Medium INTEGER +4 / -1 PYQ · JEE Mains 2023

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 (integer) 4

Solution

<p>$\mathrm{A:C_{10}H_{12}O_2}$</p> <p>DU of A $=\frac{22-12}{2}=5$</p> <p>1 DU is due to Ring (Benzene ring)</p> <p>4 $\pi$-bonds will be there</p> <p>(3 $\pi$-bonds in ring and 1 $\pi$-bond outside ring) as it decolorises alkaline KMnO$_4$.</p>

About this question

Subject: Chemistry · Chapter: Aldehydes, Ketones and Carboxylic Acids · Topic: Nucleophilic Addition Reactions

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