Q. Consider the following reaction: ethanol —(PBr₃)→ X —(alc. KOH)→ Y —((i) H₂SO₄, room temperature; (ii) H₂O, heat)→ Z The product Z is:
A) CH₃CH₂–O–CH₂–CH₃
B) CH₃–CH₂–O–SO₃H
C) CH₃CH₂OH
D) CH₂=CH₂
Explanation
[Solution from source] PBr₃ converts ethanol to bromoethane (X); alcoholic KOH eliminates HBr to give ethene (Y). Cold conc. H₂SO₄ adds across the double bond giving ethyl hydrogen sulphate, which on boiling with water is hydrolysed back to ethanol (Z). Ethyl hydrogen sulphate is only the intermediate. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.