Properties of Ether

ChemistryAlcohols, Phenols And EthersFor JEE aspirants

The properties of ethers follow from an oxygen with two lone pairs and no O-H. Ethers cannot hydrogen bond to one another, so they boil like alkanes of the same mass, yet they accept hydrogen bonds from water and dissolve about as well as alcohols. Chemically they are unreactive, except at the oxygen (oxonium salts), at the C-H next to oxygen (peroxides) and at the C-O bond once protonated (cleavage by HI). These properties of ethers, plus anisole's ring reactions, are core NEET and JEE Main material.

On this page1Physical2Lewis base3Peroxides4Halogenation5Cleavage by HX6Anisole ring7Epoxides8Examples

Key Formulas - Quick Reference

b.p.: butan-1-ol (390 K) $≫$ ethoxyethane (308 K) $\approx$ n-pentane (309 K)
★ Must learnR-O-R' + HX $\to$ R-X + R'-OH; reactivity HI $>$ HBr $>$ HCl
★ Must learnMethyl or 1° groups: S $_{N}$ 2, the smaller group becomes R-X
★ Must learn3° (or benzylic) group: S $_{N}$ 1, that group becomes R-X
★ Must learnAnisole + HI $\to$ $C_{6} H_{5} OH$ + $C H_{3} I$ (the O-aryl bond never breaks)
Anisole is o/p directing: $B r_{2}$ / $C H_{3} COOH$ $\to$ 4-bromoanisole (major)
★ Must learnEpoxide opening: acid at the more substituted C, base at the less hindered C
Air + light: ethers form explosive hydroperoxides at the $α$ -C-H

1. Physical Properties

Dimethyl ether and ethyl methyl ether are gases; higher ethers are colourless, volatile, highly flammable liquids with a pleasant smell, lighter than water. The oxygen is $s p^{3}$ hybridised and the C-O-C angle is about 111.7° in dimethyl ether (the source quotes 110°), so the molecule is bent and polar, with a dipole moment of about 1.2 D.

Figure 1: At the same molar mass an ether boils like an alkane (308 vs 309 K), about 80 K below the isomeric alcohol, because ether molecules cannot hydrogen bond to each other.

Two opposite effects are at work. Ether molecules have no O-H hydrogen, so they cannot hydrogen bond to one another: boiling points stay close to those of alkanes. Their oxygen, however, accepts hydrogen bonds from water, so ethers with up to about four carbons are fairly soluble, much like alcohols of the same size.

Figure 2: One oxygen, two effects. With no O-H, ethers cannot H-bond to each other (low b.p.); their O lone pairs accept H-bonds from water (solubility like alcohols: 7.5 vs 9 g per 100 mL).

Exam Trick

"Boils like an alkane, dissolves like an alcohol." No O-H to give, two lone pairs to accept: that single fact explains both physical properties of ethers.

Key idea

Ethers cannot donate hydrogen bonds (low b.p.) but can accept them (water solubility).

2. Ethers as Lewis Bases

Ethers dissolve in cold concentrated mineral acids by forming oxonium salts; adding water reverses the reaction. The lone pairs also bind Lewis acids: boron trifluoride forms a stable etherate, and ether molecules coordinate to the magnesium of a Grignard reagent, which is why Grignard reagents are made in dry ether.

(C_{2} H_{5})_{2} O + HCl ⇌ [(C_{2} H_{5})_{2} OH]^{+} C l^{-}

Figure 3: The ether oxygen is a weak Lewis base: it accepts a proton from cold conc. acids (oxonium salt, reversed by adding water) and donates a lone pair to BF

_{3}

or to the Mg of a Grignard reagent.

3. Peroxide Formation and Halogenation

In air and light, ethers slowly form hydroperoxides by a free-radical attack at the carbon next to oxygen, whose radical is stabilised by the oxygen lone pair. Peroxides are explosive when an ether is distilled to dryness. They are detected with acidified potassium iodide (iodine is liberated) and removed by washing with iron(II) sulphate solution. Ethers are stored in dark bottles for this reason.

Figure 4: Old ether bottles are dangerous. Air and light attack the C-H next to oxygen and form hydroperoxides, which explode when the ether is distilled to dryness. Test: KI gives I

_{2}

; remedy: wash with FeSO

_{4}

solution.

Halogenation also starts at the $α$ -carbon. In the dark chlorine gives 1-chloro and then 1,1'-dichlorodiethyl ether; in light every hydrogen is replaced.

Figure 5: Halogenation is also radical and starts at the α-carbon, whose radical is stabilised by the oxygen lone pair. Dark: two α-H replaced. Light: every H replaced.

