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State Board · Class 12 · Chemistry

Lesson 11: Alcohols, Phenols and Ethers

7 concepts · 56 practice questions with answers · Free to read

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Classification and nomenclature

From daily life

Ethanol in sanitiser, phenol disinfectant and the old ether anaesthetic are all oxygen-containing organic compounds.

Simple definition

Alcohols have –OH on an sp³ carbon (R–OH); phenols have –OH on a benzene ring; ethers are R–O–R′. Alcohols are 1°, 2° or 3° by the carbons attached to the C–OH.

ClassificationAlcohols, phenols, ethers: classes and names1°CH₃CH₂CH₂OHpropan-1-ol2°(CH₃)₂CHOHpropan-2-ol3°(CH₃)₃COH2-methylpropan-2-olPhenolC₆H₅OHphenolEtherCH₃OCH₂CH₃methoxyethaneEtherC₆H₅OCH₃methoxybenzene (anisole)1°: –CH₂OH · 2°: >CHOH · 3°: ≥C–OH (carbons on the C–OH)Name: alkane + 'ol' (OH); ethers: alkoxy + alkane

In the figure : Classes and names.

Rule 1: Classifying alcohols — 1°: –CH₂OH; 2°: >CHOH; 3°: ≥C–OH. Mono-, di-, trihydric: 1, 2, 3 OH groups.

  • Easy : Ethanol is 1°.
  • Medium : Propan-2-ol is 2°.
  • Exam-type : Glycerol is trihydric.

Rule 2: IUPAC naming — Alcohols: replace the final e of the alkane by 'ol', lowest locant to OH. Ethers: alkoxy + alkane (smaller alkyl = alkoxy).

  • Easy : CH₃CH(OH)CH₃ = propan-2-ol.
  • Medium : CH₃OCH₂CH₃ = methoxyethane.
  • Exam-type : C₆H₅OCH₃ = methoxybenzene.

Rule 3: Naming phenols — C₆H₅OH = phenol; methyl derivatives are cresols (2-methylphenol); two OH ⇒ benzene-1,2-diol (catechol).

  • Easy : o-cresol = 2-methylphenol.
  • Medium : Catechol is 1,2.
  • Exam-type : Resorcinol is 1,3.

Reasoning: why is (CH₃)₃COH tertiary?

  • Step 1: Find the carbon bearing OH.
  • Step 2: It carries three carbon groups (3 CH₃).
  • Step 3: So it is 3°; name 2-methylpropan-2-ol.

CET level (MCQ)

  • CET 1) Name of CH₃CH(OH)CH₃? 1) propan-1-ol 2) propan-2-ol 3) ethanol 4) methoxyethane — Answer: 2).
  • CET 2) A 3° alcohol? 1) (CH₃)₃COH 2) CH₃CH₂OH 3) CH₃CH(OH)CH₃ 4) CH₃OH — Answer: 1).

Real-life use: Sanitisers, disinfectants, solvents.

Remember

1° –CH₂OH; 2° >CHOH; 3° ≥C–OH; phenol = ring + OH.

Common mistake

Calling benzyl alcohol (C₆H₅CH₂OH) a phenol.

CET trap

In phenol OH is on the ring itself; in benzyl alcohol a CH₂ intervenes.

Shortcut

Memory clue: "1°-2°-3° = carbons on the C–OH".

Spot the difference

TypeGroupExample
AlcoholR–OHC₂H₅OH
PhenolAr–OHC₆H₅OH
EtherR–O–R′CH₃OCH₃

Practice questions

Easy
  1. 1What are 1°, 2°, 3°?
  2. 2Name (CH₃)₃COH.
  3. 3Phenol vs benzyl alcohol.
Medium
  1. 4What type is glycerol?
  2. 5Name CH₃OC₂H₅.
  3. 6What is catechol?
Challenging
  1. 7Class of C₂H₅OH?
  2. 8Formula of methoxybenzene.
Show answers (try them yourself first)
  1. 11, 2 or 3 carbons on the C–OH.
  2. 22-methylpropan-2-ol.
  3. 3OH on ring vs on CH₂.
  4. 4Trihydric.
  5. 5Methoxyethane.
  6. 6Benzene-1,2-diol.
  7. 7A 1° alcohol.
  8. 8C₆H₅OCH₃.

