Alcohols, Phenols and Ethers • Topic 3 of 3

Ethers

An ether has two alkyl or aryl groups joined to the same oxygen, R–O–R′. If both groups are identical (CH3OCH3) it is a symmetrical (simple) ether; if different (CH3OC2H5) it is an unsymmetrical (mixed) ether.

Nomenclature

Common names list the two groups alphabetically followed by “ether” (ethyl methyl ether). In IUPAC the smaller R–O is treated as an alkoxy substituent on the larger chain: CH3OCH3 is methoxymethane, CH3OC2H5 is methoxyethane, and C6H5OCH3 is methoxybenzene (anisole).

Preparation

Williamson synthesis is the general method: an alkoxide (or phenoxide) reacts with a primary alkyl halide by SN2 to give an ether, R–ON a + R′–X → R–O–R′ + NaX. It works well for symmetrical and unsymmetrical ethers; for unsymmetrical ethers always pair the bulkier alkoxide with the less hindered (1°) halide, because 3° halides undergo elimination (alkene) rather than substitution. Dehydration of alcohols by conc. H2SO4 at a moderate temperature (about 413 K) gives a symmetrical ether (2 ROH → R–O–R + H2O); higher temperatures favour the alkene instead, and the method suits only 1° alcohols.

Physical properties

The C–O–C bond angle is ~111° and the molecule has a small net dipole. Ethers cannot H-bond to each other (no O–H), so they have low boiling points, close to those of comparable alkanes and far below isomeric alcohols. They can accept H-bonds from water, so lower ethers are somewhat water-soluble. Ethers are good, fairly inert organic solvents (e.g. for Grignard reagents).

Chemical reactions

Cleavage of the C–O bond by HX. Hot concentrated HI or HBr cleaves ethers. With an alkyl alkyl ether and excess HX, both fragments end up as alkyl halides. The mechanism (and which C–O bond breaks) depends on the groups: with a 1°/2° group cleavage is SN2 and HX attacks the smaller/less hindered alkyl group, giving the smaller alkyl halide plus the alcohol (which can react further). With a 3° group it is SN1, so the 3° halide forms. For aryl alkyl ethers (anisole), the strong C(aryl)–O bond is never cleaved, so the products are an alkyl halide and a phenol: C6H5OCH3 + HI → C6H5OH + CH3I.

Electrophilic substitution in aromatic ethers. In anisole the –OCH3 group is activating and o/p-directing (oxygen lone pair conjugates with the ring). Halogenation, nitration and Friedel–Crafts acylation therefore occur readily at the ortho and para positions; e.g. anisole + CH3COCl/anhyd. AlCl3 gives mainly p-methoxyacetophenone.

Williamson ether synthesis: sodium ethoxide reacts with chloromethane by SN2 to give methoxyethaneCH3CH2O− Na++CH3—ClSN2CH3CH2—O—CH3methoxyethane + NaClalkoxide (nucleophile)1° alkyl halideRule: pair the bulkier alkoxide with the less hindered (1°) halide; 3° halides eliminate to alkenes.
Solved Examples
1
Worked Example
Give the IUPAC name of CH3OCH2CH2CH3 and state if it is simple or mixed.
Solution
  1. The larger group is propyl (3 C) → parent chain propane.
  2. The smaller CH3O– is a methoxy substituent on C-1.
  3. The two groups differ, so it is a mixed (unsymmetrical) ether.

Answer: 1-methoxypropane; an unsymmetrical (mixed) ether.

2
Worked Example
Show the best Williamson synthesis of tert-butyl methyl ether and explain the choice of reactants.
Solution
  1. Two pairings are possible: (a) t-BuO– + CH3I, or (b) CH3O– + (CH3)3CBr.
  2. The halide must be the less hindered one for SN2; a 3° halide gives elimination instead.
  3. So choose (a): potassium tert-butoxide + methyl iodide.

Answer: (CH3)3COK + CH3I → (CH3)3COCH3; use the bulky alkoxide with the methyl halide so SN2 succeeds.

3
Worked Example
What are the products when ethoxyethane (diethyl ether) is heated with excess HI?
Solution
  1. HI cleaves the C–O bond; both groups are 1° ethyl.
  2. One C–O bond breaks (SN2) giving iodoethane + ethanol.
  3. With excess HI the ethanol reacts further to give a second iodoethane.

Answer: Two molecules of iodoethane, CH3CH2I (with excess HI).

4
Worked Example
Why does cleavage of anisole (C6H5OCH3) with HI give phenol and methyl iodide rather than iodobenzene and methanol?
Solution
  1. The C(aryl)–O bond has partial double-bond character (oxygen lone pair conjugates with the ring) and is very strong.
  2. So HI cannot break the aryl–O bond.
  3. Instead I– attacks the methyl carbon (SN2), breaking the O–CH3 bond.
  4. This gives CH3I and the phenoxide/phenol.

Answer: The strong aryl–O bond is not cleaved, so HI breaks the O–CH3 bond to give phenol + CH3I.

5
Worked Example
Predict the major product of nitration of anisole and explain the regiochemistry.
Solution
  1. –OCH3 is an activating, o/p-directing group (lone pair conjugates with the ring).
  2. Electrophile NO2+ attacks ortho and para positions.
  3. The para product is favoured (less steric hindrance).

Answer: Mainly p-nitroanisole (with some o-nitroanisole); directed o/p by the methoxy group.

6
Worked Example
Explain why diethyl ether boils at 35°C while butan-1-ol (similar mass) boils at 118°C.
Solution
  1. Both have similar molar mass (~74 g/mol).
  2. Butan-1-ol has an O–H group and forms intermolecular hydrogen bonds.
  3. Diethyl ether has no O–H, so it cannot self-associate (only weak dipole forces).
  4. Far less energy is needed to vaporise the ether, giving the lower boiling point.

Answer: Absence of intermolecular hydrogen bonding in the ether (present in the alcohol) gives its much lower boiling point.

Key Points

  • Ethers are R–O–R′ (simple if R=R′, mixed if not); IUPAC names the smaller group as an alkoxy substituent (methoxymethane, anisole = methoxybenzene).
  • Williamson synthesis (alkoxide + 1° alkyl halide, SN2) is the general method; pair the bulky alkoxide with the less hindered halide to avoid elimination.
  • Dehydration of (1°) alcohols with conc. H2SO4 at ~413 K gives symmetrical ethers; higher temperature gives the alkene.
  • No O–H means low boiling points (near alkanes, well below isomeric alcohols); ethers accept H-bonds from water, so lower members are slightly soluble.
  • Reactions: HX cleaves C–O (anisole → phenol + alkyl halide, never aryl halide); –OR activates aromatic rings, so anisole undergoes o/p electrophilic substitution.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The best route to ethyl methyl ether by Williamson synthesis uses:
Explanation: Williamson uses an alkoxide with a 1° halide; ethoxide + methyl iodide (both unhindered) gives the ether cleanly by SN2.
Q2.Cleavage of anisole with hot HI gives:
Explanation: The strong aryl–O bond is not broken; HI cleaves the O–CH3 bond to give phenol and CH3I.
Q3.Compared with their isomeric alcohols, ethers have:
Explanation: Ethers lack O–H, so they cannot self H-bond and boil much lower than isomeric alcohols.
Q4.In anisole, electrophilic substitution occurs mainly at the:
Explanation: The –OCH3 group is activating and o/p-directing, so electrophiles attack ortho and para.
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