Which one of the following reactions will not lead to the desired ether formation in major proportion? (Given labels: textiso-Bu Rightarrow textisobutyl$\text{iso-Bu} \Rightarrow \text{isobutyl}$, textsec-Bu Rightarrow textsec-butyl$\text{sec-Bu} \Rightarrow \text{sec-butyl}$, textn-Pr Rightarrow textn-propyl$\text{n-Pr} \Rightarrow \text{n-propyl}$, ^ttextBu Rightarrow texttert-butyl${}^t\text{Bu} \Rightarrow \text{tert-butyl}$, textEt Rightarrow textethyl$\text{Et} \Rightarrow \text{ethyl}$)

### Core Logic Williamson Ether Synthesis operates strictly via an **S_N2$S_N2$ mechanism**. To optimize ether formation yields, the alkyl halide component MUST be primary (1^circ$1^\circ$) or methyl to evade spatial shielding restrictions. Let us review the alkyl halide across the options: * Option 1: Ethyl bromide (textEtBr$\text{EtBr}$) is 1^circ$1^\circ$ rightarrow$\rightarrow$ Excellent ether yield via substitution. * Option 2: Methyl bromide (textCH_3textBr$\text{CH}_3\text{Br}$) is a unhindered methyl halide rightarrow$\rightarrow$ High substitution efficiency. * Option 3: n-Propyl bromide (textn-PrBr$\text{n-PrBr}$) is 1^circ$1^\circ$ rightarrow$\rightarrow$ Clean substitution pathway. * Option 4: sec-Butyl bromide (textsec-BuBr$\text{sec-BuBr}$) is a secondary (2^circ$2^\circ$) alkyl halide. When a sterically hindered 2^circ$2^\circ$ halide is treated with a strongly basic alkoxide reagent like isobutoxide, **alkene elimination (E2$E2$) competes aggressively and dominates** as the major pathway over nucleophilic substitution (S_N2$S_N2$).

### Pattern Recognition Williamson Synthesis Rule: Alkoxide can be as massive and complex as desired (3^circ$3^\circ$ or branched), but the Alkyl Halide MUST be unhindered (1^circ$1^\circ$ or methyl). If the halide is 2^circ$2^\circ$ or 3^circ$3^\circ$, elimination wins, forming an alkene instead of an ether. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Alcohols, Phenols and Ethers