NCERT Plus Two (Class 12) Chemistry • Chapter 7

Alcohols, Phenols and Ethers

Q: While separating a mixture of ortho and para -nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Steam-volatile isomer: o -Nitrophenol . Reason: o -Nitrophenol exhibits intramolecular hydrogen bonding (forming an internal chelate ring), so molecules do not associate strongly and evaporate easily with steam. In contrast, p -nitrophenol exhibits intermolecular hydrogen bonding , causing molecular association and lower volatility.

Q: Name the product obtained when vapours of 2-methylpropan-2-ol (a tertiary alcohol) are passed over heated copper at 573 K.

Product: 2-Methylpropene (an alkene). Reaction: (CH 3 ) 3 C–OH --(Cu, 573 K)--> (CH 3 ) 2 C=CH 2 + H 2 O (Tertiary alcohols undergo dehydration instead of dehydrogenation).

Q: What is Lucas reagent? Which class of alcohol gives immediate turbidity with Lucas reagent at room temperature?

Lucas reagent: A mixture of concentrated HCl and anhydrous ZnCl 2 . Tertiary (3°) alcohols produce immediate white turbidity at room temperature due to rapid formation of stable 3° carbocations and insoluble alkyl chloride.

Q: Give the equations for the industrial preparation of phenol from cumene. Name the valuable by-product obtained.

Step 1 (Air oxidation): Cumene [C 6 H 5 –CH(CH 3 ) 2 ] + O 2 → Cumene hydroperoxide [C 6 H 5 –C(CH 3 ) 2 –O–O–H]. Step 2 (Acid hydrolysis): Cumene hydroperoxide --(H + /H 2 O)--> Phenol (C 6 H 5 OH) + Acetone (CH 3 COCH 3 ) . By-product: Acetone (Propanone) .

Q: Explain why phenol is much more acidic than ethanol.

1. sp 2 Carbon & O–H Polarity: In phenol, the –OH group is attached to a more electronegative sp 2 hybridised aromatic carbon, increasing O–H bond polarity. 2. Resonance Stabilisation of Phenoxide Ion: After losing H + , the phenoxide ion (C 6 H 5 O − ) delocalises its negative charge over the benzene ring without charge separation, making it far more stable than phenol (which has charge-separated resonance forms). In ethoxide ion (C 2 H 5 O − ), the negative charge remains localised on oxygen and is destabilised by the +I ethyl group.

Q: Write the chemical equations involved in: (i) Kolbe's reaction and (ii) Reimer–Tiemann reaction.

(i) Kolbe's Reaction: Phenol --(NaOH)--> Sodium phenoxide --((i) CO 2 , (ii) H + )--> 2-Hydroxybenzoic acid (Salicylic acid) . (ii) Reimer–Tiemann Reaction: Phenol + CHCl 3 + aq. NaOH (340 K) → Intermediate (–ONa, –CHCl 2 ) --(NaOH, then H + )--> Salicylaldehyde (2-Hydroxybenzaldehyde) .

Q: Account for the following: (a) The C–O–C bond angle in ethers (111.7°) is slightly greater than the tetrahedral angle (109.5°). (b) The C–O bond length in phenol (136 pm) is shorter than in methanol (142 pm).

(a) In ethers, strong steric repulsion between the two bulky alkyl (–R) groups widens the C–O–C angle to 111.7° despite two lone pairs on oxygen. (b) In phenol, the C–O bond is shorter ( 136 pm ) due to (i) partial double-bond character arising from conjugation of oxygen's lone pair with the aromatic ring, and (ii) the sp 2 hybridised state of the ring carbon attached to oxygen.

Q: Give reason for the higher boiling point of ethanol (351 K) compared to methoxymethane (248 K), even though both have the same molecular formula C 2 H 6 O. Why is ethoxyethane still soluble in water?

Boiling point: Ethanol molecules associate strongly through intermolecular O–H···O hydrogen bonding , requiring much more thermal energy to vaporise. Methoxymethane lacks an O–H bond and cannot H-bond with itself. Solubility of ethoxyethane: Just like alcohols, the oxygen atom of ethoxyethane can accept hydrogen bonds from water molecules (R 2 O···H–OH), making it miscible (7.5 g/100 mL).

Preparation, Grignard Synthesis, H-Bonding, Lucas Test, Dehydration, Phenol Acidity, Kolbe, Reimer-Tiemann & Williamson Ether Synthesis

📄 Official NCERT Textbook PDF

Topic 7.1 – 7.3: Classification, IUPAC Nomenclature & Functional Group Structures

Look around your study desk right now! The spirit polishing wooden furniture contains ethanol, the paper in your textbook is cellulose packed with countless –OH groups, hospital antiseptics rely on substituted phenols, and classic surgical anaesthetics and fragrances belong to ethers. One tiny oxygen atom—bonded to an aliphatic carbon, an aromatic ring, or sandwiched between two carbons—creates three powerhouse organic families!

