Question

The major products U and V in the following reaction are

• A. Fig A
• B. Fig B
• C. Fig C
• D. Fig D

Hint:

To remember this rule: "Polar Aprotic Solvents promote O-alkylation (Ether formation), while Non-polar Solvents with coordinating bases favor C-alkylation (Phenol ring substitution)." This is a classic advanced organic chemistry concept frequently tested in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given two different reaction pathways starting from phenol to form major products U and V. This tests the regioselectivity (O-alkylation vs. C-alkylation) of phenoxide ions under different solvent and counter-ion conditions.

Step 2: Key Formula or Approach:
The phenoxide ion is an ambident nucleophile that can attack electrophiles via its oxygen atom (O-alkylation) or its ortho/para carbon atoms (C-alkylation):
- Polar Aprotic Solvents (like acetone): The phenoxide oxygen is weakly solvated and highly nucleophilic. Thus, reaction with \( \text{K}_2\text{CO}_3 \) in acetone leads to rapid O-alkylation.
- Non-polar Solvents (like benzene) with coordinating cations: The alkali metal phenoxide (formed by \( \text{KOH} \)) forms tight ion-pairs in non-polar solvents. The oxygen is shielded by the metal cation, directing the electrophile to attack the nucleophilic ortho-carbon, leading to C-alkylation.

Step 3: Detailed Explanation:
1. Reaction to form U:
Phenol is treated with 1-bromo-3-methylbut-2-ene in the presence of \( \text{K}_2\text{CO}_3 \) in acetone. Since acetone is a polar aprotic solvent, the free phenoxide oxygen nucleophilically attacks the allylic halide via an \( \text{S}_{\text{N}}2 \) pathway, resulting in O-alkylation. The major product U is the corresponding ether:
\[ \text{C}_6\text{H}_5\text{-O-CH}_2\text{-CH=C(CH}_3)_2 \]
2. Reaction to form V:
Phenol is treated with \( \text{KOH} \) and allyl bromide in benzene. In a non-polar solvent like benzene, the potassium phenoxide forms tight ion pairs, shielding the oxygen atom. This promotes electrophilic attack at the ortho-carbon of the benzene ring, resulting in C-alkylation. The major product V is:
\[ o\text{-allylphenol} \quad (2\text{-allylphenol}) \]
This matches the structures shown in option (C).

Step 4: Final Answer:
The major products U and V are given in option (C).