CBSE Important Questions Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes

Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes is part of the CBSE Class 12 Chemistry syllabus. These are compounds in which one or more hydrogen atoms of a hydrocarbon are replaced by halogen atoms. Students learn their classification and nomenclature, methods of preparation, physical properties, the nucleophilic substitution reactions SN1 and SN2, elimination reactions, the behaviour of haloarenes, and the uses and environmental effects of some important halogen compounds.

These CBSE important questions for Class 12 Chemistry Chapter 6 cover the reaction of 2-chloropentane with alcoholic KOH and Saytzeff's rule, the nature of the C-X bond, reactivity order towards nucleophilic substitution, the Wurtz reaction, why vinyl chloride is least reactive, a case study on the inductive and resonance effects of chlorine in chlorobenzene, and an assertion-reason question on nitro groups. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. (a) Show the reaction between 2-chloropentane and alcoholic KOH. Identify the major product. (b) Which rule governs the formation of this product?

Answer: (a) Alcoholic KOH brings about dehydrohalogenation (an elimination reaction). A hydrogen atom from a β-carbon and the chlorine atom are removed to give an alkene.
CH₃CH₂CH₂CHClCH₃ + alcoholic KOH → pent-2-ene + pent-1-ene + KCl + H₂O
Two alkenes are possible because there are two different β-carbons. The major product is pent-2-ene.
(b) The product is decided by Saytzeff's rule, which states that in a dehydrohalogenation reaction, the more substituted alkene is the major product. Pent-2-ene has more alkyl groups attached to the double-bonded carbons, so it is more stable and forms in greater amount.

Q2. What is the nature of the C-X bond in haloalkanes?

Answer: The C-X bond is polar (polarised).
Halogen atoms are more electronegative than carbon, so the shared pair of electrons is pulled towards the halogen. This gives the carbon a partial positive charge (δ+) and the halogen a partial negative charge (δ−).
Because the carbon carries a partial positive charge, it is open to attack by nucleophiles. This polarity is the reason haloalkanes readily undergo nucleophilic substitution reactions.

Q3. Which of the following is least reactive in a nucleophilic substitution reaction?
(a) (CH₃)₃C-Cl
(b) CH₂=CHCl
(c) CH₃CH₂Cl
(d) CH₂=CHCH₂Cl

Answer: (b) CH₂=CHCl (vinyl chloride)
In vinyl chloride, the lone pair on chlorine goes into resonance with the double bond. This gives the C-Cl bond a partial double bond character, making it shorter and much stronger.
A stronger C-Cl bond is far harder to break, so vinyl chloride is the least reactive towards nucleophilic substitution.
Allyl chloride (option d) and tertiary butyl chloride (option a) are in fact very reactive, because they form stable carbocations.

Q4. An organic compound A (C₄H₉Cl) on reaction with Na in dry ether gives a hydrocarbon which on monochlorination gives only one chloro derivative. Identify A.
(a) t-butyl chloride
(b) sec-butyl chloride
(c) isobutyl chloride
(d) n-butyl chloride

Answer: (a) t-butyl chloride
This is the Wurtz reaction, in which two molecules of alkyl halide join to form a symmetrical alkane:
2(CH₃)₃C-Cl + 2Na → (CH₃)₃C-C(CH₃)₃ + 2NaCl
The product is 2,2,3,3-tetramethylbutane. All 18 hydrogen atoms in this molecule are equivalent, since they all sit on identical methyl groups.
So replacing any one of them gives exactly the same single product, which matches the condition in the question.

Q5. Read the passage and answer the question.
Chlorine withdraws electrons through the inductive effect and releases electrons through resonance. Through the inductive effect chlorine destabilises the intermediate carbocation formed during electrophilic substitution. Through resonance the halogen stabilises the carbocation, and this effect is more pronounced at the ortho and para positions.
Which of the following is the correct statement about chlorobenzene?
(a) Acetylation of chlorobenzene gives 2-chloroacetophenone as the major product
(b) Chlorobenzene is less reactive towards electrophilic aromatic substitution than benzene
(c) Nitration of chlorobenzene gives the meta product
(d) Orientation is controlled by the inductive effect and reactivity by the resonance effect

Answer: (b) Chlorobenzene is less reactive towards electrophilic aromatic substitution than benzene
Chlorine is a special case: it is deactivating but ortho-para directing.
• The inductive effect (electron withdrawing) is stronger overall, so it pulls electron density out of the ring and makes chlorobenzene less reactive than benzene.
• The resonance effect (electron releasing) puts extra electron density specifically at the ortho and para positions, so the substitution that does happen goes mainly to those positions.
So reactivity is controlled by the inductive effect and orientation by the resonance effect — the opposite of what option (d) says.

