Definition: Compounds where one hydrogen atom in alkanes is replaced by a halogen atom (F, Cl, Br, I).
General Formula: R-X (alkyl halides), Ar-X (aryl halides)
Classification:
Halogenation of Benzene:
Benzene + X₂ (Cl₂ or Br₂) + Fe or FeX₃ catalyst → Halogenobenzene
Mechanism: Electrophilic substitution
Catalyst Role: FeX₃ polarizes X-X bond, generating X⁺ electrophile
Example: C₆H₆ + Cl₂ → C₆H₅Cl + HCl
Halogenoalkanes vs Halogenoarenes:
Chloroethane (R-Cl): Polar C-Cl bond, undergoes nucleophilic substitution
Chlorobenzene (Ar-Cl): Resonance stabilization, C-Cl bond has partial double bond character, undergoes electrophilic substitution
Reason: In chlorobenzene, lone pair on Cl delocalizes into benzene ring
General Reaction: R-X + Nu⁻ → R-Nu + X⁻
Key Components:
SN1 (Unimolecular):
SN2 (Bimolecular):
E1 (Unimolecular Elimination):
E2 (Bimolecular Elimination):
Halogenoarenes (aryl halides) are prepared by electrophilic substitution of benzene with halogens in the presence of catalysts.
General Reaction: C₆H₆ + X₂ → C₆H₅X + HX
Catalysts: Fe, FeCl₃, FeBr₃, AlCl₃
X₂ + FeX₃ → X⁺ + FeX₄⁻
C₆H₆ + Cl₂ → C₆H₅Cl + HCl (in presence of FeCl₃)
Conditions: Room temperature, anhydrous conditions
Product: Chlorobenzene
C₆H₆ + Br₂ → C₆H₅Br + HBr (in presence of FeBr₃)
Conditions: Room temperature, anhydrous conditions
Product: Bromobenzene
Structure: CH₃-CH₂-Cl
Bond Character: Polar covalent bond
Electron Distribution: Carbon δ⁺, Chlorine δ⁻
Reactivity: High towards nucleophiles
Reaction Type: Nucleophilic substitution
CH₃-CH₂-Cl + OH⁻ → CH₃-CH₂-OH + Cl⁻
Structure: C₆H₅-Cl
Bond Character: Partial double bond due to resonance
Electron Distribution: Delocalized in ring
Reactivity: Low towards nucleophiles
Reaction Type: Electrophilic substitution
C₆H₅-Cl + NO₂⁺ → o-NO₂-C₆H₄-Cl + p-NO₂-C₆H₄-Cl
The C-Cl bond in chlorobenzene has partial double bond character due to resonance:
Cl donates lone pair to benzene ring → C-Cl bond gains double bond character
This makes the bond shorter and stronger than in alkyl chlorides
Steps:
R-X → R⁺ + X⁻ (rate-determining)
R⁺ + Nu⁻ → R-Nu
Rate Law: Rate = k[R-X]
Favored by: 3° alkyl halides, polar protic solvents, weak nucleophiles
Stereochemistry: Racemic mixture (both enantiomers)
Steps:
Nu⁻ + R-X → [Nu—R—X]⁻ → R-Nu + X⁻
Rate Law: Rate = k[Nu⁻][R-X]
Favored by: 1° alkyl halides, polar aprotic solvents, strong nucleophiles
Stereochemistry: Inversion of configuration (backside attack)
The stability order of carbocations determines SN1 reaction rates:
Stability Order: 3° > 2° > 1° > methyl
Reason: Hyperconjugation and inductive effects from alkyl groups
| Alkyl Halide Type | Preferred Mechanism | Reason |
|---|---|---|
| Primary (1°) | SN2 | Less steric hindrance, unstable carbocation |
| Secondary (2°) | Both SN1 and SN2 | Moderate steric hindrance, moderately stable carbocation |
| Tertiary (3°) | SN1 | High steric hindrance, stable carbocation |
Steps:
R-X → R⁺ + X⁻
R⁺ + B⁻ → Alkene + H-B
Rate Law: Rate = k[R-X]
Favored by: 3° alkyl halides, weak bases, polar protic solvents
Regioselectivity: Follows Zaitsev’s rule (more substituted alkene)
Steps:
B⁻ + R-X → Alkene + H-B + X⁻
Rate Law: Rate = k[B⁻][R-X]
Favored by: Strong bases, polar aprotic solvents
Stereochemistry: Anti-periplanar arrangement required
| Alkyl Halide | Strong Base/Nucleophile | Weak Base/Nucleophile |
|---|---|---|
| Primary (1°) | SN2 (major), E2 (minor) | SN2 |
| Secondary (2°) | E2 (major), SN2 (minor) | SN1/E1 mixture |
| Tertiary (3°) | E2 | E1/SN1 mixture |
A common test to assess halogenoalkane reactivity uses aqueous silver nitrate (AgNO₃).