Replacing a hydrogen atom of an aliphatic hydrocarbon by –OH gives an alcohol (R–OH); attaching –OH directly to an aromatic ring carbon gives a phenol (Ar–OH); replacing the hydroxyl hydrogen with an alkyl/aryl group yields an ether (R–O–R′).

1. Classification of Alcohols & Phenols

By number of –OH groups: Monohydric (1 –OH), Dihydric (e.g., ethylene glycol, ethane-1,2-diol), Trihydric (e.g., glycerol, propane-1,2,3-triol), and Polyhydric.
Compounds with sp³ C–OH bond:
- Primary (1°), Secondary (2°), Tertiary (3°): Based on whether –OH is on a 1°, 2°, or 3° carbon.
- Allylic alcohols: –OH on an sp³ carbon next to a C=C double bond (e.g., CH₂=CH–CH₂–OH, prop-2-en-1-ol).
- Benzylic alcohols: –OH on an sp³ carbon attached directly to a benzene ring (e.g., C₆H₅–CH₂–OH).
Compounds with sp² C–OH bond: Vinylic alcohols (CH₂=CH–OH) and Phenols (Ar–OH).
Ethers: Simple / Symmetrical (R = R′, e.g., C₂H₅OC₂H₅) vs Mixed / Unsymmetrical (R ≠ R′, e.g., CH₃OC₂H₅, C₆H₅OCH₃).

2. Key IUPAC Names (NCERT Tables 7.1 & 7.2)

Formula / Common Name	IUPAC Name	Notes
(CH₃)₃C–OH (tert-Butyl alcohol)	2-Methylpropan-2-ol	3° Alcohol
HO–CH₂–CH₂–OH (Ethylene glycol)	Ethane-1,2-diol	Retain final ‘e’ of alkane!
HO–CH₂–CH(OH)–CH₂–OH (Glycerol)	Propane-1,2,3-triol	Trihydric alcohol
o-, m-, p-Cresols (CH₃–C₆H₄–OH)	2-, 3-, 4-Methylphenol	1,2- / 1,3- / 1,4-Benzenediol = Catechol / Resorcinol / Hydroquinone
C₆H₅OCH₃ (Anisole)	Methoxybenzene	Anisole is IUPAC-accepted; smaller group is alkoxy!
C₆H₅OC₂H₅ (Phenetole)	Ethoxybenzene	CH₃OCH₂CH₂CH₃ → 1-Methoxypropane

NCERT Example 7.1 (IUPAC Practice):
• CH₃–CH(Cl)–CH(CH₃)–CH(CH₃)–CH₂OH → 4-Chloro-2,3-dimethylpentan-1-ol (number from –OH end!).
• CH₃–CH(CH₃)–O–CH₂CH₃ → 2-Ethoxypropane.
• 1-Ethoxy-2-nitrocyclohexane & 2,6-Dimethylphenol.

3. Structures & Bond Angles (NCERT Fig. 7.1)

Methanol (CH₃–OH): ∠C–O–H = 108.9° (< 109.5° due to lone pair–lone pair repulsion on O) | C–O = 142 pm
Phenol (C₆H₅–OH): ∠C–O–H = 109° | C–O = 136 pm (shorter: partial double-bond character + sp² C)
Methoxymethane (CH₃–O–CH₃): ∠C–O–C = 111.7° (> 109.5° due to steric repulsion of two bulky –CH₃ groups) | C–O = 141 pm

Remember two huge exam favorites! (1) The bond angle follows Alcohol (108.9°) < Phenol (109°) < Ether (111.7°). (2) Phenol's C–O bond (136 pm) is SHORTER and stronger than methanol's (142 pm) because oxygen's lone pair conjugates with the aromatic ring and bonds to an sp² carbon—so phenol's C–O bond refuses to cleave in normal substitution!

Topic 7.4.1: Preparation of Alcohols: Alkene Hydration, Hydroboration, Reduction & Grignard

Suppose you need to synthesize an exact primary, secondary, or tertiary alcohol in the lab. If you start from an alkene like propene, standard acid hydration puts –OH on the middle carbon—but H.C. Brown won the 1979 Nobel Prize for showing how boron flips the –OH right onto the end carbon! Better still, Grignard reagents let you snap two carbon pieces together like molecular Lego bricks.

1. From Alkenes

(i) Acid-Catalysed Hydration (Markovnikov): Alkenes react with water + dil. H₂SO₄ to form alcohols via a carbocation intermediate.
3-Step Mechanism (Ethene → Ethanol):
Step 1: Protonation of alkene by H₃O⁺ to form carbocation (slow/electrophilic attack).
Step 2: Nucleophilic attack of H₂O on carbocation → protonated alcohol.
Step 3: Deprotonation (loss of H⁺) → Alcohol + H₃O⁺ regenerated.
(ii) Hydroboration–Oxidation (Anti-Markovnikov): Propene reacts with diborane (BH₃)₂ to give trialkylborane (CH₃CH₂CH₂)₃B, which on oxidation with H₂O₂ / aq. NaOH gives Propan-1-ol in excellent yield! Boron attaches to the sp² carbon carrying more hydrogens.

