Williamson synthesis: which pairing gives the ether?

Choose the alkoxide and the halide for a Williamson synthesis; SN2 succeeds on methyl/1° halides while a 2° or 3° halide gives elimination.

Make tert-butyl methyl ether with no 2-methylpropene by-product.

Sodium methoxide + tert-butyl bromide gives the ether.

• Protic acids (H₂SO₄, H₃PO₄) dehydrate alcohols, and the conditions decide the product: ethanol with sulphuric acid gives ethene at 443 K but mainly ethoxyethane at 413 K.
• Ether formation from an alcohol is an SN2 reaction: one alcohol molecule attacks a protonated alcohol, water leaves, and the protonated ether loses H⁺.
• This route suits only primary alcohols with unhindered alkyl groups at low temperature. With secondary and tertiary alcohols, elimination to the alkene wins over substitution.
• Williamson synthesis: an alkyl halide reacts with a sodium alkoxide to give an ether, R–X + R′–O⁻Na⁺ → R–O–R′ + NaX. It makes both symmetrical and unsymmetrical ethers.
• The alkoxide attacks the alkyl halide by SN2, so primary alkyl halides work well. With secondary and tertiary halides elimination competes, and a tertiary halide gives only the alkene.
• Example: sodium methoxide with (CH₃)₃C–Br gives only 2-methylpropene, so tert-butyl methyl ether must be made from the tertiary alkoxide and a methyl halide instead.
• Phenols are converted to ethers the same way: the phenoxide ion is the nucleophile that attacks the alkyl halide.
• Diethyl ether was once widely used as an inhalation anaesthetic, but its slow action and unpleasant recovery led to its replacement by other compounds.

Teaching this idea? Put the simulation on your own page or LMS with this code.