2. In Williamson ether synthesis, which of the following pairs of reactants will directly give ethyl phenyl ether (${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{O}{\mathrm{C}}_6{\mathrm{H}}_5$)?

Step-by-step solution

1. Recall the Williamson ether synthesis: an alkoxide ion reacts with a primary alkyl halide via an $S_{N}2$ mechanism to form an ether.
2. Identify the alkoxide that provides the phenoxy part: phenoxide ion (${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}$).
3. Identify the primary alkyl halide that will provide the ethyl group: ethyl bromide (${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}$).
4. Check other options – any that contain a secondary/tertiary halide or use phenyl halide – would lead to side reactions or no $S_{N}2$.
5. Conclude that the correct pair is phenoxide ion and ethyl bromide.

Concept tested: Williamson ether synthesis reagents

Published 8 September 2026 · How this solution was prepared and checked