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Question
An aromatic compound ‘A’ (Molecular formula \[\ce{C8H8O}\]) gives 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 Tollen’s or Fehling’s test. On drastic oxidation with potassium permanganate it forms a carboxylic acid ‘C’ (Molecular formula \[\ce{C7H6O2}\]), which is also formed along with the yellow compound in the above reaction. Identify A, B and C and write all the reactions involved.
Solution
Molecular formula of compound is \[\ce{C8HgO}\]. As ‘A’ does not give Tollens’ or Fehling’s test. It must be a ketone. It gives positive test with 2, 4-DNP and iodoform test. It means it is methyl ketone.
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| Column I (Example) |
Column II (Reaction) |
||
| (i) | \[\begin{array}{cc} \phantom{...}\ce{O}\phantom{..............................}\ce{O}\phantom{}\\ \phantom{...}||\phantom{..............................}||\phantom{}\\ \ce{CH3 - C - Cl + H2 ->[Pd - C/BasO4] CH3 - C - H} \end{array}\] |
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| (ii) |  |
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| (iii) |  |
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| (iv) | \[\begin{array}{cc} \ce{R - CH2 - COOH ->[Br/Red P] R - CH - COOH}\\ \phantom{.....................}|\\ \phantom{.......................}\ce{Br} \end{array}\] |
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| (v) | \[\ce{CH3 - CN ->[(i) SnCl2/HCl][(ii) H2O/H+] CH3CHO}\] | (e) | Rosenmund’s reductio |
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