2. Reduction of Carbonyl Compounds & Carboxylic Acids

Aldehydes → 1° Alcohols; Ketones → 2° Alcohols: Using H₂/Pd (or Pt/Ni), NaBH₄, or LiAlH₄.
Carboxylic Acids → 1° Alcohols: R–COOH requires strong (i) LiAlH₄, (ii) H₂O (expensive, lab use). Commercially, acids are esterified (RCOOR′) and reduced via H₂ / catalyst.

3. From Grignard Reagents (R–Mg–X) — 1°, 2° & 3° Alcohols

Nucleophilic addition of R^δ−–Mg^δ+X to the polar C=O bond forms an adduct, which upon hydrolysis (H₂O / H⁺) yields alcohols:

• Methanal (HCHO) + RMgX → Adduct --(H₂O)--> R–CH₂–OH (Primary, 1° Alcohol)
• Other Aldehydes (R′CHO) + RMgX → R′–CH(OH)–R (Secondary, 2° Alcohol)
• Ketones (R′COR″) + RMgX → R′R″C(OH)R (Tertiary, 3° Alcohol)

NCERT Example 7.2:
(a) Catalytic reduction of butanal (CH₃CH₂CH₂CHO + H₂/Pd) → Butan-1-ol.
(b) Acid hydration of propene (CH₃CH=CH₂ + H₂O/H⁺) → Propan-2-ol (Markovnikov).
(c) Propanone + CH₃MgBr followed by H₂O → 2-Methylpropan-2-ol (3° alcohol).

Watch two classic exam traps! (1) NaBH₄ reduces aldehydes and ketones cleanly, but is too mild to reduce carboxylic acids or esters—only LiAlH₄ reduces RCOOH directly to RCH₂OH. (2) From propene, dil. H₂SO₄ gives Propan-2-ol, whereas (i) B₂H₆, (ii) H₂O₂/OH⁻ gives Propan-1-ol!

Topic 7.4.2: Preparation of Phenols: Dow's Process, Sulphonic Acid, Diazonium & Cumene

A century ago, phenol—then called carbolic acid—was painstakingly extracted from coal tar. Today, global industry needs millions of tonnes for medicines, resins, and dyes. The undisputed world champion? The Cumene Hydroperoxide Process, which turns simple isopropylbenzene and ordinary air into pure phenol AND bonus acetone simultaneously!

Four classic NCERT synthetic routes convert benzene derivatives into phenol:

Method	Starting Material & Reagents	Key Conditions & Products
1. From Haloarenes (Dow's Process)	Chlorobenzene + NaOH → Sodium phenoxide --(HCl)--> Phenol	623 K, 320 atm (drastic T & P due to partial C=Cl bond!)
2. From Benzene Sulphonic Acid	Benzene --(Oleum, H₂S₂O₇)--> C₆H₅SO₃H --((i) molten NaOH, Δ (ii) H⁺)--> Phenol	Two-step alkali fusion
3. From Diazonium Salts	Aniline + NaNO₂ + HCl → C₆H₅N₂⁺Cl⁻ --(H₂O, warm / dil. acid)--> Phenol + N₂↑	Diazotisation at 273–278 K (0–5 °C)
4. From Cumene (Industrial!)	Cumene [C₆H₅–CH(CH₃)₂] --(O₂/air)--> Cumene hydroperoxide --(H₃O⁺, dil. acid)--> Phenol + Acetone	Produces Acetone (CH₃COCH₃) as valuable by-product!

Cumene Process Equations (NCERT Ex 7.9 — Most Repeated!):
C₆H₅–CH(CH₃)₂ + O₂ → C₆H₅–C(CH₃)₂–O–O–H (Cumene hydroperoxide)
C₆H₅–C(CH₃)₂–O–O–H --(H⁺ / H₂O)--> C₆H₅–OH (Phenol) + CH₃–CO–CH₃ (Acetone)

Two high-frequency traps! (1) During diazotisation of aniline (NaNO₂ + HCl), keep temperature strictly at 273–278 K; only in the SECOND step do you warm the diazonium salt with water to liberate N₂ gas and phenol. (2) In exams asking "Name the by-product of industrial phenol manufacture from cumene," always write Acetone (Propanone)!

Topic 7.4.3 & 7.6.2: Physical Properties: Hydrogen Bonding, Boiling Points & Steam Distillation

Compare three molecules that weigh almost identically (~46 g/mol): Ethanol boils at a hot 351 K, Methoxymethane boils at a frigid 248 K, and Propane boils at 231 K! What gives ethanol a 103-degree leap over its own ether isomer? Intermolecular Hydrogen Bonding—molecules holding hands tightly so they refuse to escape into vapour!