An ether with no hydrogen on the carbons next to oxygen, such as tert-butyl phenyl ether, cannot form peroxides this way.

Quick Recall: tap to check

Where does oxygen attack an ether in air and light?

At the C-H next to the ether oxygen.

How are peroxides detected in an ether sample?

Acidified KI liberates iodine (blue with starch).

Why are ethers solvents for Grignard reagents?

Their lone pairs coordinate to and stabilise the Mg.

4. Cleavage by Hydrogen Halides

The C-O bond of an ether is broken only by strong acids. Heated with concentrated HI (or HBr), an ether gives an alkyl halide and an alcohol; with excess acid the alcohol is also converted to a halide. Reactivity follows HI $>$ HBr $>$ HCl, because HI is the strongest acid (easiest protonation) and iodide the best nucleophile.

R - O - R^{'} + HI R - I + R^{'} - OH

First the oxygen is protonated, which turns the alkoxy group into a neutral alcohol leaving group. Then iodide attacks. With methyl or primary groups the attack is S $_{N}$ 2 and goes to the smaller, less hindered group.

Figure 6: Ether cleavage by HI. Protonation turns the alkoxy group into a neutral alcohol leaving group; iodide, a strong nucleophile, attacks the smaller alkyl group (S

_{N}

2).

If one group is tertiary (or benzylic), the protonated ether splits by S $_{N}$ 1 to the stable carbocation, which becomes the halide. In an aryl alkyl ether the O-aryl bond has partial double-bond character and phenyl cations cannot form, so the products are always a phenol and an alkyl halide.

Figure 7: Three cleavage patterns. With a 3° (or benzylic) group the C-O bond to that group breaks (S

_{N}

1); with an aryl group the O-aryl bond never breaks, so phenol and an alkyl halide form.

Methyl ethyl ether + HIS

_{N}

2 at the smaller group

C H_{3} I

C_{2} H_{5} OH

tert-Butyl methyl ether + HIS

_{N}

1 at the 3° group

(C H_{3})_{3} CI

C H_{3} OH

JEE Advanced

The mechanism can switch with the solvent. With anhydrous HI in a non-polar medium, tert-butyl methyl ether gives $C H_{3} I$ and tert-butyl alcohol (S $_{N}$ 2 at methyl); with concentrated aqueous HI, the polar medium favours S $_{N}$ 1 and gives tert-butyl iodide and methanol. NCERT expects the S $_{N}$ 1 answer for a 3° ether unless the question says otherwise.

Key idea

Find the weakest C-O bond: aryl-O never breaks, 3° and benzylic C-O break by S

_{N}

1, otherwise the smaller group takes the halide.

5. Electrophilic Substitution in Aromatic Ethers

The $- OCH_{3}$ group of anisole releases electrons to the ring by resonance, so it activates the ring and directs incoming groups to the ortho and para positions, like $- OH$ but less strongly. Bromine in ethanoic acid reacts without a Lewis acid, and the para isomer dominates because the methoxy group crowds the ortho positions.

Figure 8: The OCH

_{3}

group activates the ring and directs ortho and para, like OH but a little less strongly: bromination needs no FeBr

_{3}

, and the para product dominates because OCH

_{3}

is bulky.

Reaction	Reagent	Major product
Halogenation	$B r_{2}$ in ethanoic acid	4-bromoanisole
Friedel-Crafts alkylation	$C H_{3} Cl$ , anhydrous $AlC l_{3}$	4-methylanisole (+ 2-)
Friedel-Crafts acylation	$C H_{3} COCl$ , anhydrous $AlC l_{3}$	4-methoxyacetophenone (+ 2-)
Nitration	conc. $HN O_{3}$ + conc. $H_{2} S O_{4}$	4-nitroanisole (+ 2-)

Exam Trick

"Methoxy is OH's calmer cousin." Same o/p direction, less activation: anisole needs no Lewis acid for bromination, but it does not give a tribromo product at once the way phenol does.

6. Epoxides: the Reactive Ethers

Three-membered cyclic ethers are strained, so their C-O bonds open with nucleophiles that ordinary ethers ignore. In acid the oxygen is protonated and the more substituted carbon carries more positive charge, so the nucleophile attacks there (S $_{N}$ 1-like). In base a strong nucleophile such as an alkoxide attacks the less hindered carbon (S $_{N}$ 2).

Figure 9: Epoxides are the reactive ethers (ring strain). In acid the nucleophile goes to the more substituted carbon (S

_{N}

1-like); in base it goes to the less hindered carbon (S

_{N}

2).

