Preparation of Ether

ChemistryAlcohols, Phenols And EthersFor JEE aspirants

The preparation of ethers rests on two ideas. Most routes are S $_{N}$ 2 reactions: an alcohol attacks a protonated alcohol (acid dehydration at 413 K), or an alkoxide attacks a methyl or primary alkyl halide (Williamson synthesis). The rest add RO to an alkene: alkoxymercuration-demercuration, or a peroxy acid for epoxides. Getting the preparation of ethers right means choosing the partner that can survive S $_{N}$ 2. A favourite JEE Main and NEET question on reagent choice.

On this page1Four routes2Dehydration3Williamson4Aryl ethers5From alkenes6Epoxides7Choose a route8Examples

Key Formulas - Quick Reference

★ Must learnDehydration: 2 $C_{2} H_{5} OH$ $conc. H_{2} SO_{4}, 413 K$ $C_{2} H_{5} O C_{2} H_{5}$ + $H_{2} O$
At 443 K the same mixture gives ethene; 2° and 3° alcohols give alkenes
★ Must learnWilliamson: RONa + R'X $\to$ ROR' + NaX (S $_{N}$ 2; R'X = $C H_{3}$ X or 1°)
★ Must learntert-Butyl methyl ether: $(C H_{3})_{3} CONa$ + $C H_{3} Br$ , never $C H_{3} ONa$ + $(C H_{3})_{3} CBr$
★ Must learnAryl ethers: ArONa + RX or $(C H_{3})_{2} S O_{4}$ ; ArX + RONa does not react
Alkoxymercuration: alkene + ROH, $Hg (OOCC F_{3})_{2}$ , then $NaB H_{4}$ : Markovnikov ether
Epoxide: alkene + RCO $_{3}$ H; ethylene oxide from ethene + $O_{2}$ over Ag

1. Four Ways to Make an Ether

An ether R-O-R' has an oxygen joined to two carbon groups. It can be made by joining two alcohol molecules (dehydration), by joining an alkoxide to an alkyl halide (Williamson), or by adding an alcohol across a C=C bond. Cyclic three-membered ethers, epoxides, come from alkenes and peroxy acids.

Figure 1: Four routes. Dehydration suits only simple symmetrical ethers from 1° alcohols; Williamson makes almost any ether; alkoxymercuration adds RO- Markovnikov to an alkene; peroxy acids make epoxides.

2. Dehydration of Alcohols

Primary alcohols heated with concentrated sulphuric acid at 413 K, with the alcohol in excess, give symmetrical ethers. The source calls this intra-molecular dehydration, but it is intermolecular: water is removed from two molecules.

2 C_{2} H_{5} OH conc. H_{2} S O_{4}, 413 K C_{2} H_{5} O C_{2} H_{5} + H_{2} O

The mechanism is S $_{N}$ 2. The acid protonates one ethanol so that water can leave; a second ethanol, acting as the nucleophile, attacks the carbon from the back; the protonated ether loses a proton.

Figure 2: Dehydration to an ether is an S

_{N}

2 reaction: one alcohol molecule is the nucleophile, the other (protonated) is the substrate, and water is the leaving group.

Conditions matter. With excess acid at 443 K elimination wins and ethene is formed. Secondary and tertiary alcohols form carbocations easily and eliminate, so they give alkenes, not ethers. Two different alcohols give a mixture of three ethers, so the method is useless for unsymmetrical ethers.

Figure 3: Same reagents, different product. At 413 K with excess ethanol, S

_{N}

2 gives the ether; at 443 K elimination wins. 2° and 3° alcohols give alkenes, so this route works only for simple ethers from 1° alcohols.

Exam Trick

"140 for ether, 170 for alkene." In °C: 413 K (140 °C) with excess alcohol gives the ether, 443 K (170 °C) with excess acid gives the alkene.

Key idea

Acid dehydration gives ethers only from 1° alcohols, only symmetrical ones, and only at the lower temperature.

3. Williamson Ether Synthesis

An alkyl halide heated with a sodium alkoxide gives an ether. It is the most general method and makes both simple and mixed ethers.

