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State Board · Class 12 · Chemistry

Lesson 10: Halogen Derivatives

7 concepts · 56 practice questions with answers · Free to read

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Classification and nomenclature of halogen derivatives

From daily life

Fire extinguishers, chloroform and refrigerant gases are all halogen derivatives.

Simple definition

Haloalkanes are R–X (X on an sp³ C); haloarenes are Ar–X (X on an sp² C); 1°, 2°, 3° depend on the carbons attached to the C–X carbon.

ClassificationClassification and naming of halogen derivativesTypeFormulaIUPAC1° (primary)CH₃CH₂CH₂Cl1-chloropropane2° (secondary)(CH₃)₂CHBr2-bromopropane3° (tertiary)(CH₃)₃CCl2-chloro-2-methylpropaneAllylicCH₂=CH–CH₂Cl3-chloroprop-1-eneBenzylicC₆H₅CH₂Cl(chloromethyl)benzeneVinylicCH₂=CHClchloroetheneArylC₆H₅ClchlorobenzeneC–X on sp³ C ⇒ haloalkane; on sp² C ⇒ haloarene / vinylicDihalides: geminal (same C) and vicinal (adjacent C)C₄H₉Cl has 4 structural isomers

In the figure : Classification and names.

Rule 1: Classification — 1°, 2°, 3° alkyl; allylic (next to C=C); benzylic (next to ring); vinylic and aryl (sp² C).

  • Easy : CH₃CH₂Cl is 1°.
  • Medium : (CH₃)₂CHCl is 2°.
  • Exam-type : (CH₃)₃CCl is 3°.

Rule 2: IUPAC naming — Longest chain; halo prefixes (chloro, bromo, iodo, fluoro); alphabetical order; lowest locants.

  • Easy : CH₃CHBrCH₃ = 2-bromopropane.
  • Medium : (CH₃)₃CCl = 2-chloro-2-methylpropane.
  • Exam-type : C₆H₅Cl = chlorobenzene.

Rule 3: Dihalides and isomerism — Geminal (both X on one C) and vicinal (adjacent C); C₄H₉Cl has 4 structural isomers.

  • Easy : CH₃CHCl₂ is geminal.
  • Medium : CH₂ClCH₂Cl is vicinal.
  • Exam-type : Four: 1-, 2-, 2-methyl-1-, 2-methyl-2-.

Calculation: chlorine % in C₄H₉Cl

  • Step 1: M = 4×12 + 9×1 + 35.5 = 92.5.
  • Step 2: Cl % = 35.5/92.5 × 100.
  • Step 3: Cl % = 38.4.

CET level (MCQ)

  • CET 1) Name of (CH₃)₃CCl? 1) 2-chloro-2-methylpropane 2) 1-chlorobutane 3) 2-chlorobutane 4) tert-pentyl chloride — Answer: 1).
  • CET 2) Vinylic halide? 1) CH₂=CHCl 2) C₆H₅CH₂Cl 3) CH₃Cl 4) C₆H₅Cl — Answer: 1).

Real-life use: Solvents, drugs, refrigerants, pesticides.

Remember

sp³ C–X = haloalkane; sp² C–X = haloarene/vinylic; geminal vs vicinal.

Common mistake

Calling a benzylic halide a haloarene.

CET trap

In C₆H₅CH₂Cl the X is on an sp³ C ⇒ alkyl type.

Shortcut

Memory clue: "geminal = twins at one house, vicinal = neighbours".

Spot the difference

TypeExample
1°CH₃CH₂Cl
2°(CH₃)₂CHBr
3°(CH₃)₃CCl
ArylC₆H₅Cl

Practice questions

Easy
  1. 1Haloalkane vs haloarene.
  2. 2Identify 1°, 2°, 3°.
  3. 3Name (CH₃)₂CHBr.
Medium
  1. 4Geminal and vicinal.
  2. 5Isomers of C₄H₉Cl.
  3. 6Allylic and benzylic.
Challenging
  1. 7Formula of vinyl chloride.
  2. 8Cl % in C₄H₉Cl.
Show answers (try them yourself first)
  1. 1sp³ vs sp² C.
  2. 2Count attached carbons.
  3. 32-bromopropane.
  4. 4Same C vs adjacent C.
  5. 54.
  6. 6Next to C=C / ring.
  7. 7CH₂=CHCl.
  8. 838.4 %.

