JEE Advanced 2022 Paper 2 Q13 Chemistry Organic Reactions & Mechanisms Carboxylic Acids & Derivatives Hard

JEE Advanced 2022 Paper 2 · Q13 · Carboxylic Acids & Derivatives

Considering the following reaction sequence, the correct statement(s) is(are):

Starting material: benzene + succinic anhydride (cyclic anhydride O=C–CH$_2$–CH$_2$–C=O joined through O).

Reagents and intermediates: - Benzene $\xrightarrow{\text{succinic anhydride, AlCl}_3}$ P (Friedel-Crafts acylation gives Ph-CO-CH$_2$-CH$_2$-COOH; the keto acid). - P $\xrightarrow{\text{Zn/Hg, HCl (Clemmensen)}}$ Q (Ph-CH$_2$-CH$_2$-CH$_2$-COOH). - Q $\xrightarrow{\text{SOCl}_2}$ R (Ph-CH$_2$-CH$_2$-CH$_2$-COCl). - R $\xrightarrow{\text{AlCl}_3}$ S (intramolecular Friedel-Crafts acylation gives α-tetralone, the bicyclic cyclic ketone). - S $\xrightarrow{\text{Zn/Hg, HCl}}$ 'a hydrocarbon' (tetrahydronaphthalene = tetralin).

  1. A. Compounds P and Q are carboxylic acids.
  2. B. Compound S decolorizes bromine water.
  3. C. Compounds P and S react with hydroxylamine to give the corresponding oximes.
  4. D. Compound R reacts with dialkylcadmium to give the corresponding tertiary alcohol.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C

Solution

Identify each intermediate: P = Ph-CO-CH$_2$-CH$_2$-COOH (3-benzoylpropanoic acid — keto carboxylic acid). Q = Ph-CH$_2$-CH$_2$-CH$_2$-COOH (4-phenylbutanoic acid — carboxylic acid). R = Ph-CH$_2$-CH$_2$-CH$_2$-COCl (acyl chloride). S = α-tetralone (bicyclic aromatic ketone; no isolated C=C alkene).

(A) P has -COOH (with ketone); Q is a -COOH. BOTH are carboxylic acids. CORRECT.

(B) S = α-tetralone has only an aromatic ring + C=O, no isolated alkene. Does NOT decolourise bromine water. WRONG.

(C) P (ketone group) and S (ketone) react with NH$_2$OH to give oximes. CORRECT.

(D) R is an acyl chloride. Acyl chloride + R'$_2$Cd gives a KETONE, not a tertiary alcohol (this is the value of dialkyl cadmium). WRONG.

Correct options: A, C.

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