Preparation of alcohols and phenols

From daily life

Ethanol is made by fermenting sugar or by hydrating ethene — two routes.

Simple definition

Alcohols: hydration, hydroboration–oxidation, carbonyl reduction, Grignard. Phenols: cumene, diazonium salt, haloarene and sulphonic acid routes.

PreparationPreparation of alcohols and phenolsRCH=CH₂ + H₂O (H⁺)Markovnikov ⇒ 2° alcoholRCH=CH₂ ⇒ B₂H₆, H₂O₂/OH⁻anti-Markovnikov ⇒ 1°RCHO / R₂CO + NaBH₄ / LiAlH₄reduction ⇒ 1° / 2°RMgX + HCHO / RCHO / R₂COGrignard ⇒ 1° / 2° / 3°cumene + O₂ ⇒ hydroperoxide + H₃O⁺phenol + acetoneC₆H₅N₂⁺Cl⁻ + H₂O, warmphenol + N₂Cumene process: industrial; acetone is the co-product

In the figure : Preparation routes.

Rule 1: From alkenes — H₃O⁺ hydration follows Markovnikov. B₂H₆ then H₂O₂/OH⁻ gives the anti-Markovnikov 1° alcohol.

  • Easy : Propene + H₂O ⇒ propan-2-ol.
  • Medium : Propene, hydroboration ⇒ propan-1-ol.
  • Exam-type : Ethene ⇒ ethanol.

Rule 2: Carbonyls and Grignard — RCHO + NaBH₄ ⇒ 1°; R₂CO ⇒ 2°. RMgX + HCHO ⇒ 1°; RCHO ⇒ 2°; R₂CO ⇒ 3° (then H₃O⁺).

  • Easy : CH₃MgBr + HCHO ⇒ ethanol.
  • Medium : CH₃MgBr + CH₃CHO ⇒ propan-2-ol.
  • Exam-type : CH₃MgBr + (CH₃)₂CO ⇒ 3°.

Rule 3: Preparing phenol — Cumene: O₂ ⇒ hydroperoxide ⇒ H₃O⁺ ⇒ phenol + acetone. Diazonium salt + H₂O ⇒ phenol + N₂.

  • Easy : Cumene = isopropylbenzene.
  • Medium : Acetone is a co-product.
  • Exam-type : C₆H₅N₂⁺Cl⁻ + H₂O ⇒ phenol.

Reasoning: propan-1-ol from propene?

  • Step 1: H₃O⁺ gives propan-2-ol (Markovnikov), not the 1° alcohol.
  • Step 2: So use B₂H₆ (hydroboration).
  • Step 3: Then H₂O₂/OH⁻ ⇒ propan-1-ol.

CET level (MCQ)

  • CET 1) CH₃MgBr + HCHO then H₃O⁺? 1) ethanol 2) propan-2-ol 3) methanol 4) 3° — Answer: 1).
  • CET 2) Co-product of the cumene process? 1) acetone 2) CO₂ 3) benzene 4) ethane — Answer: 1).

Real-life use: Ethanol industry, phenol plastics.

Remember

Markovnikov = hydration; anti = hydroboration; Grignard + HCHO = 1°.

Common mistake

Treating hydroboration as Markovnikov.

CET trap

Grignard: HCHO ⇒ 1°, other aldehydes ⇒ 2°, ketones ⇒ 3°.

Shortcut

Memory clue: "HCHO–1°, RCHO–2°, R₂CO–3°".

Spot the difference

RouteProduct
H₃O⁺ + alkene2°
Hydroboration1°
Cumenephenol + acetone

Practice questions

Easy
  1. 1Hydration vs hydroboration.
  2. 2Grignard + HCHO?
  3. 3Grignard + ketone?
Medium
  1. 4Describe the cumene process.
  2. 5Role of NaBH₄?
  3. 6Phenol from diazonium salt?
Challenging
  1. 7Conditions: propene ⇒ propan-2-ol.
  2. 8What is cumene?
Show answers (try them yourself first)
  1. 1Markovnikov vs anti-Markovnikov.
  2. 2A 1° alcohol.
  3. 3A 3° alcohol.
  4. 4O₂ ⇒ hydroperoxide ⇒ H₃O⁺ ⇒ phenol + acetone.
  5. 5Carbonyl reduction.
  6. 6Warm with water.
  7. 7H₃O⁺.
  8. 8Isopropylbenzene.