Methods of preparation of haloalkanes and haloarenes

From daily life

As a dish can be cooked several ways, R–X has several routes.

Simple definition

Main routes: from alcohols (SOCl₂/HX), from alkenes (HX), from alkanes (X₂/hν) and halogen exchange (Finkelstein, Swarts).

PreparationPreparing haloalkanesR–XhaloalkaneR–OH + SOCl₂ → R–Cl + SO₂ + HClR–OH + HX / ZnCl₂ (Lucas)C=C + HX (Markovnikov)RH + X₂ / hν (free radical)Finkelstein: RCl + NaI (acetone) → RI + NaClSwarts: RBr + AgF → RF + AgBrAryl: C₆H₆ + Cl₂/FeCl₃; ArN₂⁺ + CuCl (Sandmeyer)Ex. 7.4 g butan-1-ol ⇒ 9.25 g C₄H₉Cl (100 %); at 80 % ⇒ 7.4 g

In the figure : Routes of preparation.

Rule 1: From alcohols — R–OH + SOCl₂ ⇒ R–Cl + SO₂ + HCl (pure product); R–OH + HX/ZnCl₂ (Lucas); 3° reacts fastest; PCl₅, PCl₃ also work.

  • Easy : SOCl₂ gives gaseous by-products.
  • Medium : HX reactivity HI > HBr > HCl.
  • Exam-type : 7.4 g butanol ⇒ 9.25 g (100 %).

Rule 2: From alkanes and alkenes — Alkane + X₂/hν: radical, mixture; alkene + HX: Markovnikov, the 2° product is major.

  • Easy : CH₃CH=CH₂ + HBr ⇒ 2-bromopropane.
  • Medium : Alkanes: poor selectivity.
  • Exam-type : Peroxide gives anti-Markovnikov.

Rule 3: Exchange and arenes — Finkelstein: RCl + NaI ⇒ RI (acetone); Swarts: RBr + AgF ⇒ RF; arenes: X₂/FeCl₃; Sandmeyer: ArN₂⁺ + CuCl.

  • Easy : NaCl is insoluble in acetone, driving the reaction.
  • Medium : Direct ring iodination is hard.
  • Exam-type : Fluorination: Balz–Schiemann.

Calculation: C₄H₉Cl (M = 92.5) from 7.4 g butan-1-ol (M = 74), 80 % yield

  • Step 1: n = 7.4/74 = 0.1 mol.
  • Step 2: theoretical mass = 0.1 × 92.5 = 9.25 g.
  • Step 3: 80 % ⇒ 9.25 × 0.8 = 7.4 g.

CET level (MCQ)

  • CET 1) Gives pure R–Cl? 1) SOCl₂ 2) Cl₂/hν 3) HCl 4) NaCl — Answer: 1).
  • CET 2) Finkelstein reaction? 1) RCl + NaI ⇒ RI 2) RBr + AgF ⇒ RF 3) RX + Mg 4) RX + Na — Answer: 1).

Real-life use: Drug synthesis, solvent production.

Remember

R–OH + SOCl₂ (clean); alkene + HX (Markovnikov); NaI/acetone ⇒ RI; AgF ⇒ RF.

Common mistake

Mixing up Finkelstein and Swarts.

CET trap

Finkelstein ⇒ I; Swarts ⇒ F.

Shortcut

Memory clue: "Finkelstein = iodine, Swarts = fluorine".

Spot the difference

ReactionProduct
FinkelsteinR–I
SwartsR–F
SandmeyerAr–Cl

Practice questions

Easy
  1. 1Three routes from alcohols to R–Cl.
  2. 2Advantage of SOCl₂.
  3. 3Explain Finkelstein.
Medium
  1. 4Explain Swarts.
  2. 5Why does alkane chlorination give mixtures?
  3. 6Sandmeyer reaction.
Challenging
  1. 77.4 g butanol ⇒ product (80 %).
  2. 8Product of propene + HBr.
Show answers (try them yourself first)
  1. 1SOCl₂, HX/ZnCl₂, PCl₅.
  2. 2SO₂ and HCl are gases.
  3. 3RCl + NaI ⇒ RI.
  4. 4RBr + AgF ⇒ RF.
  5. 5Radical, many sites.
  6. 6ArN₂⁺ + CuCl.
  7. 77.4 g.
  8. 82-bromopropane.

