Class 12 Chemistry Chapter 6 Important Questions – Haloalkanes and Haloarenes

Haloalkanes and haloarenes contain halogen atoms attached to aliphatic or aromatic carbon atoms. These questions cover nomenclature, preparation, physical properties, reactions, mechanisms, stereochemistry and organic conversions.

Class 12 Chemistry Chapter 6 Important Questions help students revise Haloalkanes and Haloarenes through theory, reactions and conversions. The chapter explains how the carbon-halogen bond affects the preparation, properties and reactions of halogen-containing compounds.

Students should understand the difference between SN1 and SN2 mechanisms. They should also revise elimination, named reactions, haloarene reactivity and organic conversions. Every reaction answer should mention the reagent, condition and main product.

Key Takeaways

• Haloalkanes contain halogens bonded to sp³-hybridised carbon atoms.
• Haloarenes contain halogens bonded directly to aromatic sp² carbon atoms.
• Carbon-halogen bond strength decreases from C–F to C–I.
• Primary haloalkanes generally favour SN2 reactions.
• Tertiary haloalkanes generally favour SN1 reactions.
• Aqueous KOH forms alcohols, while alcoholic KOH generally forms alkenes.
• Haloarenes resist nucleophilic substitution under ordinary conditions.
• SN2 reactions involve inversion of configuration.
• SN1 reactions can produce racemisation.

Important Reactions from Haloalkanes and Haloarenes

Preparation from Alcohols

R–OH + HX → R–X + H₂O

R–OH + SOCl₂ → R–Cl + SO₂ + HCl

Finkelstein Reaction

R–Cl + NaI → R–I + NaCl

The reaction is carried out in dry acetone.

Swarts Reaction

R–Cl + AgF → R–F + AgCl

Nucleophilic Substitution

R–X + KOH(aq) → R–OH + KX

R–X + KCN → R–CN + KX

R–X + AgCN → R–NC + AgX

Elimination

R–CH₂–CHX–R′ + KOH(alc) → Alkene + KX + H₂O

Wurtz Reaction

2R–X + 2Na → R–R + 2NaX

Grignard Reagent Formation

R–X + Mg → R–MgX

The reaction is carried out in dry ether.

Wurtz–Fittig Reaction

Ar–X + R–X + 2Na → Ar–R + 2NaX

Fittig Reaction

2Ar–X + 2Na → Ar–Ar + 2NaX

Sandmeyer Reaction

ArN₂⁺Cl⁻ + CuCl → ArCl + N₂

ArN₂⁺Cl⁻ + CuBr → ArBr + N₂

Reactivity Orders

Leaving-group reactivity:

R–I > R–Br > R–Cl >> R–F

SN1 reactivity:

Tertiary > Secondary > Primary > Methyl

SN2 reactivity:

Methyl > Primary > Secondary >> Tertiary

Boiling points for the same alkyl group:

RI > RBr > RCl > RF

Access Class 12 Chemistry Chapter 6 Important Questions in 30 Minutes

First 10 minutes: Revise classification, nomenclature and preparation methods.

Next 10 minutes: Review SN1, SN2, elimination and named reactions.

Final 10 minutes: Practise conversions, reasoning questions and haloarene reactions.

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Important Questions for Class 12 Chemistry Chapter 6

Q1. Differentiate between haloalkanes and haloarenes.

Answer:

Q2. Give the IUPAC names of the following compounds.

(i) CH₃CH₂CH(Cl)CH₃
(ii) (CH₃)₃CCH₂Br
(iii) CH₂=CHCH₂Br
(iv) CH₃CHBrCH₂CH₃

Answer:

(i) 2-Chlorobutane
(ii) 1-Bromo-2,2-dimethylpropane
(iii) 3-Bromoprop-1-ene
(iv) 2-Bromobutane

Q3. Write the structures of the following compounds.

(i) 2-Chloro-3-methylpentane
(ii) 1,4-Dibromobut-2-ene
(iii) 2-Bromo-2-methylpropane

Answer:

(i) CH₃–CH(Cl)–CH(CH₃)–CH₂–CH₃

(ii) BrCH₂–CH=CH–CH₂Br

(iii) (CH₃)₃CBr

Q4. Write all structural isomers of C₄H₉Br and classify them.

