Alcohols, Phenols & Ethers | Class 12 Unit 7

7.1 Classification

Three Key Classes

Alcohol: –OH group directly attached to an sp³ hybridised carbon of an aliphatic system. Example: CH₃OH (Methanol)

Phenol: –OH group directly attached to an sp² hybridised carbon of an aromatic ring. Example: C₆H₅OH (Phenol)

Ether: Oxygen atom bonded to two alkyl or aryl groups (R–O–R', Ar–O–R, Ar–O–Ar). Example: CH₃OCH₃ (Dimethyl ether)

7.1.1 Classification of Alcohols

By number of –OH groups:

Monohydric

C₂H₅OH

One –OH group

Dihydric

HOCH₂CH₂OH

Ethylene glycol — Two –OH groups

Trihydric

HOCH₂CH(OH)CH₂OH

Glycerol — Three –OH groups

Monohydric alcohols further classified by hybridisation of C bearing –OH:

sp³ C–OH Alcohols (Most Important)

Primary (1°)

–OH on a carbon bonded to only one other carbon. Example: CH₃CH₂OH (Ethanol), CH₂=CHCH₂OH (Allyl alcohol — also allylic)

Secondary (2°)

–OH on a carbon bonded to two other carbons. Example: CH₃CH(OH)CH₃ (Propan-2-ol)

Tertiary (3°)

–OH on a carbon bonded to three other carbons. Example: (CH₃)₃COH (2-Methylpropan-2-ol / tert-Butyl alcohol)

Allylic alcohols: –OH on an sp³ carbon adjacent to C=C. Can be 1°, 2°, or 3°.

Benzylic alcohols: –OH on an sp³ carbon adjacent to an aromatic ring. Example: C₆H₅CH₂OH (Benzyl alcohol — primary benzylic)

Vinylic alcohols (sp² C–OH): –OH directly on a C=C carbon. Example: CH₂=CH–OH. These are unstable and tautomerise to carbonyl compounds.

7.1.2 Classification of Phenols

Classified by number of –OH groups on the aromatic ring: monohydric (phenol, cresols), dihydric (catechol, resorcinol, hydroquinone), trihydric (pyrogallol).

7.1.3 Classification of Ethers

Simple / Symmetrical

Both groups on oxygen are the same. Example: C₂H₅–O–C₂H₅ (Diethyl ether)

Mixed / Unsymmetrical

Two different groups on oxygen. Example: CH₃–O–C₂H₅ (Ethyl methyl ether), C₂H₅–O–C₆H₅ (Ethyl phenyl ether)

7.2 Nomenclature

(a) Alcohols

IUPAC Rules for Alcohols

Select the longest carbon chain containing the –OH group as the parent alkane.
Replace the terminal 'e' of alkane with 'ol'. (e.g., ethane → ethanol)
Number the chain from the end nearest to the –OH group to give it the lowest locant.
For polyhydric alcohols: retain the 'e' of alkane and add 'diol', 'triol', etc. (e.g., ethane-1,2-diol)
For cyclic alcohols: use prefix 'cyclo', –OH attached to C-1.

Common and IUPAC Names — Master Table

Structure	Common Name	IUPAC Name	Type
CH₃OH	Methyl alcohol / Wood spirit	Methanol	1°
CH₃CH₂OH	Ethyl alcohol / Grain alcohol	Ethanol	1°
CH₃CH₂CH₂OH	n-Propyl alcohol	Propan-1-ol	1°
CH₃CH(OH)CH₃	Isopropyl alcohol	Propan-2-ol	2°
CH₃CH₂CH₂CH₂OH	n-Butyl alcohol	Butan-1-ol	1°
CH₃CH₂CH(OH)CH₃	sec-Butyl alcohol	Butan-2-ol	2°
(CH₃)₂CHCH₂OH	Isobutyl alcohol	2-Methylpropan-1-ol	1°
(CH₃)₃COH	tert-Butyl alcohol	2-Methylpropan-2-ol	3°
HOCH₂CH₂OH	Ethylene glycol	Ethane-1,2-diol	Dihydric
HOCH₂CH(OH)CH₂OH	Glycerol	Propane-1,2,3-triol	Trihydric

(b) Phenols

The simplest hydroxy derivative of benzene is phenol — both common and accepted IUPAC name. Substituted phenols use ortho/meta/para (common) or numerical locants (IUPAC).

Phenol

C₆H₅OH

o-Cresol

2-Methylphenol

Catechol

Benzene-1,2-diol

Hydroquinone

Benzene-1,4-diol (Quinol)

(c) Ethers

Common names: Name both alkyl/aryl groups alphabetically + "ether". Example: CH₃OC₂H₅ = Ethyl methyl ether.

