Cumene undergoes air oxidation followed by acid hydrolysis to give phenol and acetone:
\[
\text{C}_6\text{H}_5\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{O}_2, \, \text{H}^+ / \text{H}_2\text{O}} \text{C}_6\text{H}_5\text{OH}
\]
Thus, C = phenol.
Step 4: Bromination of phenol in CS$_2$.
Phenol is a strongly activating ortho/para-directing group due to resonance donation of lone pair electrons from oxygen.
In non-polar solvent (CS$_2$), controlled monobromination occurs. Both ortho and para products are possible, but steric hindrance at ortho position favors para substitution as the major product:
\[
\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{Br}_2 / \text{CS}_2} p\text{-bromophenol}
\]
Conclusion:
The final product $Z$ is $p$-bromophenol. Hence, Option (3) is correct.
