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Ethers Preparation Reactions and Commercial Uses

🎓 Class 12 Chemistry CBSE Theory Ch 7 – Alcohols, Phenols and Ethers ⏱ ~14 min
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Ethers — Preparation, Reactions and Commercial Uses

7.9 Some Commercial Uses of Alcohols

7.9.1 Methanol (wood spirit)

Methanol was once made by destructive distillation of wood — hence the old name "wood spirit". Today it is produced industrially by catalytic hydrogenation of carbon monoxide over a Cu/ZnO/Al₂O₃ catalyst:

CO + 2 H₂  Cu/ZnO/Al₂O₃, 573K, 200 atm→ CH₃OH

Uses: solvent for paints, varnishes and shellac; feedstock for formaldehyde (→ bakelite), acetic acid and methyl tert-butyl ether (petrol additive); antifreeze; emerging as a clean liquid fuel for direct-methanol fuel cells. Toxicity: methanol is metabolised to formaldehyde and formic acid; as little as 10 mL can cause blindness, 30 mL can be fatal.

7.9.2 Ethanol

Made industrially by (i) acid-catalysed hydration of ethylene, and (ii) fermentation of sugars — zymase enzymes of yeast convert glucose into ethanol:

C₆H₁₂O₆  zymase, 298-303K→ 2 C₂H₅OH + 2 CO₂

Uses: alcoholic beverages; organic solvent for tinctures and perfumes; industrial feedstock for diethyl ether, ethyl acetate, acetaldehyde; automotive fuel additive (gasohol); hand sanitisers at 60–70% v/v (denatures viral envelope proteins and bacterial membranes). Ordinary ethanol is denatured with methanol or pyridine to make it non-potable for industrial purposes.

7.10 Ethers

7.10.1 Structure Recap

An ether — R–O–R' — has a bent oxygen (C–O–C ≈ 112°) and is sp³-hybridised. Because there is no O–H, ethers cannot hydrogen-bond to themselves; they can, however, accept hydrogen bonds from water. Diethyl ether is therefore slightly soluble in water (≈ 8 g per 100 g at 20 °C) but boils at only 35 °C — lower than an alcohol of the same molar mass.

7.10.2 Preparation of Ethers

(a) Dehydration of an alcohol

Heating ethanol with conc. H₂SO₄ at 413 K forms diethyl ether (condensation — two alcohol molecules + acid); at 443 K the acid drives the reaction further to ethylene + water. So temperature is the key switch between "ether" and "alkene".

2 CH₃CH₂OH  conc. H₂SO₄, 413K→ CH₃CH₂–O–CH₂CH₃ + H₂O

The mechanism: one alcohol is protonated, a second alcohol attacks its carbon by SN2 and expels water, giving a protonated ether that deprotonates to the neutral product. This method is practical only for symmetrical ethers of primary alcohols. For 2° and 3° alcohols, elimination to the alkene dominates even at 413 K.

(b) Williamson synthesis

This is the general laboratory method — it works for both symmetrical and unsymmetrical ethers. An alkoxide ion displaces halide from an alkyl halide in an SN2 step:

R–O⁻Na⁺ + R'–X → R–O–R' + NaX

Because the step is SN2, the alkyl halide partner must be primary — with a 2° or 3° halide the alkoxide acts as a base instead of a nucleophile and elimination (E2) dominates.

Target: t-Bu–O–CH₃ (methyl tert-butyl ether) CORRECT (CH₃)₃C–O⁻Na⁺ + CH₃–I → (CH₃)₃C–O–CH₃ SN2 on CH₃ (1°) works cleanly WRONG CH₃–O⁻Na⁺ + (CH₃)₃C–Br → (CH₃)₂C=CH₂ + MeOH + NaBr methoxide is basic; E2 on 3° halide dominates Golden rule: always put the alkoxide on the more substituted side, the alkyl halide on the less substituted (primary) side.
Fig 7.11: Williamson synthesis — pick the right alkyl halide partner.

7.10.3 Physical Properties of Ethers

  • Boiling points much lower than those of isomeric alcohols (dimethyl ether –25 °C vs ethanol 78 °C) because no O–H means no H-bond donation.
  • Miscibility with water comparable to that of the isomeric alcohol for low molar mass (ether O accepts an H-bond from water).
  • Excellent solvents for Grignard reagents, LiAlH₄ and many organic reactions — inert, volatile and weakly coordinating.

