In the following reaction sequence, the compound J is an intermediate. \[ \mathbf{I} \xrightarrow[\mathrm{CH}_{3} \mathrm{COONa}]{\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}} \mathbf{J} \xrightarrow[\text { (ii) } \mathrm{SOCl}_{2}]{\text { (i) } \mathrm{H}_{2}, \mathrm{Pd} / \mathrm{C}} \mathbf{K} \] (iii) anhyd. \(\mathrm{AlCl}_{3}\) J \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) gives effervescence on treatment with \(\mathrm{NaHCO}_{3}\) and a positive Baeyer's test. The compound I is

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Aldehydes Ketones and Carboxylic Acids

In the following reaction sequence, the compound J is an intermediate.

I (CH_{3} CO)_{2} O CH_{3} COONa J (i) H_{2}, Pd / C (ii) SOCl_{2} K

(iii) anhyd.

AlCl_{3}

(C_{9} H_{8} O_{2})

gives effervescence on treatment with

NaHCO_{3}

and a positive Baeyer's test.

The compound I is

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

J (C₉H₈O₂) gives effervescence with NaHCO₃ (indicating a carboxylic acid) and a positive Baeyer's test (indicating a C=C double bond). Working backward, J is cinnamic acid (C₆H₅-CH=CH-COOH) formed by Perkin condensation. Therefore, I must be benzaldehyde (C₆H₅CHO), which undergoes Perkin reaction with acetic anhydride and sodium acetate. The answer is (A).

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Aldehydes Ketones and Carboxylic Acids

Solution

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