Chapter 16: Alcohols, Phenols and Ethers

Introduction

Alcohols and Phenols

• Alcohols: Hydroxyl (—OH) group replaces hydrogen in aliphatic hydrocarbons.
• Phenols: Hydroxyl (—OH) group directly attached to aromatic benzene ring (C₆H₅OH).
• Simple phenol (C₆H₅OH) is also called carbolic acid.

Ethers

• Ethers: Alkoxy (R–O) or aryloxy (Ar–O) group replaces hydrogen in hydrocarbons.
• General formula: C_nH_2n+2O, where n > 1.

Solved Examples

• Ex 1: Identify the compound type: C₂H₅OH.
  Answer: Alcohol (ethyl alcohol) with one —OH group attached to aliphatic carbon.
• Ex 2: Identify C₆H₅OH.
  Answer: Phenol, —OH attached to benzene ring.
• Ex 3: Identify C₂H₅OCH₃.
  Answer: Ether (unsymmetrical), alkoxy group between two different alkyl groups.

Practice Set

• Q1: Which compound is a phenol?
  A. C₂H₅OH
  B. C₆H₅OH
  C. C₂H₅OCH₃
  D. CH₃OH
• Q2: General formula of ethers is:
  A. C_nH_2n+1OH
  B. C_nH_2nO
  C. C_nH_2n+2O
  D. C_nH_2n+1O
• Q3: Which compound is an ether?
  A. C₆H₅OH
  B. C₂H₅OH
  C. C₂H₅OCH₃
  D. CH₃OH

Answer Key

• Q1: B - Phenol has —OH attached to benzene ring.
• Q2: C - Ethers have formula C_nH_2n+2O.
• Q3: C - C₂H₅OCH₃ is an ether with two different alkyl groups.

Memory Aids

• Alcohol = Hydroxyl (—OH) on aliphatic carbon.
• Phenol = Hydroxyl (—OH) on benzene ring.
• Ethers = Oxygen linking two alkyl/aryl groups.

Classification of Alcohols, Phenols and Ethers

Alcohols

• Classified by number of —OH groups: Monohydric (1 —OH), Polyhydric (multiple —OH).
• Monohydric alcohols classified by carbon attachment:
  • Primary (1°): —OH on carbon attached to 1 other carbon (e.g., ethyl alcohol).
  • Secondary (2°): —OH on carbon attached to 2 other carbons (e.g., isopropyl alcohol).
  • Tertiary (3°): —OH on carbon attached to 3 other carbons (e.g., tert-butyl alcohol).
• Special monohydric types:
  • Vinylic alcohol (unstable)
  • Allylic alcohol (—OH near double bond)
  • Benzylic alcohol (—OH attached to benzene ring carbon)
• Polyhydric alcohols:
  • Dihydric (2 —OH), e.g., ethylene glycol
  • Trihydric (3 —OH), e.g., glycerol
  • Hexahydric (6 —OH), e.g., sorbitol

Phenols

• Phenols have —OH directly attached to benzene ring.
• Classified by number of —OH groups:
  • Monohydric phenols: 1 —OH (e.g., phenol, cresol)
  • Dihydric phenols: 2 —OH groups
  • Trihydric phenols: 3 —OH groups

Ethers

• Classified by alkyl/aryl groups attached to oxygen:
• Simple/Symmetrical: Both groups same (e.g., diethyl ether C₂H₅OC₂H₅).
• Mixed/Unsymmetrical: Different groups (e.g., C₂H₅OCH₃).

Solved Examples

• Ex 1: Classify CH₃CH₂OH.
  Answer: Monohydric primary alcohol.
• Ex 2: Classify C₆H₄(OH)₂.
  Answer: Dihydric phenol.
• Ex 3: Classify C₂H₅OC₆H₅.
  Answer: Unsymmetrical ether.

Practice Set

• Q1: Which is a tertiary alcohol?
  A. CH₃CH₂OH
  B. (CH₃)₃COH
  C. CH₃CHOHCH₃
  D. C₆H₅OH
• Q2: Which phenol has two —OH groups?
  A. Phenol
  B. Catechol
  C. Cresol
  D. Resorcinol
• Q3: Which ether is symmetrical?
  A. C₂H₅OC₆H₅
  B. CH₃OCH₃
  C. C₂H₅OCH₃
  D. C₆H₅OCH₃

Answer Key

• Q1: B - Tertiary alcohol has —OH on carbon attached to 3 carbons.
• Q2: B - Catechol is a dihydric phenol with 2 —OH groups.
• Q3: B - CH₃OCH₃ has identical alkyl groups, symmetrical ether.

