Class 12 Chemistry | CBSE 2025-26
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KEY CONCEPTS
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1. CLASSIFICATION

• Alcohols — OH on sp³ carbon (R-OH)
• 1°: CH₃CH₂OH (ethanol)
• 2°: (CH₃)₂CHOH (propan-2-ol)
• 3°: (CH₃)₃COH (2-methylpropan-2-ol)
• Phenols — OH directly on benzene ring (C₆H₅OH)
• Ethers — oxygen between two alkyl/aryl groups (R-O-R')

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2. IUPAC NOMENCLATURE

• CH₃OH → Methanol
• CH₃CH₂OH → Ethanol
• CH₃CH(OH)CH₃ → Propan-2-ol
• (CH₃)₃COH → 2-methylpropan-2-ol
• C₆H₅OH → Phenol
• CH₃-O-CH₃ → Methoxymethane
• C₂H₅-O-CH₃ → Methoxyethane

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3. PREPARATION OF ALCOHOLS

From alkenes (hydration):
CH₂=CH₂ + H₂O ==H⁺==> CH₃CH₂OH

From carbonyl compounds (reduction):
RCHO + 2[H] ==LiAlH₄==> RCH₂OH (1° alcohol)
RCOR' + 2[H] ==LiAlH₄==> RCHOHR' (2° alcohol)

From Grignard reagent:
HCHO + RMgX ---> RCH₂OH (1° alcohol)
RCHO + R'MgX ---> RR'CHOH (2° alcohol)
RCOR' + R''MgX ---> RR'R''COH (3° alcohol)

From haloalkanes:
RX + KOH(aq) ---> ROH + KX

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4. PREPARATION OF PHENOLS

Dow's process:
C₆H₅Cl + NaOH ==623K, 300atm==> C₆H₅ONa ==H⁺==> C₆H₅OH

From diazonium salt:
C₆H₅N₂⁺Cl⁻ + H₂O ==heat==> C₆H₅OH + N₂ + HCl

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5. PREPARATION OF ETHERS

Williamson Synthesis:
C₂H₅ONa + C₂H₅Br ---> C₂H₅-O-C₂H₅ + NaBr
(Use 1° alkyl halide only — 3° gives elimination)

Dehydration of alcohols:
2ROH ==conc H₂SO₄, 413K==> R-O-R + H₂O
(413K → ether; 443K → alkene)

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6. PHYSICAL PROPERTIES

• BP: alcohols > ethers > alkanes of same MW
• Reason: alcohols have O-H intermolecular H-bonding
• BP of alcohols: 1° > 2° > 3° (H-bonding decreases)
• Lower alcohols miscible with water; solubility decreases with chain length
• Ethers have low BP (no H-bonding between ether molecules)
• Phenols — colourless solid, slightly soluble in water

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7. CHEMICAL REACTIONS OF ALCOHOLS

A. O-H bond reactions

With Na:
2ROH + 2Na ---> 2RONa + H₂

Acidity order:
Phenol > H₂O > 1° > 2° > 3° alcohol

With PCl₅:
ROH + PCl₅ ---> RCl + POCl₃ + HCl

With SOCl₂:
ROH + SOCl₂ ---> RCl + SO₂ + HCl

Esterification:
ROH + RCOOH ==conc H₂SO₄==> RCOOR' + H₂O

B. C-O bond reactions (dehydration)

At 413K: 2C₂H₅OH ---> C₂H₅OC₂H₅ + H₂O
At 443K: C₂H₅OH ---> CH₂=CH₂ + H₂O

C. Oxidation

1° alcohol:
RCH₂OH ==KMnO₄==> RCHO ---> RCOOH

2° alcohol:
R₂CHOH ==KMnO₄==> R₂C=O (ketone)

3° alcohol:
No oxidation under mild conditions

D. Lucas Test (ZnCl₂ + conc HCl)

• 3° → turbidity immediately
• 2° → turbidity in 5 min
• 1° → no turbidity at room temp

E. Victor Meyer Test

• 1° → red
• 2° → blue
• 3° → colourless

F. Iodoform Test (CH₃CH(OH)- group):
CH₃CH₂OH + I₂ + NaOH ---> CHI₃↓ + HCOONa + NaI + H₂O
(CHI₃ = yellow ppt = iodoform)

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8. CHEMICAL REACTIONS OF PHENOLS

A. Acidic nature
C₆H₅OH + NaOH ---> C₆H₅ONa + H₂O
C₆H₅OH + Na₂CO₃ ---> C₆H₅ONa + NaHCO₃
Phenol does NOT react with NaHCO₃
(weaker acid than carbonic acid)

