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The WILLIAMSON ETHER SYNTHESIS prepares ethers by reacting

Aan alkyl halide with an alkoxide ion (SN2 mechanism)
Btwo alcohols with hot conc. $\mathrm{H_2SO_4}$
Can alcohol with an aldehyde and ZnCl$_2$
Dan aldehyde with hydrogen and Pd
Answer & Solution
Correct answer: A. an alkyl halide with an alkoxide ion (SN2 mechanism)
1. Your syllabus: the Williamson synthesis is the standard preparative route for unsymmetric ethers. 2. Generic reaction: $\mathrm{R{-}O^-Na^+} + \mathrm{R'{-}X} \to \mathrm{R{-}O{-}R'} + \mathrm{NaX}$. 3. The alkoxide (a strong nucleophile) attacks the alkyl halide via SN2 — back-side displacement of the leaving halide. 4. For best yield, use METHYL or PRIMARY alkyl halides (better SN2 substrates). Secondary alkyl halides give mixtures of substitution and elimination tertiary halides give mostly elimination. 5. Example: $\mathrm{CH_3O^-Na^+ + CH_3CH_2Br \to CH_3{-}O{-}CH_2CH_3 + NaBr}$. 6. Option B is INTRAMOLECULAR/INTERMOLECULAR dehydration (gives symmetric ethers only, lower yield). Options C and D are unrelated. _Source: syllabus Chemistry Part 2, Ch 7, §7.8 (Williamson ether synthesis), p. 21._
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