JEE Main 2021 — Hydrocarbons Question with Solution
From: JEE Main 2021 (Online) 25th February Morning Shift
Question
Identify A in the given chemical reaction.
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
Mo2O3 at 773 K temperature and 10-20-atm pressure is aromatising agent.
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This is a previous-year question from JEE Main 2021, covering the Hydrocarbons chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.