C H_{3} C H_{2} Br + C H_{3} ONa C H_{3} C H_{2} OC H_{3} + NaBr

Figure 4: Williamson synthesis is S

_{N}

2: the alkoxide is the nucleophile and the alkyl halide must be methyl or 1°, because the attack comes from the back of the C-X bond.

Because the step is S $_{N}$ 2, the halide must be methyl or primary. Alkoxides are also strong bases: with a tertiary halide they remove a $β$ -hydrogen and give an alkene (E2). The rule is to put the bulky or tertiary group on the alkoxide and the small group on the halide.

Figure 5: Put the bulky group on the alkoxide, never on the halide. A 3° halide meets a strong base and eliminates (E2), giving an alkene instead of the ether.

Good pairing

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

C H_{3} Br

_{N}

2 at an unhindered carbon
ether in good yield

Bad pairing

C H_{3} O^{-}

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

E2 at a crowded carbon
2-methylpropene

Exam Trick

"Big O, small X." The bulky group rides on the oxygen, the small group carries the halogen. If both halves are bulky, the Williamson route cannot make that ether.

JEE Advanced

Reactivity of primary halides in the Williamson synthesis falls $C H_{3} X > C H_{3} C H_{2} X > C H_{3} C H_{2} C H_{2} X$ because S $_{N}$ 2 is sensitive to crowding, while the tendency to eliminate rises $3^{\circ} > 2^{\circ} > 1^{\circ}$ . For a chiral 2° halide the ether forms with inversion, a direct proof of backside attack.

Quick Recall: tap to check

Best reagents for tert-butyl ethyl ether?

Sodium tert-butoxide and bromoethane.

What does sodium ethoxide give with tert-butyl bromide?

2-Methylpropene (elimination), not the ether.

Mechanism of the Williamson synthesis?

_{N}

4. Aryl Alkyl Ethers

Phenols are acidic enough to be converted to phenoxide by aqueous NaOH. The phenoxide ion then attacks an alkyl halide or dimethyl sulphate, which is cheaper than iodomethane and has a better leaving group.

C_{6} H_{5} ONa + C H_{3} I C_{6} H_{5} OC H_{3} + NaI

C_{6} H_{5} ONa + (C H_{3})_{2} S O_{4} C_{6} H_{5} OC H_{3} + C H_{3} OS O_{3} Na

Figure 6: For aryl alkyl ethers the aryl group must be on the oxygen (phenoxide) and the alkyl group on the halide. Aryl halides do not undergo S

_{N}

2, so the reverse pairing fails.

The reverse pairing, an aryl halide with an alkoxide, fails: the C-X bond of a haloarene has partial double-bond character and the

s p^{2}

carbon cannot be attacked from the back. So anisole is made from phenoxide and a methylating agent, never from bromobenzene and methoxide.

Key idea

In any Williamson question the aryl group goes on the oxygen and the alkyl group on the halide.

5. Ethers from Alkenes

Alkoxymercuration-demercuration works like oxymercuration with an alcohol in place of water. The alkene reacts with the alcohol and mercury(II) trifluoroacetate; sodium borohydride then replaces the mercury by hydrogen. The RO group adds to the more substituted carbon (Markovnikov) and, because no free carbocation forms, there is no rearrangement.

Figure 7: Alkoxymercuration: the same as oxymercuration, with an alcohol in place of water. RO adds Markovnikov and there is no carbocation, so no rearrangement.

5.1 Epoxides

An alkene treated with a peroxy acid, such as m-chloroperoxybenzoic acid (mCPBA), gives an epoxide (oxirane) in one step. Industrially, ethylene oxide is made from ethene and oxygen over a silver catalyst at about 523 K.

Figure 8: Epoxides (three-membered cyclic ethers) are made by oxidising an alkene: a peroxy acid in the laboratory,

O_{2}

over silver for ethylene oxide in industry.

6. Choosing a Route

The flowchart turns the rules into four questions. The mind map after it collects every condition.

Figure 9: Flowchart: in a Williamson question, always ask which half can survive S

_{N}

2 as the halide: methyl or 1°, never 3°, never aryl.

Quick Recall: tap to check

Why can't dehydration make ethyl methyl ether cleanly?