IUPAC names: Larger group = parent hydrocarbon; smaller group = alkoxy (–OR) or aryloxy (–OAr) substituent. Example: CH₃OC₂H₅ = Methoxyethane.

Structure	Common Name	IUPAC Name
CH₃OCH₃	Dimethyl ether	Methoxymethane
C₂H₅OC₂H₅	Diethyl ether	Ethoxyethane
C₆H₅OCH₃	Methyl phenyl ether (Anisole)	Methoxybenzene
C₆H₅OC₂H₅	Ethyl phenyl ether (Phenetole)	Ethoxybenzene
CH₃OCH₂CH₂CH₃	Methyl n-propyl ether	1-Methoxypropane

7.3 Structures of Functional Groups

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Bond Angles and Lengths — Methanol, Phenol, Methoxymethane

NCERT p.198 (Fig. 7.1)

Shows bond lengths and angles: Methanol (C–O = 142 pm, O–H = 96 pm, angle = 108.9°); Phenol (C–O = 136 pm, O–C–H angle = 109°); Methoxymethane (C–O = 141 pm, C–O–C = 111.7°)

AI Image Prompt

Professional chemistry diagram showing molecular structures of three compounds side by side: (1) Methanol with bond lengths C–O=142pm, O–H=96pm and bond angle 108.9°, (2) Phenol with benzene ring showing C–O=136pm and O–H angle=109°, (3) Methoxymethane with C–O=141pm and C–O–C angle=111.7°. Show lone pairs on oxygen atoms as dots. Fully white background, clean minimalistic style, labeled bond lengths and angles, no background patterns.

Fig. 7.1 — Bond parameters in methanol, phenol and methoxymethane. Note phenol has the shortest C–O bond (136 pm) due to partial double bond character.

Bond Structure Analysis

Alcohols: O is sp³ hybridised. Bond angle slightly less than 109.5° (actual ~108.9°) due to lone pair–lone pair repulsion on oxygen compressing the C–O–H angle.
Phenol: O–H attached to sp² carbon. C–O bond = 136 pm (shorter than in methanol 142 pm). Reason: (i) partial double bond character due to conjugation of O's lone pair with the aromatic ring; (ii) higher electronegativity of sp² carbon holds bonding electrons tighter.
Ethers: O has two bond pairs + two lone pairs, arranged tetrahedrally. Bond angle slightly greater than 109.5° (actual ~111.7° in dimethyl ether) due to repulsion between bulky R groups. C–O bond = 141 pm.

7.4a Preparation of Alcohols

1. From Alkenes

(i) Acid-Catalysed Hydration (Markovnikov Addition)

Alkenes react with water in presence of dilute H₂SO₄ or H₃PO₄. For unsymmetrical alkenes, follows Markovnikov's rule → OH adds to the more substituted carbon (via more stable carbocation).

>C=C< + H₂O ⇌^H⁺ >C–C<
| |
H OH

Example (Markovnikov):
CH₃CH=CH₂ + H₂O —^H⁺→ CH₃CH(OH)CH₃ (major: propan-2-ol)

3-step mechanism: (1) Protonation of alkene → carbocation; (2) Nucleophilic attack of H₂O on carbocation; (3) Deprotonation to give alcohol. Step 2 is rate-determining.

(ii) Hydroboration–Oxidation (Anti-Markovnikov) — H.C. Brown, Nobel 1979

Diborane (BH₃)₂ adds to alkene → trialkyl borane → oxidised with H₂O₂/NaOH → alcohol. Boron attaches to the less substituted sp² carbon (carrying more H atoms) → gives anti-Markovnikov product with excellent yield.

CH₃CH=CH₂ + (H–BH₂)₂ → (CH₃CH₂CH₂)₃B
(CH₃CH₂CH₂)₃B —^{H₂O₂, NaOH}→ 3CH₃CH₂CH₂OH (Propan-1-ol, anti-Markovnikov)

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Key Contrast Acid hydration → Markovnikov → OH on more substituted C → secondary/tertiary alcohol.
Hydroboration–oxidation → Anti-Markovnikov → OH on less substituted C → primary alcohol. No carbocation intermediate → no rearrangement.

2. From Carbonyl Compounds

(i) Reduction of Aldehydes and Ketones

Reagents: H₂/Pd or Pt (catalytic hydrogenation), NaBH₄, or LiAlH₄

RCHO + H₂ —^Pd→ RCH₂OH (1° alcohol)
RCOR' —^NaBH₄→ R–CH(OH)–R' (2° alcohol)

Aldehydes → primary alcohols; Ketones → secondary alcohols

(ii) Reduction of Carboxylic Acids and Esters

Carboxylic acids → reduced by LiAlH₄ (strong; expensive) → primary alcohol

RCOOH —^{(i) LiAlH₄ (ii) H₂O}→ RCH₂OH

Commercial route: RCOOH —^{R'OH, H⁺}→ RCOOR' —^H₂/cat.→ RCH₂OH

3. From Grignard Reagents — KEY SYNTHETIC METHOD

Grignard reagent (RMgX) adds nucleophilically to the carbonyl group of an aldehyde or ketone. Hydrolysis of the adduct gives alcohol. This is an excellent method for constructing new C–C bonds.

Grignard + Carbonyl → Alcohol

Carbonyl Compound	Reaction	Product Type
Methanal (HCHO)	HCHO + RMgX → RCH₂OMgX → ^H₂O RCH₂OH	Primary alcohol (R–CH₂OH)
Other aldehydes (RCHO)	RCHO + R'MgX → R–CH(OMgX)–R' → ^H₂O R–CH(OH)–R'	Secondary alcohol
Ketone (RCOR')	RCOR' + R''MgX → R–C(OMgX)(R')(R'') → ^H₂O R–C(OH)(R')(R'')	Tertiary alcohol

Mechanism: Step 1 — Nucleophilic addition of R⁻ (carbanion character) to δ+ carbonyl carbon → forms adduct with C–O–MgX. Step 2 — Hydrolysis with H₂O → releases alcohol + Mg(OH)X.

📝 Solved Example

(a) Catalytic reduction of butanal: CH₃CH₂CH₂CHO + H₂ →^Pd CH₃CH₂CH₂CH₂OH (Butan-1-ol)

(b) Hydration of propene (dilute H₂SO₄): CH₃CH=CH₂ + H₂O →^H⁺ CH₃CH(OH)CH₃ (Propan-2-ol, Markovnikov)

7.4b Preparation of Phenols

4 Methods of Preparation

1. From Haloarenes (Dow Process)

C₆H₅Cl + NaOH(aq) —^{623 K, 300 atm}→ C₆H₅ONa (sodium phenoxide) —^HCl→ C₆H₅OH (Phenol)
Requires extreme conditions — C–Cl bond in haloarene is very strong (sp² carbon, resonance).

2. From Benzenesulphonic Acid

C₆H₆ —^Oleum→ C₆H₅SO₃H —^{(i) NaOH (molten) (ii) H⁺}→ C₆H₅OH

3. From Diazonium Salts

C₆H₅NH₂ —^{NaNO₂ + HCl, 273–278 K}→ C₆H₅N₂⁺Cl^– —^{H₂O, warm}→ C₆H₅OH + N₂↑ + HCl

4. From Cumene — Industrial Method (Most Important)

Most of the world's phenol is produced by this method. Acetone (CH₃COCH₃) is the valuable by-product.

Step 1 — Oxidation of Cumene (Isopropylbenzene):
C₆H₅–CH(CH₃)₂ —^{O₂ (air)}→ C₆H₅–C(CH₃)₂–O–O–H (Cumene hydroperoxide)

Step 2 — Acid cleavage:
Cumene hydroperoxide —^{H⁺, H₂O (dilute acid)}→ C₆H₅OH + CH₃COCH₃
Phenol + Acetone (both commercially valuable products from one reaction)

7.4c Physical Properties of Alcohols and Phenols

Boiling Points

Why Alcohols Have Abnormally High Boiling Points

The –OH group forms intermolecular hydrogen bonds with other –OH groups. Breaking these H-bonds requires significant energy → high boiling points compared to hydrocarbons, ethers, and haloalkanes of comparable molecular mass.

Compound	Molecular Mass	Boiling Point (K)	Reason
Ethanol (C₂H₅OH)	46	351	Intermolecular H-bonding
Methoxymethane (CH₃OCH₃)	46	248	No H-bonding (no O–H)
Propane (C₃H₈)	44	231	Only van der Waals

Trend in alcohols: bp increases with chain length (↑ van der Waals); bp decreases with branching (↓ surface area → weaker van der Waals).

Solubility in Water

Alcohols and phenols are soluble in water because they form hydrogen bonds with water molecules via their –OH groups. Solubility decreases as the size of the hydrophobic alkyl/aryl group increases. Lower molecular mass alcohols (methanol, ethanol, propanol) are miscible with water in all proportions.

Boiling Point Order (Important for Exams)

Increasing boiling point: Methanol < Ethanol < Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol

For same MW group: n-Butane < Ethoxyethane < Pentanal < Pentan-1-ol (alkane < ether < aldehyde < alcohol)

7.4d Chemical Reactions of Alcohols

Alcohols are versatile — they can act as nucleophiles (O–H bond broken) or as electrophiles (C–O bond broken after protonation). Reaction type depends on conditions.