Physical properties of halogen derivatives

From daily life

Chloroform sinks in water — dense and immiscible.

Simple definition

The C–X bond is polar: Cδ⁺–Xδ⁻; b.p. RI > RBr > RCl > RF; insoluble in water; density rises.

PhysicalPhysical properties and the C–X bondBondlength pmenthalpy kJb.p. °CCH₃–F139452—CH₃–Cl178351–24CH₃–Br1932934CH₃–I21423442b.p.: RI > RBr > RCl > RF; 1° > 2° > 3° (branching lowers)Longer bond ⇒ weaker bond ⇒ reactivity RI > RBr > RClInsoluble in water (no H-bonds); soluble in organic solventsDensity rises RCl < RBr < RI; CHCl₃ sinks in waterPolar bond Cδ⁺–Xδ⁻; electronegativity F > Cl > Br > I

In the figure : Properties and the C–X bond.

Rule 1: Boiling points — Larger halogen ⇒ higher b.p.; 1° > 2° > 3° (branching reduces surface area); CH₃Cl –24, CH₃Br 4, CH₃I 42 °C.

  • Easy : Butyl chloride isomers: n- > iso- > tert-.
  • Medium : Higher b.p. than the parent alkane.
  • Exam-type : Dipole–dipole plus dispersion.

Rule 2: Solubility and density — Insoluble in water (no H-bonds with water); soluble in organic solvents; density RCl < RBr < RI.

  • Easy : CHCl₃ sinks in water.
  • Medium : CH₃Cl is lighter.
  • Exam-type : Iodides are heaviest.

Rule 3: The C–X bond — Bond length F 139, Cl 178, Br 193, I 214 pm; enthalpy 452, 351, 293, 234 kJ/mol; reactivity RI > RBr > RCl.

  • Easy : C–F is strongest.
  • Medium : C–I is weakest.
  • Exam-type : Longer ⇒ weaker.

Reasoning: why is tert-butyl chloride's b.p. lower than n-butyl chloride's

  • Step 1: The tert- isomer is nearly spherical with less surface area.
  • Step 2: Weaker dispersion forces.
  • Step 3: Lower b.p. (about 51 °C versus 78 °C).

CET level (MCQ)

  • CET 1) Highest b.p.? 1) CH₃I 2) CH₃Br 3) CH₃Cl 4) CH₃F — Answer: 1).
  • CET 2) Longest C–X bond? 1) C–I 2) C–Br 3) C–Cl 4) C–F — Answer: 1).

Real-life use: Solvents, stain removal, dry cleaning.

Remember

b.p. RI > RBr > RCl > RF; 1° > 2° > 3°; insoluble; C–I is weak.

Common mistake

Thinking RF is most reactive.

CET trap

C–F is strongest ⇒ least reactive.

Shortcut

Memory clue: "iodine is big, boils late".

Spot the difference

BondLength pm
C–F139
C–Cl178
C–Br193
C–I214

Practice questions

Easy
  1. 1State the b.p. order.
  2. 2Effect of branching on b.p.
  3. 3Why insoluble in water?
Medium
  1. 4Order of density.
  2. 5Order of C–X bond enthalpy.
  3. 6Reactivity order.
Challenging
  1. 7A feature of CHCl₃.
  2. 8Bond polarity.
Show answers (try them yourself first)
  1. 1RI > RBr > RCl > RF.
  2. 2Branching ⇒ lower b.p.
  3. 3No H-bonding.
  4. 4RCl < RBr < RI.
  5. 5C–F > C–Cl > C–Br > C–I.
  6. 6RI > RBr > RCl.
  7. 7Sinks in water.
  8. 8Cδ⁺–Xδ⁻.

Optical isomerism and chirality

From daily life

Left and right hands are mirror images that cannot be superimposed — so are chiral molecules.

Simple definition

A chiral molecule is non-superimposable on its mirror image; a chiral C* has four different groups; enantiomers are mirror-image pairs.

