Organic Chemistry: Haloalkanes and Haloarenes

1. Introduction

Haloalkanes and haloarenes contain halogen atoms (F, Cl, Br, I) bonded to sp³ and sp² carbon atoms respectively. They are versatile intermediates in organic synthesis.

2. Nomenclature

Common name: Alkyl halide (methyl chloride). IUPAC: Halo as prefix (chloromethane).

The halogen is named as fluoro, chloro, bromo, iodo. For compounds with functional groups, halogen is a substituent.

3. Preparation

3.1 Haloalkanes

  1. From alcohols: R-OH + HX → R-X + H₂O (using Lucas reagent for distinction).
  2. From alkenes: Addition of HX or X₂ (Markovnikov's rule).
  3. Free radical halogenation of alkanes.
  4. From alkenes via hydroboration (anti-Markovnikov addition of HBr with peroxide).

3.2 Haloarenes

  1. Direct halogenation of benzene (electrophilic substitution: FeCl₃ catalyst).
  2. Sandmeyer reaction: Diazonium salt → Aryl halide (CuCl/CuBr).
  3. Balz-Schiemann reaction: Diazonium tetrafluoroborate → Aryl fluoride.

4. SN1 and SN2 Mechanisms

4.1 SN2 (Substitution Nucleophilic Bimolecular)

  • One step, concerted. Inversion of configuration.
  • Favoured by: Primary substrate, strong nucleophile, aprotic solvent (polar).
  • Rate = k[RX][Nu⁻]

4.2 SN1 (Substitution Nucleophilic Unimolecular)

  • Two steps. Carbocation intermediate. Racemization.
  • Favoured by: Tertiary substrate, weak nucleophile, protic solvent.
  • Rate = k[RX]

5. Grignard Reagent

R-X + Mg → RMgX (in dry ether). Highly reactive, used in:

  • Synthesis of alcohols (with carbonyl compounds).
  • Synthesis of alkanes (with water).
  • Synthesis of carboxylic acids (with CO₂).

'Grignard reagents must be prepared in absolutely dry ether. Even traces of water destroy them.'

5. Comparison of SN1 and SN2 Reactions

FeatureSN1SN2
Full nameUnimolecular nucleophilic substitutionBimolecular nucleophilic substitution
StepsTwo (carbocation intermediate)One (concerted)
KineticsRate = k[RX]Rate = k[RX][Nu⁻]
Substrate3° > 2° >> 1°1° > 2° >> 3°
NucleophileWeak favouredStrong favoured
SolventPolar proticPolar aprotic
StereochemistryRacemisationInversion (Walden inversion)
RearrangementPossible (carbocation rearrangement)Not possible
Leaving groupGood leaving group neededGood leaving group needed

'SN1 reactions may involve carbocation rearrangement (hydride or alkyl shift) to form a more stable carbocation. SN2 reactions never involve rearrangement.'

6. Polyhalogen Compounds

Chloroform (CHCl₃): Anaesthetic, solvent. Forms phosgene (COCl₂) on oxidation. Carbon tetrachloride (CCl₄): Fire extinguisher, solvent. DDT: Insecticide.

7. Worked Problems

Problem 1: Arrange 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane in order of SN1 reactivity. Solution: More stable carbocation → faster SN1. 3° > 2° > 1°. So: 2-bromo-2-methylpropane > 2-bromobutane > 1-bromobutane.

Problem 2: Write the product of CH₃CH₂Br + aq. KOH. Solution: CH₃CH₂OH (ethyl alcohol). Aqueous KOH gives substitution (SN2). Alcoholic KOH gives elimination (alkene).

8. Common Mistakes

'Students often confuse whether alcoholic or aqueous KOH is used. Aqueous KOH gives substitution (OH), alcoholic KOH gives elimination (alkene).'

9. ISC Exam Focus

TopicTheory MarksPractical Marks
Nomenclature21
SN1/SN2 mechanisms42
Grignard reagent31
Polyhalogen compounds21

10. Self-Test Questions

  1. Distinguish between SN1 and SN2 reactions with examples.
  2. How is Grignard reagent prepared? Give two synthetic applications.
  3. Explain why haloarenes are less reactive than haloalkanes towards nucleophilic substitution.
  4. Write the mechanism for the preparation of bromoethane from ethene.
  5. Complete: C₆H₅N₂⁺Cl⁻ + CuCl → ?

11. Additional Reactions and Distinction Tests

Distinction Between Alkyl and Aryl Halides

TestAlkyl HalideAryl Halide
AgNO₃ (alcoholic)Immediate precipitate of AgXNo precipitate (even on heating)
Aq. KOH (boiling)Hydrolyses readilyNo reaction
NaOH (high pressure, 300°C)-Forms phenol (Dow process)

'Lack of reactivity in haloarenes is due to resonance delocalisation of the lone pair on halogen into the benzene ring, giving partial double bond character to the C-X bond. This makes the C-X bond stronger and less susceptible to nucleophilic attack.'

Additional Worked Problems

Problem A: Write the product of CH₃CH₂Br + alc. KOH.

Solution: Alcoholic KOH favours elimination. CH₂=CH₂ (ethene) is formed along with KBr and H₂O.

Problem B: Complete the reaction: C₆H₅Cl + NaOH (high pressure, 300°C) → ?

Solution: C₆H₅OH (phenol) is formed by the Dow process. The reaction proceeds through a benzyne intermediate.

Problem C: Which alkyl halide reacts faster in SN2 — CH₃CH₂Br or (CH₃)₂CHBr? Explain.

Solution: CH₃CH₂Br (primary) reacts faster. In SN2, the nucleophile attacks the carbon from the back, and bulky alkyl groups hinder this approach (steric hindrance). (CH₃)₂CHBr is secondary and has more steric hindrance.

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