(a) Reactions Involving Cleavage of O–H Bond

1. Acidity of Alcohols — Reaction with Active Metals

2R–OH + 2Na → 2R–O–Na (Sodium alkoxide) + H₂↑
6(CH₃)₃COH + 2Al → 2[(CH₃)₃CO]₃Al (Aluminium tert-butoxide) + 3H₂↑

Acid strength of alcohols: Primary > Secondary > Tertiary

Reason: More alkyl groups (electron-donating, +I effect) → increase electron density on O → decrease polarity of O–H → harder to lose H⁺. Tertiary alcohols have three +I groups → most electron-rich O → weakest acid.

Alcohols vs Water: Alcohols are weaker acids than water. R–O⁻ + H–OH → R–OH + ⁻OH. Water is a better proton donor. Alkoxides (RO⁻) are stronger bases than hydroxide (HO⁻).

2. Esterification — Reaction with Carboxylic Acids / Acid Chlorides / Anhydrides

R/Ar–OH + R'–COOH ⇌^H⁺ R/Ar–O–CO–R' + H₂O (reversible — remove water)

R/Ar–OH + (R'CO)₂O ⇌^H⁺ R/Ar–O–CO–R' + R'COOH

R/Ar–OH + R'COCl —^pyridine→ R/Ar–O–CO–R' + HCl
Pyridine neutralises HCl, shifting equilibrium right for acid chloride reactions.

Acetylation: Introduction of CH₃CO– (acetyl) group using acetic anhydride or acetyl chloride.

★ Aspirin Synthesis (Acetylation of Salicylic Acid):
Salicylic acid (2-hydroxybenzoic acid) + (CH₃CO)₂O —^H⁺→ Acetylsalicylic acid (Aspirin) + CH₃COOH
Aspirin = analgesic + anti-inflammatory + antipyretic

(b) Reactions Involving Cleavage of C–O Bond

1. Reaction with Hydrogen Halides (→ Alkyl Halides)

R–OH + HX → R–X + H₂O (HI > HBr > HCl in reactivity)

Lucas Test: Conc. HCl + anhydrous ZnCl₂ (Lucas reagent). Distinguishes 1°, 2°, 3° alcohols:

Tertiary alcohols: Turbidity (cloudiness) appears immediately at room temperature.
Secondary alcohols: Turbidity appears after ~5 minutes at room temperature.
Primary alcohols: No turbidity at room temperature (requires heating).

Reason: Tertiary carbocations are most stable → form most easily via SN1 → react fastest. Primary must go through SN2 (slower).

2. Dehydration — Formation of Alkenes (C–O Bond Cleavage)

Alcohols dehydrate (lose H₂O) to form alkenes with conc. H₂SO₄ or H₃PO₄ at high temperature, or with Al₂O₃.

C₂H₅OH —^{conc. H₂SO₄, 443 K}→ CH₂=CH₂ + H₂O
CH₃CHOHCH₃ —^{85% H₃PO₄, 440 K}→ CH₃CH=CH₂ + H₂O
(CH₃)₃COH —^{20% H₃PO₄, 358 K}→ (CH₃)₂C=CH₂ + H₂O

Ease of dehydration: Tertiary > Secondary > Primary — because tertiary carbocations are more stable.

3-step Mechanism (for ethanol → ethene):

Step 1 (Fast): Protonation by H₂SO₄ → formation of ethyl oxonium ion (CH₃CH₂–OH₂⁺)
Step 2 (Slow, rate-determining): Carbocation formation — loss of water → ethyl carbocation (CH₃CH₂⁺)
Step 3 (Fast): Loss of proton from β-carbon → ethene + H⁺ (acid regenerated)

At lower temperature (413 K), ethanol + conc. H₂SO₄ gives ether (ethoxyethane) not alkene — SN2 mechanism.

3. Oxidation — Dehydrogenation Reactions (Most Important)

Oxidation of alcohols involves breaking O–H and C–H bonds to form C=O double bond. Also called dehydrogenation.

Alcohol Type	Oxidising Agent	Product
Primary (RCH₂OH)	Mild: CrO₃ (anhydrous), PCC	Aldehyde (RCHO)
Primary (RCH₂OH)	Strong: KMnO₄/H⁺, K₂Cr₂O₇/H⁺	Carboxylic acid (RCOOH)
Secondary (R–CHOH–R')	CrO₃, K₂Cr₂O₇/H⁺	Ketone (RCOR')
Tertiary (R₃COH)	Mild oxidising agents	No reaction (resistant)
Tertiary (R₃COH)	Strong: KMnO₄, high T	Mixture of carboxylic acids (C–C cleavage)

★ PCC (Pyridinium Chlorochromate) — Best for 1° alcohol → Aldehyde:
CH₃CH=CH–CH₂OH —^PCC→ CH₃CH=CH–CHO (no over-oxidation to acid)

Vapour phase dehydrogenation over heated Cu (573 K):
RCH₂OH —^{Cu, 573 K}→ RCHO + H₂
R–CHOH–R' —^{Cu, 573 K}→ R–CO–R' + H₂
(CH₃)₃COH —^{Cu, 573 K}→ (CH₃)₂C=CH₂ + H₂O (tertiary → dehydration, not dehydrogenation)

⚠️ Methanol Poisoning — Bio-oxidation

Methanol → (oxidised in body) → Methanal (HCHO) → Methanoic acid (HCOOH) → blindness and death.