7.10.4 Chemical Reactions of Ethers

(a) Cleavage by HX

Ethers are otherwise inert, but strong hydrogen halides (HI > HBr >> HCl) cleave the C–O–C linkage at high temperature to give an alcohol and an alkyl halide. HI is most reactive because I⁻ is the best nucleophile and HI is the strongest acid. HF does not cleave ethers.

R–O–R' + HX → R–X + R'–OH   (or the reverse pairing, depending on structure)

Mechanism & regioselectivity:

  • Ether of only 1°/2° alkyl groups: proton adds to ether O; halide attacks by SN2 on the less hindered carbon, expelling the more-substituted alcohol. Example: CH₃–O–C₂H₅ + HI → CH₃I + C₂H₅OH (I⁻ prefers the less hindered methyl).
  • Ether with a 3° or benzylic group: protonation then SN1 — the tertiary C–O cleaves to give a stable 3° carbocation and the smaller alcohol; halide then traps the cation. Example: (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH.
  • Aryl alkyl ethers (e.g., anisole, C₆H₅–O–CH₃ + HI): cleavage always gives phenol + alkyl halide, never aryl iodide + methanol. Reason: the C(aryl)–O bond has partial double-bond character and cannot undergo SN1 or SN2; only the alkyl–O bond breaks. Product: C₆H₅OH + CH₃I.
SN2 (1°/2° ether) CH₃–O–C₂H₅ + HI → CH₃–I + C₂H₅OH I⁻ attacks less-hindered C SN1 (3° ether) (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH 3° carbocation intermediate Aryl alkyl ether C₆H₅–O–CH₃ + HI → C₆H₅OH + CH₃I C(aryl)–O is unbreakable Reactivity of HX: HI > HBr >> HCl (HF no reaction) In aryl alkyl ethers, phenol never converts to aryl iodide because Ar–O has partial double-bond character.
Fig 7.12: Three regiochemical patterns for ether cleavage by HX.

(b) Electrophilic aromatic substitution on anisole

The –OCH₃ group is a strong ortho/para director (like –OH) because its oxygen lone pair donates into the ring. Anisole undergoes:

  • Halogenation: with Br₂ in ethanoic acid → p-bromoanisole (major) + o-bromoanisole.
  • Nitration: with dilute HNO₃ → o- and p-nitroanisole.
  • Friedel–Crafts alkylation with R–X / anhyd. AlCl₃ → o/p-alkylanisole.
  • Friedel–Crafts acylation with CH₃COCl / AlCl₃ → p-methoxyacetophenone (major).
C₆H₅OCH₃ + CH₃COCl  anhyd. AlCl₃→ p-CH₃CO–C₆H₄–OCH₃ (p-methoxyacetophenone)

Worked Examples

Example 7.16 — Plan a Williamson synthesis

Design a Williamson synthesis of tert-butyl methyl ether, (CH₃)₃C–O–CH₃.

Option A: (CH₃)₃C–Br + CH₃O⁻Na⁺ → mostly isobutylene + methanol (E2 on 3° halide). BAD.

Option B: (CH₃)₃C–O⁻K⁺ + CH₃I → (CH₃)₃C–O–CH₃ + KI. SN2 on a 1° halide works cleanly. GOOD.

Example 7.17 — Predict ether cleavage products

Write the products when each ether is heated with excess HI: (i) CH₃OCH₃; (ii) CH₃OC(CH₃)₃; (iii) C₆H₅OCH₃.

(i) 1°/1° ether: CH₃OH + CH₃I (SN2). With excess HI, the methanol also converts: two moles CH₃I in total.

(ii) 1°/3° ether: SN1 — I⁻ attaches to the 3° carbon. Products: (CH₃)₃C–I + CH₃OH.

(iii) Aryl alkyl ether: only the alkyl–O bond breaks. Products: C₆H₅OH + CH₃I.

Example 7.18 — Explain the boiling-point gap

Why does ethanol (b.p. 78 °C) boil ~100 °C higher than dimethyl ether (b.p. –25 °C) although they have the same molecular formula (C₂H₆O)?

