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 \(\mathbf{K}\) is

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Hydrocarbons

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

K

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

Answer: (C)

Solution

J is cinnamic acid. Step (i) hydrogenates the double bond to give 3-phenylpropanoic acid. Step (ii) converts it to the acid chloride with SOCl₂. Step (iii) performs intramolecular Friedel-Crafts acylation with AlCl₃ to form a six-membered ring ketone: α-tetralone (1-tetralone). The answer is (C).

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

Practise more: all JEE Advanced PYQs on Hydrocarbons →

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Hydrocarbons

Solution

More JEE Advanced questions on Hydrocarbons