General Reaction: R-X + AgNO₃ → R-ONO₂ + AgX↓
The reactivity follows the order: RI > RBr > RCl > RF
Reason: Decreasing C-X bond strength down the group
| Halogenoalkane | Precipitate Color | Solubility in NH₄OH | Reactivity |
|---|---|---|---|
| R-F | No precipitate | – | Very low |
| R-Cl | White | Soluble | Low |
| R-Br | Cream | Partially soluble | Medium |
| R-I | Yellow | Insoluble | High |
“1-2-3 Rule”:
1° alkyl halides → SN2
3° alkyl halides → SN1
2° alkyl halides → Can do both
“The More, The Merrier”:
More alkyl groups = More stable carbocation
Order: 3° > 2° > 1° > CH₃⁺
Reason: Hyperconjugation and inductive effects
Silver Nitrate Test:
RI > RBr > RCl > RF
Mnemonic: “I Bring Cool Friends”
Remember: Bond strength decreases down the group
Protic vs Aprotic:
Protic (H-bond donors): Favor SN1/E1
Aprotic (no H-bond donors): Favor SN2/E2
Examples: Water/alcohols (protic) vs Acetone/DMSO (aprotic)
Good Leaving Groups: I⁻, Br⁻, Cl⁻, HSO₄⁻
Poor Leaving Groups: OH⁻, OR⁻, NH₂⁻
Rule: Weak bases make good leaving groups
Halogenoalkane: C-X bond polar, nucleophilic substitution
Halogenoarene: C-X bond partial double bond, electrophilic substitution
Memory: “Arenes like Electrophiles, Alkanes like Nucleophiles”
i. Which catalyst is commonly used in the halogenation of benzene with Cl₂ or Br₂?
Answer: b) FeCl₃
Explanation: FeCl₃ or FeBr₃ are commonly used as Lewis acid catalysts in the halogenation of benzene. They polarize the halogen molecule, generating the electrophile needed for the reaction.
ii. In the reaction of benzene with Cl₂ in the presence of a catalyst, the product is:
Answer: b) Chlorobenzene
Explanation: Benzene undergoes electrophilic substitution with chlorine in the presence of FeCl₃ catalyst to form chlorobenzene (C₆H₅Cl).
iii. What is the major product of the reaction between benzene and Br₂ with FeBr₃ as a catalyst?
Answer: b) Bromobenzene
Explanation: The reaction produces bromobenzene (C₆H₅Br) through electrophilic aromatic substitution. Further bromination requires more vigorous conditions.
iv. Which mechanism do primary halogenoalkanes typically follow in nucleophilic substitution reactions?
Answer: b) SN2
Explanation: Primary halogenoalkanes prefer SN2 mechanism due to less steric hindrance and the instability of primary carbocations for SN1.
v. What type of bond is broken in the nucleophilic substitution reaction of a halogenoalkane?
Answer: c) C-X (where X is a halogen)
Explanation: In nucleophilic substitution reactions, the carbon-halogen bond (C-X) is broken as the halogen is replaced by a nucleophile.
vi. Which factor does NOT significantly affect the rate of SN1 reactions?
Answer: c) Strength of the nucleophile
Explanation: SN1 reactions are unimolecular and the rate depends only on the concentration of the substrate. The nucleophile concentration doesn’t affect the rate since it attacks in the fast second step.
vii. Which halogenoalkane is more reactive in an SN2 reaction?