Treatment: Intravenous infusions of diluted ethanol. The enzyme oxidising HCHO to acid becomes saturated with ethanol, allowing kidneys to excrete methanol before it becomes fully toxic.

7.4e Acidity of Phenols & Chemical Reactions

Acidity of Phenols — Why More Acidic than Alcohols?

PHENOL vs ALCOHOL ACIDITY

Phenol is approximately 10⁶ times (1 million times) more acidic than ethanol. (pKa: phenol ≈ 10.0; ethanol ≈ 15.9)

Reason 1 — sp² C of ring: The ring carbon is sp² hybridised (more electronegative than sp³ of alkyl C). This electron withdrawal reduces electron density on O → increases O–H polarity → easier to ionise.

Reason 2 — Resonance stabilisation of phenoxide ion: In alkoxide (RO⁻), negative charge is localised on O. In phenoxide ion (C₆H₅O⁻), the negative charge is delocalised over the ring (5 resonance structures — charge at O, ortho, and para positions). More stable ion → more favourable ionisation.

🔬

Resonance Structures of Phenoxide Ion (Structures I–V)

NCERT p.207

Shows 5 resonance structures of phenoxide ion: negative charge on O (I), delocalised to ortho positions (II, IV) and para position (III), and fully in ring (V). Demonstrates why phenoxide is more stable than alkoxide.

AI Image Prompt

Professional chemistry diagram showing 5 resonance structures of phenoxide ion (C₆H₅O⁻) connected by double-headed resonance arrows. Fully white background, clean minimalistic. Structure I: O⁻ on oxygen, benzene ring neutral. Structure II: O neutral (double bond to ring), negative charge at ortho carbon. Structure III: charge at para carbon. Structure IV: charge at other ortho carbon. Structure V: delocalized. Show each as a benzene ring with oxygen group. No background patterns, label each structure I through V below.

Resonance structures of phenoxide ion — negative charge delocalised over 5 positions → high stability → favours ionisation of phenol.

Effect of Substituents on Phenol Acidity

Substituent	Position	Effect on Acidity	Reason
–NO₂ (electron withdrawing)	ortho/para	Increases acid strength (↓ pKa)	Withdraws electrons → destabilises phenol more than phenoxide → more ionisation. Negative charge in phenoxide delocalised onto N of NO₂
–NO₂	meta	Small increase (inductive only)	Resonance stabilisation not possible at meta; only inductive effect
–CH₃, alkyl (electron releasing)	any	Decreases acid strength (↑ pKa)	+I effect pushes electrons onto ring → increases electron density on O → harder to release H⁺

Acidity order (increasing): Propan-1-ol < Cresols < Phenol < o/p-Nitrophenol < o/p-Dinitrophenol < 2,4,6-Trinitrophenol (Picric acid)

Reactions Unique to Phenols

1. Electrophilic Aromatic Substitution — Nitration

–OH group is a strong activating and ortho/para-directing group (via resonance). Phenol reacts with EAS more readily than benzene.

With dilute HNO₃ (298 K):
C₆H₅OH —^{dil. HNO₃, 298 K}→ o-Nitrophenol (2-nitrophenol) + p-Nitrophenol (4-nitrophenol)
Separable by steam distillation: o-isomer is steam volatile (intramolecular H-bonding); p-isomer is not (intermolecular H-bonding, associated molecules).

With concentrated HNO₃:
C₆H₅OH —^{conc. HNO₃}→ 2,4,6-Trinitrophenol (Picric acid) (yield is poor directly)

🧠

o vs p Nitrophenol — H-Bonding Trick o-Nitrophenol: –NO₂ at ortho position → intramolecular H-bond between –OH and –NO₂ of the same molecule → does NOT associate with other molecules → lower bp → steam volatile.
p-Nitrophenol: –NO₂ at para → cannot form intramolecular H-bond → forms intermolecular H-bonds between molecules → associated → higher bp → not steam volatile.

2. Halogenation of Phenol

(a) In non-polar solvent (CHCl₃ or CS₂), low temperature:
C₆H₅OH + Br₂ —^{CS₂, 273 K}→ o-Bromophenol (minor) + p-Bromophenol (major)
No Lewis acid catalyst needed — the activating –OH group polarises Br₂ directly.

