Sigma Percentile
JEE Advanced 2018
LEVELJEE Advanced

Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product.
Question 1:

The compound Y is :-

Select Answer:

Question 2:

The compound Z is :-

Select Answer:

Visualized Solution

  • We are given a multi-step organic synthesis starting from benzene.
  • We need to identify the major products , , and .

  • Benzene reacts with and in the presence of anhydrous and to form benzaldehyde.

  • Benzaldehyde reacts with acetic anhydride () and sodium acetate () to form cinnamic acid (Compound ).

  • Compound reacts with to undergo electrophilic addition across the double bond, forming a dibromo intermediate.

  • Heating the dibromo intermediate with moist at causes double dehydrohalogenation and decarboxylation, yielding phenylacetylene (Compound ).

  • In the second path, Compound (cinnamic acid) is treated with , which selectively reduces the alkene double bond to form 3-phenylpropanoic acid.

  • Treatment of 3-phenylpropanoic acid with generates an acyl cation, which undergoes intramolecular electrophilic aromatic substitution to form 1-indanone (Compound ).

  • Compound is phenylacetylene (Option C for Q7).
  • Compound is 1-indanone (Option A for Q8).

The Sigma Insight: Carbonyl Compounds

Solution Diagram
Welcome to this beautiful organic synthesis problem! We are going on a journey starting from a simple benzene ring, transforming it step-by-step into complex molecules. Let's break down this sequence and uncover the identities of compounds X, Y, and Z.

The Starting Point

Gattermann-Koch Formylation
Look closely at our first step. We treat benzene with carbon monoxide () and hydrogen chloride (), using anhydrous aluminum chloride () and copper chloride () as catalysts.
Do you remember this named reaction? Yes, it's the classic Gattermann-Koch formylation! It introduces a formyl group directly onto the benzene ring, giving us benzaldehyde.

The Perkin Condensation

Forging Compound X
Next, we take our benzaldehyde and react it with acetic anhydride () and sodium acetate (). This is the Perkin condensation!
The acetate ion acts as a base, generating an enolate from acetic anhydride, which then attacks the aldehyde. After condensation and hydrolysis, we get an -unsaturated acid. This is cinnamic acid, which is our Compound X!

Path 1

The Dramatic Transformation to Compound Y
Now, let's follow the first path to find Compound Y. We treat cinnamic acid with bromine () and sodium carbonate (). The double bond undergoes electrophilic addition, adding two bromine atoms across it. We get a dibromo intermediate. The sodium carbonate just keeps the medium slightly basic.
There is a catch here, watch carefully. We heat this dibromo compound strongly with moist potassium hydroxide () at . This is a harsh condition! It triggers a double dehydrohalogenation—removing two molecules of hydrogen bromide ()—and simultaneously, a decarboxylation occurs, removing the carboxylic acid group as carbon dioxide ().
The result? A terminal alkyne! We get phenylacetylene. This is our Compound Y.

Path 2

Reduction and Cyclization to Compound Z
Let's go back to Compound X and explore the second path to find Compound Z. We treat cinnamic acid with hydrogen gas over a palladium-carbon catalyst (). This is a catalytic hydrogenation. It selectively reduces the carbon-carbon double bond without touching the benzene ring or the carboxylic acid. We get 3-phenylpropanoic acid.
Finally, we treat this saturated acid with phosphoric acid (). Phosphoric acid protonates the carboxylic acid , water leaves, and we generate a highly reactive acyl cation.
Because this cation is tethered to the benzene ring, it bends around and attacks the ring intramolecularly! This is an intramolecular Friedel-Crafts acylation. It forms a new five-membered ring fused to the benzene. This beautiful bicyclic molecule is 1-indanone, our Compound Z!

The Final Verdict

And there we have it! We've successfully navigated the entire reaction map.
Compound Y is phenylacetylene, which matches option (C) for the first question. Compound Z is 1-indanone, matching option (A) for the second question.
A brilliant exercise in recalling named reactions and intramolecular cyclizations!

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