JEE Challenger
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Identification of Major Product Z in Multistep Benzene Synthesis

Comprehension Passage

Treatment of benzene with CO/HCl\text{CO}/\text{HCl} in the presence of anhydrous AlCl3/CuCl\text{AlCl}_3/\text{CuCl}, followed by reaction with Ac2O/NaOAc\text{Ac}_2\text{O}/\text{NaOAc}, yields compound X\mathbf{X} as the principal product. Compound X\mathbf{X}, when reacted with Br2/Na2CO3\text{Br}_2/\text{Na}_2\text{CO}_3 and subsequently heated at 473 K473\text{ K} with moist KOH\text{KOH}, furnishes Y\mathbf{Y} as the primary product. Catalytic hydrogenation of X\mathbf{X} over H2/Pd-C\text{H}_2/\text{Pd-C}, followed by cyclization using H3PO4\text{H}_3\text{PO}_4, gives Z\mathbf{Z} as the major product.

(There are two questions based on PARAGRAPH "X", the question given below is one of them)

Identify the structure of compound Z\mathbf{Z}.

Options

A
Option A
Correct
B
Option B
C
Option C
D
Option D

Step-by-Step Solution

To determine the structure of compound Z\mathbf{Z}, we trace the reaction sequence step-by-step:

  1. Formation of Benzaldehyde (Gattermann-Koch Reaction): Benzene reacts with carbon monoxide (CO\text{CO}) and hydrogen chloride (HCl\text{HCl}) in the presence of anhydrous AlCl3\text{AlCl}_3 and CuCl\text{CuCl} to yield benzaldehyde: C6H6+CO+HCl→anh. AlCl3/CuClC6H5CHO\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{anh. } \text{AlCl}_3 / \text{CuCl}} \text{C}_6\text{H}_5\text{CHO}

  2. Formation of Compound X\mathbf{X} (Perkin Condensation): Benzaldehyde undergoes a Perkin reaction upon treatment with acetic anhydride (Ac2O\text{Ac}_2\text{O}) in the presence of sodium acetate (NaOAc\text{NaOAc}) followed by hydrolysis to give cinnamic acid as the major product: C6H5CHO+(CH3CO)2O→ΔCH3COONaC6H5−CH=CH−COOH(X)\text{C}_6\text{H}_5\text{CHO} + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow[\Delta]{\text{CH}_3\text{COONa}} \text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{COOH} \quad (\mathbf{X})

  3. Catalytic Hydrogenation of Compound X\mathbf{X}: Catalytic hydrogenation of cinnamic acid over H2/Pd-C\text{H}_2/\text{Pd-C} selectively reduces the aliphatic carbon-carbon double bond, yielding 3-phenylpropanoic acid (hydrocinnamic acid): C6H5−CH=CH−COOH→H2/Pd-CC6H5−CH2−CH2−COOH\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{COOH} \xrightarrow{\text{H}_2/\text{Pd-C}} \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{COOH}

  4. Intramolecular Cyclization to Form Compound Z\mathbf{Z}: Treatment of 3-phenylpropanoic acid with phosphoric acid (H3PO4\text{H}_3\text{PO}_4), a dehydrating and acid catalyst, induces an intramolecular Friedel-Crafts acylation. The acylium ion intermediate undergoes electrophilic aromatic substitution onto the adjacent position of the phenyl ring to form the fused bicyclic ketone, 1-indanone: C6H5−CH2−CH2−COOH→H3PO41-Indanone(Z)\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{COOH} \xrightarrow{\text{H}_3\text{PO}_4} \text{1-Indanone} \quad (\mathbf{Z})

The structure of 1-indanone consists of a benzene ring fused to a five-membered ring containing a carbonyl group (C=O\text{C}=\text{O}) adjacent to the benzene ring. This corresponds to the structure shown in option (A).

Correct Answer: (A)