*Class 12 Chemistry | CBSE 2025-26* ━━━━━━━━━━━━━━━━━━━━

KEY CONCEPTS
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1. CLASSIFICATION

• Haloalkanes → halogen on sp³ carbon (R-X)
• Haloarenes → halogen on benzene ring (Ar-X)
• Types by carbon: 1° (primary), 2° (secondary), 3° (tertiary)
• Types by halogen count: mono, di, tri, tetra

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2. C-X BOND NATURE

• Polar covalent bond (C is δ+, X is δ-)
• Bond length: C-F < C-Cl < C-Br < C-I
• Bond strength: C-F > C-Cl > C-Br > C-I
• Reactivity: C-I > C-Br > C-Cl > C-F

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3. IUPAC NOMENCLATURE

• Halogen = prefix: fluoro / chloro / bromo / iodo
• CH₃CH₂Cl → Chloroethane
• CH₃CHBrCH₃ → 2-Bromopropane
• CH₃CHClCH₂CH₃ → 2-Chlorobutane
• C₆H₅Cl → Chlorobenzene

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4. METHODS OF PREPARATION

From Alkanes:
CH₄ + Cl₂ ==hv==> CH₃Cl + HCl

From Alkenes:
CH₂=CH₂ + HBr ---> CH₃CH₂Br

From Alcohols:
ROH + PCl₅ ---> RCl + POCl₃ + HCl
ROH + SOCl₂ ---> RCl + SO₂ + HCl
(SOCl₂ preferred — pure product)

Finkelstein:
RCl + NaI ==acetone==> RI + NaCl↓

Sandmeyer's:
C₆H₅N₂⁺Cl⁻ + CuCl ==heat==> C₆H₅Cl + N₂
C₆H₅N₂⁺Cl⁻ + CuBr ==heat==> C₆H₅Br + N₂

Wurtz:
2RX + 2Na ==dry ether==> R-R + 2NaX

Grignard:
RX + Mg ==dry ether==> RMgX

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5. PHYSICAL PROPERTIES

• BP order: RI > RBr > RCl > RF
• BP increases with chain length
• Branching decreases BP
• Density: haloalkanes > parent alkane
• CHCl₃ and CCl₄ are heavier than water
• Slightly soluble in water

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6. CHEMICAL REACTIONS

SN2

• One step, concerted
• Walden inversion (configuration flips)
• Rate = k[RX][Nu⁻]
• Favoured by: 1° halide + strong Nu + polar aprotic solvent
• Reactivity: 1° > 2° > 3°

SN1

• Two steps via carbocation
• Racemisation
• Rate = k[RX]
• Favoured by: 3° halide + weak Nu + polar protic solvent
• Reactivity: 3° > 2° > 1°

Elimination with KOH/alc:
CH₃CH₂Br + KOH ==alc==> CH₂=CH₂ + KBr + H₂O

Saytzeff's Rule → more substituted alkene = major product

CH₃CHBrCH₂CH₃ + KOH ==alc==>
Major: CH₃CH=CHCH₃ (but-2-ene)
Minor: CH₂=CHCH₂CH₃ (but-1-ene)

Haloarenes — Electrophilic Substitution (easy):

• Halogen = ortho/para director
• Halogenation, Nitration, Sulphonation on ring

Haloarenes — Nucleophilic Substitution (hard):
C₆H₅Cl + NaOH ==623K, 300atm==> C₆H₅OH + NaCl

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7. OPTICAL ACTIVITY

• Chiral carbon = 4 different groups attached
• Enantiomers = non-superimposable mirror images
• Racemic mixture = equal d + l = optically inactive
• SN1 → racemisation
• SN2 → inversion (Walden inversion)
• CIP rules → R/S configuration

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8. POLYHALOGEN COMPOUNDS

Chloroform (CHCl₃) — solvent, anaesthetic
Carbon tetrachloride (CCl₄) — fire extinguisher
Freon (CCl₂F₂) — refrigerant, causes ozone depletion
DDT (C₁₄H₉Cl₅) — insecticide, banned
Iodoform (CHI₃) — antiseptic

CHCl₃ oxidation in air + light:
2CHCl₃ + O₂ ==light==> 2COCl₂ + 2HCl
(COCl₂ = phosgene = toxic)
Store in dark amber bottle + add 1% ethanol

DDT = non-biodegradable → biomagnification → banned

Freons destroy ozone:
CCl₂F₂ ==UV==> Cl• + •CClF₂
Cl• + O₃ ---> ClO• + O₂

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Q&A — 20 QUESTIONS
CBSE + JEE Mains Mix
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Q1 (CBSE 1M)
Arrange in increasing order of BP:
CH₃F, CH₃Cl, CH₃Br, CH₃I

Ans: CH₃F < CH₃Cl < CH₃Br < CH₃I
BP increases with molecular mass → stronger van der Waals forces.

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Q2 (JEE MCQ)
Finkelstein reaction uses:
(a) KI/water
(b) NaI/acetone
(c) NaI/ethanol
(d) KI/acetone

Ans: (b) NaI/acetone
RCl + NaI ==acetone==> RI + NaCl↓
NaCl insoluble in acetone → equilibrium shifts forward.

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Q3 (CBSE 2M)
Why is C-X bond in haloarenes shorter and stronger than in haloalkanes?

Ans: In haloarenes, lone pair of halogen delocalises into benzene ring through resonance → C-X gets partial double bond character → shorter and stronger. No such resonance in haloalkanes.

