Chemistry · 2023
JEE Main · 30 January 2023, Shift 1 · Q60
A trisubstituted compound ' A ', C_10 H_12 O_2 gives neutral FeCl_3 test positive. Treatment of compound 'A' with NaOH and CH_3 Br gives C_11 H_14…
A trisubstituted compound ' A ', $\displaystyle \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_2$ gives neutral $\displaystyle \mathrm{FeCl}_3$ test positive. Treatment of compound 'A' with NaOH and $\displaystyle \mathrm{CH}_3 \mathrm{Br}$ gives $\displaystyle \mathrm{C}_{11} \mathrm{H}_{14} \mathrm{O}_2$, with hydroiodic acid gives methyl iodide and with hot conc. NaOH gives a compound $\displaystyle \mathrm{B}, \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_2$. Compound ' A ' also decolorises alkaline $\displaystyle \mathrm{KMnO}_4$. The number of $\displaystyle \pi$ bond/s present in the compound ' A ' is $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
4
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JEE Main 2023 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.