Williamson Ether Synthesis: The Williamson synthesis is the principal method for preparing ethers:

$\text{R-O}^-\text{Na}^+ + \text{R'-X} \xrightarrow{\text{SN2}} \text{R-O-R'} + \text{NaX}$

The alkoxide (RO^{-}$$Na^{+}) acts as the nucleophile. The alkyl halide (R'X) acts as the electrophile. The mechanism is SN2 (bimolecular nucleophilic substitution). This has a critical implication: R'X must be a primary (1°) alkyl halide. If a secondary or tertiary alkyl halide is used, the alkoxide (which is also a strong base) will abstract a β-hydrogen from the halide in an E2 elimination reaction, producing an alkene rather than an ether. The rule: "pair alkoxide with 1° halide, never 2° or 3°."

Example — Synthesis of Diethyl Ether: CH_{3}CH_{2}O^{-}$$Na^{+} + $CH_{3}CH_{2}Br$ → $CH_{3}CH_{2}OCH_{2}CH_{3}$ + NaBr SMILES of diethyl ether: SMILES:CCOCC

Example — Anisole Synthesis (methyl phenyl ether): PhO^{-}$$Na^{+} + $CH_{3}Br$ (1°) → $PhOCH_{3}$ (SMILES:COc1ccccc1) + NaBr

Cleavage of Ethers with HI: Ethers are generally unreactive (no good leaving group, no easily ionizable H). However, HI is the strongest hydrohalic acid and can cleave ethers:

• Step 1: R-O-R' + HI → RI + R'OH (protonation of O, then SN2 attack by $I^{-}$ on smaller alkyl group)
• Step 2: R'OH + HI → R'I + $H_{2}O$ (if excess HI present)
• Net with excess HI: R-O-R' → RI + R'I (both fragments become alkyl iodides)

For anisole (SMILES:COc1ccccc1) + excess HI: the methyl group (smaller, 1°) leaves as $CH_{3}I$; the phenyl group remains as phenol then $C_{6}H_{5}I$.