Memory Aids

• Primary alcohol: —OH on carbon with 1 carbon neighbor.
• Phenols: Number of —OH groups changes reactivity.
• Ethers: Symmetry depends on alkyl/aryl groups.

Nomenclature

Alcohols

• Common system: Named as alkyl alcohol (e.g., methyl alcohol).
• IUPAC system: Named as alkanols by replacing ‘e’ of alkane with ‘-ol’ (e.g., methanol from methane).
• Cyclic alcohols: Use prefix ‘cyclo’ and number —OH group as attached to C-1 (e.g., cyclohexanol).

Solved Examples

• Ex 1: Name CH₃CH₂OH.
  • Step 1: Identify alkane: Ethane.
  • Step 2: Replace ‘e’ with ‘ol’: Ethanol.

Practice Set

• Q1: What is the IUPAC name of CH₃CH₂OH?
  • A. Ethanol
  • B. Ethyl alcohol
  • C. Methanol
  • D. Methyl alcohol
• Q2: How are cyclic alcohols named?
  • A. Using prefix ‘cyclo’ and numbering —OH at C-1
  • B. As alkyl alcohols
  • C. Using ‘diol’ suffix
  • D. Using ‘phenol’
• Q3: Which of the following is a correct IUPAC name?
  • A. Propyl alcohol
  • B. Propanol
  • C. Propanediol
  • D. Cyclopropanol

Answer Key

• Q1: A. Ethanol - IUPAC replaces ‘e’ of ethane with ‘-ol’.
• Q2: A. Using prefix ‘cyclo’ and numbering —OH at C-1 - standard for cyclic alcohols.
• Q3: B. Propanol - correct IUPAC name for 3-carbon alcohol.

Memory Aids

• Alcohols: Alkane ‘e’ → ‘ol’
• Cyclic alcohols: ‘cyclo’ + —OH at C-1

Phenols

Phenols

• Phenol: Hydroxy derivative of benzene, common and IUPAC name.
• Substituted phenols use ortho (1,2-), meta (1,3-), para (1,4-) for positions.
• Dihydroxy benzenes named as 1,2-, 1,3-, 1,4-benzenediol.

Solved Examples

• Ex 1: Name 1,4-dihydroxybenzene.
  • Step 1: Identify benzene ring with two —OH groups at 1 and 4.
  • Step 2: Name as 1,4-benzenediol.

Practice Set

• Q1: What is the common name of hydroxybenzene?
  • A. Phenol
  • B. Benzene diol
  • C. Cyclohexanol
  • D. Benzenol
• Q2: What does ‘ortho’ indicate in substituted phenols?
  • A. 1,2-disubstituted positions
  • B. 1,3-disubstituted positions
  • C. 1,4-disubstituted positions
  • D. No substitution
• Q3: How are dihydroxy derivatives of benzene named?
  • A. Benzenediol with position numbers
  • B. Alkyl alcohols
  • C. Cycloalkanes
  • D. Phenols without numbers

Answer Key

• Q1: A. Phenol - simplest hydroxybenzene name.
• Q2: A. 1,2-disubstituted positions - ortho means adjacent positions.
• Q3: A. Benzenediol with position numbers - indicates dihydroxy substitution.

Memory Aids

• Phenol = hydroxybenzene
• Ortho = adjacent (1,2-), Meta = 1,3-, Para = 1,4-
• Dihydroxy benzene = benzenediol

Ethers

Ethers

• Common names: Alkyl/aryl groups in alphabetical order + ‘ether’ (e.g., ethyl methyl ether).
• If alkyl groups same, prefix ‘di’ added (e.g., diethyl ether).
• IUPAC: Hydrocarbon derivative with H replaced by —OR or —OAr.
• Parent hydrocarbon chosen as larger R group.

Solved Examples

• Ex 1: Name CH₃OCH₂CH₃.
  • Step 1: Identify alkyl groups: CH₃ (methyl), CH₂CH₃ (ethyl).
  • Step 2: Alphabetical order: ethyl methyl ether.

