Q.21 The major product formed in the following reaction is (ignore product stereochemistry)
Question Overview
The given substrate contains:
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A cyclohexene ring (alkene)
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A carboxylic acid side chain
The reaction conditions are:
I₂, NaHCO₃ (ignore stereochemistry)
We are asked to identify the major product.
Reaction Type: Iodolactonization
This reaction is a textbook example of iodolactonization, which occurs when:
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An alkene and a carboxylic acid are present in the same molecule
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Treated with halogen (I₂) in mild base (NaHCO₃)
Step-by-Step Mechanism
Step 1: Iodonium Ion Formation
The alkene reacts with I₂ to form a cyclic iodonium ion.
Step 2: Intramolecular Nucleophilic Attack
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NaHCO₃ deprotonates the –COOH group
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The resulting carboxylate (–COO⁻) attacks the more substituted carbon of the iodonium ion
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This occurs via 5-exo-trig cyclization (Baldwin-allowed)
Step 3: Lactone Formation
A five-membered lactone ring forms with:
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One iodine atom
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One cyclic ester (lactone)
Why 5-Membered Lactone?
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5-membered rings are kinetically favored
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6-membered lactones are less favorable under these conditions
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Intramolecular attack is faster than intermolecular reactions
Evaluation of Options
Option (A): ❌ Incorrect
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Shows a strained bicyclic anhydride-like structure
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No reasonable pathway via iodolactonization
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Not consistent with alkene + COOH chemistry
Option (B): ✅ Correct
✔ Contains:
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A 5-membered lactone ring
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Iodine attached to the carbon adjacent to oxygen
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Matches iodolactonization mechanism
✔ Follows:
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Baldwin’s rules
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Regioselective opening of iodonium ion
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Intramolecular cyclization
👉 This is the major product
Option (C): ❌ Incorrect
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Shows a different regiochemistry
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Places iodine at a less favorable position
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Violates preferred iodonium opening pathway
Option (D): ❌ Incorrect
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Represents a carboxylate salt
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No lactone ring formation
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Reaction conditions strongly favor cyclization, not salt stabilization
Final Answer
🎯 The major product formed is Option (B)
This product results from iodolactonization, involving:
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Alkene → iodonium ion
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Intramolecular nucleophilic attack by –COO⁻
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Formation of a five-membered iodolactone
Exam Tip (GATE/JAM/NET)
Whenever you see:
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Alkene + COOH
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I₂ / NaHCO₃
Immediately think:
🔥 Iodolactonization → 5-membered lactone