Physical properties: H-bonding, boiling point, solubility

From daily life

Ethanol mixes freely with water, but petrol does not — because of hydrogen bonding.

Simple definition

Alcohols form H-bonds with each other and with water ⇒ high boiling points and water solubility. Ethers cannot H-bond with each other ⇒ low boiling points.

H-bond bpBoiling points (K): effect of H-bonding185C₂H₆231C₃H₈249CH₃OCH₃308C₂H₅OC₂H₅338CH₃OH351C₂H₅OH370C₃H₇OH391C₄H₉OHSimilar size: alcohol > ether > alkane (H-bonding)More carbons ⇒ less soluble; up to 3 C miscible with water

In the figure : Boiling points.

Rule 1: Boiling point — Similar size: alcohol > ether > alkane. More carbons ⇒ higher b.p.; branching ⇒ lower.

  • Easy : C₂H₅OH 351 K vs CH₃OCH₃ 249 K.
  • Medium : 1-butanol 391 K vs tert-butanol lower.
  • Exam-type : Phenol 455 K.

Rule 2: Solubility — Alcohols up to 3 C are miscible with water; a larger hydrocarbon part lowers solubility. Ethers are slightly soluble.

  • Easy : Methanol and ethanol are miscible.
  • Medium : 1-hexanol is poorly soluble.
  • Exam-type : C₂H₅OC₂H₅ ≈ 7 g/100 mL.

Rule 3: H-bond strength — O–H···O bonds are strong; an ether oxygen accepts H-bonds from water but cannot donate its own.

  • Easy : Intermolecular H-bonds ⇒ higher b.p.
  • Medium : o-nitrophenol (intramolecular) has a lower b.p.
  • Exam-type : p-nitrophenol is higher.

Reasoning: why is ethanol's b.p. above CH₃OCH₃'s?

  • Step 1: Both have M = 46.
  • Step 2: Ethanol has O–H ⇒ intermolecular H-bonds.
  • Step 3: The ether has none ⇒ 351 K vs 249 K.

CET level (MCQ)

  • CET 1) Highest b.p.? 1) C₂H₆ 2) CH₃OCH₃ 3) C₂H₅OH 4) C₃H₈ — Answer: 3).
  • CET 2) Water solubility falls when? 1) carbons increase 2) carbons decrease 3) OH increases 4) temp falls — Answer: 1).

Real-life use: Choice of solvent, drug solubility.

Remember

b.p.: alcohol > ether > alkane; up to 3 C miscible.

Common mistake

Assuming ethers form intermolecular H-bonds.

CET trap

o-Nitrophenol has intramolecular H-bonding ⇒ lower b.p., steam volatile.

Shortcut

Memory clue: "OH = high b.p. + water-loving".

Spot the difference

CompoundMb.p. (K)
C₂H₅OH46351
CH₃OCH₃46249
C₃H₈44231

Practice questions

Easy
  1. 1Why high b.p. for alcohols?
  2. 2Why low b.p. for ethers?
  3. 3Which alcohols are miscible?
Medium
  1. 4Solubility as C increases?
  2. 5o- vs p-nitrophenol.
  3. 6Effect of branching on b.p.?
Challenging
  1. 7Compare C₂H₅OH and C₃H₈.
  2. 8Why are ethers slightly soluble?
Show answers (try them yourself first)
  1. 1Intermolecular H-bonds.
  2. 2No H-bonds.
  3. 3Up to 3 C.
  4. 4It decreases.
  5. 5Intramolecular vs intermolecular H-bonds.
  6. 6Branching lowers b.p.
  7. 7351 K vs 231 K.
  8. 8O accepts H-bonds from water.