Practice Set

• Q1: What is the common name of CH₃OCH₃?
  • A. Methyl ether
  • B. Dimethyl ether
  • C. Ethyl methyl ether
  • D. Diethyl ether
• Q2: According to IUPAC, ethers are derivatives where H is replaced by?
  • A. —OH
  • B. —OR or —OAr
  • C. —COOH
  • D. —NH₂
• Q3: How is the parent hydrocarbon chosen in IUPAC naming of ethers?
  • A. Smaller R group
  • B. Larger R group
  • C. Alphabetically first group
  • D. Based on molecular weight

Answer Key

• Q1: B. Dimethyl ether - both alkyl groups same, prefix ‘di’ used.
• Q2: B. —OR or —OAr - ethers have H replaced by these groups.
• Q3: B. Larger R group - chosen as parent hydrocarbon.

Memory Aids

• Ethers: Alphabetical alkyl/aryl + ‘ether’
• Same alkyl groups: prefix ‘di’
• IUPAC: H replaced by —OR or —OAr

Structures of Functional Groups

Alcohols

• O atom in hydroxyl (–OH) group attached to C-atom by sigma (σ) bond.
• Bond formed by overlap of sp³ hybridised orbital of C with sp³ hybridised orbital of O.

Phenols

• –OH group attached to sp² hybridised C-atom of aromatic ring.

Ethers

• O atom has four electron pairs: two bond pairs and two lone pairs.
• Electron pairs arranged approximately tetrahedrally around O atom.

Solved Examples

• Ex 1: Identify hybridisation of O in methanol.
  Solution: O in methanol is sp³ hybridised due to tetrahedral electron pair arrangement.
• Ex 2: Compare bond angles in phenol and methoxymethane.
  Solution: Phenol O–C bond angle ~109°, methoxymethane O–C–C bond angle ~111.7° due to different hybridisation and bonding.

Practice Set

• Q1: What is the hybridisation of oxygen in alcohols?
  A. sp²
  B. sp³
  C. sp
  D. dsp²
• Q2: In phenol, the –OH group is attached to which type of carbon?
  A. sp³ carbon
  B. sp² carbon
  C. sp carbon
  D. None of these
• Q3: Which arrangement best describes electron pairs around oxygen in ethers?
  A. Linear
  B. Trigonal planar
  C. Tetrahedral
  D. Octahedral

Answer Key

• Q1: B. sp³ – Oxygen in alcohols is sp³ hybridised due to tetrahedral electron pair geometry.
• Q2: B. sp² – Phenol’s –OH is attached to sp² hybridised carbon of aromatic ring.
• Q3: C. Tetrahedral – Oxygen in ethers has two bond pairs and two lone pairs arranged tetrahedrally.

Memory Aids

• Alcohol O: sp³ hybridised, sigma bond with C.
• Phenol O: attached to sp² aromatic C.
• Ethers: O electron pairs tetrahedral (2 bond + 2 lone pairs).

Alcohols and Phenols

Introduction

• Alcohols contain one or more hydroxyl (-OH) groups attached to saturated carbon atoms.
• Phenols have hydroxyl groups attached directly to an aromatic benzene ring.
• Both classes show distinct physical and chemical properties due to their structure.

Solved Examples

• Ex 1: Identify the functional group in ethanol and phenol.
  Answer: Ethanol - hydroxyl group on alkane; Phenol - hydroxyl group on benzene ring.

Practice Set

• Q1: Which group is present in phenol?
  A. Alkyl group
  B. Hydroxyl attached to benzene
  C. Carbonyl group
  D. Carboxyl group
• Q2: Alcohols differ from phenols in that alcohols have -OH attached to?
  A. Aromatic ring
  B. Saturated carbon
  C. Unsaturated carbon
  D. Nitrogen atom
• Q3: Which compound is a phenol?
  A. CH3CH2OH
  B. C6H5OH
  C. CH3COOH
  D. C6H5CH3

Answer Key

• Q1: B - Phenol has hydroxyl group attached to benzene ring.
• Q2: B - Alcohols have -OH on saturated carbon atoms.
• Q3: B - C6H5OH is phenol.