Two alcohols give three ethers (dimethyl, diethyl, ethyl methyl).

Which reagent adds RO Markovnikov to an alkene without rearrangement?

Hg (OOCC F_{3})_{2}

and ROH, then

NaB H_{4}

Reagent that converts propene into methyloxirane?

A peroxy acid such as mCPBA.

Figure 10: Mind map: every route is either S

_{N}

2 (dehydration, Williamson) or addition to an alkene (alkoxymercuration, epoxidation).

7. Solved Examples

Solved Example 1

Identify the reagent(s) for:

(C H_{3})_{2} C = C H_{2}

\to

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

Solution:

$B r_{2}$ in ethanol. Bromine forms a bromonium ion; ethanol, the solvent, attacks the more substituted carbon (as in halohydrin formation), giving the bromo ether. This is ethoxybromination of the $π$ bond.

Solved Example 2

Identify the reagents for:

Ph - OH

\to

Ph - O - C H_{2} - CH = C H_{2}

Solution:

(1) NaOH (1 equivalent) to form sodium phenoxide; (2) $Br - C H_{2} - CH = C H_{2}$ (allyl bromide), a reactive primary halide, in a Williamson synthesis.

Solved Example 3

The end product Y of

(C_{6} H_{5})_{2} CHC H_{2} OH Na X C H_{3} I Y

is
(A)

(C_{6} H_{5})_{2} C = C H_{2}

(B)

(C_{6} H_{5})_{2} CHC H_{2} O^{-} N a^{+}

(C)

(C_{6} H_{5})_{2} CHC H_{2} OCOSNa

(D)

(C_{6} H_{5})_{2} CHC H_{2} OC H_{3}

Solution:

Answer: (D). Sodium gives the alkoxide X, which is methylated by iodomethane in an S $_{N}$ 2 Williamson step.

Solved Example 4

Which gives the best yield of tert-butyl ethyl ether?
(A)

(C H_{3})_{3} CBr + C_{2} H_{5} ONa

(B)

(C H_{3})_{3} CONa + C_{2} H_{5} Br

(C)

(C H_{3})_{3} COH + C_{2} H_{5} OH

, conc.

H_{2} S O_{4}

(D)

(C H_{3})_{3} CCl + C_{2} H_{5} OH

Solution:

Answer: (B). S $_{N}$ 2 at the primary carbon of bromoethane. (A) eliminates to 2-methylpropene; (C) dehydrates the 3° alcohol; (D) gives mainly elimination and some ether by S $_{N}$ 1.

Solved Example 5

Suggest a synthesis of 1,4-dioxane from ethene.

Solution:

Cold dilute alkaline $KMn O_{4}$ converts ethene to ethane-1,2-diol. Acid-catalysed dehydration of two glycol molecules gives $HOC H_{2} C H_{2} OC H_{2} C H_{2} OH$ (diethylene glycol), which cyclises with loss of a second water molecule to 1,4-dioxane, a cyclic diether.

Solved Example 6

Why is acid dehydration not used to make ethyl methyl ether?

Solution:

Heating methanol and ethanol with acid lets either alcohol act as nucleophile and substrate, so a mixture of $C H_{3} OC H_{3}$ , $C_{2} H_{5} O C_{2} H_{5}$ and $C H_{3} O C_{2} H_{5}$ forms and is hard to separate. The Williamson synthesis ( $C H_{3} ONa + C_{2} H_{5} Br$ or $C_{2} H_{5} ONa + C H_{3} I$ ) gives one product.

Solved Example 7

9.41 g of phenol is converted to anisole with NaOH and dimethyl sulphate. What mass of anisole forms at 100% yield?

Solution:

$M$ (phenol) = 94.11, $M$ (anisole) = 108.14 g mol $^{- 1}$ ; 0.100 mol of phenol gives 0.100 mol of anisole = 10.8 g.

Solved Example 8

Why cannot bromobenzene and sodium ethoxide be used to make phenetole?

Solution:

Aryl halides do not undergo S $_{N}$ 2: the C-Br bond has partial double-bond character, the carbon is $s p^{2}$ and backside attack is blocked by the ring. Phenetole is made from sodium phenoxide and ethyl iodide.