Answer:

1. CH₃CH₂CH₂CH₂Br
   1-Bromobutane — Primary
2. CH₃CH(Br)CH₂CH₃
   2-Bromobutane — Secondary
3. (CH₃)₂CHCH₂Br
   1-Bromo-2-methylpropane — Primary
4. (CH₃)₃CBr
   2-Bromo-2-methylpropane — Tertiary

Q5. Arrange the following compounds in increasing order of boiling point.

(i) Chloromethane, bromomethane and dibromomethane
(ii) Isopropyl chloride, 1-chloropropane and 1-chlorobutane

Answer:

(i)

Chloromethane < Bromomethane < Dibromomethane

Boiling point increases with molecular mass and intermolecular forces.

(ii)

Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

Branching lowers the surface area and boiling point. A longer chain increases intermolecular attraction.

Q6. Explain why haloalkanes are only slightly soluble in water.

Answer: Water molecules are held together through strong hydrogen bonds.

Energy is required to break these hydrogen bonds. The attractions formed between water and haloalkane molecules do not release enough energy to compensate.

Therefore, haloalkanes are only slightly soluble in water.

Q7. Why is thionyl chloride preferred for converting alcohols into alkyl chlorides?

Answer: Thionyl chloride converts alcohols into alkyl chlorides according to the reaction:

R–OH + SOCl₂ → R–Cl + SO₂ + HCl

Both SO₂ and HCl escape as gases. The alkyl chloride is therefore obtained without difficult separation from solid or liquid by-products.

Q8. Explain the Finkelstein and Swarts reactions with equations.

Answer:

Finkelstein reaction

An alkyl chloride or bromide reacts with sodium iodide in dry acetone.

R–Cl + NaI → R–I + NaCl

The precipitation of NaCl helps the reaction proceed forward.

Swarts reaction

An alkyl chloride or bromide is heated with a metallic fluoride.

R–Cl + AgF → R–F + AgCl

The reaction is used to prepare alkyl fluorides.

Q9. Why does KCN form an alkyl cyanide while AgCN forms an isocyanide?

Answer: Cyanide is an ambident nucleophile because it can attack through carbon or nitrogen.

KCN is mainly ionic and provides free CN⁻ ions. Attack occurs mainly through carbon:

R–X + KCN → R–CN + KX

AgCN is mainly covalent. Nitrogen is more available for bonding:

R–X + AgCN → R–NC + AgX

Q10. Explain the SN2 mechanism using the hydrolysis of methyl chloride.

Answer: Hydroxide ion attacks the carbon atom from the side opposite to chlorine.

CH₃Cl + OH⁻ → CH₃OH + Cl⁻

The carbon-oxygen bond forms while the carbon-chlorine bond breaks. Both processes occur simultaneously in one step.

The rate law is:

Rate = k[CH₃Cl][OH⁻]

The reaction rate depends on both reactants. Therefore, it is a bimolecular nucleophilic substitution reaction.

Q11. Explain the SN1 mechanism using tert-butyl bromide.

Answer: The reaction takes place in two main steps.

Step 1: Carbocation formation

This is the slow step.

(CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻

Step 2: Nucleophilic attack

(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH

The rate law is:

Rate = k[(CH₃)₃CBr]

The rate depends only on the concentration of tert-butyl bromide. Therefore, the reaction is unimolecular.

Q12. Distinguish between SN1 and SN2 reactions.

Answer:

Q13. Arrange the following compounds in decreasing order of SN2 reactivity.

Methyl bromide, ethyl bromide, isopropyl bromide and tert-butyl bromide.

Answer:

Methyl bromide > Ethyl bromide > Isopropyl bromide > Tert-butyl bromide

SN2 reactions require back-side attack. Increasing alkyl substitution creates greater steric hindrance and slows the reaction.

Q14. Why are haloarenes less reactive towards nucleophilic substitution than haloalkanes?

Answer: Haloarenes resist nucleophilic substitution because:

1. Resonance gives the carbon-halogen bond partial double-bond character.
2. The carbon bonded to halogen is sp² hybridised.
3. The carbon-halogen bond is shorter and stronger.
4. Formation of a phenyl carbocation is highly unstable.
5. Back-side attack on the aromatic carbon is difficult.

Q15. Why is chlorine deactivating but ortho- and para-directing in chlorobenzene?

Answer: Chlorine withdraws electron density through its negative inductive effect. This reduces the overall reactivity of the benzene ring.