1. Boiling Points: Why Alcohols Beat Ethers & Alkanes

Intermolecular H-bonding: Alcohols and phenols form extensive intermolecular O–H···O hydrogen bonds, requiring extra thermal energy to break before boiling. Ethers and hydrocarbons lack O–H bonds and rely mainly on weaker dipole or van der Waals forces.
Number of Carbons vs Branching: Boiling point increases with carbon chain length (greater surface area → stronger van der Waals forces) and decreases with branching (compact shape → smaller surface area).

Compound (~Comparable Molar Mass)	Boiling Point	Water Miscibility (per 100 mL)
n-Pentane (CH₃CH₂CH₂CH₂CH₃, 72 u)	309.1 K	Immiscible (no H-bonds)
Ethoxyethane (C₂H₅OC₂H₅, 74 u)	307.6 K	7.5 g (accepts H-bonds from water!)
Butan-1-ol (CH₃CH₂CH₂CH₂OH, 74 u)	390 K	9.0 g (forms H-bonds with water!)

NCERT Example 7.3 (Increasing Boiling Point):
(a) Methanol < Ethanol < Propan-1-ol < Butan-2-ol (branched) < Butan-1-ol (straight) < Pentan-1-ol.
(b) n-Butane < Ethoxyethane < Pentanal < Pentan-1-ol.

2. Solubility & Steam Distillation of o- and p-Nitrophenol

Water Solubility: Alcohols, phenols, AND ethers can form hydrogen bonds with water molecules—explaining why ethoxyethane and butan-1-ol have comparable water solubility! Solubility drops as the hydrophobic hydrocarbon tail grows larger.
Steam Distillation (o- vs p-Nitrophenol):
- o-Nitrophenol exhibits intramolecular H-bonding (internal 6-membered chelate ring between –OH and adjacent –NO₂), so separate molecules do not associate → steam volatile (distils with steam)!
- p-Nitrophenol exhibits intermolecular H-bonding (molecules link into chains) → higher boiling point, less volatile, remains behind!

Two super-common conceptual traps! (1) Ethers have low boiling points like alkanes because they have NO O–H bond to H-bond with each other—YET they are surprisingly soluble in water (7.5 g/100 mL for diethyl ether) because ether oxygen CAN accept H-bonds from water molecules! (2) Always pair ortho-nitrophenol = intra = steam volatile and para-nitrophenol = inter = less volatile!

Topic 7.4.4(a)–(b) & 7.5: Reactions of Alcohols: Acidity, Esterification, Lucas Test, Dehydration & Oxidation

An alcohol molecule plays a clever double game: break the O–H bond and it acts as a nucleophile (reacting with active metals or forming fragrant esters and Aspirin!); break the C–O bond after protonation and it acts as an electrophile—letting Lucas reagent tell 1°, 2°, and 3° alcohols apart by simple cloudiness timing!

(A) Reactions Involving Cleavage of O–H Bond

1. Acidity of Alcohols (1° > 2° > 3°): React with Na, K, Al to release H₂ gas (2 R–OH + 2 Na → 2 R–ONa + H₂↑). Electron-releasing alkyl groups (+I) increase electron density on oxygen, reducing O–H polarity. Crucially, alcohols are weaker acids than water (R–O⁻ + H₂O → R–OH + OH⁻, showing alkoxides are stronger bases than OH⁻).
2. Esterification & Acetylation: R–OH + R′–COOH ⇌ R′–COOR + H₂O (cat. conc. H₂SO₄, reversible). With R′–COCl, pyridine is added to neutralise HCl and drive equilibrium forward. Acetylation of salicylic acid with acetic anhydride gives Aspirin!

(B) Reactions Involving Cleavage of C–O Bond (Alcohols Only!)

1. Lucas Test (conc. HCl + anhydrous ZnCl₂): Alcohols dissolve in Lucas reagent, while formed alkyl chlorides are immiscible and create turbidity. Ease follows carbocation stability (3° > 2° > 1°):
- 3° Alcohol: Immediate turbidity!
- 2° Alcohol: Turbidity appears in ~5 minutes!
- 1° Alcohol: No turbidity at room temperature (only on heating)!
2. Acid-Catalysed Dehydration (3° > 2° > 1°):
Temperature Switch with Ethanol + conc. H₂SO₄:
• At 443 K: CH₃CH₂OH → CH₂=CH₂ (Ethene) + H₂O
• At 413 K: 2 CH₃CH₂OH → C₂H₅–O–C₂H₅ (Ethoxyethane) + H₂O
Mechanism (443 K): (1) Protonation → CH₃CH₂OH₂⁺; (2) Slow RDS loss of H₂O → CH₃CH₂⁺; (3) Loss of H⁺ → Ethene.
Milder conditions work for 2° (85% H₃PO₄, 440 K) and 3° (20% H₃PO₄, 358 K) because 3° carbocations are most stable!
3. Oxidation & Heated Copper (573 K):
- 1° Alcohol: With PCC (pyridinium chlorochromate) or anhyd. CrO₃ → Aldehyde (RCHO); with strong KMnO₄/H⁺ or K₂Cr₂O₇/H⁺ → Carboxylic acid.
- 2° Alcohol: With CrO₃ → Ketone.
- Vapours over Cu at 573 K: 1° → Aldehyde (dehydrogenation); 2° → Ketone (dehydrogenation); 3° → Alkene (dehydration!).