Memory Aids

• Phenol = benzene + OH
• Alcohol = alkane + OH

Preparation of Alcohols

From Alkenes

• Acid catalysed hydration: Alkene + H₂O (acid catalyst) → Alcohol
• Markovnikov's rule: -OH attaches to carbon with fewer H atoms in unsymmetrical alkenes.
• Hydroboration-oxidation: Alkene + B₂H₆ → trialkylborane → oxidation with H₂O₂/OH^- → alcohol (anti-Markovnikov product)

From Carbonyl Compounds

• Reduction of aldehydes and ketones: Aldehydes → primary alcohols; Ketones → secondary alcohols using H₂/metal or NaBH₄/LiAlH₄.
• Reduction of carboxylic acids and esters: LiAlH₄ reduces acids/esters to primary alcohols.
• From Grignard reagents: Aldehydes/ketones + R-MgX → adduct → hydrolysis → alcohols (1°, 2°, 3° depending on substrate).

Solved Examples

• Ex 1: Acid catalysed hydration of propene gives propan-2-ol (Markovnikov addition).
• Ex 2: Hydroboration-oxidation of propene gives propan-1-ol (anti-Markovnikov addition).
• Ex 3: Reduction of acetaldehyde with NaBH₄ gives ethanol (primary alcohol).

Practice Set

• Q1: Acid catalysed hydration of ethene produces?
  A. Ethanol
  B. Ethane
  C. Ethanal
  D. Ethyne
• Q2: Hydroboration-oxidation of propene yields?
  A. Propan-2-ol
  B. Propan-1-ol
  C. Propene oxide
  D. Propanal
• Q3: Grignard reagent reaction with formaldehyde followed by hydrolysis gives?
  A. Primary alcohol
  B. Secondary alcohol
  C. Tertiary alcohol
  D. Ketone

Answer Key

• Q1: A - Ethene + H₂O (acid) → Ethanol.
• Q2: B - Hydroboration-oxidation adds -OH anti-Markovnikov.
• Q3: A - Formaldehyde + R-MgX → primary alcohol after hydrolysis.

Memory Aids

• Markovnikov: -OH to carbon with fewer H.
• Hydroboration: anti-Markovnikov addition.
• Grignard + formaldehyde → 1° alcohol.

Preparation of Phenol

From Haloarenes (Dow's Process)

• Chlorobenzene + NaOH (623 K, 300 atm) → sodium phenoxide → acidification → phenol.

From Benzenesulphonic Acid

• Benzenesulphonic acid + molten NaOH → sodium phenoxide → acidification → phenol.
• Benzene + oleum → benzenesulphonic acid.

From Diazonium Salts

• Benzene diazonium chloride + warm water/dilute acid → phenol + N₂ + HCl.
• Preparation: Benzene → nitration → nitrobenzene → reduction → aniline → diazotisation → diazonium salt.

From Cumene (Industrial Method)

• Cumene + O₂ → cumene hydroperoxide → acid treatment → phenol + acetone.
• Cumene prepared by Friedel-Crafts alkylation of benzene with propyl chloride or propene.

Solved Examples

• Ex 1: Dow's process converts chlorobenzene to phenol using NaOH at high T and P.
• Ex 2: Diazonium salt hydrolysis produces phenol and nitrogen gas.

Practice Set

• Q1: Dow's process uses which reagent?
  A. NaCl
  B. NaOH
  C. HCl
  D. H₂SO₄
• Q2: Benzene diazonium chloride on hydrolysis gives?
  A. Aniline
  B. Phenol
  C. Nitrobenzene
  D. Benzene
• Q3: Industrial phenol production uses?
  A. Benzene sulfonation
  B. Cumene oxidation
  C. Diazotisation
  D. Friedel-Crafts alkylation

Answer Key

• Q1: B - NaOH used in Dow's process.
• Q2: B - Hydrolysis of diazonium salt yields phenol.
• Q3: B - Cumene oxidation is industrial method.

Memory Aids

• Dow's: Haloarene + NaOH → phenol.
• Diazonium salt → phenol + N₂.
• Cumene → phenol + acetone industrially.

Physical Properties

Alcohols

• Lower alcohols: colourless liquids with characteristic smell.
• Higher alcohols: colourless, odourless waxy solids.
• Solubility: Decreases with alkyl size; increases with branching due to hydrogen bonding.
• Boiling points: Increase with carbon number; decrease with branching (van der Waals forces).

Phenols

• Physical state: Colourless crystalline solids or liquids with phenolic odour.
• Boiling points: Higher than aromatic hydrocarbons due to intermolecular hydrogen bonding.
• Solubility: Less soluble than alcohols due to large hydrophobic benzene ring.