Q6. Read the passage above. Identify product X when chlorobenzene undergoes nitration.
(a) 2-chlorobenzenesulphonic acid
(b) 4-chlorobenzenesulphonic acid
(c) 1-chloro-2-nitrobenzene
(d) 1-chloro-4-nitrobenzene

Answer: (d) 1-chloro-4-nitrobenzene (the para product is the major one)
Nitration introduces an −NO₂ group, not a sulphonic acid group, so options (a) and (b) are ruled out straight away.
Since chlorine is ortho-para directing, both 1-chloro-2-nitrobenzene and 1-chloro-4-nitrobenzene form. The para product is the major one, because the ortho position suffers from steric hindrance from the chlorine atom sitting right next to it.

Q7. Read the Assertion and Reason and choose the correct option.
Assertion (A): p-nitrochlorobenzene is more reactive towards nucleophilic substitution than chlorobenzene.
Reason (R): The −NO₂ group is electron withdrawing, so it decreases the electron density on the benzene ring.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (a) Both A and R are true, and R is the correct explanation of A
Haloarenes are normally very unreactive towards nucleophilic substitution. But an electron-withdrawing group such as −NO₂ at the ortho or para position changes this.
The nitro group pulls electron density away from the ring, which makes the carbon bearing the chlorine more open to nucleophilic attack. It also stabilises the negatively charged intermediate formed during the reaction, by spreading the charge onto its own oxygen atoms.
With more nitro groups the effect grows stronger: 2,4-dinitrochlorobenzene and 2,4,6-trinitrochlorobenzene react more and more easily.

Q8. Distinguish between SN1 and SN2 reactions.

SN1 SN2
Two steps, through a carbocation intermediate One step, through a transition state
First order kinetics: rate depends only on the halide Second order kinetics: rate depends on halide and nucleophile
Favoured by tertiary halides Favoured by primary halides
Gives a racemic mixture Gives inversion of configuration (Walden inversion)
Favoured by polar protic solvents such as water Favoured by polar aprotic solvents such as acetone

The order of reactivity is therefore reversed: for SN1 it is 3° > 2° > 1°, while for SN2 it is 1° > 2° > 3°.

Q9. Why are haloarenes much less reactive than haloalkanes towards nucleophilic substitution?

Answer: There are three reasons.
1. Resonance: The lone pair on the halogen goes into resonance with the benzene ring. This gives the C-X bond partial double bond character, making it shorter and stronger, so it is much harder to break.
2. Hybridisation: In haloarenes the carbon bonded to the halogen is sp² hybridised, while in haloalkanes it is sp³. The sp² carbon is more electronegative and holds the shared electrons more tightly, again strengthening the bond.
3. Repulsion: The approaching nucleophile is repelled by the electron-rich π cloud of the benzene ring, so it cannot get close enough to attack.

Q10. What are the uses and environmental effects of chloroform, DDT and freons?

Answer: Chloroform (CHCl₃): Used as a solvent for fats, waxes and rubber, and earlier as an anaesthetic.
Harm: It is slowly oxidised in air and light to the highly poisonous phosgene (COCl₂), which is why it is stored in dark bottles filled to the brim. It also damages the liver and central nervous system.
DDT: Was used widely as an insecticide, especially against malaria-carrying mosquitoes.
Harm: It is not biodegradable and builds up in the fatty tissue of animals and in the food chain. It is now banned in most countries.
Freons (chlorofluorocarbons): Used as refrigerants and in aerosol sprays.
Harm: They rise into the stratosphere and destroy the ozone layer, which allows harmful UV radiation to reach the earth. Their use is now restricted under the Montreal Protocol.

Q11. Why is the boiling point of a haloalkane higher than that of the corresponding alkane?

Answer: Two reasons work together.
1. Higher molecular mass: A halogen atom is much heavier than the hydrogen atom it replaces. Heavier molecules have stronger van der Waals forces between them.
2. