Haloalkanes and Haloarenes — Previous Year Questions (Class 12 Chemistry)
Haloalkanes and haloarenes are organic compounds containing halogens. Understanding their synthesis, reactions, and properties is essential for organic che
TL;DR: Haloalkanes and haloarenes are organic compounds containing halogens. Understanding their synthesis, reactions, and properties is essential for organi…
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Haloalkanes and haloarenes are organic compounds containing halogens. Understanding their synthesis, reactions, and properties is essential for organic che
Haloalkanes and Haloarenes — Previous Year Questions with Solutions
Q (2023, 2 marks): Distinguish between haloalkanes and haloarenes. Give two examples of each with structural formulas.
Answer: Haloalkanes:
- Organic compounds with halogen bonded to sp3 hybridized carbon (alkane backbone).
- General formula: CnH(2n+1)X
- Example 1: Chloromethane (CH3Cl) - carbon is saturated
- Example 2: 1,2-Dichloroethane (ClCH2-CH2Cl)
Haloarenes:
- Organic compounds with halogen bonded directly to aromatic ring (benzene).
- General formula: C6H5X (for monosubstituted benzene)
- Example 1: Chlorobenzene (C6H5Cl) - halogen on benzene ring
- Example 2: 1,4-Dibromobenzene (para-dibromobenzene)
Key difference: In haloalkanes, the C-X bond is from an sp3 carbon; in haloarenes, the C-X bond is from an sp2 aromatic carbon.
Q (2022, 3 marks): Explain the preparation of haloalkanes from alcohols using PX3 or HX. Write equations.
Answer: Method 1: Using phosphorus halides (PX3 or PX5)
With PCl3:
3ROH + PCl3 → 3RCl + H3PO3
Example: 3CH3CH2OH + PCl3 → 3CH3CH2Cl + H3PO3 (Ethanol to chloroethane)
With PBr3:
3ROH + PBr3 → 3RBr + H3PO3
With SOCl2 (better method):
ROH + SOCl2 → RCl + SO2 + HCl
Example: CH3CH2OH + SOCl2 → CH3CH2Cl + SO2 + HCl
Method 2: Using hydrogen halides (HX)
ROH + HX → RX + H2O
Example: CH3OH + HCl → CH3Cl + H2O (Methanol to chloromethane)
(Requires catalyst like H2SO4 or ZnCl2; reaction rate: primary > secondary > tertiary)
Q (2023, 3 marks): Compare the reactivity of haloalkanes and haloarenes towards nucleophilic substitution. Explain the reasons.
Answer: Haloalkanes:
- Highly reactive towards nucleophilic substitution (SN1 and SN2 mechanisms).
- C-X bond: σ-bond with C-X single bond character
- Easy bond breaking and C+ formation
- Reactivity order: Iodides > Bromides > Chlorides
Haloarenes:
- Extremely unreactive towards nucleophilic substitution (cannot undergo SN1 or SN2).
- C-X bond: Has partial aromatic double bond character (resonance stabilization)
- Very strong C-X bond (bond dissociation energy ≈ 430 kJ/mol)
- Cannot form C+ (breaks aromaticity)
Reason for difference:
1. Aromatic stability of benzene ring is lost if the C-X bond breaks
2. Resonance stabilization keeps the C-X bond very strong
3. The carbon becomes sp2, making it less susceptible to nucleophilic attack
4. Very high activation energy (>200 kJ/mol) for nucleophilic substitution in haloarenes
Q (2022, 3 marks): Explain the mechanism of SN1 reaction for haloalkanes with an example. Write the reaction equation and mechanism.
