Question

Suitable reaction condition for preparation of Methyl phenyl ether is

• A. PhO- Na+ , MeBr
• B. PhO- Na+ , MeOH
• C. Ph - Br, MeO- Na+
• D. Benzene, MeBr

Hint:

The Williamson Ether Synthesis is a key reaction for preparing ethers. Remem ber that it’s an SN₂ reaction, so it works best with primary alkyl halides (or those with minimal steric hindrance). If a tertiary alkyl halide is used, an elim ination reaction is more likely to occur.

Answer 1

The Correct Option is A

To prepare Methyl phenyl ether, also known as anisole, we need to consider the suitable reaction conditions that will enhance the Williamson ether synthesis. The correct conditions include using a phenoxide ion and an alkyl halide. Let's break down the answer in detail:

1. Understanding Williamson Ether Synthesis: This method involves the reaction of a phenoxide ion with a primary alkyl halide. The reaction follows an S_N2 mechanism where the nucleophile (phenoxide ion) attacks the alkyl halide leading to the formation of the ether.

2. Analyzing the Options:

   • Option 1: PhO^- \text{ Na}^+ , \text{MeBr}. In this option, phenoxide ion is a good nucleophile and methyl bromide (MeBr) is a suitable substrate for S_N2 reactions. Therefore, it will lead to the formation of methyl phenyl ether (anisole).
   • Option 2: PhO^- \text{ Na}^+ , \text{MeOH}. Here, methanol is not an alkyl halide, so it cannot participate in an S_N2 reaction with phenoxide ion.
   • Option 3: Ph - \text{Br, MeO}^- \text{ Na}^+. While methoxide ion (MeO^-) is a good nucleophile, phenyl bromide (Ph-Br) is not a good substrate for S_N2 reactions with such a strong base since it tends to undergo elimination or a different mechanism.
   • Option 4: \text{Benzene, MeBr}. Benzene cannot act as a nucleophile because it's a stable aromatic compound and inert under these conditions.

3. Conclusion: The correct choice for generating methyl phenyl ether via the Williamson synthesis is using PhO^- Na⁺ and MeBr as given in Option 1.