Ethers: Preparation, Cleavage & Reactions

CHEEthers

Ethers are prepared mainly by Williamson synthesis and are notable for being relatively unreactive — their main reaction is cleavage by HI/HBr. Exam questions test the mechanism choice in Williamson, the boiling-point comparison with alcohols, and which C–O bond breaks on cleavage.

Williamson synthesis (

S_N2

)

R\text{-}O^-Na^+ + R'\text{-}X \rightarrow R\text{-}O\text{-}R' + NaX

Cleavage of anisole with HI

C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I

Cleavage of diethyl ether

C_2H_5OC_2H_5 + HI \rightarrow C_2H_5I + C_2H_5OH \xrightarrow{\text{excess HI}} 2\,C_2H_5I + H_2O

Oxonium salt formation

R_2O + BF_3 \rightarrow [R_2O\!\rightarrow\!BF_3]

ether acts as a Lewis base via the O lone pair

Williamson synthesis: $R\text{-}O^-Na^+ + R'\text{-}X \rightarrow R\text{-}O\text{-}R' + NaX$ , an $S_N2$ reaction; it makes both symmetrical and unsymmetrical ethers.
Uses: diethyl ether was historically used as a general anaesthetic and remains a common lab/extraction solvent (volatile, flammable — stored away from heat/sparks and tested for peroxides before distillation); THF is an important industrial solvent; anisole is used in fragrances and as a pharmaceutical intermediate; dimethyl ether serves as an aerosol propellant and fuel.
Williamson fails with a $3^\circ$ alkyl halide because $S_N2$ is hindered and the strong-base alkoxide instead causes elimination, giving an alkene — use the $3^\circ$ alkoxide with a $1^\circ$ halide instead.
Ethers have much lower boiling points than alcohols of similar mass because ether molecules cannot hydrogen-bond to each other (no $O\text{-}H$ ).
HI cleavage of dialkyl ethers: the smaller/less hindered alkyl group is attacked by $I^-$ in an $S_N2$ step, so it becomes the alkyl iodide (e.g. $CH_3OC_2H_5 + HI \rightarrow CH_3I + C_2H_5OH$ ).
Anisole $+ HI$ gives phenol $+ CH_3I$ ; the aryl C–O bond is not cleaved because that bond has partial double-bond (resonance) character and aryl carbon resists $S_N2$ attack.
Diethyl ether $+$ 1 mol HI $\rightarrow$ ethyl iodide $+$ ethanol; with excess HI the ethanol is further converted, giving $2\,C_2H_5I + H_2O$ .
Ethers are good solvents and used in Grignard chemistry because they are largely inert (unreactive toward bases, reducing agents, and dilute acids).
Industrial/lab ethers also form by acid dehydration of alcohols (conc. $H_2SO_4$ , $413\,\text{K}$ ): $2\,C_2H_5OH \rightarrow C_2H_5OC_2H_5 + H_2O$ , but this gives only symmetrical ethers and works best for $1^\circ$ alcohols.
When the ether has a $3^\circ$ alkyl group, cleavage by HI goes by $S_N1$ : the $3^\circ$ carbocation forms, so the $3^\circ$ group becomes the iodide and the other becomes the alcohol.
Reactivity of HX toward ether cleavage is $HI > HBr > HCl$ (matching nucleophilicity of the halide); HI is the standard reagent for quantitative ether cleavage.
Ethers dissolve in cold conc. $H_2SO_4$ by protonation of the oxygen lone pair to form an oxonium salt — a property used to distinguish them from alkanes (which do not dissolve).
Ethers slowly form explosive peroxides on standing in air/light, so old ether is tested and treated before distillation — a safety point sometimes asked.

Where the marks go

Cleaving the aryl–O bond of anisole — HI always breaks the $CH_3\text{-}O$ bond to give phenol $+ CH_3I$ ; the strong, partly-double aryl C–O bond stays intact.
Choosing a $3^\circ$ halide for Williamson — it gives mainly the alkene by elimination; pair the bulky group as the alkoxide with a $1^\circ$ halide instead.
Saying the larger alkyl group becomes the iodide in $S_N2$ cleavage — for primary ethers it is the SMALLER, less hindered group that is attacked by $I^-$ .
Expecting ethers to H-bond like alcohols — ethers have no $O\text{-}H$ , so they cannot H-bond to each other and boil far lower than isomeric alcohols.
Forgetting the second step with excess HI — the ethanol from diethyl ether is further converted to ethyl iodide, so the final products are $2\,C_2H_5I + H_2O$ .

How the board asks it

Give reasonsanisole cleavage and aryl c-o resonance
Account for the following: When anisole is treated with $HI$ , the products are phenol and $CH_3I$ , and not iodobenzene and methanol.
Predict the producthi cleavage of unsymmetrical dialkyl ethers
Write the products formed when ethyl methyl ether ( $CH_3OC_2H_5$ ) is heated with one mole of $HI$ , and give the balanced equation.
Give reasonswilliamson failure with tertiary halide and elimination
Account for the following: $tert$ -butyl methyl ether is prepared by reacting sodium $tert$ -butoxide with $CH_3I$ , and not by reacting sodium methoxide with $tert$ -butyl bromide.
Conversionwilliamson synthesis from alkoxide and alkyl halide
How will you prepare ethyl methyl ether by Williamson's synthesis? Give the equation involved.
Distinguishabsence of o-h bond and boiling point versus alcohols
Give a reason: Ethers have considerably lower boiling points than alcohols of comparable molecular mass.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.