Ethanol has an O–H bond → every molecule can donate AND accept H-bonds, forming an extensive network in the liquid. Dimethyl ether has no O–H → it can only accept, not donate, so it cannot self-associate. Breaking pure London-dispersion forces in liquid DME costs far less energy than breaking the H-bonded network of ethanol.

Example 7.19 — Anisole substitution

Predict the product(s) when anisole reacts with acetyl chloride in the presence of anhydrous AlCl₃.

Friedel–Crafts acylation. –OCH₃ activates the ring and directs to ortho/para. The bulky acylium ion preferentially enters para to avoid the methoxy group. Major product: p-methoxyacetophenone (4-methoxyacetophenone) with a small amount of the ortho isomer.

Activity 7.3 — Williamson Partner SelectorL4 Analyse

For each target ether below, decide which alkoxide + alkyl halide pair would give the highest yield:

  1. CH₃OCH₂CH₃ (methoxyethane)
  2. (CH₃)₃COCH₂C₆H₅ (benzyl tert-butyl ether)
  3. C₆H₅OCH₃ (anisole)
Predict: for each target, which partner is the alkyl halide and which is the alkoxide?

1. Either CH₃O⁻ + C₂H₅–Br or C₂H₅O⁻ + CH₃–I works (both partners 1°); CH₃I is slightly preferred (better leaving group).

2. Put the alkoxide on the tertiary side and the halide on the 1° side: (CH₃)₃C–O⁻Na⁺ + C₆H₅CH₂–Br → target.

3. Phenol is the alkoxide source; methyl must be the halide partner: C₆H₅O⁻Na⁺ + CH₃–I → anisole. An aryl halide would fail (sp² C resists SN2).

Interactive — Ether Cleavage Product Predictor

Pick an ether; the tool tells you the mechanism (SN1/SN2/aryl) and the products with excess HI.

Result will appear here …

Chapter 7 — Summary

  • Alcohols R–OH (sp³ C–O–H), phenols Ar–OH (sp² C–O–H) and ethers R–O–R' (sp³ C–O–C) are the three oxygen functional-group families derived formally from water.
  • IUPAC: alcohol = alkaneol; ether = alkoxy-alkane; phenol has its own parent name.
  • Alcohols are prepared from alkenes (Markovnikov hydration; anti-Markovnikov hydroboration–oxidation), from aldehydes/ketones/acids/esters (reduction with LiAlH₄, NaBH₄, H₂/Ni), and from Grignard addition to carbonyls.
  • Phenols — four routes: chlorobenzene + NaOH (Dow); benzenesulphonic acid + NaOH fusion; diazonium salt + H₂O; cumene + O₂ (industrial).
  • Alcohols acidity order: H₂O > 1° > 2° > 3°. Phenols (pKa ≈ 10) are a million times stronger than alcohols because the phenoxide is resonance-stabilised. Electron-withdrawing substituents (–NO₂) raise acidity further (picric acid pKa 0.4).
  • Key reactions of alcohols: with Na → alkoxide + H₂; with R'COOH/H₂SO₄ → ester; with HX/PCl₅/SOCl₂ → alkyl halide; dehydration (conc. H₂SO₄ or Al₂O₃) → alkene; oxidation (PCC, K₂Cr₂O₇) → aldehyde/ketone/acid depending on class; dehydrogenation (Cu, 573 K) → carbonyl.
  • Key reactions of phenols: with NaOH → phenoxide; with Br₂/H₂O → 2,4,6-tribromophenol; with HNO₃ → nitrophenols and picric acid; Kolbe → salicylic acid; Reimer–Tiemann → salicylaldehyde.
  • Ethers are made by alcohol dehydration (413 K) or — better — Williamson (alkoxide + 1° R'–X). They are cleaved by HI/HBr: SN2 on 1°/2° ethers, SN1 on ethers containing a 3° group, and selectively at the alkyl–O bond in aryl alkyl ethers. Anisole undergoes ortho/para electrophilic substitution.
  • Methanol: industrial feedstock, solvent, fuel; TOXIC. Ethanol: beverages, solvent, fuel additive, sanitiser.

