Phenols and Ethers — Part 2: Ethers: visual lesson, ISC Class 12 Chemistry

Lesson

Ethers

Part 2 — R–O–R′: structure, preparation, properties and reactions, including anisole.

In this lesson you will

Describe the structure of the ether group and name ethers
Choose reactants for Williamson’s synthesis of a given ether
Write reactions of ethers with chlorine, oxygen, HI and PCl₅
Write halogenation, nitration and Friedel–Crafts reactions of anisole, and state uses of ethers

Diagram

Structure and nomenclature

The ether oxygen is sp³ with two lone pairs; the C–O–C angle (about 111.7° in methoxymethane) is larger than tetrahedral because the alkyl groups repel. Ethers are symmetrical (R–O–R) or unsymmetrical (R–O–R′). IUPAC: alkoxyalkanes, the larger group being the parent: CH₃OC₂H₅ methoxyethane; C₂H₅OC₂H₅ ethoxyethane; C₆H₅OCH₃ methoxybenzene (anisole).

Concept

Williamson’s synthesis

An alkoxide (or phenoxide) reacts with a primary alkyl halide by SN2: R–O⁻Na⁺ + R′–X → R–O–R′ + NaX. With secondary and especially tertiary halides, the strongly basic alkoxide causes elimination, so the bulky group must be on the alkoxide. Aryl halides cannot be used (unreactive).

\ce{CH3CH2ONa + CH3CH2CH2Br -> CH3CH2OCH2CH2CH3 + NaBr}

\ce{(CH3)3CONa + CH3CH2Br -> (CH3)3COCH2CH3 + NaBr}

Table of Williamson choices
Target	Good choice	Poor choice (why)
1-Ethoxypropane	C₂H₅ONa + 1-bromopropane	—
2-Ethoxy-2-methylpropane	(CH₃)₃CONa + C₂H₅Br	(CH₃)₃CBr + C₂H₅ONa (elimination)
Ethoxybenzene	C₆H₅ONa + C₂H₅Br	C₆H₅Br + C₂H₅ONa (aryl halide inert)

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