Chemical Reactions of Ethers

Chemical Reactions of Ethers — Overview

The C-O bonds in ethers are fairly strong and unreactive, so ethers are generally inert toward bases, reducing agents and dilute acids — which is why they make good solvents. Their two important reactions are:

1. Cleavage of the C-O bond by hydrogen halides (HX), and
2. Electrophilic aromatic substitution when the ether is aromatic (e.g. anisole), because -OR activates the ring.

Key Point: ethers resist most reagents; the exam reactions are C-O cleavage by HX and aromatic substitution of anisole.

Cleavage of the C-O Bond by HX

When heated with hydrogen halides, ethers are cleaved at the C-O bond to give an alkyl halide and an alcohol (and, with excess HX, two alkyl halides):

R-O-R' + HX → R-X + R'-OH (and R'-X + H$_2$O with excess HX)

Reactivity order of the acids: HI > HBr > HCl (HI is the strongest and most commonly used).

Which fragment becomes the halide? With excess HX, the alkyl group that gives the more stable carbocation (or, for a simple case, the less hindered group via SN2) is the one that ends up as the halide.

• With primary/methyl ethers the reaction is SN2: the halide attacks the less hindered (smaller) carbon, so e.g. CH$_3$-O-C$_2$H$_5$ + HI → CH$_3$I + C$_2$H$_5$OH (the smaller methyl becomes the iodide).
• With ethers bearing a tertiary/branched group the reaction is SN1: cleavage gives the more stable (3°) carbocation, so e.g. (CH$_3$)$_3$C-O-CH$_3$ + HI → (CH$_3$)$_3$C-I + CH$_3$OH.

Key Point: HX cleaves ethers (HI > HBr > HCl). For methyl/1° ethers the smaller group becomes the halide (SN2); when a 3° group is present it forms the carbocation and becomes the halide (SN1).

Cleavage of Aryl Alkyl Ethers (e.g. Anisole)

In an aryl alkyl ether such as anisole (C$_6$H$_5$-O-CH$_3$), the C(aryl)-O bond does NOT break because of the partial double-bond character (the oxygen lone pair is delocalised into the ring) and because aryl cations/aryl-SN2 are not feasible. So cleavage always breaks the alkyl-oxygen bond and gives a phenol + an alkyl halide:

C$_6$H$_5$-O-CH$_3$ + HI → C$_6$H$_5$-OH (phenol) + CH$_3$I

The phenol is not further converted to an aryl halide.

Key Point: with aryl alkyl ethers the alkyl-O bond breaks → you always get phenol + alkyl halide. The aryl-O bond is too strong to cleave.

Electrophilic Substitution in Anisole

The -OCH$_3$ (alkoxy) group is activating and ortho/para-directing, just like -OH, because the oxygen lone pair feeds electron density into the ring. So anisole undergoes electrophilic aromatic substitution faster than benzene, mainly at the ortho and para positions.

Halogenation: anisole + Br$_2$ in ethanoic acid → mainly p-bromoanisole (para is favoured because ortho is hindered by the -OCH$_3$ group).

Nitration: anisole + conc. HNO$_3$/H$_2$SO$_4$ → ortho- and para-nitroanisole.

Friedel-Crafts reaction: anisole + CH$_3$Cl (or CH$_3$COCl) with anhydrous AlCl$_3$ → mainly the para methyl/acetyl product (ortho + para; para predominates).

Key Point: -OR activates the ring (o/p-directing). Anisole gives mainly para products in halogenation, nitration and Friedel-Crafts because the ortho position is sterically hindered.

Solved Examples

Example 1: Cleavage products of a methyl ether

What are the products when CH$_3$-O-C$_2$H$_5$ reacts with excess HI?

Solution: The reaction is SN2 and the iodide attacks the less hindered methyl group: first CH$_3$I + C$_2$H$_5$OH. With excess HI the ethanol is also converted, giving CH$_3$I + C$_2$H$_5$I.

Example 2: Reactivity of HX

Arrange HCl, HBr and HI in order of reactivity toward ether cleavage.

Solution: HI > HBr > HCl. HI is the strongest acid and best nucleophile (iodide), so it cleaves ethers most readily.

Example 3: Directing effect in anisole

Why does bromination of anisole give mainly the para product?

Solution: The -OCH$_3$ group is o/p-directing and activating. Both ortho and para are favoured electronically, but the ortho position is sterically hindered by the methoxy group, so the para product (p-bromoanisole) predominates.

Example 4: Identify the ether

An ether C$_4$H$_{10}$O on cleavage with HI gives methanol and a tertiary iodide. Identify it.

Solution: The tertiary iodide must be (CH$_3$)$_3$CI, so the ether is methyl tert-butyl ether, (CH$_3$)$_3$C-O-CH$_3$ (2-methoxy-2-methylpropane). The 3° group forms the carbocation/iodide, leaving methanol.

Example 5: Nitration of anisole

What products form when anisole is nitrated?

Solution: -OCH$_3$ is o/p-directing, so nitration gives o-nitroanisole and p-nitroanisole (para usually predominates).

Example 6: Why ethers are good solvents

Why are ethers chemically suitable as reaction solvents?

Solution: The C-O bonds are strong and ethers are inert to bases, dilute acids, oxidising and reducing agents under mild conditions, so they dissolve organic reactants without reacting themselves (caution: flammable and form explosive peroxides on storage).

Example 7: Product with limited HX

CH$_3$-O-CH$_3$ reacts with one mole of HI. Write the product.

Solution: Symmetrical dimethyl ether gives CH$_3$I + CH$_3$OH (with one mole of HI). With excess HI both halves become CH$_3$I.

Example 8: Friedel-Crafts on anisole

What is the main product of anisole with CH$_3$COCl/anhydrous AlCl$_3$?

Solution: Friedel-Crafts acylation on the activated, o/p-directing ring gives mainly p-methoxyacetophenone (4-methoxyacetophenone), the para acetyl product.