Keywords grid

Monohydricalcohol with one –OH group
Allylic alcohol–OH on sp³ C adjacent to C=C
MarkovnikovH adds to H-richer C of C=C
Hydroboration–oxidationB₂H₆ then H₂O₂/OH⁻ → 1° alcohol
Grignard reagentR–MgX, carbanionic R
Dow processC₆H₅Cl + NaOH (623K/300atm) → phenol
Cumene processindustrial phenol + acetone
EsterificationR–OH + R'COOH (H₂SO₄) → ester
Lucas reagentconc. HCl + ZnCl₂ — distinguishes 1°, 2°, 3°
Saytzeff rulemore substituted alkene preferred
DehydrogenationCu/573K removes H₂ from alcohol
Kolbe reactionphenoxide + CO₂ → salicylic acid
Reimer–Tiemannphenol + CHCl₃/NaOH → salicylaldehyde
Picric acid2,4,6-trinitrophenol, pKa ≈ 0.4
Williamson synthesisR–O⁻ + R'–X → R–O–R'
Anisolemethoxybenzene, o/p-director
Fermentationglucose → ethanol + CO₂ (zymase)
Denatured alcoholnon-drinkable industrial ethanol

Competency-Based Questions — Ethers & Applications

A research group wants to prepare the fragrance compound p-methoxyacetophenone. They have anisole, acetyl chloride, benzene, diethyl ether and anhydrous AlCl₃.

1. The correct reagent mixture to convert anisole into p-methoxyacetophenone is:

  • (a) CH₃COOH + H₂SO₄
  • (b) CH₃COCl + anhyd. AlCl₃
  • (c) CH₃COCl + H₂O
  • (d) CH₃COCl + pyridine
(b). Friedel–Crafts acylation; –OCH₃ directs the acyl cation to the para position.

2. Which pair of reagents best delivers methyl tert-butyl ether (MTBE) in high yield?

  • (a) (CH₃)₃C–Br + CH₃O⁻Na⁺
  • (b) (CH₃)₃C–O⁻K⁺ + CH₃I
  • (c) 2 (CH₃)₃C–OH + conc. H₂SO₄ at 413 K
  • (d) (CH₃)₃C–OH + CH₃OH at 443 K / H⁺
(b). Alkoxide on 3° carbon + 1° halide = clean SN2. (a) would give mostly alkene via E2.

3. True/False: The cleavage of anisole by HI at 373 K gives iodobenzene and methanol.

False. It gives phenol and CH₃I because the C(aryl)–O bond is not broken; only the alkyl–O bond cleaves.

4. Fill in the blank: Fermentation of glucose is catalysed by the enzyme ________ in yeast.

Zymase.

5. Why do ethers have lower boiling points than alcohols of comparable molar mass but higher solubilities in water than alkanes?

Ethers lack O–H, so they cannot hydrogen-bond to each other — lower b.p. than alcohols. But the ether oxygen still carries lone pairs that accept H-bonds from water — higher water solubility than alkanes.

Assertion–Reason Questions

Assertion (A): Williamson synthesis fails when the alkyl halide is tertiary.

Reason (R): A tertiary carbon is too hindered for SN2 attack; alkoxide acts as a base and elimination (E2) dominates.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Steric hindrance at the 3° carbon plus basicity of the alkoxide shift the outcome to E2.

Assertion (A): In the HI cleavage of an aryl alkyl ether, phenol (not aryl iodide) is always formed.

Reason (R): The C(sp²)–O bond has partial double-bond character due to resonance and resists both SN1 and SN2 cleavage.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Only the alkyl–O bond is available for cleavage.

Assertion (A): Diethyl ether is a useful solvent for Grignard reactions.

Reason (R): The ether oxygen coordinates the magnesium atom of RMgX and stabilises the reagent without reacting with the carbanion.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. Donation of the ether lone pair to Mg is the key — ether has no acidic H for R⁻ to pick off.
Next: Part 4 is the dedicated exercises file — all 33 NCERT end-of-chapter problems with full solutions.

Frequently Asked Questions - NCERT Exercises and Solutions: Alcohols, Phenols and Ethers

What are the key NCERT exercise types in Chapter 7 Alcohols, Phenols and Ethers?
NCERT Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in Alcohols, Phenols and Ethers?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 7?
From NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 7: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 7 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 7 (Alcohols, Phenols and Ethers) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 7 Alcohols, Phenols and Ethers solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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