Report Issue

JEE-Main_2023
30-1-23{S1}
Question
A trisubstituted compound 'A', $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_2$ gives neutral $FeCl_3$ test positive. Treatment of compound 'A' with NaOH and $\mathrm{CH}_3 \mathrm{Br}$ gives $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{14} \mathrm{O}_2$, with hydroiodic acid gives methyl iodide and with hot cone. NaOH gives a compound B, $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_2$. Compound 'A' also decolorizes alkaline KMn$O_4$. The number of n bond/s present in the compound 'A' is_______.
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Chemistry
Hard
Start Preparing for JEE with Competishun