ChiralChirality: 2-bromobutaneBrC₂H₅CH₃HC*(R)BrC₂H₅CH₃HC*(S)mirrorFour different groups ⇒ chiral C* ⇒ two mirror-image enantiomersd (+) and l (–) rotate plane-polarised light oppositely(±) racemic mixture ⇒ optically inactive (external compensation)n stereocentres ⇒ 2ⁿ stereoisomers (n = 1 ⇒ 2)

In the figure : 2-bromobutane.

Live figure

Optical isomerism and chirality

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Rule 1: Chiral C* — Four different groups ⇒ 2 enantiomers; CH₃CHBrC₂H₅ is chiral, CH₃CH₂Br is not.

  • Easy : 2-bromobutane is chiral.
  • Medium : 2-bromopropane is achiral.
  • Exam-type : n C* ⇒ 2ⁿ stereoisomers.

Rule 2: Optical activity — d (+) rotates clockwise, l (–) anticlockwise by the same angle; (±) racemic ⇒ inactive.

  • Easy : Racemic mixture is 50:50.
  • Medium : (R)/(S) by the CIP rules.
  • Exam-type : External compensation.

Rule 3: Stereochemistry in reactions — SN2 ⇒ inversion (Walden); SN1 ⇒ racemisation; destroying the chiral centre removes activity.

  • Easy : (R)-2-bromobutane + OH⁻ (SN2) ⇒ (S)-alcohol.
  • Medium : SN1 ⇒ 50:50.
  • Exam-type : Often only one enantiomer of a drug is active.

Reasoning: why CH₃CH₂CHBrCH₃ is chiral

  • Step 1: C-2 carries Br, H, CH₃ and C₂H₅ — four different groups.
  • Step 2: The mirror image is different and cannot be superimposed.
  • Step 3: Hence chiral, with 2 enantiomers (2¹).

CET level (MCQ)

  • CET 1) Chiral? 1) 2-bromobutane 2) 2-bromopropane 3) bromoethane 4) 1-bromopropane — Answer: 1).
  • CET 2) A racemic mixture is? 1) inactive 2) only d 3) only l 4) active — Answer: 1).

Real-life use: Drug design, fragrances, food chemistry.

Remember

Four different groups ⇒ chiral; enantiomers; racemic = inactive; 2ⁿ.

Common mistake

Treating a racemic mixture as active.

CET trap

(±) rotations cancel out.

Shortcut

Memory clue: "four different = a hand (chiral)".

Spot the difference

FormRotation
d (+)right
l (–)left
(±)zero

Practice questions

Easy
  1. 1What is chiral?
  2. 2What is C*?
  3. 3What are enantiomers?
Medium
  1. 4Racemic mixture.
  2. 5Why is 2-bromobutane chiral?
  3. 6n = 2 C* ⇒ isomers.
Challenging
  1. 7Walden inversion in SN2.
  2. 8Racemisation in SN1.
Show answers (try them yourself first)
  1. 1Not superimposable.
  2. 2C with four different groups.
  3. 3Mirror-image pair.
  4. 450:50, inactive.
  5. 5Four different groups.
  6. 64.
  7. 7Configuration inverted.
  8. 8Planar carbocation.

Nucleophilic substitution: SN1 and SN2

From daily life

A new guest takes the seat of one who leaves — a nucleophile replaces the leaving group.

Simple definition

SN2: one step, rate = k[RX][Nu], inversion; SN1: two steps, rate = k[RX], carbocation, racemisation.

SNSN2 आणि SN1SN2Nu⁻ ··· C ··· X[Nu···C···X]⁻ transition stateone step; inversion (Walden)SN1R–X → R⁺ + X⁻ → R–Nutwo steps; carbocationracemisationRatek[RX][Nu]k[RX]Order21SubstrateCH₃ > 1° > 2° > 3°3° > 2° > 1°Solventpolar aproticpolar proticStereoinversionracemicEx. [RX] 0.1⇒0.2, [OH⁻] 0.1⇒0.3: SN2 rate ×6; SN1 rate ×2Leaving group ability: I⁻ > Br⁻ > Cl⁻ ≫ F⁻

In the figure : SN2 vs SN1.

