Question

What are X, Y, Z in the following reaction sequence?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Friedel-Crafts Alkylation: Benzene + Alkyl Halide ($R-X$) $\xrightarrow{AlCl_3}$ Alkyl Benzene. Lucas Reagent: Alcohol $\to$ Alkyl Halide.

Answer 1

The Correct Option is D

Step 1: Initial Hydrolysis:
The starting material (likely a vinyl ether or cyclic ether based on $C_5H_8O$) hydrolyzes to form a compound. Based on the options and flow, let's look at the reagents.
Step 2: Reagent X and Transformation to Y:
$C_5H_8O$ seems to be an alcohol or ketone precursor. The reagent $X$ in Option 4 is $HCl/ZnCl_2$ (Lucas Reagent). This suggests the reactant is an alcohol, converting it to an alkyl chloride ($Y$). Let's assume the $C_5H_8O$ is Cyclopentanol or similar precursor formed. Actually, looking at the product Z (phenyl cyclopentane derivative), the reaction sequence implies: Precursor $\rightarrow$ Cyclopentyl Chloride (Y) $\rightarrow$ Phenyl Cyclopentane (Z).
Step 3: Friedel-Crafts Alkylation ($Y \rightarrow Z$):
$Y + C_6H_6$ (Benzene) in presence of anhydrous $AlCl_3$. This is a standard Friedel-Crafts alkylation. For this to yield Z (Phenylcyclopentane), Y must be Chlorocyclopentane (Cyclopentyl chloride).
Step 4: Backtracking to X:
To convert Cyclopentanol (implied) to Chlorocyclopentane, we need $HCl/ZnCl_2$ (Lucas reagent). Therefore: $X = HCl/ZnCl_2$. $Y =$ Chlorocyclopentane. $Z =$ Phenylcyclopentane.
Step 5: Verify Option 4:
Option 4 lists $HCl/ZnCl_2$, Structure Y (Chlorocyclopentane), Structure Z (Phenylcyclopentane). This fits the chemical logic perfectly.