B. Electrophilic Substitution
-OH = powerful ortho/para director

Bromination:
C₆H₅OH + 3Br₂(aq) ---> C₆H₂Br₃OH↓ + 3HBr
(2,4,6-tribromophenol — white ppt — detection test)

Nitration (dilute HNO₃):
C₆H₅OH + HNO₃ ---> o-nitrophenol + p-nitrophenol

C. Kolbe's Reaction
C₆H₅ONa + CO₂ ==400K, 4-7atm==> C₆H₄(OH)COONa ==H⁺==> salicylic acid

D. Reimer-Tiemann Reaction
C₆H₅OH + CHCl₃ + 3NaOH ==heat==> o-HO-C₆H₄-CHO + 3NaCl + H₂O
(salicylaldehyde)

E. FeCl₃ Test
Phenol + FeCl₃ ---> violet/purple colour
(detection of phenol)

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9. CHEMICAL REACTIONS OF ETHERS

Cleavage by HI:
R-O-R + HI ---> RI + ROH
Excess HI: ROH + HI ---> RI + H₂O

Cleavage of aryl ether:
C₆H₅-O-CH₃ + HI ---> C₆H₅OH + CH₃I
(aryl C-O too strong → only alkyl C-O breaks)

Electrophilic substitution on aryl ethers:
-OR = ortho/para director

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10. KEY DISTINCTIONS

o-nitrophenol vs p-nitrophenol:

• o-nitrophenol → intramolecular H-bond → more volatile → steam distillable
• p-nitrophenol → intermolecular H-bond → higher BP → less volatile

Alcohol vs Phenol:

• Phenol reacts with NaOH, alcohol does not
• Phenol gives violet with FeCl₃, alcohol does not
• Phenol gives white ppt with Br₂/water, alcohol does not

Ether vs Alcohol (same MW):

• Ether — no H-bonding → very low BP
• Alcohol — H-bonding → high BP

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Q&A — 20 QUESTIONS
CBSE + JEE Mains Mix
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Q1 (CBSE 1M)
Arrange in increasing BP:
CH₃OCH₃, CH₃CH₂OH, CH₃CH₂CH₃

Ans: CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CH₂OH
Propane (no H-bond) < ether (no H-bond between molecules) < ethanol (O-H H-bonding)

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Q2 (JEE MCQ)
Which alcohol gives immediate turbidity with Lucas reagent?
(a) 1° (b) 2° (c) 3° (d) all same

Ans: (c) 3°
3° carbocation forms instantly → immediate turbidity.

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Q3 (CBSE 2M)
Why is phenol more acidic than ethanol?

Ans:
C₆H₅O⁻ stabilised by resonance → negative charge delocalised over ring.
C₂H₅O⁻ has no resonance → less stable.
More stable conjugate base = stronger acid.
Acidity: phenol > ethanol

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Q4 (JEE)
Write Williamson synthesis of diethyl ether. Why cannot 3° alkyl halide be used?

Ans:
C₂H₅ONa + C₂H₅Br ---> C₂H₅-O-C₂H₅ + NaBr

3° alkyl halide → strong base (alkoxide) causes elimination → alkene formed instead of ether.

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Q5 (CBSE 2M)
Write Kolbe's reaction.

Ans:
C₆H₅ONa + CO₂ ==400K, 4-7atm==> C₆H₄(OH)COONa
C₆H₄(OH)COONa + HCl ---> C₆H₄(OH)COOH + NaCl
(salicylic acid — used in aspirin synthesis)

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Q6 (JEE MCQ)
Phenol reacts with:
(a) NaHCO₃
(b) Na₂CO₃
(c) NaOH
(d) Both b and c

Ans: (d) both NaOH and Na₂CO₃
Does NOT react with NaHCO₃ (phenol weaker than carbonic acid).

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Q7 (CBSE 3M)
Complete:
(i) C₂H₅OH + Na ---> ?
(ii) C₂H₅OH ==conc H₂SO₄, 443K==> ?
(iii) C₂H₅OH ==conc H₂SO₄, 413K==> ?

Ans:
(i) 2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂
(ii) C₂H₅OH ---> CH₂=CH₂ + H₂O (ethene)
(iii) 2C₂H₅OH ---> C₂H₅OC₂H₅ + H₂O (diethyl ether)

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Q8 (JEE MCQ)
Product when 2° alcohol is oxidised with K₂Cr₂O₇/H₂SO₄:
(a) Aldehyde
(b) Ketone
(c) Carboxylic acid
(d) Ester

Ans: (b) Ketone
R₂CHOH ==K₂Cr₂O₇/H₂SO₄==> R₂C=O

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Q9 (CBSE 2M)
Write Reimer-Tiemann reaction.