NCERT 7.5 Commercial Alcohols: Methanol (CH₃OH, wood spirit) is made via CO + 2H₂ (ZnO–Cr₂O₃, 573–673 K, 200–300 atm; b.p. 337 K; causes blindness/death). Ethanol (C₂H₅OH) is made by fermentation of molasses/sugars using yeast enzymes invertase (sucrose → glucose + fructose) and zymase (→ ethanol + CO₂, anaerobic, stops > 14%). Commercial ethanol is made unfit for drinking (denaturation) by adding CuSO₄ (colour) and pyridine (foul smell).

TWO MEGA EXAM TRAPS! (1) Passing 3° alcohol vapours over heated Cu at 573 K does NOT oxidise—it undergoes dehydration to an alkene (e.g., 2-methylpropan-2-ol → 2-methylpropene)! (2) Ethanol + conc. H₂SO₄ gives Ethene at 443 K, but Ethoxyethane at 413 K—just 30 K decides alkene vs ether!

Topic 7.4.4(a)(iii): Acidity of Phenols vs Alcohols & Substituent Effects (pKa Logic)

Drop ethanol into aqueous NaOH and nothing reacts. Drop phenol into aqueous NaOH and it reacts at once—phenol is a million times (10⁶) more acidic than ethanol! Why? When ethanol loses H⁺, oxygen holds the negative charge all alone like a burning coal. When phenol loses H⁺, its benzene ring passes that charge around four atoms like a team relay!

Unlike alcohols, phenols react with both active metals (Na, K, Al) AND aqueous NaOH to yield sodium phenoxide (C₆H₅ONa) + H₂O, proving phenols are stronger acids than alcohols and water.

Why Phenol (pK_a = 10.0) is Far More Acidic than Ethanol (pK_a = 15.9)

Higher Electronegativity of sp² Carbon: In phenol, –OH is bonded to an sp²-hybridised benzene carbon, which withdraws electron density from oxygen, weakening the O–H bond and easing H⁺ release (phenol resonance even puts a δ+ charge on oxygen).
Resonance Stabilisation of Phenoxide Ion vs Alkoxide Ion:
- In alkoxide ion (R–O⁻), the negative charge is completely localised on oxygen (and further destabilised by +I alkyl groups).
- In phenoxide ion (C₆H₅O⁻), the negative charge is delocalised across oxygen and the ortho and para carbons (structures I–V) with NO charge separation!
- Although neutral phenol also shows resonance, those structures require charge separation (+ on O, − on ring), making phenol less stable than the phenoxide ion—strongly driving ionisation forward!

Effect of Ring Substituents on Phenol Acidity (NCERT Table 7.3)

Electron-Withdrawing Groups (–NO₂, –CN, –X): Disperse negative charge of phenoxide ion and increase acidity (lower pK_a), strongest at ortho & para positions where the negative charge lands!
Electron-Releasing Groups (–CH₃, –OCH₃): Intensify negative charge on phenoxide ion and decrease acidity (higher pK_a; cresols are weaker acids than phenol).

Compound	pK_a Value	Acidity Comparison
2,4,6-Trinitrophenol (Picric acid)	~0.38	Very strong acid (three –NO₂ groups!)
p-Nitrophenol	7.1	Strongest mononitrophenol (−R & −I at para)
o-Nitrophenol	7.2	Slightly weaker than para (intramolecular H-bond holds H⁺)
m-Nitrophenol	8.3	Only −I operates at meta (no −R resonance!)
Phenol (C₆H₅OH)	10.0	Reference (~10⁶× more acidic than ethanol!)
Cresols (m-, o-, p-CH₃–C₆H₄OH)	10.1 – 10.2	Weaker than phenol (+I / hyperconjugation of –CH₃)
Ethanol (C₂H₅OH)	15.9	Very weak acid (localised alkoxide)

NCERT Example 7.4 (Increasing Acid Strength):
Propan-1-ol < 4-Methylphenol < Phenol < 3-Nitrophenol < 3,5-Dinitrophenol < 2,4,6-Trinitrophenol.

Three favorite exam traps! (1) Smaller pK_a = Stronger acid (do not reverse the order!). (2) p-Nitrophenol (pK_a 7.1) is slightly MORE acidic than o-Nitrophenol (7.2) because intramolecular H-bonding in ortho holds back the proton slightly! (3) Even though m-Nitrophenol doesn't experience −R resonance, it is STILL more acidic than phenol (pK_a 8.3 vs 10.0) due to the −I inductive pull of –NO₂!

