JEE MAIN - Chemistry (2023 - 30th January Morning Shift - No. 19)
A trisubstituted compound '$$\mathrm{A}$$', $$\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$$ gives neutral $$\mathrm{FeCl}_{3}$$ test positive. Treatment of compound 'A' with $$\mathrm{NaOH}$$ and $$\mathrm{CH}_{3} \mathrm{Br}$$ gives $$\mathrm{C}_{11} \mathrm{H}_{14} \mathrm{O}_{2}$$, with hydroiodic acid gives methyl iodide and with hot conc. $$\mathrm{NaOH}$$ gives a compound $$\mathrm{B}, \mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$$. Compound 'A' also decolorises alkaline $$\mathrm{KMnO}_{4}$$. The number of $$\pi$$ bond/s present in the compound '$$\mathrm{A}$$' is _____________.
Answer
4
Explanation
$$\mathrm{A:C_{10}H_{12}O_2}$$
DU of A $$=\frac{22-12}{2}=5$$
1 DU is due to Ring (Benzene ring)
4 $$\pi$$-bonds will be there
(3 $$\pi$$-bonds in ring and 1 $$\pi$$-bond outside ring) as it decolorises alkaline KMnO$$_4$$.
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