Live figure

Nucleophilic substitution: SN1 and SN2

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Rule 1: SN2 — Back-side attack; a transition state; CH₃ > 1° > 2° > 3° (steric hindrance); polar aprotic solvents.

  • Easy : CH₃Br + OH⁻ ⇒ CH₃OH.
  • Medium : Doubling both ⇒ rate ×4.
  • Exam-type : Walden inversion.

Rule 2: SN1 — R–X ⇒ R⁺ + X⁻ (slow), then + Nu; 3° > 2° > 1° (stable R⁺); polar protic solvents; racemic.

  • Easy : (CH₃)₃CBr + H₂O ⇒ (CH₃)₃COH.
  • Medium : Doubling [Nu] ⇒ same rate.
  • Exam-type : Benzylic and allylic cations are stable.

Rule 3: Leaving group — I⁻ > Br⁻ > Cl⁻ ≫ F⁻ (weaker base = better leaving group); 2° can follow both.

  • Easy : RI > RBr > RCl.
  • Medium : Solvent steers the mechanism.
  • Exam-type : Strong Nu favours SN2.

Calculation: [RX] 0.1 ⇒ 0.2 and [OH⁻] 0.1 ⇒ 0.3

  • Step 1: SN2: rate = k[RX][OH⁻] ⇒ 2 × 3 = 6 times.
  • Step 2: SN1: rate = k[RX] ⇒ 2 times.
  • Step 3: SN2 ×6; SN1 ×2.

CET level (MCQ)

  • CET 1) A 3° halide mostly reacts by? 1) SN1 2) SN2 3) neither 4) not E2 — Answer: 1).
  • CET 2) Walden inversion occurs in? 1) SN2 2) SN1 3) both 4) neither — Answer: 1).

Real-life use: Synthesis of alcohols, ethers, nitriles, drugs.

Remember

SN2: second order, inversion; SN1: first order, racemic; I⁻ is the best leaving group.

Common mistake

Assigning SN1 to 1° halides.

CET trap

1° ⇒ SN2; 3° ⇒ SN1; 2° mixed.

Shortcut

Memory clue: "SN2: order two, one step; SN1: order one, two steps".

Spot the difference

AspectSN2SN1
Ratek[RX][Nu]k[RX]
Steps12
Stereoinversionracemic

Practice questions

Easy
  1. 1SN2 mechanism.
  2. 2SN1 mechanism.
  3. 3Two differences.
Medium
  1. 4Reactivity order.
  2. 5Leaving group.
  3. 6[RX] ×2, [OH⁻] ×3 ⇒ SN2 rate.
Challenging
  1. 7What is Walden inversion?
  2. 8Role of the solvent.
Show answers (try them yourself first)
  1. 1Back-side attack.
  2. 2Carbocation.
  3. 3Order and stereochemistry.
  4. 4SN2: 1° > 2° > 3°.
  5. 5I⁻ > Br⁻ > Cl⁻.
  6. 66 times.
  7. 7Configuration inverts.
  8. 8Protic ⇒ SN1.

Elimination, organometallic reactions and haloarenes

From daily life

With a strong hot base the same halide eliminates instead of substituting.

Simple definition

Elimination: alcoholic KOH ⇒ alkene (Saytzeff); Grignard: RX + Mg ⇒ RMgX; Wurtz: 2RX + 2Na ⇒ R–R.

EliminationElimination, organometallics and haloarenesE2R–CH₂CH₂X + KOH(alc) → R–CH=CH₂ + KX + H₂OSaytzeff2-bromobutane ⇒ but-2-ene (more substituted) majorWurtz2 RX + 2 Na → R–R + 2 NaX (dry ether)GrignardRX + Mg → RMgX (dry ether)RMgX+ HCHO ⇒ 1° alcohol; + RCHO ⇒ 2°; + R₂CO ⇒ 3°Haloarenes: sp² C–Cl, resonance shortens it (169 pm) ⇒ NAS is hardEAS: o/p-directing yet deactivating (–I > +R)Nitro groups at o/p ⇒ NAS becomes easy (2,4-dinitrochlorobenzene)Fittig: 2 ArX + 2 Na ⇒ Ar–Ar; Wurtz–Fittig: ArX + RX + 2 Na ⇒ Ar–R

In the figure : Elimination, Grignard, haloarenes.