Topic 7.4.4(c): Reactions of Phenols: Kolbe, Reimer–Tiemann, Bromination, Nitration & Zn Dust

Attach an –OH group to benzene and the aromatic ring becomes super-charged—reacting rapidly with bromine even WITHOUT an FeBr₃ catalyst! Even better, two legendary reactions—Kolbe and Reimer–Tiemann—attach a –COOH or –CHO right next door to –OH, turning phenol into the parent molecule of Aspirin and the sweet scent of meadowsweet!

The –OH group strongly activates the benzene ring and is ortho, para directing because oxygen donates lone pairs into the ring (+R effect).

1. Halogenation (Dramatic Solvent Effect!)

In non-polar/low-polarity solvent (Br₂ in CS₂ or CHCl₃ at 273 K): Gives monobromophenols — 4-Bromophenol (major) + 2-Bromophenol (minor), without FeBr₃!
In Bromine Water (Br₂ / H₂O, polar): Phenoxide ion forms easily → instantaneous white precipitate of 2,4,6-Tribromophenol!

2. Nitration

Dilute HNO₃ at 298 K: Gives o-Nitrophenol + p-Nitrophenol (separated by steam distillation).
Conc. HNO₃: Converts phenol to 2,4,6-Trinitrophenol (Picric acid) (low yield directly due to oxidation; industrially prepared by sulphonating phenol to phenol-2,4-disulphonic acid first, then treating with conc. HNO₃).

3. Named Board-Exam Reactions & Special Reactions

Reaction	Reagents	Major Product & Key Idea
Kolbe's Reaction	(i) NaOH → Phenoxide (ii) CO₂, then H⁺	2-Hydroxybenzoic acid (Salicylic acid) (phenoxide is reactive enough to attack weak electrophile CO₂!)
Reimer–Tiemann Reaction	CHCl₃ + aq. NaOH (340 K), then H⁺	Salicylaldehyde (2-Hydroxybenzaldehyde) (via –CHCl₂ substituted benzal chloride intermediate)
Zn Dust Reduction	Phenol + Zn dust, Δ	Benzene (C₆H₆) + ZnO (the ONLY reaction cleaving phenol's C–O bond!)
Oxidation	Na₂Cr₂O₇ + H₂SO₄ (Chromic acid)	Benzoquinone (conjugated diketone)

NEVER mix up Kolbe and Reimer–Tiemann! Kolbe = CO₂ + NaOH → Salicylic ACID (–COOH), whereas Reimer–Tiemann = CHCl₃ + NaOH → SalicylALDEHYDE (–CHO)! Also remember: Phenol + Br₂/H₂O gives a white precipitate of 2,4,6-tribromophenol, while Br₂/CS₂ gives only monobromophenol!

Topic 7.6: Ethers: Williamson Synthesis (SN2 vs Elimination), HI Cleavage & Anisole

Imagine trying to squeeze a bulky crowd through a crowded narrow doorway—everyone gets stuck outside and starts snapping off bonds instead! That is exactly what happens in Williamson Ether Synthesis if you pair an alkoxide with a tertiary alkyl halide: elimination wins and only an alkene forms! Flip the partners—put the bulky group on the alkoxide and keep the alkyl halide primary—and the S_N2 handshake succeeds beautifully!

1. Preparation of Ethers

(i) Acid Dehydration of 1° Alcohols (413 K): 2 CH₃CH₂OH --(H₂SO₄, 413 K)--> C₂H₅OC₂H₅ + H₂O via S_N2. Works ONLY for unhindered 1° alcohols; 2° and 3° alcohols eliminate to alkenes!
(ii) Williamson Synthesis (R–X + R′–ONa → R–O–R′ + NaX):
- Involves S_N2 attack of alkoxide on alkyl halide → Alkyl halide (R–X) MUST be 1° (or methyl)!
- Critical Limitation: Because alkoxides are strong bases as well as nucleophiles, reacting a 3° alkyl halide like (CH₃)₃C–Br with CH₃ONa causes Elimination → 2-Methylpropene ONLY, NO ether!
- Correct Trick for 3° or Aryl Ethers (NCERT Ex 7.6 & 7.11): Make tert-butyl methyl ether from (CH₃)₃C–O⁻Na⁺ + CH₃Br; make Anisole from C₆H₅O⁻Na⁺ + CH₃I (never C₆H₅Br + CH₃ONa!).

2. Cleavage of C–O Bond with Hydrogen Halides (HI > HBr > HCl)

3 Golden Cleavage Rules with Conc. HI:
1. Two 1°/2° Alkyl Groups (S_N2): I⁻ attacks the smaller/less hindered group → smaller group forms alkyl iodide; larger forms alcohol (CH₃OC₂H₅ + HI → CH₃I + C₂H₅OH).
2. One 3° Alkyl Group (S_N1): Stable 3° carbocation forms! → (CH₃)₃C–O–CH₃ + HI → (CH₃)₃C–I + CH₃OH.
3. Alkyl Aryl Ether (Anisole): Partial double bond of Ar–O resists cleavage → C₆H₅–O–CH₃ + HI → C₆H₅OH (Phenol) + CH₃I (NEVER iodobenzene!).