Acidic character and laboratory tests

From daily life

Phenol burns skin because it is acidic; ethanol is not.

Simple definition

Lower pKa ⇒ stronger acid. Phenol (10.0) > water (15.7) > ethanol (16). Phenoxide is resonance-stabilised.

AcidityAcidity: pKa (lower = more acidic)C₂H₅OH16.0H₂O15.7phenol10.0p-nitrophenol7.15picric acid0.4Phenol: phenoxide is resonance-stabilised ⇒ dissolves in NaOH, not in NaHCO₃–NO₂ (withdrawing) ⇒ stronger acid; –CH₃ ⇒ weakerFeCl₃: phenol ⇒ violet colour

In the figure : pKa comparison.

Rule 1: Acidity order — 1° > 2° > 3° alcohols (alkyl +I). Phenol > alcohols. –NO₂ increases acidity, –CH₃ decreases it.

  • Easy : Picric acid pKa 0.4.
  • Medium : p-nitrophenol 7.15.
  • Exam-type : Cresols above 10.

Rule 2: Tests for phenol — FeCl₃ ⇒ violet; dissolves in NaOH but not in NaHCO₃ (weaker than H₂CO₃); bromine water ⇒ white precipitate.

  • Easy : A carboxylic acid with NaHCO₃ ⇒ CO₂.
  • Medium : Phenol with NaHCO₃ ⇒ nothing.
  • Exam-type : C₆H₅OH + NaOH ⇒ C₆H₅ONa.

Rule 3: Tests for alcohols — Lucas: ZnCl₂ + conc. HCl. 3° turbid at once, 2° in about 5 min, 1° none at room temperature. Iodoform: for CH₃CH(OH)– groups.

  • Easy : (CH₃)₃COH at once.
  • Medium : Propan-2-ol in about 5 min.
  • Exam-type : Ethanol iodoform ⇒ yellow CHI₃.

Reasoning: why no reaction of phenol with NaHCO₃?

  • Step 1: pKa phenol = 10.0; H₂CO₃ = 6.4.
  • Step 2: Phenol is a weaker acid than H₂CO₃.
  • Step 3: So HCO₃⁻ cannot take its proton.

CET level (MCQ)

  • CET 1) Phenol with FeCl₃? 1) violet 2) red 3) yellow 4) green — Answer: 1).
  • CET 2) 3° alcohol in the Lucas test? 1) at once 2) 5 min 3) none 4) on heating — Answer: 1).

Real-life use: Identifying phenols, classifying alcohols.

Remember

pKa: phenol 10; picric 0.4; Lucas: 3° at once.

Common mistake

Writing that phenol reacts with NaHCO₃.

CET trap

3° alcohols are least acidic yet fastest in the Lucas test.

Shortcut

Memory clue: "Lucas: tertiary = instant".

Spot the difference

TestResult
FeCl₃phenol ⇒ violet
Lucas 3°at once
NaHCO₃phenol ⇒ none

Practice questions

Easy
  1. 1Why is phenol more acidic than alcohols?
  2. 2Describe the Lucas test.
  3. 3Why is picric acid strong?
Medium
  1. 4FeCl₃ test for phenol?
  2. 5Phenol with NaOH vs NaHCO₃.
  3. 6Acidity order of 1°, 2°, 3°?
Challenging
  1. 7Iodoform test is for?
  2. 8What does a lower pKa mean?
Show answers (try them yourself first)
  1. 1Resonance-stabilised phenoxide.
  2. 2ZnCl₂/HCl; 3° at once.
  3. 3–NO₂ stabilises the anion.
  4. 4A violet colour.
  5. 5Dissolves in NaOH, not NaHCO₃.
  6. 61° > 2° > 3°.
  7. 7CH₃CH(OH)– / CH₃CO–.
  8. 8Stronger acid.

Reactions of alcohols

From daily life

Sanitiser burns (combustion) and wine turns sour (oxidation to acetic acid) — alcohol reactions.

Simple definition

Alcohols break O–H (Na, esterification) or C–O (HX, dehydration); oxidation needs an α-hydrogen.