Rule 1: Elimination — R–CH₂CH₂X + alcoholic KOH ⇒ R–CH=CH₂; Saytzeff: the more substituted alkene is major; 2-bromobutane ⇒ but-2-ene.

  • Easy : Aqueous KOH ⇒ substitution.
  • Medium : Alcoholic KOH ⇒ elimination.
  • Exam-type : High temperature favours elimination.

Rule 2: Grignard and Wurtz — RX + Mg (dry ether) ⇒ RMgX; + HCHO ⇒ 1° alcohol; + RCHO ⇒ 2°; + R₂CO ⇒ 3°; Wurtz: 2 CH₃Br + 2 Na ⇒ CH₃CH₃.

  • Easy : Moisture destroys Grignard ⇒ dry conditions.
  • Medium : RMgX + H₂O ⇒ RH.
  • Exam-type : Fittig: 2 ArX + 2 Na ⇒ Ar–Ar.

Rule 3: Haloarenes — sp² C–Cl with resonance is shorter (169 pm), so NAS is hard; EAS: o/p-directing but deactivating; o/p nitro groups make NAS easy.

  • Easy : Chlorobenzene + NaOH at 623 K, 300 atm ⇒ phenol (Dow).
  • Medium : 2,4-dinitrochlorobenzene reacts easily.
  • Exam-type : Nitration ⇒ o/p-nitrochlorobenzene.

Reasoning: why the C–Cl bond in chlorobenzene is shorter

  • Step 1: Cl's lone pair conjugates with the ring.
  • Step 2: The C–Cl bond gains partial double-bond character.
  • Step 3: With sp² C the bond is shorter (169 vs 178 pm), so NAS is hard.

CET level (MCQ)

  • CET 1) By Saytzeff's rule, the major product is? 1) more substituted alkene 2) less substituted 3) alkane 4) alcohol — Answer: 1).
  • CET 2) RMgX + HCHO ⇒ (after hydrolysis)? 1) 1° alcohol 2) 2° 3) 3° 4) aldehyde — Answer: 1).

Real-life use: Drug synthesis, polymers, dyes.

Remember

Alcoholic KOH ⇒ elimination; Saytzeff; RMgX in dry ether; haloarenes resist NAS.

Common mistake

Mixing up aqueous and alcoholic KOH.

CET trap

Aqueous ⇒ substitution; alcoholic ⇒ elimination.

Shortcut

Memory clue: "alcoholic KOH eliminates to alkene".

Spot the difference

ReagentProduct
KOH (aq)R–OH
KOH (alc)alkene
Mg / etherRMgX
Na / etherR–R

Practice questions

Easy
  1. 1State Saytzeff's rule.
  2. 2Aqueous vs alcoholic KOH.
  3. 3How is a Grignard reagent made?
Medium
  1. 4RMgX + HCHO.
  2. 5Wurtz reaction.
  3. 6Why is NAS hard in haloarenes?
Challenging
  1. 7Effect of o/p nitro groups.
  2. 8Dow process.
Show answers (try them yourself first)
  1. 1More substituted alkene.
  2. 2Substitution vs elimination.
  3. 3Mg in dry ether.
  4. 41° alcohol.
  5. 52RX + 2Na ⇒ R–R.
  6. 6sp² and resonance.
  7. 7NAS becomes easy.
  8. 8ClC₆H₅ + NaOH ⇒ phenol.

Polyhalogen compounds, uses and environmental effects

From daily life

Iodoform on a wound and DDT in a field are both halogen compounds with very different effects.

Simple definition

Polyhalogen compounds contain several halogens: CHCl₃, CCl₄, CHI₃, freons, DDT.