Solved Examples

• Ex 1: Explain why phenol has higher boiling point than benzene.
  Answer: Due to intermolecular hydrogen bonding.

Practice Set

• Q1: Lower alcohols are usually?
  A. Solids
  B. Liquids
  C. Gases
  D. Waxy solids
• Q2: Phenols have higher boiling points than?
  A. Alcohols
  B. Aromatic hydrocarbons
  C. Alkanes
  D. Ethers
• Q3: Solubility of alcohols in water decreases with?
  A. Branching
  B. Increase in alkyl size
  C. Decrease in alkyl size
  D. None

Answer Key

• Q1: B - Lower alcohols are liquids.
• Q2: B - Phenols have higher boiling points than aromatic hydrocarbons.
• Q3: B - Solubility decreases with larger alkyl groups.

Memory Aids

• Hydrogen bonding raises boiling points.
• Branching increases solubility.

Chemical Properties of Alcohols

General

• Alcohols act as nucleophiles (O-H bond cleavage) and electrophiles (C-O bond cleavage).
• Acidity order: 1° > 2° > 3° alcohols.
• Electron releasing groups decrease acidity.
• Reactivity order for C-O bond cleavage: 3° > 2° > 1°.
• 3° alcohols resist oxidation in neutral/alkaline but oxidize in acidic medium forming carboxylic acids.
• Dehydration follows Saytzeff rule: more substituted alkene is major product.
• Ease of dehydration: 3° > 2° > 1° (carbocation stability).

Tests to Distinguish Alcohols

• Victor Meyer test: Color changes identify 1°, 2°, 3° alcohols.
• Lucas test: Cloudiness time indicates alcohol type (3° immediate, 2° in 5-10 min, 1° no cloudiness).

Solved Examples

• Ex 1: Lucas test cloudiness order: 3° > 2° > 1° alcohol.
• Ex 2: Victor Meyer test color for 1° alcohol is blood red.

Practice Set

• Q1: Order of acidity in alcohols?
  A. 3° > 2° > 1°
  B. 1° > 2° > 3°
  C. 2° > 1° > 3°
  D. Equal
• Q2: Lucas test shows immediate cloudiness for?
  A. 1° alcohol
  B. 2° alcohol
  C. 3° alcohol
  D. Phenol
• Q3: Dehydration of alcohols follows?
  A. Markovnikov rule
  B. Saytzeff rule
  C. Anti-Markovnikov rule
  D. None

Answer Key

• Q1: B - Acidity decreases from 1° to 3° due to electron releasing groups.
• Q2: C - 3° alcohols react fastest in Lucas test.
• Q3: B - Dehydration follows Saytzeff rule.

Memory Aids

• Saytzeff: More substituted alkene formed.
• Lucas: Cloudiness time indicates alcohol type.
• Victor Meyer: Color test for alcohol types.

Chemical Properties of Phenols

Acidic Nature

• Phenols are more acidic than alcohols due to:
• (i) -OH attached to electronegative sp² carbon of benzene, increasing O-H bond polarity.
• (ii) Phenoxide ion stabilized by resonance, unlike phenol.
• Electron withdrawing groups increase acidity; electron releasing groups decrease acidity.

Chemical Reactions

• Bromination: Phenol + Br₂ → 2,4,6-tribromophenol (white precipitate).
• Nitration: Phenol + HNO₃ → nitrophenols, picric acid.
• Sulfonation: Phenol + H₂SO₄ → phenol sulfonic acid.
• Neutralization: Phenol + Na/NaOH → sodium phenoxide.
• Oxidation: Phenol + Na₂Cr₂O₇/H₂SO₄ → p-benzoquinone.
• Kolbe’s reaction: Sodium phenoxide + CO₂ → salicylic acid.
• Hydrogenation: Phenol + H₂/Ni → cyclohexanol.
• Reimer-Tiemann: Sodium phenoxide + CHCl₃/NaOH → salicylaldehyde.
• Alkylation: Phenol + CH₃Cl/AlCl₃ → cresols.

Identification Tests

• Ferric chloride test: Phenol + FeCl₃ → violet color.
• Bromine water test: Phenol + Br₂ water → white precipitate (2,4,6-tribromophenol).

Solved Examples

• Ex 1: Phenol + Br₂ → 2,4,6-tribromophenol (white precipitate).
• Ex 2: Ferric chloride test gives violet color with phenol.