3. Electrophilic Substitution of Anisole (–OCH₃ is +R, o/p-directing)

Bromination: Anisole + Br₂ in CH₃COOH (no FeBr₃ needed!) → p-Bromoanisole (90%) + o-Bromoanisole.
Nitration: Conc. HNO₃ + conc. H₂SO₄ → o- and p-Nitroanisole (p-major).
Friedel–Crafts: CH₃Cl / anhyd. AlCl₃ (Alkylation) or CH₃COCl / anhyd. AlCl₃ (Acetylation) → o- and p-products (p-major).

NCERT Example 7.6 & 7.7:
• Ex 7.6: CH₃CH₂ONa + (CH₃)₃C–Br yields 2-methylprop-1-ene (elimination!). Correct route to tert-butyl ethyl ether: (CH₃)₃C–ONa + C₂H₅Br.
• Ex 7.7 (HI cleavage): C₆H₅CH₂–O–C₂H₅ + HI → benzyl carbocation is resonance-stabilised (S_N1) → C₆H₅CH₂I + C₂H₅OH!

Watch BOTH classic ether traps! (1) In Williamson synthesis, NEVER pick a 3° halide or aryl halide—always put the 3° or phenyl group on the –O⁻Na⁺ partner and keep the halide 1°! (2) In HI cleavage of tert-butyl methyl ether, the product switches from CH₃I to (CH₃)₃C–I + CH₃OH because the 3° carbocation forms via S_N1! And remember benzyl ethyl ether + HI also follows S_N1 to give benzyl iodide + ethanol!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (NCERT Ex 7.8) • 1 Marks): While separating a mixture of ortho and para-nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.

Steam-volatile isomer: o-Nitrophenol.

Reason: o-Nitrophenol exhibits intramolecular hydrogen bonding (forming an internal chelate ring), so molecules do not associate strongly and evaporate easily with steam. In contrast, p-nitrophenol exhibits intermolecular hydrogen bonding, causing molecular association and lower volatility.

Q2 (Kerala HSE SAY • 1 Marks): Name the product obtained when vapours of 2-methylpropan-2-ol (a tertiary alcohol) are passed over heated copper at 573 K.

Product: 2-Methylpropene (an alkene).

Reaction: (CH₃)₃C–OH --(Cu, 573 K)--> (CH₃)₂C=CH₂ + H₂O (Tertiary alcohols undergo dehydration instead of dehydrogenation).

Q3 (Kerala HSE Model • 1 Marks): What is Lucas reagent? Which class of alcohol gives immediate turbidity with Lucas reagent at room temperature?

Lucas reagent: A mixture of concentrated HCl and anhydrous ZnCl₂.

Tertiary (3°) alcohols produce immediate white turbidity at room temperature due to rapid formation of stable 3° carbocations and insoluble alkyl chloride.

Q4 (Kerala HSE March (NCERT Ex 7.9) • 2 Marks): Give the equations for the industrial preparation of phenol from cumene. Name the valuable by-product obtained.

Step 1 (Air oxidation): Cumene [C₆H₅–CH(CH₃)₂] + O₂ → Cumene hydroperoxide [C₆H₅–C(CH₃)₂–O–O–H].

Step 2 (Acid hydrolysis): Cumene hydroperoxide --(H⁺/H₂O)--> Phenol (C₆H₅OH) + Acetone (CH₃COCH₃).

By-product: Acetone (Propanone).

Q5 (Kerala HSE March (NCERT Ex 7.14) • 2 Marks): Explain why phenol is much more acidic than ethanol.

1. sp² Carbon & O–H Polarity: In phenol, the –OH group is attached to a more electronegative sp² hybridised aromatic carbon, increasing O–H bond polarity.

2. Resonance Stabilisation of Phenoxide Ion: After losing H⁺, the phenoxide ion (C₆H₅O⁻) delocalises its negative charge over the benzene ring without charge separation, making it far more stable than phenol (which has charge-separated resonance forms). In ethoxide ion (C₂H₅O⁻), the negative charge remains localised on oxygen and is destabilised by the +I ethyl group.

Q6 (Kerala HSE March (Repeated Favorite) • 2 Marks): Write the chemical equations involved in: (i) Kolbe's reaction and (ii) Reimer–Tiemann reaction.

(i) Kolbe's Reaction: Phenol --(NaOH)--> Sodium phenoxide --((i) CO₂, (ii) H⁺)--> 2-Hydroxybenzoic acid (Salicylic acid).

(ii) Reimer–Tiemann Reaction: Phenol + CHCl₃ + aq. NaOH (340 K) → Intermediate (–ONa, –CHCl₂) --(NaOH, then H⁺)--> Salicylaldehyde (2-Hydroxybenzaldehyde).

Q7 (Kerala HSE Model • 2 Marks): Account for the following: (a) The C–O–C bond angle in ethers (111.7°) is slightly greater than the tetrahedral angle (109.5°). (b) The C–O bond length in phenol (136 pm) is shorter than in methanol (142 pm).