Alcohol reactionsReactions of alcohols (CH₃CH₂OH)C₂H₅OHNa ⇒ C₂H₅ONa + ½H₂PCC ⇒ CH₃CHOK₂Cr₂O₇ ⇒ CH₃COOH443 K, H₂SO₄ ⇒ C₂H₄413 K ⇒ C₂H₅OC₂H₅CH₃COOH ⇒ ester1°: PCC gives aldehyde; K₂Cr₂O₇ gives acid2° ⇒ ketone · 3° ⇒ resists oxidationLucas: 3° at once · 2° ≈ 5 min · 1° none at room temperature

In the figure : Alcohol reactions.

Rule 1: O–H bond reactions — 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O (H⁺).

  • Easy : 4.6 g ethanol ⇒ 1.12 L H₂ (STP).
  • Medium : CH₃COOH + C₂H₅OH ⇒ ethyl acetate.
  • Exam-type : 60 g + 46 g ⇒ 88 g ester.

Rule 2: C–O bond reactions — HX / PCl₅ / SOCl₂ ⇒ alkyl halide. Dehydration: 443 K ⇒ alkene; 413 K ⇒ ether. Zaitsev's rule.

  • Easy : C₂H₅OH + SOCl₂ ⇒ C₂H₅Cl.
  • Medium : 92 g ethanol (413 K) ⇒ 74 g ether.
  • Exam-type : 443 K ⇒ C₂H₄.

Rule 3: Oxidation — 1°: PCC ⇒ aldehyde; K₂Cr₂O₇/KMnO₄ ⇒ carboxylic acid. 2° ⇒ ketone. 3° has no α-H and resists oxidation.

  • Easy : 4.6 g C₂H₅OH ⇒ 6.0 g CH₃COOH.
  • Medium : Propan-2-ol ⇒ acetone.
  • Exam-type : (CH₃)₃COH ⇒ no change.

Calculation: H₂ (STP) from 4.6 g ethanol + Na

  • Step 1: n = 4.6 ÷ 46 = 0.1 mol.
  • Step 2: 2 C₂H₅OH ⇒ 1 H₂ ⇒ 0.05 mol.
  • Step 3: V = 0.05 × 22.4 = 1.12 L.

CET level (MCQ)

  • CET 1) Oxidation of a 2° alcohol gives? 1) aldehyde 2) ketone 3) acid 4) ether — Answer: 2).
  • CET 2) 443 K with conc. H₂SO₄ gives? 1) ether 2) alkene 3) acid 4) ester — Answer: 2).

Real-life use: Fuel, drug synthesis, esters.

Remember

443 K alkene; 413 K ether; PCC = aldehyde; 3° resists.

Common mistake

Thinking PCC gives the acid; it stops at the aldehyde.

CET trap

Temperature decides the dehydration product.

Shortcut

Memory clue: "443 = alkene, 413 = ether (cooler ⇒ ether)".

Spot the difference

Reagent1° product
PCCaldehyde
K₂Cr₂O₇acid
Naalkoxide

Practice questions

Easy
  1. 1Ethanol + Na?
  2. 2H₂ from 4.6 g in L?
  3. 3Two dehydration conditions.
Medium
  1. 4PCC oxidation of 1°.
  2. 5Why 3° resists oxidation?
  3. 6Write esterification.
Challenging
  1. 7Ether from 92 g ethanol?
  2. 8State Zaitsev's rule.
Show answers (try them yourself first)
  1. 1Ethoxide + H₂.
  2. 21.12 L.
  3. 3443 K alkene; 413 K ether.
  4. 4An aldehyde.
  5. 5No α-hydrogen.
  6. 6RCOOH + R′OH ⇌ RCOOR′ + H₂O.
  7. 774 g.
  8. 8The more substituted alkene is major.

Reactions of phenols

From daily life

Phenol gives picric acid (an old dye and explosive) and salicylic acid (the base of aspirin).

Simple definition

In phenol –OH donates electrons to the ring ⇒ an activating o/p director ⇒ easy electrophilic substitution; O–H is acidic.