Practice Set

• Q1: Phenol is more acidic than alcohol because?
  A. sp³ carbon
  B. Resonance stabilization
  C. Less electronegative carbon
  D. No resonance
• Q2: Bromine water test with phenol produces?
  A. Yellow solution
  B. White precipitate
  C. No reaction
  D. Red color
• Q3: Kolbe’s reaction produces?
  A. Salicylic acid
  B. Benzoic acid
  C. Phenol
  D. Benzene

Answer Key

• Q1: B - Phenoxide ion resonance stabilizes phenol's acidity.
• Q2: B - 2,4,6-tribromophenol precipitates as white solid.
• Q3: A - Kolbe’s reaction forms salicylic acid.

Memory Aids

• Phenol acidity due to resonance and sp² carbon.
• Bromination gives white precipitate.
• Kolbe’s reaction forms salicylic acid.

Some Commercially Important Alcohols

Methanol Production

• Methanol (CH₃OH) is produced by catalytic hydrogenation of carbon monoxide (CO) using hydrogen gas (H₂).
• Catalyst: Zinc oxide-chromium oxide (ZnO-Cr₂O₃).
• Conditions: High temperature (573-673 K) and high pressure (200-300 atm).
• Reaction: CO + 2H₂ → CH₃OH (methanol).

Methanol Properties and Uses

• Colourless liquid, highly poisonous.
• Small ingestion causes blindness; large ingestion causes death.
• Used as solvent in paints, varnishes, and for making formaldehyde.

Ethanol Production

• Produced by fermentation of sugars in molasses, sugarcane, or fruits (e.g., grapes).
• Enzymes involved: Invertase breaks sucrose into glucose and fructose; zymase converts glucose into ethanol and CO₂.
• Reaction: C₁₂H₂₂O₁₁ + H₂O → glucose + fructose → ethanol + CO₂.

Ethanol Properties and Uses

• Colourless liquid used as solvent in paint industry and preparation of carbon compounds.
• Commercial alcohol is denatured by adding copper sulphate and pyridine to make it unfit for drinking.

Ethylene Glycol Uses

• Used as antifreeze in car radiators.
• 50% aqueous solution freezes at -34°C.
• Used as de-icing fluid for airplane wings.

Types of Alcohol

• Absolute alcohol: 100% ethanol.
• Rectified spirit: 95% alcohol + 5% water.
• Denatured spirit: 95% ethanol + 5% methanol.
• Methylated spirit: 90% ethanol + 9% methanol + 1% poisonous substances.
• Power alcohol: 80% petrol + 20% ethanol.
• Different types have different uses and purity levels.

Solved Examples

• Ex 1: Write the reaction for methanol production from CO and H₂.
  Answer: CO + 2H₂ 13 ZnO-Cr₂O₃, 573-673 K, 200-300 atm 13 CH₃OH
• Ex 2: Write the fermentation reaction of glucose to ethanol.
  Answer: C₆H₁₂O₆ 13 zymase 13 2C₂H₅OH + 2CO₂

Practice Set

• Q1: Which catalyst is used in methanol production?
  A. Cu-Cr₂O₃
  B. ZnO-Cr₂O₃
  C. Fe₂O₃
  D. Al₂O₃
• Q2: What is the main enzyme that converts glucose to ethanol in fermentation?
  A. Invertase
  B. Zymase
  C. Amylase
  D. Catalase
• Q3: Which type of alcohol is made unfit for drinking by adding copper sulphate and pyridine?
  A. Absolute alcohol
  B. Rectified spirit
  C. Denatured alcohol
  D. Methylated spirit

Answer Key

• Q1: B - ZnO-Cr₂O₃ is the catalyst used for methanol production.
• Q2: B - Zymase converts glucose into ethanol and CO₂ during fermentation.
• Q3: C - Denatured alcohol is made unfit for drinking by adding copper sulphate and pyridine.

Memory Aids

• Methanol production catalyst: ZnO-Cr₂O₃ (Zinc-Chrome combo).
• Fermentation enzymes: Invertase breaks sucrose; Zymase forms ethanol.
• Denatured alcohol = ethanol + copper sulphate + pyridine (unfit for drinking).

