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 involves the reaction between an alkyl halide and an alkoxide ion to produce an ether: \[ \text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-. \]
However, when tertiary alkyl halides are used: 1. The reaction mechanism favors elimination (E2) rather than substitution (SN2). 2. This occurs because the steric hindrance in tertiary alkyl halides hinders nucleophilic substitution. For instance: \[ \text{(CH}_3\text{)}_3\text{C-Br} + \text{CH}_3\text{ONa} \rightarrow \text{(CH}_3\text{)}_2\text{C=CH}_2 + \text{CH}_3\text{OH} + \text{NaBr}. \] Thus, the primary product of the reaction is the alkene (\( \text{(CH}_3\text{)}_2\text{C=CH}_2 \)), and no ether is produced. Final Answer: \[ \boxed{\text{Alkene is the only reaction product.}} \]