Answer: SN1 mechanism (Unimolecular nucleophilic substitution):
Applies to: Tertiary and secondary haloalkanes (less common)
Example: 2-Bromo-2-methylpropane + Water (aqueous NaOH)
(CH3)3CBr + H2O → (CH3)3COH + HBr
Mechanism (two steps):
Step 1 (Rate-determining): Ionization
(CH3)3CBr → (CH3)3C+ + Br⁻ (slow)
Bonds break slowly, forming carbocation
Step 2 (Fast): Nucleophilic attack
(CH3)3C+ + H2O → (CH3)3COH + H+ (fast)
Nuclei attacks carbocation, forms product
Characteristics:
1. Rate = k[(CH3)3CBr] (first-order)
2. Rate independent of nucleophile concentration
3. Forms carbocation intermediate (prone to rearrangement)
4. Racemization occurs (stereospecificity lost)
5. Favored in polar solvents, high temperature
Q (2023, 3 marks): Explain the mechanism of SN2 reaction for haloalkanes. What is the role of solvent?
Answer: SN2 mechanism (Bimolecular nucleophilic substitution):
Applies to: Primary and secondary haloalkanes
Example: CH3CH2Br + KOH → CH3CH2OH + KBr
Mechanism (single step, concerted):
OH⁻ attacks from back side (opposite to Br⁻)
As O-C bond forms, C-Br bond breaks simultaneously
Tsuchida state: [OH···C···Br]‡ (transition state with partial bonds)
Characteristics:
1. Rate = k[RX][Nu⁻] (second-order, bimolecular)
2. Rate depends on both substrate and nucleophile concentration
3. No carbocation intermediate
4. Inversion of configuration (Walden inversion) - stereochemistry inverts
5. Favored by strong nucleophile, polar aprotic solvent, primary substrate
Role of solvent:
1. Polar aprotic solvents (DMSO, DMF) increase rate:
- Do not form H-bonds with nucleophile
- Nucleophile remains unhydrated, highly reactive
2. Polar protic solvents (water, alcohol) decrease rate:
- Form H-bonds with nucleophile
- Nucleophile becomes solvated, less reactive
3. Non-polar solvents: Reaction does not occur (ions not solvated)
Q (2021, 3 marks): How are haloarenes prepared? Describe the preparation from benzene using benzene diazonium chloride.
Answer: Haloarenes are primarily prepared using benzene diazonium chloride through diazo coupling.
Procedure:
Step 1: Aniline to diazonium chloride
C6H5NH2 + NaNO2 + 2HCl → C6H5N2⁺Cl⁻ + 2H2O + NaCl
(At 0-5°C, cold conditions)
Step 2: Sandmeyer reaction (halogen substitution)
C6H5N2⁺Cl⁻ + CuCl → C6H5Cl + N2 + CuCl2 (Chlorobenzene)
C6H5N2⁺Cl⁻ + CuBr → C6H5Br + N2 + CuBr2 (Bromobenzene)
C6H5N2⁺Cl⁻ + KI → C6H5I + N2 + KCl (Iodobenzene)
Alternative for fluorine:
C6H5N2⁺Cl⁻ + HBF4 → C6H5F + N2 + BF4⁻ (Fluorobenzene)
Other methods:
1. Direct halogenation: Benzene + X2 (with AlX3 catalyst)
C6H6 + Br2 (AlBr3) → C6H5Br + HBr
2. Reaction with halogen carriers: Benzene + Cl2 (CrO2Cl2)
Advantage of diazonium method: Good control, can introduce halogen at specific positions using directing effects
Frequently Asked Questions
Why does fluorobenzene not undergo nucleophilic aromatic substitution easily?
Fluorine is the most electronegative element and forms the strongest C-F bond with carbon (~485 kJ/mol). Although fluorine is small and increases electron density through resonance, the extremely strong C-F bond and the sp2 nature of the aromatic carbon make nucleophilic substitution very difficult. Fluorobenzene requires harsh conditions (high temperature, high pressure) for nucleophilic substitution.
What is the Sandmeyer reaction and why is it important?
The Sandmeyer reaction is the replacement of the diazonium group (-N2+) by halogens using cuprous halides (CuX). It is important because it allows the introduction of halogens (especially Cl, Br) into aromatic rings via aniline, which is easily prepared. This provides a route to haloarenes that are otherwise difficult to synthesize. It also allows placement of halogens at different positions on benzene through controlling the directing effects of the aniline amino group.
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