JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Aldehydes Ketones and Carboxylic Acids

The compound has to meet two tests at once: react with ethylmagnesium bromide and decolourise bromine water. A phenolic $-\mathrm{OH}$ provides the acidic hydrogen that reacts with the Grignard reagent, liberating ethane, while a vinyl group ($-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$) adds bromine and so decolourises bromine water. The compound carrying both groups is $o$-vinylphenol (2-ethenylphenol): its phenolic hydroxyl reacts with ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgBr}$ and its side-chain double bond decolourises bromine water, so it alone satisfies both requirements as intended by the official key.