However, chlorine donates its lone pair through resonance. This increases electron density at the ortho and para positions.

The inductive effect causes deactivation, while resonance controls orientation.

Q16. What happens when 2-bromobutane is heated with alcoholic KOH? Name the rule involved.

Answer: 2-Bromobutane undergoes beta-elimination.

CH₃CHBrCH₂CH₃ + KOH(alc)

→ CH₃CH=CHCH₃ + KBr + H₂O

But-2-ene is the major product. But-1-ene is formed in a smaller amount.

The reaction follows Zaitsev’s rule. The more substituted alkene is formed as the major product.

Q17. Explain the Wurtz, Wurtz–Fittig and Fittig reactions.

Answer:

Wurtz reaction

Two alkyl halide molecules react with sodium in dry ether.

2CH₃Br + 2Na → C₂H₆ + 2NaBr

Wurtz–Fittig reaction

An alkyl halide and an aryl halide react with sodium.

C₆H₅Br + CH₃Br + 2Na

→ C₆H₅CH₃ + 2NaBr

Fittig reaction

Two aryl halide molecules react with sodium.

2C₆H₅Br + 2Na

→ C₆H₅–C₆H₅ + 2NaBr

All three reactions are carried out in dry ether.

Q18. Complete the following conversions.

(i) 1-Bromopropane to 2-bromopropane
(ii) Ethyl chloride to propanoic acid
(iii) Aniline to chlorobenzene
(iv) Tert-butyl bromide to isobutyl bromide

Answer:

(i) 1-Bromopropane to 2-bromopropane

CH₃CH₂CH₂Br + KOH(alc)

→ CH₃CH=CH₂ + KBr + H₂O

CH₃CH=CH₂ + HBr

→ CH₃CHBrCH₃

(ii) Ethyl chloride to propanoic acid

CH₃CH₂Cl + KCN

→ CH₃CH₂CN + KCl

CH₃CH₂CN + 2H₂O + H⁺

→ CH₃CH₂COOH + NH₄⁺

(iii) Aniline to chlorobenzene

C₆H₅NH₂ + NaNO₂ + 2HCl

→ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

C₆H₅N₂⁺Cl⁻ + CuCl

→ C₆H₅Cl + N₂

(iv) Tert-butyl bromide to isobutyl bromide

(CH₃)₃CBr + KOH(alc)

→ (CH₃)₂C=CH₂ + KBr + H₂O

(CH₃)₂C=CH₂ + HBr

→ (CH₃)₂CHCH₂Br

The second reaction is carried out in the presence of peroxide.

Q19. Explain inversion and racemisation in nucleophilic substitution reactions.

Answer:

Inversion

In an SN2 reaction, the nucleophile attacks from the side opposite to the leaving group.

The spatial arrangement around the chiral carbon becomes inverted. This is called inversion of configuration.

Racemisation

In an SN1 reaction, a planar carbocation is formed.

The nucleophile can attack from either side. Both enantiomers are therefore produced, forming an approximately racemic mixture.

Q20. Answer the following case-based question.

A student reacts bromoethane separately with aqueous KOH and alcoholic KOH. The two reactions form different products.

(a) Name the product formed with aqueous KOH.
(b) Name the product formed with alcoholic KOH.
(c) Explain the difference.

Answer:

(a) Aqueous KOH forms ethanol.

CH₃CH₂Br + KOH(aq)

→ CH₃CH₂OH + KBr

(b) Alcoholic KOH forms ethene.

CH₃CH₂Br + KOH(alc)

→ CH₂=CH₂ + KBr + H₂O

(c) In aqueous solution, OH⁻ mainly acts as a nucleophile and replaces bromine.

In alcoholic solution, the reagent acts mainly as a base and removes a beta-hydrogen. This produces an alkene through elimination.

How to Prepare Haloalkanes and Haloarenes

Begin with classification and nomenclature because these concepts appear throughout the chapter. Then prepare a separate table for SN1 and SN2 mechanisms.

Create a reaction map for substitution, elimination, Finkelstein, Swarts, Wurtz, Fittig and Sandmeyer reactions. Mention the reagent and condition above every arrow.

Practise conversions by identifying whether the carbon chain must increase, decrease or remain unchanged. KCN increases the chain by one carbon atom, while elimination forms a double bond.

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