Answer: a) Chloromethane
Explanation: Chloromethane (CH₃Cl) is the most reactive in SN2 reactions because it has the least steric hindrance around the carbon bearing the chlorine.
viii. The C-Cl bond in chlorobenzene is less reactive towards nucleophilic substitution because:
Answer: c) Resonance effect makes the bond partially double
Explanation: In chlorobenzene, the lone pair on chlorine delocalizes into the benzene ring through resonance, giving the C-Cl bond partial double bond character, making it stronger and less reactive.
ix. When chloroethane reacts with aqueous silver nitrate, the precipitate formed is:
Answer: a) Silver chloride
Explanation: Chloroethane reacts slowly with silver nitrate to form silver chloride (AgCl) precipitate, which is white and soluble in ammonia.
i. Describe the halogenation of benzene.
Answer: Halogenation of benzene involves the electrophilic substitution of a hydrogen atom by a halogen atom (Cl or Br) in the presence of a Lewis acid catalyst like FeCl₃ or FeBr₃.
Mechanism:
Example: C₆H₆ + Cl₂ → C₆H₅Cl + HCl (with FeCl₃ catalyst)
ii. Explain why chlorobenzene is less reactive than chloroethane in nucleophilic substitution reactions.
Answer: Chlorobenzene is less reactive than chloroethane in nucleophilic substitution due to:
In contrast, chloroethane has a polar C-Cl bond with no resonance stabilization, making it more susceptible to nucleophilic attack.
iii. What is the major product when benzene reacts with Br₂ in the presence of FeBr₃?
Answer: The major product is bromobenzene (C₆H₅Br).
Reaction: C₆H₆ + Br₂ → C₆H₅Br + HBr (with FeBr₃ catalyst)
Mechanism: Electrophilic aromatic substitution where Br⁺ attacks the benzene ring.
Note: Further bromination to form dibromobenzene requires more vigorous conditions as the first bromine deactivates the ring towards further electrophilic substitution.
i. Explain the mechanisms of SN1 and SN2 reactions in detail, including the factors that affect each mechanism and examples of substrates that prefer each pathway.
Answer:
Steps:
Rate Law: Rate = k[R-X] (first order)
Stereochemistry: Racemic mixture (both enantiomers formed equally)
Steps:
Rate Law: Rate = k[Nu⁻][R-X] (second order)
Stereochemistry: Inversion of configuration (backside attack)
1. Structure of Alkyl Halide:
2. Nature of Nucleophile:
3. Solvent Effects:
4. Leaving Group Ability:
SN2 Preferred: CH₃-Br + OH⁻ → CH₃-OH + Br⁻ (methyl bromide)
SN1 Preferred: (CH₃)₃C-Br + H₂O → (CH₃)₃C-OH + H⁺ + Br⁻ (tert-butyl bromide)
ii. Compare and contrast the reactivity of halogenoalkanes and halogenoarenes with examples. Discuss the factors contributing to their differing reactivities.
Answer:
| Aspect | Halogenoalkanes (e.g., Chloroethane) | Halogenoarenes (e.g., Chlorobenzene) |
|---|---|---|
| General Formula | R-X | Ar-X |
| Bond Character | Polar covalent bond | Partial double bond due to resonance |
| Reactivity towards Nucleophiles | High | Very low |
| Reactivity towards Electrophiles | Low | High (ring substitution) |
| Preferred Reaction Type | Nucleophilic substitution | Electrophilic substitution |
| Bond Strength (C-X) | Weaker | Stronger |
| Hybridization of Carbon | sp³ | sp² |
1. Resonance Effect:
2. Bond Strength:
3. Electronic Effects:
4. Steric Effects:
Chloroethane: CH₃-CH₂-Cl + OH⁻ → CH₃-CH₂-OH + Cl⁻ (nucleophilic substitution)
Chlorobenzene: C₆H₅-Cl + HNO₃ → o-NO₂-C₆H₄-Cl + p-NO₂-C₆H₄-Cl (electrophilic substitution)