(a) In ethers, strong steric repulsion between the two bulky alkyl (–R) groups widens the C–O–C angle to 111.7° despite two lone pairs on oxygen.

(b) In phenol, the C–O bond is shorter (136 pm) due to (i) partial double-bond character arising from conjugation of oxygen's lone pair with the aromatic ring, and (ii) the sp² hybridised state of the ring carbon attached to oxygen.

Q8 (Kerala HSE SAY (NCERT Ex 7.22) • 2 Marks): Give reason for the higher boiling point of ethanol (351 K) compared to methoxymethane (248 K), even though both have the same molecular formula C₂H₆O. Why is ethoxyethane still soluble in water?

Boiling point: Ethanol molecules associate strongly through intermolecular O–H···O hydrogen bonding, requiring much more thermal energy to vaporise. Methoxymethane lacks an O–H bond and cannot H-bond with itself.

Solubility of ethoxyethane: Just like alcohols, the oxygen atom of ethoxyethane can accept hydrogen bonds from water molecules (R₂O···H–OH), making it miscible (7.5 g/100 mL).

Q9 (Kerala HSE March (NCERT Ex 7.19) • 3 Marks): Write the three-step mechanism of acid-catalysed dehydration of ethanol to yield ethene at 443 K. What product is obtained if ethanol is heated with conc. H₂SO₄ at 413 K?

Mechanism (Ethanol → Ethene at 443 K):
Step 1 (Fast protonation): CH₃CH₂OH + H⁺ ⇌ CH₃CH₂–OH₂⁺ (ethyl oxonium ion).
Step 2 (Slow, RDS — carbocation formation): CH₃CH₂–OH₂⁺ → CH₃CH₂⁺ + H₂O.
Step 3 (Deprotonation): CH₃CH₂⁺ → CH₂=CH₂ + H⁺ (acid catalyst regenerated).

At 413 K: Nucleophilic bimolecular substitution (S_N2) yields Ethoxyethane (C₂H₅–O–C₂H₅).

Q10 (Kerala HSE March (NCERT Ex 7.6 & 7.25) • 3 Marks): Explain the limitation of Williamson synthesis when preparing ethers containing tertiary alkyl groups. Write the appropriate reagents to prepare (i) 2-methoxy-2-methylpropane and (ii) ethoxybenzene.

Limitation: Williamson synthesis proceeds via S_N2 attack of alkoxide on alkyl halide. If a 3° alkyl halide is used, alkoxides (strong bases) cause elimination instead of substitution, yielding an alkene exclusively (e.g., CH₃ONa + (CH₃)₃C–Br → 2-methylpropene).

Correct Reagents:
(i) 2-Methoxy-2-methylpropane: Sodium tert-butoxide [(CH₃)₃C–ONa] + Methyl bromide (CH₃Br).
(ii) Ethoxybenzene: Sodium phenoxide (C₆H₅ONa) + Bromoethane (C₂H₅Br).

Q11 (Kerala HSE SAY (NCERT Ex 7.7 & 7.28) • 3 Marks): Write the major products formed when each of the following ethers is heated with concentrated HI: (i) 1-Propoxypropane, (ii) tert-Butyl methyl ether, (iii) Methoxybenzene (Anisole). Explain the pathway for (ii) and (iii).

(i) 1-Propoxypropane: CH₃CH₂CH₂–I + CH₃CH₂CH₂–OH (with excess HI → 2 moles CH₃CH₂CH₂I).

(ii) tert-Butyl methyl ether: (CH₃)₃C–I + CH₃OH. Follows S_N1 mechanism because departure of CH₃OH forms the stable 3° carbocation (CH₃)₃C⁺.

(iii) Methoxybenzene (Anisole): C₆H₅OH (Phenol) + CH₃I. The O–C₆H₅ bond is stronger due to sp² aryl carbon and partial double-bond character, so I⁻ exclusively cleaves the weaker O–CH₃ bond.

Q12 (Kerala HSE March • 4 Marks): (a) How will you prepare primary, secondary, and tertiary alcohols using Grignard reagent (CH₃MgBr)? Give equations. (b) Distinguish the products formed when phenol reacts with (i) Br₂ in H₂O and (ii) Br₂ in CS₂ at 273 K.

(a) From CH₃MgBr (followed by H₂O/H⁺):
• With Methanal (HCHO) → CH₃CH₂OH (Ethanol, 1° alcohol).
• With Ethanal (CH₃CHO) → CH₃CH(OH)CH₃ (Propan-2-ol, 2° alcohol).
• With Propanone (CH₃COCH₃) → (CH₃)₃C–OH (2-Methylpropan-2-ol, 3° alcohol).

(b) Bromination of Phenol:
• (i) With Br₂/H₂O (polar): Phenoxide ion forms → 2,4,6-Tribromophenol (white precipitate).
• (ii) With Br₂ in CS₂ at 273 K (low polarity): Monobromination → 4-Bromophenol (major) + 2-Bromophenol (minor), without FeBr₃.