(b) With bromine water:
C₆H₅OH + 3Br₂(aq) → 2,4,6-Tribromophenol↓ (white precipitate) + 3HBr
This is a qualitative test for phenol — immediate white precipitate confirms presence of phenol.

★ 3. Kolbe's Reaction (Kolbe–Schmitt Reaction)

Phenol treated with NaOH → sodium phenoxide (phenoxide ion is even more reactive than phenol). Sodium phenoxide reacts with CO₂ (weak electrophile) under pressure at 400 K → electrophilic substitution at the ortho position → gives sodium salicylate → acidify → salicylic acid.

C₆H₅OH + NaOH → C₆H₅ONa (sodium phenoxide)
C₆H₅ONa —^{(i) CO₂, 400 K, pressure (ii) H⁺}→ 2-HO–C₆H₄–COOH
Product = 2-Hydroxybenzoic acid (Salicylic acid) — precursor for aspirin synthesis

★ 4. Reimer–Tiemann Reaction

Phenol + CHCl₃ in presence of aqueous NaOH → –CHO group introduced at the ortho position → salicylaldehyde (2-hydroxybenzaldehyde).

C₆H₅OH + CHCl₃ —^{aq. NaOH}→ [Intermediate: C₆H₄(ONa)(CHCl₂)] —^NaOH→ 2-HO–C₆H₄–CHO
Salicylaldehyde (2-hydroxybenzaldehyde)

Mechanism: CHCl₃ + NaOH → :CCl₂ (dichlorocarbene, an electrophile) → attacks ortho position of phenoxide → benzal dichloride intermediate → hydrolysis by NaOH → salicylaldehyde + H⁺.

5. Reaction with Zinc Dust

Phenol is reduced (C–O bond cleaved) to benzene when heated with zinc dust. –OH is removed.

C₆H₅OH + Zn —^heat→ C₆H₆ + ZnO

6. Oxidation of Phenol

Phenol is oxidised by Na₂Cr₂O₇/H₂SO₄ to give benzoquinone (a conjugated diketone).

C₆H₅OH —^{Na₂Cr₂O₇/H₂SO₄}→ C₆H₄O₂ (Benzoquinone)

Phenols slowly oxidise in air (dark coloured mixtures of quinones).

7.5 Commercially Important Alcohols

Methanol (CH₃OH) — Wood Spirit

Industrial production: Catalytic hydrogenation of CO

CO + 2H₂ —^{ZnO–Cr₂O₃, 200–300 atm, 573–673 K}→ CH₃OH

bp: 337 K. Colourless liquid.

Uses: Solvent in paints & varnishes; manufacture of formaldehyde (plastics).

Highly poisonous: Even small amounts cause blindness; large amounts → death. Body oxidises it to HCHO → HCOOH.

Ethanol (C₂H₅OH) — Grain/Spirit Alcohol

Commercial production: Fermentation of sugars (molasses, starch, grapes)

C₁₂H₂₂O₁₁ + H₂O —^invertase→ C₆H₁₂O₆ + C₆H₁₂O₆
C₆H₁₂O₆ —^zymase→ 2C₂H₅OH + 2CO₂↑

bp: 351 K. Colourless liquid.

Denaturation: Making ethanol unfit for drinking by adding CuSO₄ (colour) + pyridine (foul smell) = denatured alcohol.

Fermentation limit: Zymase inhibited above 14% ethanol. CO₂ released makes process anaerobic.

7.6a Preparation of Ethers

1. Dehydration of Alcohols (Intermolecular)

Primary alcohols with conc. H₂SO₄ at lower temperature (413 K) give ethers via intermolecular dehydration (SN2).

Example — Ethoxyethane synthesis:
CH₃CH₂OH —^{H₂SO₄, 413 K}→ CH₃CH₂–O–CH₂CH₃ + H₂O (Ethoxyethane)

SN2 Mechanism:
Step 1: C₂H₅OH + H⁺ → C₂H₅–OH₂⁺ (Protonated alcohol)
Step 2: C₂H₅OH + C₂H₅–OH₂⁺ → C₂H₅–O⁺(H)–C₂H₅ → C₂H₅–O–C₂H₅ + H⁺

Limitations of Dehydration Method

Suitable only for primary alkyl groups — must be unhindered and temperature kept low.
Secondary and tertiary alcohols undergo elimination (alkene formation) rather than ether formation at higher temperatures.
Cannot be used to make unsymmetrical ethers like ethyl methyl ether — would give a mixture of three ethers.

★ 2. Williamson Synthesis — Best Laboratory Method

An alkyl halide (preferably primary) reacts with a sodium alkoxide or sodium aryloxide via SN2 mechanism. Works for both symmetrical and unsymmetrical ethers, including aryl ethers.