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Q4 (JEE)
(R)-2-bromobutane reacts with NaOH via SN2. Product configuration?

Ans: SN2 = backside attack = Walden inversion
(R)-2-bromobutane → (S)-2-butanol

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

Ans:
C₆H₅N₂⁺Cl⁻ + CuCl ==heat==> C₆H₅Cl + N₂
C₆H₅N₂⁺Cl⁻ + CuBr ==heat==> C₆H₅Br + N₂

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Q6 (JEE MCQ)
SN1 is favoured by:
(a) 1° halide + strong Nu
(b) 3° halide + weak Nu + polar protic solvent
(c) 1° halide + polar aprotic solvent
(d) 2° halide + strong base

Ans: (b)
3° carbocation most stable. Polar protic solvent stabilises carbocation.

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Q7 (CBSE 3M)
Complete:
(i) CH₃CH₂Br + KOH(aq) ---> ?
(ii) CH₃CH₂Br + KOH(alc) ---> ?
(iii) CH₃Br + Mg ==dry ether==> ?

Ans:
(i) CH₃CH₂Br + KOH(aq) ---> CH₃CH₂OH + KBr
(ii) CH₃CH₂Br + KOH(alc) ---> CH₂=CH₂ + KBr + H₂O
(iii) CH₃Br + Mg ==dry ether==> CH₃MgBr

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Q8 (JEE MCQ)
Reactivity toward SN2:
(a) 3° > 2° > 1°
(b) 1° > 2° > 3°
(c) 2° > 1° > 3°
(d) All same

Ans: (b) 1° > 2° > 3°
More branching = more steric hindrance = less SN2 reactivity.

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Q9 (CBSE 2M)
Why is chloroform stored in dark amber bottles?

Ans:
2CHCl₃ + O₂ ==light==> 2COCl₂ + 2HCl
Dark bottle prevents photochemical oxidation to toxic phosgene.
1% ethanol added to convert any COCl₂ formed to harmless diethyl carbonate.

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Q10 (JEE Numerical)
Molecular formula C₃H₇Cl — how many structural isomers?

Ans: 2 isomers

1. CH₃CH₂CH₂Cl → 1-Chloropropane
2. CH₃CHClCH₃ → 2-Chloropropane

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Q11 (CBSE 2M)
What is Wurtz reaction?

Ans:
2C₂H₅Br + 2Na ==dry ether==> C₄H₁₀ + 2NaBr
Limitation: Cannot prepare odd-carbon alkanes efficiently.

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Q12 (JEE MCQ)
Most reactive toward nucleophilic substitution:
(a) Chlorobenzene
(b) 2-chloro-2-methylpropane
(c) Chloromethane
(d) 2-chloropropane

Ans: (c) Chloromethane
Least steric hindrance → best SN2.
Chlorobenzene = least reactive due to resonance.

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Q13 (CBSE 3M)
Why are haloarenes less reactive than haloalkanes toward nucleophilic substitution?

Ans:

1. Resonance → C-X partial double bond → stronger, harder to break
2. Electron-rich benzene ring repels incoming Nu⁻
3. sp² carbon less electron-deficient than sp³
   Reaction requires 623K + 300 atm.

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Q14 (JEE)
2-bromobutane + KOH/alc → major product and rule?

Ans: Saytzeff's Rule → more substituted alkene is major product.
CH₃CHBrCH₂CH₃ + KOH ==alc==>
Major: CH₃CH=CHCH₃ (but-2-ene)
Minor: CH₂=CHCH₂CH₃ (but-1-ene)

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Q15 (CBSE 2M)
Decreasing SN1 reactivity:
(CH₃)₃CBr, CH₃Br, (CH₃)₂CHBr

Ans: (CH₃)₃CBr > (CH₃)₂CHBr > CH₃Br
SN1 depends on carbocation stability: 3° > 2° > 1°

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Q16 (JEE MCQ)
IUPAC name of CH₃-CHBr-CH₂-CH₃:
(a) 1-bromobutane
(b) 2-bromobutane
(c) 3-bromobutane
(d) 1-bromo-2-methylpropane

Ans: (b) 2-bromobutane

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Q17 (CBSE 2M)
What happens when chlorobenzene reacts with dilute NaOH?

Ans: No reaction under normal conditions.
Resonance makes C-Cl strong + ring repels OH⁻.
Reacts only under extreme conditions:
C₆H₅Cl + NaOH ==623K, 300atm==> C₆H₅OH + NaCl

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Q18 (JEE Assertion-Reason)
A: BP of n-butyl chloride > tert-butyl chloride
R: Branching decreases surface area → weaker van der Waals forces.

Ans: Both A and R correct. R correctly explains A.
n-BuCl (78°C) > t-BuCl (51°C)

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Q19 (CBSE 3M)
Prepare 1-iodobutane from:
(i) 1-butanol
(ii) 1-chlorobutane

Ans:
(i) CH₃CH₂CH₂CH₂OH + HI ---> CH₃CH₂CH₂CH₂I + H₂O

(ii) Finkelstein:
CH₃CH₂CH₂CH₂Cl + NaI ==acetone==> CH₃CH₂CH₂CH₂I + NaCl↓

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Q20 (JEE MCQ)
Which is correct about DDT?
(a) It is biodegradable
(b) It causes biomagnification in food chains
(c) Contains only chlorine atoms
(d) Currently used as pesticide in India

Ans: (b)
DDT non-biodegradable → accumulates in fat → concentration increases up food chain → banned.