Preparation from Alcohols (by Dehydration)

Explanation

• Primary alcohols heated with protic acid (H₂SO₄) at 413 K undergo dehydration to form ethers.
• At higher temperature (443 K), the same reaction forms alkenes or alcohols instead.
• Suitable only for simple ethers with primary, unhindered alkyl groups at low temperature.
• Mechanism: Acid-catalyzed nucleophilic bimolecular substitution (S_N2).
• At low temperature, substitution (ether formation) dominates; at high temperature, elimination (alkene formation) dominates.
• Secondary and tertiary alcohols mainly undergo elimination to form alkenes, not ethers.

Solved Examples

• Ex 1: Ethanol + H₂SO₄ at 413 K → Diethyl ether + H₂O
• Ex 2: Ethanol + H₂SO₄ at 443 K → Ethene + H₂O

Practice Set

• Q1: Which alcohol gives ether on dehydration at low temperature?
  A. Primary
  B. Secondary
  C. Tertiary
  D. All
• Q2: What is the major product when ethanol is heated with H₂SO₄ at 443 K?
  A. Diethyl ether
  B. Ethanol
  C. Ethene
  D. Water
• Q3: The mechanism of ether formation from alcohol dehydration is:
  A. S_N1
  B. S_N2
  C. E1
  D. E2

Answer Key

• Q1: A - Primary alcohols form ethers at low temperature by substitution.
• Q2: C - Ethene is formed at higher temperature by elimination.
• Q3: B - Ether formation follows S_N2 mechanism.

Memory Aids

• Primary alcohol + H₂SO₄ at low temp → ether (S_N2)
• At high temp → alkene (elimination)
• Secondary/tertiary alcohols → alkene only

Preparation from Alkyl Halides (Williamson Synthesis)

Explanation

• Primary alkyl halide reacts with sodium alkoxide to form ether via S_N2 mechanism.
• R and R' groups can be same or different, allowing preparation of simple or mixed ethers.
• Secondary and tertiary alkyl halides are unsuitable due to elimination competing with substitution.
• Alkoxide ion is a stronger base than nucleophile, causing elimination in secondary/tertiary halides.
• Phenols react with NaOH to form phenoxide ion, which then reacts with alkyl halides to form aryl ethers.

Solved Examples

• Ex 1: CH₃CH₂Cl + CH₃ONa → CH₃CH₂OCH₃ + NaCl
• Ex 2: C₆H₅OH + NaOH → C₆H₅O^-Na⁺; then C₆H₅O^-Na⁺ + R-X → C₆H₅-O-R + NaX

Practice Set

• Q1: Williamson synthesis is suitable for:
  A. Primary alkyl halides only
  B. Secondary alkyl halides only
  C. Tertiary alkyl halides only
  D. Primary and secondary alkyl halides
• Q2: Phenol reacts with NaOH to form:
  A. Phenol sodium salt
  B. Sodium phenoxide
  C. Sodium hydroxide
  D. Phenyl halide
• Q3: In Williamson synthesis, elimination occurs mainly with:
  A. Primary alkyl halides
  B. Secondary alkyl halides
  C. Tertiary alkyl halides
  D. Aryl halides

Answer Key

• Q1: A - Only primary alkyl halides give ethers via S_N2.
• Q2: B - Phenol + NaOH forms sodium phenoxide.
• Q3: C - Tertiary alkyl halides favor elimination, no ether formed.

Memory Aids

• Williamson synthesis: Primary alkyl halide + sodium alkoxide → ether (S_N2)
• Secondary/tertiary alkyl halides → elimination (alkene)
• Phenol + NaOH → sodium phenoxide → reacts with alkyl halide to form aryl ether

Physical Properties

Explanation

• Dimethyl and diethyl ethers are gases at room temperature; others are colorless liquids with ethereal smell.
• Ethers have bent structure, giving them polarity (net dipole moment).
• Boiling points are lower than alcohols due to absence of hydrogen bonding.
• Soluble in water to some extent due to hydrogen bonding; solubility decreases with molecular mass.
• Soluble in organic solvents like alcohol, chloroform, benzene.
• Density less than water (lighter than water).

Solved Examples

• Ex 1: Diethyl ether is liquid at room temperature with characteristic smell.
• Ex 2: Boiling point of ether < alcohol of similar molecular weight.