Quick Recall: tap to check

Products of anisole with HI?

Phenol and iodomethane.

Which carbon of 2,2-dimethyloxirane does methoxide attack?

The primary (

C H_{2}

) carbon.

Major product of anisole with

B r_{2}

in ethanoic acid?

4-Bromoanisole.

Figure 10: Flowchart: find the bond that breaks. Aryl-O never; 3° or benzylic C-O by S

_{N}

1; otherwise the smaller group takes the halide by S

_{N}

Figure 11: Mind map: ethers are unreactive except at three places: the oxygen lone pairs, the α-C-H, and the C-O bond once the oxygen is protonated.

7. Solved Examples

Solved Example 1

Predict the products: (a)

Ph - O - C H_{2} - Ph

+ excess HI (b)

(C H_{3})_{2} CH - O - C (C H_{3})_{3}

+ 1 equiv. HBr

Solution:

(a) Phenol + benzyl iodide ( $C_{6} H_{5} C H_{2} I$ ): the protonated ether cleaves by S $_{N}$ 1 at the benzylic carbon; the O-phenyl bond cannot break. (b) Propan-2-ol + tert-butyl bromide: S $_{N}$ 1 at the more substituted carbon.

Solved Example 2

Styrene oxide (phenyloxirane) is treated (i) with

Ph O^{-}

, then

H^{+}

; (ii) with phenol and catalytic acid. Give the products.

Solution:

(i) S $_{N}$ 2 at the less hindered $C H_{2}$ : $Ph - CH (OH) - C H_{2} - OPh$ . (ii) In acid the benzylic carbon carries more positive charge and is attacked: $Ph - CH (OPh) - C H_{2} OH$ .

Solved Example 3

2,2-Dimethyloxirane reacts with

H_{2}^{18} O

and

H^{+}

to give A, and with

C H_{3} ONa

in methanol to give B. Identify A and B.

Solution:

A: $(C H_{3})_{2} C (^{18} OH) C H_{2} OH$ , labelled oxygen on the tertiary carbon (acid: more substituted carbon). B: $(C H_{3})_{2} C (OH) C H_{2} OC H_{3}$ (base: less hindered carbon).

Solved Example 4

(C H_{3})_{3} C - O - C H_{3}

with anhydrous HI gives A; with concentrated aqueous HI it gives B. Then
(A) A =

C H_{3} I + (C H_{3})_{3} COH

, B =

C H_{3} OH + (C H_{3})_{3} CI

(B) both are

C H_{3} I + (C H_{3})_{3} COH

C H_{3} OH + (C H_{3})_{3} CI

(D) none

Solution:

Answer: (A). In a non-polar medium iodide attacks the unhindered methyl carbon (S $_{N}$ 2); in a polar aqueous medium the tertiary carbocation forms (S $_{N}$ 1) and becomes tert-butyl iodide.

Solved Example 5

Why does S

_{N}

2 cleavage of ethers occur faster with HI than with HCl?

Solution:

HI is the stronger acid, so it protonates the ether more completely in step 1; and in step 2 iodide is a better nucleophile than chloride, so the displacement is faster.

Solved Example 6

Ethylene oxide with acidic water gives ethylene glycol, but at high concentration with a trace of

O H^{-}

it gives a polymer. Suggest its structure.

Solution:

$O H^{-}$ opens one ring to give $HOC H_{2} C H_{2} O^{-}$ , which opens the next ring, and so on (anionic polymerisation). The product is polyethylene glycol, $HO - (C H_{2} C H_{2} O)_{n} - H$ , whose chain length depends on the conditions.

Solved Example 7

Which ether is least likely to form peroxides in air?
(A) diethyl ether
(B) tetrahydrofuran
(C) tert-butyl phenyl ether
(D) diisopropyl ether

Solution:

Answer: (C). Peroxides form at a C-H next to oxygen. The tert-butyl carbon has no H and the phenyl carbon is aromatic, so there is no reactive $α$ -H. (D) is especially dangerous.

Solved Example 8

Arrange in increasing boiling point: butan-1-ol, ethoxyethane, n-pentane.
(A) ethoxyethane

<

n-pentane

<

butan-1-ol
(B) n-pentane

<

ethoxyethane

<

butan-1-ol
(C) butan-1-ol

<

ethoxyethane

<

n-pentane
(D) all equal

Solution:

Answer: (A) using data: ethoxyethane 308 K, n-pentane 309 K, butan-1-ol 390 K. The ether and alkane are almost equal because neither hydrogen bonds; the alcohol is far higher.