Phenol reactionsReactions of phenolBr₂/H₂O2,4,6-tribromophenol (white ppt)dil. HNO₃o- + p-nitrophenolconc. HNO₃2,4,6-trinitrophenol (picric acid)NaOH, CO₂ (398 K, 6 atm) ; H⁺Kolbe ⇒ salicylic acidCHCl₃, NaOH ; H⁺Reimer–Tiemann ⇒ salicylaldehydeZn dust, heatbenzene–OH is an activating o/p director ⇒ easy electrophilic substitution

In the figure : Phenol reactions.

Rule 1: Halogenation and nitration — Br₂/H₂O ⇒ 2,4,6-tribromophenol (white). Dilute HNO₃ ⇒ o- and p-nitrophenol; concentrated ⇒ picric acid.

  • Easy : 9.4 g phenol ⇒ 33.1 g tribromophenol.
  • Medium : 9.4 g ⇒ 22.9 g picric acid.
  • Exam-type : Br₂ in CS₂ ⇒ o-/p-bromo.

Rule 2: Kolbe and Reimer–Tiemann — Kolbe: C₆H₅ONa + CO₂ (398 K, 6 atm) ⇒ H⁺ ⇒ salicylic acid. Reimer–Tiemann: CHCl₃ + NaOH ⇒ H⁺ ⇒ salicylaldehyde.

  • Easy : 9.4 g phenol ⇒ 13.8 g salicylic acid.
  • Medium : Salicylic acid ⇒ aspirin.
  • Exam-type : Salicylaldehyde.

Rule 3: Reduction and oxidation — Zn dust + heat ⇒ benzene. Na₂Cr₂O₇/H₂SO₄ ⇒ benzoquinone.

  • Easy : C₆H₅OH + Zn ⇒ C₆H₆ + ZnO.
  • Medium : Oxidation ⇒ yellow quinone.
  • Exam-type : The ring C–OH bond is hard to break.

Calculation: tribromophenol from 9.4 g phenol

  • Step 1: n = 9.4 ÷ 94 = 0.1 mol.
  • Step 2: 1 : 1 ⇒ 0.1 mol product.
  • Step 3: M = 330.8 ⇒ 33.1 g.

CET level (MCQ)

  • CET 1) Phenol + bromine water ⇒ ? 1) white ppt 2) yellow 3) red 4) none — Answer: 1).
  • CET 2) Product of the Kolbe reaction? 1) salicylic acid 2) benzene 3) aniline 4) toluene — Answer: 1).

Real-life use: Drugs (aspirin), dyes, explosives.

Remember

Br₂ water = tri; dilute HNO₃ = o,p; Kolbe = salicylic acid.

Common mistake

Calling the Reimer–Tiemann product salicylic acid; it is salicylaldehyde.

CET trap

Br₂/CS₂ gives mono-bromo; Br₂/water gives tri.

Shortcut

Memory clue: "Kolbe = CO₂, Reimer = CHCl₃".

Spot the difference

ReactionReagentProduct
KolbeCO₂salicylic acid
Reimer–TiemannCHCl₃salicylaldehyde
Br₂/H₂OBr₂tribromo

Practice questions

Easy
  1. 1Why is phenol o/p directing?
  2. 2Write the Kolbe reaction.
  3. 3Write Reimer–Tiemann.
Medium
  1. 4How is picric acid made?
  2. 5Tribromophenol from 9.4 g?
  3. 6Phenol with Zn?
Challenging
  1. 7Phenol to quinone.
  2. 8Product with Br₂/CS₂?
Show answers (try them yourself first)
  1. 1OH donates electrons.
  2. 2C₆H₅ONa + CO₂ ⇒ salicylic acid.
  3. 3CHCl₃ + NaOH ⇒ salicylaldehyde.
  4. 4Nitration with conc. HNO₃.
  5. 533.1 g.
  6. 6Benzene.
  7. 7Benzoquinone.
  8. 8o-/p-bromophenol.

Ethers: preparation, reactions and uses

From daily life

Diethyl ether once anaesthetised surgical patients; today it is mostly a solvent.

Simple definition

Ethers (R–O–R′) are less reactive and low boiling. Williamson synthesis: RONa + R′X.