Practice Set

• Q1: Which ether is gaseous at room temperature?
  A. Diethyl ether
  B. Dimethyl ether
  C. Dipropyl ether
  D. Di-n-butyl ether
• Q2: Ethers are polar because:
  A. They have hydrogen bonding
  B. They have bent structure
  C. They are ionic
  D. They have linear structure
• Q3: Ethers have lower boiling points than alcohols because:
  A. They have weaker intermolecular forces
  B. They have hydrogen bonding
  C. They are ionic
  D. They have higher molecular weight

Answer Key

• Q1: B - Dimethyl ether is gaseous at room temperature.
• Q2: B - Bent structure causes net dipole moment.
• Q3: A - Absence of hydrogen bonding lowers boiling point.

Memory Aids

• Small ethers (dimethyl, diethyl) are gases; others liquids.
• Bent structure → polarity.
• No hydrogen bonding → lower boiling points than alcohols.

Chemical Reactions

Explanation

• Ethers react with hydrogen halides (HX) to cleave C–O bond, forming alcohol and alkyl halide.
• Reaction follows S_N2 mechanism for primary and secondary alkyl groups.
• Reactivity order of HX: HCl < HBr < HI (strongest acid).
• Alkyl aryl ethers cleave at R–O bond; Ar–O bond is more stable due to partial double bond character.
• In mixed ethers, smaller alkyl group forms alkyl halide if 1° or 2°; tertiary alkyl halide forms via S_N1 mechanism.
• Anisole forms methylphenyl oxonium ion on protonation; O–CH₃ bond breaks preferentially.
• Phenols do not form halides due to sp² hybridized carbon preventing nucleophilic substitution.
• Electrophilic substitution on aromatic ethers: OR group activates ring, directs ortho/para; para product major.

Solved Examples

• Ex 1: C₂H₅-O-C₂H₅ + HI → C₂H₅I + C₂H₅OH
• Ex 2: Mixed ether cleavage: smaller alkyl group forms alkyl halide.
• Ex 3: Tertiary alkyl halide formed via S_N1 mechanism in ether cleavage.

Practice Set

• Q1: Order of reactivity of hydrogen halides in ether cleavage is:
  A. HI < HBr < HCl
  B. HCl < HBr < HI
  C. HBr < HI < HCl
  D. HCl < HI < HBr
• Q2: In ether cleavage, which bond breaks in alkyl aryl ethers?
  A. Ar–O
  B. R–O
  C. C–C
  D. C–H
• Q3: In mixed ethers, which alkyl halide forms when one group is tertiary?
  A. Primary alkyl halide
  B. Secondary alkyl halide
  C. Tertiary alkyl halide
  D. No alkyl halide

Answer Key

• Q1: B - Reactivity increases from HCl to HI.
• Q2: B - R–O bond breaks; Ar–O bond is stable.
• Q3: C - Tertiary alkyl halide forms via S_N1.

Memory Aids

• HX reactivity: HCl < HBr < HI
• Alkyl aryl ethers break R–O bond, not Ar–O
• Mixed ethers: smaller alkyl halide forms unless tertiary group present
• Tertiary alkyl halide forms via S_N1

Crown Ethers

Explanation

• Cyclic polyethers with at least four -OCH₂CH₂- units.
• Named crown ethers due to crown-like ring shape.
• Example: 18-Crown-6 (18 atoms in ring, 6 oxygen atoms).
• Can complex metal ions inside the ring, useful in ion transport and phase transfer catalysis.

Solved Examples

• Ex 1: 18-Crown-6 binds K⁺ ion selectively.
• Ex 2: 12-Crown-4 binds smaller ions like Na⁺.

Practice Set

• Q1: Crown ethers are named based on:
  A. Number of carbon atoms
  B. Number of oxygen atoms
  C. Ring size and number of oxygen atoms
  D. Number of hydrogen atoms
• Q2: 18-Crown-6 has:
  A. 18 oxygen atoms
  B. 6 carbon atoms
  C. 18 atoms in ring and 6 oxygen atoms
  D. 6 atoms in ring and 18 oxygen atoms
• Q3: Crown ethers are useful because they:
  A. React with metals
  B. Bind metal ions inside ring
  C. Are acidic
  D. Are gases

Answer Key

• Q1: C - Named by ring size and oxygen count.
• Q2: C - 18 atoms total, 6 oxygen atoms.
• Q3: B - Bind metal ions inside ring.

Memory Aids

• Crown ethers = cyclic polyethers with repeating -OCH₂CH₂- units
• 18-Crown-6 = 18 atoms ring, 6 oxygen atoms
• Bind metal ions selectively