Solved Example 9

Why does anisole with HI give phenol and iodomethane and not iodobenzene and methanol?

Solution:

After protonation, iodide attacks the methyl carbon (S $_{N}$ 2). Attack at the ring carbon is impossible: the O-aryl bond has partial double-bond character, the carbon is $s p^{2}$ , and a phenyl cation is far too unstable for S $_{N}$ 1.

Practice Questions

Give the products of ethoxyethane with hot excess HI.Answer: two moles of iodoethane (and water).
Name the products of methoxybenzene with HBr.Answer: phenol and bromomethane.
Why is ether stored in dark bottles?Answer: light and air form explosive hydroperoxides.
Product of anisole with $C H_{3} COCl$ / $AlC l_{3}$ (major)?Answer: 4-methoxyacetophenone.
Which ether gives tert-butyl iodide with HI?Answer: any tert-butyl alkyl ether, e.g. tert-butyl methyl ether.
Suggest the mechanism for benzene + ethylene oxide with $AlC l_{3}$ .Answer: $AlC l_{3}$ opens the epoxide to a carbocation-like electrophile; Friedel-Crafts gives 2-phenylethanol.
Why do small ethers dissolve in water though they cannot H-bond to each other?Answer: their oxygen accepts H-bonds from water molecules.

Common Mistakes to Avoid

Watch out

Saying ethers hydrogen bond to each other. They have no O-H; they only accept H-bonds from water.
Writing iodobenzene from anisole and HI. The O-aryl bond never breaks: phenol + $C H_{3} I$ .
Putting the halogen on the larger group in SN2 cleavage. The smaller, less hindered group becomes R-X.
Using SN2 for a tert-butyl ether with aqueous HI. The 3° group leaves as a carbocation (SN1) and becomes R-X.
Forgetting that excess HX converts the alcohol product into a second alkyl halide.
Writing meta products for anisole. The $OC H_{3}$ group is an o/p director.
Adding $AlC l_{3}$ or $FeB r_{3}$ to brominate anisole. $B r_{2}$ in ethanoic acid is enough.
Opening epoxides at the same carbon in acid and base. Acid: more substituted C; base: less hindered C.

Frequently Asked Questions

Why do ethers have lower boiling points than isomeric alcohols?

Ether molecules have no hydrogen attached to oxygen, so they cannot form hydrogen bonds with one another. Only weak dipole and dispersion forces hold them together, so ethoxyethane boils at 308 K, like n-pentane, while its isomer butan-1-ol boils at 390 K.

Why are ethers soluble in water?

The lone pairs on the ether oxygen accept hydrogen bonds from water molecules. Ethers with up to about four carbons are therefore fairly soluble: ethoxyethane dissolves to about 7.5 g per 100 mL, close to butan-1-ol at 9 g per 100 mL.

What happens when an ether reacts with HI?

The ether oxygen is protonated and iodide then attacks a carbon next to oxygen, breaking the C-O bond to give an alkyl iodide and an alcohol. Methyl and primary groups react by SN2, the smaller group becoming the iodide; tertiary groups react by SN1. Excess HI converts both groups to iodides.

Why does anisole give phenol and methyl iodide with HI?

Iodide attacks the methyl carbon of protonated anisole by SN2. The bond between oxygen and the benzene ring has partial double-bond character and a phenyl cation cannot form, so that bond never breaks and iodobenzene is not produced.

Why is old diethyl ether dangerous?

In air and light ethers form hydroperoxides at the carbon next to oxygen by a radical reaction. These peroxides explode when the ether is distilled to dryness. They are detected with acidified potassium iodide and removed by washing with iron(II) sulphate.

Is anisole ortho-para directing?

Yes. The methoxy group donates electron density to the ring by resonance, activating it and directing electrophiles to the ortho and para positions. Bromination, Friedel-Crafts reactions and nitration all give the para product as the major isomer.

What does NEET ask about the properties of ethers?

NEET most often asks the products of anisole or a mixed ether with HI, the major product of bromination or Friedel-Crafts reaction of anisole, and why ethers boil lower than alcohols. The smaller-group rule and the unbreakable O-aryl bond cover most questions.

How does JEE Main test ether cleavage?

JEE Main gives ethers with primary, tertiary, benzylic or aryl groups and asks which bond breaks and by SN1 or SN2, sometimes with labelled oxygen or with epoxides in acid versus base. Identifying the most stable carbocation or the least hindered carbon decides the answer.

Previous year questions on Properties of Ether

5 questions from past papers, each with a step-by-step solution.

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