EthersEthers: preparation and reactionsWilliamson: R–ONa + R′–X → R–O–R′ + NaX (SN2)Use a 1° halide; 3° gives elimination (alkene)Alcohol at 413 K, H₂SO₄: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂OCleavage by HI: R–O–R′ + HI → R–I + R′OHAnisole + HI → phenol + CH₃I (aryl C–O bond stays)The halide forms on the smaller alkyl group (SN2)Uses: diethyl ether ⇒ old anaesthetic, solventHazard: peroxides on air exposure ⇒ explosive; flammableLower b.p. and less reactive than alcohols

In the figure : Ethers.

Rule 1: Preparation — Alcohol at 413 K + H₂SO₄ ⇒ symmetrical ether. Williamson: unsymmetrical; 1° halide via SN2.

  • Easy : 2C₂H₅OH ⇒ C₂H₅OC₂H₅ + H₂O.
  • Medium : C₂H₅ONa + CH₃I ⇒ C₂H₅OCH₃.
  • Exam-type : 3° halide ⇒ alkene.

Rule 2: Reactions — HI cleavage: R–O–R′ + HI ⇒ RI + R′OH. Anisole + HI ⇒ phenol + CH₃I. Electrophilic substitution on the ring (o/p).

  • Easy : C₂H₅OCH₃ + HI ⇒ CH₃I + C₂H₅OH.
  • Medium : Anisole nitration ⇒ o/p.
  • Exam-type : Anisole + HI ⇒ phenol.

Rule 3: Uses and hazards — Solvent (extraction), former anaesthetic; forms explosive peroxides in air; highly flammable.

  • Easy : Test peroxides with FeSO₄.
  • Medium : Keep away from flames.
  • Exam-type : Store in dark bottles.

Reasoning: products of C₂H₅OCH₃ + HI

  • Step 1: O is protonated to an oxonium ion.
  • Step 2: I⁻ attacks the smaller alkyl (CH₃) by SN2.
  • Step 3: CH₃I + C₂H₅OH.

CET level (MCQ)

  • CET 1) Williamson synthesis uses? 1) RONa + R′X 2) ROH + HX 3) RCHO 4) RMgX — Answer: 1).
  • CET 2) Anisole + HI ⇒ ? 1) phenol + CH₃I 2) benzene 3) toluene 4) CH₃OH — Answer: 1).

Real-life use: Solvents, drug synthesis.

Remember

Williamson: SN2, 1° halide; HI: iodide goes to the smaller alkyl.

Common mistake

Using a 3° halide in Williamson synthesis.

CET trap

In anisole + HI the aryl C–O bond does not break.

Shortcut

Memory clue: "ether = no flame".

Spot the difference

MethodProduct
413 Ksymmetrical
Williamsonunsymmetrical

Practice questions

Easy
  1. 1What is Williamson synthesis?
  2. 2Why avoid 3° halides?
  3. 3Ether + HI?
Medium
  1. 4Anisole + HI?
  2. 5Why are ethers hazardous?
  3. 62C₂H₅OH at 413 K?
Challenging
  1. 7State uses.
  2. 8Why are ethers unreactive?
Show answers (try them yourself first)
  1. 1RONa + R′X ⇒ ROR′.
  2. 2Elimination occurs.
  3. 3RI + R′OH.
  4. 4Phenol + CH₃I.
  5. 5Peroxides and flammable.
  6. 6Diethyl ether.
  7. 7Solvent, former anaesthetic.
  8. 8Stable C–O–C, no O–H.

Quick revision of the whole lesson

ConceptKey formula / fact
1°/2°/3°carbons on C–OH
b.p.alcohol > ether > alkane
pKaphenol 10; picric 0.4
Lucas3° at once
Dehydration443 K alkene; 413 K ether
Kolbesalicylic acid
Reimer–Tiemannsalicylaldehyde
WilliamsonRONa + R′X
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🎯 Most important points for the exam

  • Classes and names.
  • Preparation.
  • H-bonding, b.p.
  • Acidity, tests.
  • Alcohol reactions.
  • Phenol reactions.
  • Ethers.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.