The major products U and V in the following reaction are
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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.
Step 1: Understanding the Concept:
This problem centers on the regioselectivity of the phenoxide ion, which is a classic example of an ambident nucleophile.
An ambident nucleophile possesses two different nucleophilic sites through which it can attack an electrophile.
In the phenoxide ion, the negative charge is delocalized via resonance from the oxygen atom to the ortho and para positions of the benzene ring.
The fundamental question in these reactions is whether the electrophile (an alkyl halide) will react at the oxygen atom (O-alkylation) to form an ether, or at the carbon atom of the ring (C-alkylation) to form a substituted phenol.
The outcome is strictly governed by the reaction environment, specifically the solvent polarity, the nature of the base (and its counter-ion), and the degree of solvation of the nucleophilic centers. Step 2: Key Formula or Approach:
We evaluate the reaction conditions based on the solvent-directed regioselectivity principle:
1. Polar Aprotic Solvents (e.g., Acetone): These solvents solvate cations well but leave anions relatively "naked" and highly reactive. This favors attack by the more electronegative and more nucleophilic oxygen atom (O-alkylation).
2. Non-polar Solvents (e.g., Benzene): These solvents do not solvate ions effectively, leading to the formation of "tight ion pairs" between the metal cation (like $K^+$) and the phenoxide oxygen. This shielding of the oxygen directs the attack toward the nucleophilic carbon atoms of the ring (C-alkylation). Step 3: Detailed Explanation: Reaction 1 (Formation of U):
Phenol reacts with $K_2CO_3$ in acetone. The base $K_2CO_3$ deprotonates phenol to generate the potassium phenoxide salt.
Acetone is a polar aprotic solvent with a significant dielectric constant. In this medium, the potassium cation is solvated, but the phenoxide anion remains relatively free.
The oxygen atom is the most nucleophilic site because it carries the primary negative charge density.
The nucleophilic oxygen attacks the 1-bromo-3-methylbut-2-ene in an $S_N2$ fashion.
This results in the formation of an allylic phenyl ether. Thus, product U is $C_6H_5-O-CH_2-CH=C(CH_3)_2$. Reaction 2 (Formation of V):
Phenol reacts with KOH in benzene with allyl bromide.
Benzene is a non-polar solvent with a very low dielectric constant. In benzene, potassium phenoxide exists as a tight ion-pair where the $K^+$ cation is closely coordinated with the phenoxide oxygen.
This coordination effectively "masks" or shields the oxygen atom, preventing it from acting as the nucleophilic center.
The electron density concentrated at the ortho position of the benzene ring is more available for nucleophilic attack.
The ortho carbon attacks the allyl bromide. After the initial C-alkylation, the system undergoes rapid rearomatization to restore the stability of the benzene ring.
The resulting product V is 2-allylphenol (ortho-allylphenol).
Comparing these logical conclusions to the provided figures, Figure C correctly depicts the ether for U and the ortho-substituted phenol for V. Step 4: Final Answer:
By analyzing the solvent effects on the ambident phenoxide ion, we conclude that U is an ether and V is an ortho-alkylated phenol. This matches Option (C).
