Decoding Multi-Step Syntheses: The Intramolecular Friedel-Crafts Reaction
One of the hallmarks of a difficult JEE Main Organic Chemistry question is a multi-step reaction (A → B → C) where the molecule literally folds in on itself to form a new ring.
In this brilliant PYQ from 2024, examiners combined three distinct concepts: Anhydride Ring Opening, Selective Reduction, and Intramolecular Acylation. Let's break down exactly how Compound A, B, and C are formed.
Video Tutorial: Finding the Major Product (C)
Watch Abhishek Sengar sir from CHEMCA expertly map out the reaction intermediates and demonstrate how the molecule undergoes ring closure.
Step-by-Step Reaction Breakdown
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Formation of A (Intermolecular Friedel-Crafts Acylation):
Benzene reacts with a cyclic anhydride (like Succinic Anhydride) in the presence of AlCl3. The Lewis acid breaks open the anhydride ring, generating an acylium ion electrophile. This attaches to the benzene ring.
The other end of the opened anhydride ring becomes a Carboxylic Acid.
Compound A: Ph-C(=O)-CH2-CH2-COOH (β-aroylpropionic acid). -
Formation of B (Clemmensen Reduction):
Compound A is treated with Zinc amalgam and Hydrochloric acid (Zn(Hg) / HCl). This is a highly selective reduction!
It reduces the ketone carbonyl (C=O) all the way down to a methylene group (-CH2-), but it completely ignores the carboxylic acid.
Compound B: Ph-CH2-CH2-CH2-COOH (γ-phenylbutyric acid). -
Formation of C (Intramolecular Ring Closure):
Compound B is heated with concentrated H2SO4. The strong acid protonates the -OH of the carboxylic acid, which leaves as water. This generates a reactive acylium ion (-C+=O) at the end of the alkyl chain.
Because this chain is highly flexible, it bends around and attacks the ortho position of its own benzene ring! This forms a new, highly stable fused 6-membered ring containing a ketone.
Compound C: α-Tetralone.
Whenever an electrophile is generated on a flexible chain attached to a benzene ring, always check if it can bend around to form a 5- or 6-membered ring. Intramolecular reactions are kinetically favored over reacting with a separate molecule!
Fig: Tracing the exact sequence from A to C. Note the chemoselectivity of Clemmensen reduction in Step 2.
Practice Questions for JEE & NEET
Ensure you understand the specific reagents used in this reaction sequence.
Question 1: In Step 2, we used the Clemmensen Reduction (Zn(Hg) / HCl). Why didn't this reagent also reduce the Carboxylic Acid (-COOH) group at the end of the chain?
Answer: Clemmensen is Chemoselective.
Reasoning:
The Clemmensen reduction is specific to Aldehydes and Ketones. The carbonyl carbon in a carboxylic acid is highly stabilized by resonance from the adjacent hydroxyl oxygen, making it significantly less electrophilic and totally unreactive towards Zinc amalgam and HCl.
If we wanted to reduce the carboxylic acid, we would need a much more powerful reagent, such as Lithium Aluminum Hydride (LiAlH4).
Question 2: In the final step (B → C), what alternative reagent sequence could be used instead of Concentrated H2SO4 to achieve the exact same ring closure?
Answer: PCl5 (or SOCl2) followed by AlCl3.
Reasoning:
While concentrated H2SO4 generates an acylium ion directly from the acid (by dehydrating it), another standard approach is to first convert the carboxylic acid into an acid chloride using PCl5 or SOCl2.
Once the acid chloride is formed (-CH2COCl), you add anhydrous AlCl3. The Lewis acid rips off the chlorine to generate the exact same acylium ion electrophile, leading to identical intramolecular ring closure!
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