JEE Mains Chemistry · Alcohols, Phenols and Ethers

Ethers: Williamson Synthesis and Cleavage

Ethers are made by the Williamson synthesis, an SN2 reaction of an alkoxide or phenoxide with a methyl or primary halide, and are split by HI or HBr: iodide attacks the smaller alkyl group, a tertiary group leaves as a cation, and the aryl–oxygen bond never breaks.

All Alcohols, Phenols and Ethers notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, seventeen of them multiple choice, and two from 2026. Six are about making ethers: which halide and alkoxide to pair, SN2 on an allylic bromide, and ethers formed from enol ethers and alcohols in acid. Nine ask which C–O bond HI or HBr breaks and what the fragments become. Four follow an aryl ether through ring substitution or a structure proof, and two of those ask for a number.

Concept 1 of 3: Making ethers: Williamson synthesis and acid routes

In the Williamson synthesis an alkoxide or phenoxide ion attacks the carbon of an alkyl halide from the back (SN2). That carbon must be easy to reach, so the halide should be methyl or primary. Put any bulky group on the alkoxide side. A tertiary halide would lose HBr instead, and an aryl halide does not react at all.

Definition

$\mathrm{RO^-Na^+ + R'X \to ROR' + NaX}$ , with $\mathrm{R'}$ methyl or primary.
Aryl alkyl ethers: phenoxide + alkyl halide, e.g. $\mathrm{C_6H_5O^-Na^+ + CH_3I}$ → anisole. Never aryl halide + alkoxide.
A tertiary halide with an alkoxide gives an alkene (elimination).
Symmetrical ethers: a primary alcohol with conc. $\mathrm{H_2SO_4}$ at 413 K (ethanol gives ethoxyethane); at 443 K the alkene forms instead.
SN2 on an allylic halide keeps the double bond where it was: phenoxide + $\mathrm{(CH_3)_2C{=}CHCH_2Br}$ gives $\mathrm{C_6H_5OCH_2CH{=}C(CH_3)_2}$ .
In acid, an alcohol adds to an enol ether: 3,4-dihydro-2H-pyran + ROH gives a THP ether (an acetal). $\mathrm{Br_2}$ in methanol on an enol ether puts OCH₃ on the carbon next to the ring oxygen, trans to Br. An OH in the same molecule can trap a carbocation and close a cyclic ether.

Williamson synthesis

\mathrm{R{-}O^-Na^+ + R'{-}X \xrightarrow{S_N2} R{-}O{-}R' + NaX} \qquad (\mathrm{R'} = \mathrm{CH_3} \text{ or primary; never aryl})

Worked example

Plan a Williamson synthesis of 2-ethoxy-2-methylpropane,

\mathrm{(CH_3)_3C{-}O{-}CH_2CH_3}

, and say which pairing fails.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q49Moderate

Example 1 · Alcohols, Phenols and Ethers · Ethers: Williamson Synthesis and Cleavage

Suitable reaction condition for preparation of Methyl phenyl ether is

The halide side must be simple

The alkoxide can be as bulky as you like; the halide cannot. A tertiary halide eliminates and an aryl halide does not undergo SN2.

SN2 does not move the double bond

With an allylic bromide, the SN2 product has oxygen on the carbon that carried Br. The product with oxygen on the far end of the old C=C comes from SN1 or SN2′ and is the usual wrong option.

Drill 5 more on making ethers: williamson synthesis and acid routes

Concept 2 of 3: Cleaving ethers with HI and HBr

The acid protonates the ether oxygen, turning an alcohol into a leaving group. Then iodide attacks. If both groups are methyl or primary, it attacks the less hindered one (SN2). If one group is tertiary, that group leaves as a stable cation (SN1) and ends up as the iodide. An aryl–oxygen bond is never broken.

Definition

Reactivity: HI > HBr > HCl.
Methyl or primary groups: SN2 at the smaller one. $\mathrm{CH_3OCH_2CH_3 + HI \to CH_3I + CH_3CH_2OH}$ .
A tertiary (or other stable-cation) group: SN1, the halide goes to that carbon. $\mathrm{(CH_3)_3COCH_3 + HI \to (CH_3)_3CI + CH_3OH}$ .
Aryl alkyl ethers give a phenol and an alkyl halide: $\mathrm{C_6H_5OCH_3 + HI \to C_6H_5OH + CH_3I}$ . No iodobenzene forms, and the phenol reacts no further.
Excess hot HI converts the alcohol fragment into a second alkyl iodide as well.
In a molecule with another reactive group, such as a C=C, excess HBr reacts there too.

Which bond HI breaks

\mathrm{Ar{-}O{-}R + HI \to Ar{-}OH + R{-}I} \qquad \mathrm{R_3C{-}O{-}R' + HI \to R_3C{-}I + R'{-}OH}

Worked example

Give the products of one equivalent of hot HI with (a) methoxyethane, (b) ethoxybenzene and (c) 2-ethoxy-2-methylpropane.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q131Moderate

Example 2 · Alcohols, Phenols and Ethers · Ethers: Williamson Synthesis and Cleavage

Major product formed in the following reaction is a mixture of:

A tertiary group reverses the 'smaller group' rule

With a methyl or primary partner, iodide ends on the smaller group. When one group is tertiary, the iodide ends on the tertiary carbon, because that bond breaks first to give the cation.