PolyhaloPolyhalogen compounds and the environmentCHCl₃Chloroform: old anaesthetic; light + O₂ ⇒ COCl₂ (phosgene), so dark bottlesCCl₄Carbon tetrachloride: fire extinguisher, solvent; harms ozoneCHI₃Iodoform: antiseptic, yellow, characteristic smellCCl₂F₂Freon: refrigerant; Cl· destroys ozoneDDTInsecticide; non-biodegradable, accumulates in food chainsO₃ + Cl· → ClO· + O₂ ; ClO· + O → Cl· + O₂One Cl· can destroy thousands of O₃ molecules (catalytic cycle)Iodoform test: CH₃CO– + I₂/NaOH ⇒ yellow CHI₃

In the figure : Polyhalogen compounds.

Rule 1: CHCl₃ and CCl₄ — Chloroform: solvent, old anaesthetic; light + air ⇒ toxic COCl₂, so dark bottles with a little ethanol; CCl₄: solvent, fire extinguisher, harms ozone.

  • Easy : 2CHCl₃ + O₂ ⇒ 2COCl₂ + 2HCl.
  • Medium : CCl₄ damages the liver.
  • Exam-type : CCl₄ is non-flammable.

Rule 2: Iodoform and freons — CHI₃: antiseptic, yellow solid; CCl₂F₂: refrigerant and aerosol propellant; stable but releases Cl· in the stratosphere.

  • Easy : Haloform test.
  • Medium : Freons are non-flammable.
  • Exam-type : Montreal Protocol bans them.

Rule 3: DDT and ozone — DDT: insecticide, non-biodegradable, accumulates in fat; O₃ + Cl· ⇒ ClO· + O₂; ClO· + O ⇒ Cl· + O₂ (one Cl· destroys thousands of O₃).

  • Easy : DDT thins bird eggshells.
  • Medium : The ozone layer blocks UV.
  • Exam-type : HFCs are alternatives.

Reasoning: how freons deplete ozone

  • Step 1: UV breaks CCl₂F₂ to release Cl·.
  • Step 2: Cl· + O₃ ⇒ ClO· + O₂; ClO· + O ⇒ Cl· + O₂.
  • Step 3: Cl· is regenerated ⇒ chain continues (catalytic).

CET level (MCQ)

  • CET 1) Why is chloroform kept in dark bottles? 1) to avoid COCl₂ 2) for cold 3) smell 4) density — Answer: 1).
  • CET 2) Depletes the ozone layer? 1) CCl₂F₂ 2) CH₄ 3) CO₂ 4) N₂ — Answer: 1).

Real-life use: Solvents, cooling, antiseptics, pesticides.

Remember

CHCl₃ ⇒ phosgene; CHI₃ antiseptic; freons: Cl· depletes O₃; DDT non-biodegradable.

Common mistake

Thinking freons harm because they are toxic.

CET trap

Freons are stable; the harm comes from Cl·.

Shortcut

Memory clue: "one chlorine, a thousand ozones".

Spot the difference

CompoundUse
CHCl₃solvent, anaesthetic
CHI₃antiseptic
CCl₂F₂refrigerant
DDTinsecticide

Practice questions

Easy
  1. 1Uses and storage of CHCl₃.
  2. 2Uses and risks of CCl₄.
  3. 3Iodoform.
Medium
  1. 4Freons and ozone.
  2. 5Drawbacks of DDT.
  3. 6Write the ozone reactions.
Challenging
  1. 7Montreal Protocol.
  2. 8How is COCl₂ formed?
Show answers (try them yourself first)
  1. 1Solvent; dark bottle.
  2. 2Solvent; liver damage.
  3. 3Antiseptic, yellow.
  4. 4Cl· destroys ozone.
  5. 5Non-biodegradable, accumulates.
  6. 6Cl· + O₃ ⇒ ClO· + O₂.
  7. 7Ban on CFCs.
  8. 8CHCl₃ + O₂ (light).

Quick revision of the whole lesson

ConceptKey formula / fact
SN2 ratek[RX][Nu]
SN1 ratek[RX]
Stereoisomers2ⁿ
Saytzeffmore substituted alkene
Wurtz2RX + 2Na → R–R
GrignardRX + Mg → RMgX
FinkelsteinRCl + NaI → RI
Ozone lossCl· + O₃ → ClO· + O₂
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🎯 Most important points for the exam

  • Classification, naming.
  • Preparation.
  • Physical properties.
  • Chirality.
  • SN1, SN2.
  • Elimination, RMgX.
  • Polyhalogen compounds.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.