Question:
What is the end product Z in the given reaction sequence? $Phenol \xrightarrow[Zn/\Delta]{(i)} X \xrightarrow[Cl_2/FeCl_3, \Delta]{(ii)} Y \xrightarrow{CH_3COCl/Anh. AlCl_3} Z \xrightarrow[Glycol, \Delta]{N_2H_4/OH^-} Z?$
What is the end product Z in the given reaction sequence? $Phenol \xrightarrow[Zn/\Delta]{(i)} X \xrightarrow[Cl_2/FeCl_3, \Delta]{(ii)} Y \xrightarrow{CH_3COCl/Anh. AlCl_3} Z \xrightarrow[Glycol, \Delta]{N_2H_4/OH^-} Z?$
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Wolff-Kishner reduction turns a ketone/aldehyde group into a hydrocarbon chain.
Updated On: Oct 5, 2026
- Structure 1
- Structure 2
- Structure 3
- Structure 4
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The Correct Option is C
Solution and Explanation
Step 1: Concept
$Phenol \rightarrow Benzene \rightarrow Chlorobenzene \rightarrow$ Acetophenone (Friedel-Crafts) $\rightarrow$ Ethylbenzene (Wolff-Kishner).
Step 2: Analysis
Benzene is formed (X). Chlorobenzene is Y (substitution). Friedel-Crafts acylation gives Acetophenone. Wolff-Kishner reduces the carbonyl group to ethyl.
Step 3: Conclusion
The end product is ethylbenzene (Structure 3).
Final Answer: (C)
$Phenol \rightarrow Benzene \rightarrow Chlorobenzene \rightarrow$ Acetophenone (Friedel-Crafts) $\rightarrow$ Ethylbenzene (Wolff-Kishner).
Step 2: Analysis
Benzene is formed (X). Chlorobenzene is Y (substitution). Friedel-Crafts acylation gives Acetophenone. Wolff-Kishner reduces the carbonyl group to ethyl.
Step 3: Conclusion
The end product is ethylbenzene (Structure 3).
Final Answer: (C)
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