Phenol is the end of the line

An aryl alkyl ether gives a phenol and an alkyl halide. Even with excess HI the phenol is not turned into an aryl iodide.

Drill 8 more on cleaving ethers with hi and hbr

Concept 3 of 3: Aryl ethers on the ring and in structure proofs

An alkoxy group on a ring pushes electrons into it, like OH but less strongly. So anisole reacts with electrophiles at ortho and para, mostly para because the group is bulky. In a structure proof each reagent reports one feature: FeCl₃ a free phenolic OH, HI a methoxy group, KMnO₄ a C=C.

Definition

$\mathrm{OCH_3}$ and $\mathrm{OC_2H_5}$ are activating, ortho/para directing; para is the major product.
Anisole + $\mathrm{Br_2}$ in ethanoic acid (no $\mathrm{FeBr_3}$ ) → mainly p-bromoanisole.
Nitration (conc. $\mathrm{HNO_3/H_2SO_4}$ ) → o- and p-nitroanisole, para major.
Friedel–Crafts: $\mathrm{CH_3Cl/AlCl_3}$ → 2- and 4-methoxytoluene; $\mathrm{CH_3COCl/AlCl_3}$ → 2- and 4-methoxyacetophenone.
When the para position is taken, the next group goes ortho to the alkoxy group, which is the stronger director.
Clues in a structure proof: neutral $\mathrm{FeCl_3}$ colour, a free phenolic OH; NaOH + $\mathrm{CH_3Br}$ adding $\mathrm{CH_2}$ to the formula, one such OH; $\mathrm{CH_3I}$ from hot HI, a methoxy group; decolouring alkaline $\mathrm{KMnO_4}$ , a C=C; hot alkali giving an isomer, an allyl side chain moving into conjugation.
An aldehyde with no α-hydrogen + HCHO in conc. NaOH (crossed Cannizzaro) gives $\mathrm{ArCH_2OH}$ , which can then be made into an ether.

Degree of unsaturation (rings + π bonds)

\text{DU} = \dfrac{2C + 2 - H}{2} \qquad (\text{O atoms do not count})

Worked example

Compound A,

\mathrm{C_7H_8O}

, does not dissolve in NaOH and gives no colour with

\mathrm{FeCl_3}

. Hot HI converts it into

\mathrm{CH_3I}

and B, which gives a violet colour with

\mathrm{FeCl_3}

. A with

\mathrm{Br_2}

in ethanoic acid gives C. Identify A, B and C, and count the π bonds in A.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q56Moderate

Example 3 · Alcohols, Phenols and Ethers · Ethers: Williamson Synthesis and Cleavage

A trisubstituted compound '

A

C_{10}H_{12}O_{2}

gives neutral

FeCl_{3}

test positive. Treatment of compound 'A' with

NaOH

and

CH_{3}

Br gives

C_{11}H_{14}O_{2}

, with hydroiodic acid gives methyl iodide and with hot conc.

NaOH

gives a compound

B,C_{10}H_{12}O_{2}

. Compound 'A' also decolorises alkaline

KMnO_{4}

. The number of

\pi

bond/s present in the compound '

A

' is

The ring is not a π bond

The degree of unsaturation counts rings and π bonds together. A benzene ring uses four: one for the ring and three for its π bonds.

Para blocked means ortho to the alkoxy group

In 4-nitroanisole the next bromine goes ortho to

\mathrm{OCH_3}

, not ortho to

\mathrm{NO_2}

. The activating group wins the directing contest.

Drill 3 more on aryl ethers on the ring and in structure proofs

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Making ethers: Williamson synthesis and acid routes
Williamson synthesis
$\mathrm{R{-}O^-Na^+ + R'{-}X \xrightarrow{S_N2} R{-}O{-}R' + NaX} \qquad (\mathrm{R'} = \mathrm{CH_3} \text{ or primary; never aryl})$
Cleaving ethers with HI and HBr
Which bond HI breaks
$\mathrm{Ar{-}O{-}R + HI \to Ar{-}OH + R{-}I} \qquad \mathrm{R_3C{-}O{-}R' + HI \to R_3C{-}I + R'{-}OH}$
Aryl ethers on the ring and in structure proofs
Degree of unsaturation (rings + π bonds)
$\text{DU} = \dfrac{2C + 2 - H}{2} \qquad (\text{O atoms do not count})$

Watch out for (6)

The halide side must be simple
→ Making ethers: Williamson synthesis and acid routes
SN2 does not move the double bond
→ Making ethers: Williamson synthesis and acid routes
A tertiary group reverses the 'smaller group' rule
→ Cleaving ethers with HI and HBr
Phenol is the end of the line
→ Cleaving ethers with HI and HBr
The ring is not a π bond
→ Aryl ethers on the ring and in structure proofs
Para blocked means ortho to the alkoxy group
→ Aryl ethers on the ring and in structure proofs

Test yourself on Alcohols, Phenols and Ethers

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