Question

In Williamson synthesis, if tertiary alkyl halide is used, then:

• A. \( \text{Ether is obtained in good yield.} \)
• B. \( \text{Ether is obtained in poor yield.} \)
• C. \( \text{Alkene is the only reaction product.} \)
• D. \( \text{A mixture of alkene as a major product and ether as a minor product forms.} \)

Hint:

Tertiary alkyl halides are not suitable for the Williamson synthesis as they primarily undergo elimination reactions, producing alkenes instead of ethers.

Answer 1

The Correct Option is C

The Williamson synthesis, a reaction between an alkyl halide and an alkoxide ion yielding an ether (\[\text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-.\]), deviates when tertiary alkyl halides are employed. In such cases, the reaction mechanism shifts from substitution (SN2) to elimination (E2), a consequence of steric hindrance in tertiary alkyl halides impeding nucleophilic attack. For example, \[\text{(CH}_3\text{)}_3\text{C-Br} + \text{CH}_3\text{ONa} \rightarrow \text{(CH}_3\text{)}_2\text{C=CH}_2 + \text{CH}_3\text{OH} + \text{NaBr}.\] Consequently, the predominant product formed is an alkene (\( \text{(CH}_3\text{)}_2\text{C=CH}_2 \)), with ether formation being negligible. Final Answer: \[\boxed{\text{Alkene is the only reaction product.}}\]