Ethers

Williamson's ether synthesis: an alkyl halide is heated with a sodium alkoxide, and the halide is replaced by the alkoxy group ($S_N2$). Example: sodium ethoxide and bromoethane give ethoxyethane (diethyl ether).

$$C_2H_5ONa + C_2H_5Br \xrightarrow{\Delta} C_2H_5\text{-}O\text{-}C_2H_5 + NaBr$$

Preparation of ethoxyethane from $C_2H_5Br$ (Williamson's synthesis)

First, part of the bromoethane is converted into sodium ethoxide:

$$C_2H_5Br + KOH(aq) \rightarrow C_2H_5OH + KBr$$ $$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$$

Then sodium ethoxide is heated with bromoethane:

$$C_2H_5ONa + C_2H_5Br \xrightarrow{\Delta} C_2H_5OC_2H_5 + NaBr$$

i) Heating with excess HI

The ether is cleaved and both alkyl groups end up as iodoethane:

$$C_2H_5OC_2H_5 + HI \rightarrow C_2H_5I + C_2H_5OH$$ $$C_2H_5OH + HI \rightarrow C_2H_5I + H_2O$$ $$\text{Overall: } C_2H_5OC_2H_5 + 2HI \xrightarrow{\Delta} 2C_2H_5I + H_2O$$

ii) Why old ether is not distilled to dryness

On standing in air and light, ether is slowly oxidised to ether peroxide, e.g. $CH_3CH(OOH)OC_2H_5$:

$$C_2H_5OC_2H_5 + O_2 \xrightarrow{\text{light}} CH_3CH(OOH)OC_2H_5$$

The peroxide is less volatile than ether, so it collects in the residue during distillation. Near dryness it is concentrated and heated, and it explodes violently. Old ether is therefore tested for peroxides (KI–starch, or $FeSO_4$/KCNS) and never distilled to dryness.

iii) Phenol to anisole

Phenol is converted into sodium phenoxide, which then undergoes Williamson's reaction with methyl iodide:

$$C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$$ $$C_6H_5ONa + CH_3I \xrightarrow{\Delta} \underset{\text{anisole (methoxybenzene)}}{C_6H_5OCH_3} + NaI$$

Dimethyl sulphate, $(CH_3)_2SO_4$, can also be used in place of $CH_3I$.

(b) Ethanoic acid

$C_7H_8O$ here is anisole, $C_6H_5OCH_3$. HI cleaves the alkyl–oxygen bond to give phenol (A) and methyl iodide (B):

$$C_6H_5OCH_3 + HI \rightarrow \underset{(A)}{C_6H_5OH} + \underset{(B)}{CH_3I}$$ $$CH_3I \xrightarrow{alc.\ KCN} \underset{(C)}{CH_3CN} \xrightarrow{H_2O/conc.\ HCl} \underset{(D)}{CH_3COOH}$$

(b) Williamson's synthesis

An alkoxide reacting with an alkyl halide gives an ether. This is Williamson's synthesis:

$$C_2H_5ONa + C_2H_5Cl \rightarrow C_2H_5OC_2H_5\ (C_4H_{10}O) + NaCl$$

The carbon chain must be shortened from $C_2$ to $C_1$, and then the methyl groups joined by Williamson's synthesis.

1. Hydrolyse the ether to ethanol:

$$C_2H_5OC_2H_5 + H_2O \xrightarrow[\text{pressure}]{dil.\ H_2SO_4} 2C_2H_5OH$$

2. Oxidise ethanol to ethanoic acid, then decarboxylate it to methane:

$$C_2H_5OH \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH \xrightarrow{NaOH} CH_3COONa \xrightarrow{NaOH/CaO,\ \Delta} CH_4$$

3. Make methyl chloride and methanol:

$$CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$$ $$CH_3Cl + KOH(aq) \rightarrow CH_3OH + KCl$$ $$2CH_3OH + 2Na \rightarrow 2CH_3ONa + H_2$$

4. Williamson's synthesis:

$$CH_3ONa + CH_3Cl \xrightarrow{\Delta} CH_3OCH_3 + NaCl$$

The product is methoxymethane.

(b) 2

The ethers of formula $C_4H_{10}O$ are:

• $CH_3OCH_2CH_2CH_3$ (1-methoxypropane) — unsymmetrical
• $CH_3OCH(CH_3)_2$ (2-methoxypropane) — unsymmetrical
• $C_2H_5OC_2H_5$ (ethoxyethane) — symmetrical

So only two of them are unsymmetrical (mixed) ethers.

(b) 2-Methoxypropane

HBr adds to propene by Markovnikov's rule to give 2-bromopropane. Sodium methoxide then replaces Br (Williamson's synthesis):

$$CH_3CH=CH_2 \xrightarrow{HBr} \underset{(B)}{CH_3CHBrCH_3} \xrightarrow[\Delta]{CH_3ONa} \underset{(C)}{CH_3CH(OCH_3)CH_3} + NaBr$$

(d) Anisole

This is Williamson's synthesis. The phenoxide ion displaces bromide from methyl bromide to form methyl phenyl ether:

$$C_6H_5ONa + CH_3Br \rightarrow C_6H_5OCH_3 + NaBr$$