Ans:
C₆H₅OH + CHCl₃ + 3NaOH ==heat==> o-HO-C₆H₄-CHO + 3NaCl + H₂O
(major product = salicylaldehyde)

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Q10 (JEE Numerical)
How many alcohols are possible for C₄H₁₀O?

Ans: 4 alcohols

1. CH₃CH₂CH₂CH₂OH → 1-butanol (1°)
2. CH₃CH₂CH(OH)CH₃ → 2-butanol (2°)
3. (CH₃)₂CHCH₂OH → 2-methylpropan-1-ol (1°)
4. (CH₃)₃COH → 2-methylpropan-2-ol (3°)

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Q11 (CBSE 2M)
How is phenol detected?

Ans:

1. FeCl₃ test → violet/purple colour
2. Bromine water test:
   C₆H₅OH + 3Br₂(aq) ---> C₆H₂Br₃OH↓ + 3HBr
   (white ppt of 2,4,6-tribromophenol)

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Q12 (JEE MCQ)
Anisole (C₆H₅OCH₃) with excess HI gives:
(a) C₆H₅I + CH₃OH
(b) C₆H₅OH + CH₃I
(c) C₆H₅I + CH₃I
(d) C₆H₅OH + CH₃OH

Ans: (b) C₆H₅OH + CH₃I
Aryl C-O bond too strong → only alkyl C-O cleaves.

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Q13 (CBSE 3M)
Explain:
(i) Phenol does not react with NaHCO₃
(ii) o-nitrophenol more volatile than p-nitrophenol
(iii) Diethyl ether has much lower BP than n-butanol

Ans:
(i) Phenol (pKa ~10) weaker than carbonic acid (pKa ~6.4) → cannot displace CO₂.

(ii) o-nitrophenol → intramolecular H-bond → no intermolecular H-bonding → lower BP → more volatile.
p-nitrophenol → intermolecular H-bond → higher BP → less volatile.

(iii) Ether has no O-H → no intermolecular H-bonding → BP -24°C.
n-butanol has O-H → strong H-bonding → BP 118°C.

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Q14 (JEE)
1° alcohol oxidised with KMnO₄ — products in sequence?

Ans:
RCH₂OH ==mild==> RCHO ==further==> RCOOH
CH₃CH₂OH ---> CH₃CHO ---> CH₃COOH

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Q15 (CBSE 2M)
Which compound gives iodoform test? Give equation.

Ans: Compounds with CH₃CH(OH)- group.
Ethanol gives positive iodoform test.
CH₃CH₂OH + I₂ + NaOH ---> CHI₃↓ + HCOONa + NaI + H₂O
(yellow ppt with antiseptic smell)

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Q16 (JEE MCQ)
Dehydration of (CH₃)₃COH with conc H₂SO₄ gives mainly:
(a) 1-butene
(b) 2-methylpropene
(c) 2-butene
(d) 2-methylpropan-1-ol

Ans: (b) 2-methylpropene
(CH₃)₃COH ==H₂SO₄==> (CH₃)₂C=CH₂ + H₂O
(Saytzeff's rule → more substituted alkene)

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Q17 (CBSE 2M)
Why does phenol undergo electrophilic substitution more easily than benzene?

Ans: -OH donates lone pair into ring through resonance → ring becomes electron-rich → more reactive toward E⁺. Substitution at ortho and para positions.

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Q18 (JEE Assertion-Reason)
A: Alcohols have higher BP than ethers of same molecular mass.
R: Alcohols form intermolecular H-bonds; ethers do not.

Ans: Both A and R correct. R correctly explains A.
C₂H₅OH (78°C) >> CH₃OCH₃ (-24°C)

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Q19 (CBSE 3M)
Reactions of phenol with:
(i) Na metal
(ii) Br₂/water
(iii) Conc HNO₃

Ans:
(i) 2C₆H₅OH + 2Na ---> 2C₆H₅ONa + H₂

(ii) C₆H₅OH + 3Br₂(aq) ---> C₆H₂Br₃OH↓ + 3HBr

(iii) C₆H₅OH + dil HNO₃ ---> o-nitrophenol + p-nitrophenol
With fuming HNO₃ ---> 2,4,6-trinitrophenol (picric acid)

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Q20 (JEE MCQ)
Victor Meyer test — colour for 2° alcohol:
(a) Red
(b) Blue
(c) Colourless
(d) Violet

Ans: (b) Blue
1° → red
2° → blue
3° → colourless