R–X + NaO–R' → R–O–R' + NaX

Examples:
C₂H₅Br + NaOC₂H₅ → C₂H₅–O–C₂H₅ (Diethyl ether) + NaBr
CH₃Br + NaOC₂H₅ → CH₃–O–C₂H₅ (Methyl ethyl ether) + NaBr

For aryl ethers (phenol used as phenoxide):
C₆H₅OH + NaOH → C₆H₅ONa + R–X → C₆H₅–O–R (Aryl alkyl ether) + NaX

Critical Limitation of Williamson Synthesis

Alkoxide ions are both nucleophiles AND strong bases. With secondary or tertiary alkyl halides, elimination (E2) competes with substitution (SN2) and usually wins → alkene is the major product, not ether.

CH₃ONa + (CH₃)₃C–Cl → (CH₃)₂C=CH₂ (2-Methylpropene, major) + NaCl + CH₃OH

Solution for t-butyl ethyl ether: Use t-butoxide + primary halide (not primary alkoxide + t-butyl halide).

(CH₃)₃CONa + C₂H₅Cl → (CH₃)₃C–O–C₂H₅ ✓ (primary halide → SN2 prevails)

7.6b Physical Properties of Ethers

Key Physical Properties

Boiling points: Comparable to alkanes of same MW (not to alcohols). Much lower than corresponding alcohols because ethers cannot form intermolecular H-bonds with each other (no O–H bond). Example: Ethoxyethane (MW 74) bp = 307.6 K vs Butan-1-ol (MW 74) bp = 390 K.
Solubility in water: Comparable to alcohols of same MW — oxygen in ether can form H-bonds with water (R–O: ···H–O–H). Ethoxyethane: 7.5 g/100 mL; Butan-1-ol: 9 g/100 mL (similar).
Diethyl ether (common ether) was widely used as anaesthetic but replaced due to slow effect and unpleasant recovery.
C–O bonds are polar → ethers have a net dipole moment.

7.6c Chemical Reactions of Ethers

Ethers are the least reactive of all organic functional groups. Their main chemical reactions:

1. Cleavage of C–O Bond by Hydrogen Halides

C–O bond in ethers is cleaved by concentrated HI or HBr at high temperature. Reactivity order: HI > HBr > HCl

Dialkyl ether + HX:
R–O–R + HX → R–OH + R–X
R–OH + HX → R–X + H₂O (if HX in excess)

Mixed ether + HX:
R–O–R' + HX → R–OH + R'–X (smaller alkyl group → alkyl halide)

Mechanism of Ether Cleavage — 3 Steps

Protonation: Lone pair on O accepts H⁺ from HI → forms oxonium ion (R–O⁺(H)–R')
Nucleophilic attack: I⁻ (good nucleophile) attacks the least substituted carbon of oxonium ion by SN2 → displaces alcohol → gives alkyl iodide + alcohol
If excess HI: The alcohol formed also reacts with another molecule of HI → second alkyl iodide

Special case — tertiary group present: If one alkyl group is tertiary, the tertiary carbocation [(CH₃)₃C⁺] forms in step 2 → SN1 mechanism → tertiary halide formed.

(CH₃)₃C–O–CH₃ + HI → CH₃OH + (CH₃)₃C–I (tert-butyl iodide)

Cleavage of Aryl Alkyl Ethers (Anisole type)

In aryl alkyl ethers (Ar–O–R), cleavage always occurs at the alkyl–oxygen bond, NOT the aryl–oxygen bond.

Reason: The Ar–O bond has partial double bond character (sp² carbon + resonance). I⁻ attacks the –CH₃ (alkyl side) via SN2 → gives CH₃I + phenol. Phenol cannot react further (sp² C cannot undergo nucleophilic substitution).

C₆H₅–O–CH₃ + HI → C₆H₅OH + CH₃I

2. Electrophilic Aromatic Substitution of Aryl Ethers (Anisole)

The alkoxy group (–OR) is a strong ortho/para director and activating group — lone pairs on O conjugate with the ring, increasing electron density at ortho and para positions.

Reaction	Conditions	Products	Notes
Halogenation	Br₂ in ethanoic acid (no catalyst)	p-Bromoanisole (90%, major) + o-Bromoanisole (minor)	–OCH₃ activates ring so much that no FeBr₃ needed
Friedel-Crafts Alkylation	CH₃Cl / anhyd. AlCl₃, CS₂	4-Methoxytoluene (major) + 2-Methoxytoluene (minor)	Para product predominates
Friedel-Crafts Acylation	CH₃COCl / anhyd. AlCl₃	4-Methoxyacetophenone (major) + 2-Methoxyacetophenone (minor)	Para major product
Nitration	Conc. H₂SO₄ + conc. HNO₃	4-Nitroanisole (major) + 2-Nitroanisole (minor)	Milder than benzene nitration

✏️ Practice Questions

Arrange in increasing order of boiling points and give reason: Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.

