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Preparation of Ethers

Synthesis via Dehydration, Williamson Synthesis, and Dry Silver Oxide.

By chemca Team • Updated Jan 2026

Ethers ($R-O-R'$) are organic compounds where an oxygen atom is bonded to two alkyl or aryl groups. They can be prepared by two main methods: Dehydration of Alcohols and Williamson Synthesis.

1. Dehydration of Alcohols

Acid Catalyzed Dehydration

Alcohols undergo dehydration in the presence of protic acids ($H_2SO_4, H_3PO_4$). The product depends critically on the Temperature.

At 413 K (140°C): Formation of Ether (Substitution).

$$ 2C_2H_5OH \xrightarrow{H_2SO_4, 413K} \underset{\text{Diethyl Ether}}{C_2H_5-O-C_2H_5} + H_2O $$

At 443 K (170°C): Formation of Alkene (Elimination).

$$ C_2H_5OH \xrightarrow{H_2SO_4, 443K} CH_2=CH_2 + H_2O $$

Limitation: This method is suitable ONLY for preparing symmetrical ethers from primary alcohols. Secondary and tertiary alcohols prefer elimination to form alkenes.

2. Williamson Synthesis

Best Laboratory Method

Reaction of an alkyl halide with sodium alkoxide yields ether. This follows an $S_N2$ mechanism.

$$ R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX $$

Condition for Best Yield: The alkyl halide ($R-X$) must be Primary ($1^\circ$). The alkoxide ($R'-O^-$) can be primary, secondary, or tertiary.

Example: Preparation of tert-Butyl Methyl Ether

Correct: $1^\circ$ Halide + $3^\circ$ Alkoxide

$$ CH_3Br + (CH_3)_3C-ONa \rightarrow (CH_3)_3C-O-CH_3 + NaBr $$

Wrong: $3^\circ$ Halide + $1^\circ$ Alkoxide

If tertiary halide is used, elimination dominates, forming an alkene.

$$ (CH_3)_3C-Br + CH_3ONa \rightarrow \underset{\text{Isobutylene}}{(CH_3)_2C=CH_2} + NaBr + CH_3OH $$

3. Reaction with Dry Silver Oxide

From Alkyl Halides

Heating alkyl halides with dry silver oxide ($Ag_2O$) produces ethers.

$$ 2R-X + Ag_2O \text{ (dry)} \xrightarrow{\Delta} R-O-R + 2AgX $$

Note: Using moist $Ag_2O$ yields Alcohols.

4. Preparation of Phenolic Ethers

Preparation of Anisole

Phenol is converted to sodium phenoxide, which then reacts with methyl halide (Williamson Synthesis).

$$ \underset{\text{Phenol}}{C_6H_5OH} \xrightarrow{NaOH} \underset{\text{Sodium Phenoxide}}{C_6H_5ONa} \xrightarrow{CH_3I} \underset{\text{Anisole}}{C_6H_5OCH_3} + NaI $$

Important: Aryl halides ($Ar-X$) cannot be used as the substrate because they do not undergo nucleophilic substitution easily. The aryl group must come from the phenoxide.

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