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An aromatic compound ‘A’ (Molecular formula \mathrm{C}_8 \mathrm{H}_8 \mathrm{O}) gives a positive 2, 4-DNP test. It gives a yellow precipitate of compound ‘B’ on treatment with iodine and sodium hydroxide solution. Compound ‘A’ does not give Tollens or Fehling’s test. On drastic oxidation with potassium permanganate, it forms a carboxylic acid ‘C’ (Molecular formula \mathrm{C}_7 \mathrm{H}_6 \mathrm{O}_2), which is also formed along with the yellow compound in the above reaction. Identify A, B, and C and write all reactions involved.

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The molecular formula of the compound in question is \mathrm{C}_8 \mathrm{H}_8 \mathrm{O}. It is given that ‘A’ does not give Tollens’ or Fehling’s test, making it a ketone. Since we get positive tests with 2, 4-DN, P, and iodoform test, It means it is methyl ketone.

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