Why is phenol a much stronger acid than ethanol? Explain with reference to the stability of their conjugate bases.

What is the Lucas test? How does it distinguish between primary, secondary and tertiary alcohols? Give the mechanism.

o-Nitrophenol is more volatile (steam volatile) than p-nitrophenol. Explain.

Arrange in order of increasing acid strength and give complete reasoning: Propan-1-ol, 4-methylphenol (p-cresol), phenol, 3-nitrophenol, 3,5-dinitrophenol, 2,4,6-trinitrophenol (picric acid).

Write the reactions involved in (i) Kolbe's reaction (ii) Reimer–Tiemann reaction. What products are formed?

How is 2-methylpropan-2-ol dehydrated differently from propan-1-ol? Give the conditions and explain why.

Why does the reaction of (CH₃)₃CBr with CH₃ONa give 2-methylpropene (an alkene) rather than (CH₃)₃C–O–CH₃ (the expected ether)?

What product is obtained when anisole (methoxybenzene) is treated with HI? Show the mechanism.

Q10

Predict the major product of hydroboration–oxidation of (i) 1-methylcyclohexene (ii) propene, and explain how this differs from acid hydration.

Q11

Starting from suitable Grignard reagent(s) and carbonyl compounds, show how you would synthesise: (a) 1-phenylethanol (b) 2-methylpropan-2-ol

Q12

Outline the reactions to synthesise phenol starting from (i) chlorobenzene (ii) benzene + conc. H₂SO₄ + NaOH (iii) cumene.

🎯 Important Exam Points — Quick Reference

CONCEPTAlcohol vs Phenol: Alcohol has –OH on sp³ C (aliphatic); Phenol has –OH on sp² C of aromatic ring. C–O bond in phenol (136 pm) shorter than in methanol (142 pm) due to resonance.

CONCEPTAcidity order: Picric acid >> p-Nitrophenol >> Phenol >> p-Cresol >> Cyclohexanol >> Ethanol. Phenol is ~10⁶ more acidic than ethanol due to resonance stabilisation of phenoxide ion.

CONCEPTAcid strength of alcohols: Primary > Secondary > Tertiary (fewer alkyl groups → less electron donation → more O–H polarity). Alcohols are weaker acids than water.

CONCEPTEase of dehydration: Tertiary > Secondary > Primary (stability of carbocation intermediate). Ease of esterification (with HX): Tertiary > Secondary > Primary (Lucas test principle).

CONCEPTHydroboration–oxidation: Anti-Markovnikov, syn addition, no carbocation, no rearrangement, gives primary alcohol from terminal alkene. (H.C. Brown, Nobel 1979 with Wittig)

CONCEPTWilliamson synthesis limitation: Use primary alkyl halide only. Secondary/tertiary → elimination product (alkene), not ether. Alkoxide is a strong base + nucleophile.

CONCEPTEther cleavage by HI: Dialkyl ether → 2 alkyl halides. Aryl alkyl ether → always cleaves at alkyl–O bond (Ar–O stronger due to sp² C resonance) → ArOH + RX.

REACTIONKolbe's reaction: Phenol + NaOH → PhONa + CO₂ (400K) → Sodium salicylate → H⁺ → Salicylic acid. Product is precursor for aspirin.

REACTIONReimer–Tiemann reaction: Phenol + CHCl₃ + NaOH → Salicylaldehyde (2-hydroxybenzaldehyde). Electrophilic agent is dichlorocarbene (:CCl₂). –CHO goes to ortho position.

REACTIONAspirin synthesis: Salicylic acid + (CH₃CO)₂O →(H⁺) Acetylsalicylic acid (Aspirin) + CH₃COOH. Properties: analgesic, anti-inflammatory, antipyretic.

REACTIONPhenol + Br₂(aq) → 2,4,6-Tribromophenol↓ (white ppt) + 3HBr. Qualitative test for phenol — no Lewis acid needed because –OH highly activates ring.

MCQPCC (Pyridinium chlorochromate) oxidises 1° alcohol → aldehyde only (no over-oxidation to acid). KMnO₄/H⁺ gives carboxylic acid from primary alcohol.

MCQMethanol → blindness/death (oxidised to HCHO then HCOOH). Treatment: dilute ethanol IV (competes with methanol for oxidising enzyme). Methanol = wood spirit; Ethanol = grain spirit.

MCQGrignard + HCHO → 1° alcohol; Grignard + RCHO → 2° alcohol; Grignard + R₂CO → 3° alcohol. Key rule for synthesis questions.

Alcohols, Phenols & Ethers –
Complete Detailed Notes

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