Practice Questions

Give the product of ethanol with conc. $H_{2} S O_{4}$ at 413 K, and at 443 K.Answer: ethoxyethane; ethene.
Choose reagents for 2-methoxy-2-methylpropane by Williamson synthesis.Answer: sodium tert-butoxide + iodomethane.
What is formed from sodium ethoxide and 2-bromo-2-methylpropane?Answer: 2-methylpropene (E2).
Write reagents for anisole from phenol.Answer: NaOH, then $C H_{3} I$ or $(C H_{3})_{2} S O_{4}$ .
Product of 1-methylcyclohexene with $C H_{3} OH$ , $Hg (OOCC F_{3})_{2}$ , then $NaB H_{4}$ ?Answer: 1-methoxy-1-methylcyclohexane.
Name the epoxide formed from propene and mCPBA.Answer: methyloxirane (1,2-epoxypropane).
Which alcohols cannot give ethers by acid dehydration?Answer: 2° and 3° alcohols (they give alkenes).

Common Mistakes to Avoid

Watch out

Calling alcohol dehydration to ether intramolecular. Two molecules lose one water: it is intermolecular.
Heating to 443 K for an ether. That temperature gives the alkene; ethers need 413 K and excess alcohol.
Using a 3° alkyl halide in the Williamson synthesis. It eliminates to an alkene.
Using an aryl halide with an alkoxide to make an aryl ether. Aryl halides do not undergo SN2.
Making unsymmetrical ethers by dehydrating two different alcohols. A mixture of three ethers forms.
Writing anti-Markovnikov addition in alkoxymercuration. RO goes to the more substituted carbon.
Expecting rearrangement in alkoxymercuration. The bridged mercurinium ion prevents it.
Writing dehydration of tert-butyl alcohol to di-tert-butyl ether. 3° alcohols give alkenes.

Frequently Asked Questions

How are ethers prepared from alcohols?

Primary alcohols heated with concentrated sulphuric acid at 413 K, with the alcohol in excess, lose one water molecule between two molecules and give symmetrical ethers such as diethyl ether. The reaction is an SN2 attack of one alcohol on another protonated alcohol.

What is the Williamson ether synthesis?

It is the reaction of a sodium alkoxide or phenoxide with an alkyl halide to form an ether. The alkoxide attacks the carbon of the halide in one SN2 step, so the halide must be methyl or primary. It makes both simple and mixed ethers.

Why are tertiary alkyl halides not used in the Williamson synthesis?

Alkoxide ions are strong bases as well as nucleophiles. With a tertiary halide backside attack is blocked by crowding, so the alkoxide removes a beta hydrogen instead and an alkene forms by E2 elimination. The tertiary part must be on the alkoxide.

Why can't unsymmetrical ethers be made by dehydrating two alcohols?

When two different alcohols are heated with acid, each can act as both nucleophile and substrate, so three ethers form together: two symmetrical and one mixed. Separating them is difficult, so the Williamson synthesis is used instead.

How is anisole prepared?

Phenol is converted to sodium phenoxide with sodium hydroxide, and the phenoxide is treated with iodomethane or dimethyl sulphate. Bromobenzene and sodium methoxide cannot be used because aryl halides do not undergo SN2 substitution.

What is alkoxymercuration-demercuration?

It converts an alkene into an ether. The alkene reacts with an alcohol and mercury(II) trifluoroacetate, and the mercury is then replaced by hydrogen using sodium borohydride. The alkoxy group adds with Markovnikov orientation and no rearrangement occurs.

Which preparation of ethers is most asked in NEET?

NEET usually asks for the ether formed from ethanol and concentrated sulphuric acid at 413 K, the correct Williamson pairing for a tertiary ether, and why aryl halides cannot be used. The temperature 413 K and the bulky-on-oxygen rule answer most questions.

How does JEE Main test the Williamson synthesis?

JEE Main gives two possible pairings and asks which gives the ether and which gives an alkene, or asks for the product of a phenoxide with an allyl or benzyl halide. Recognising SN2 versus E2 at the halide carbon is the key skill.

Previous year questions on Preparation of Ether

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

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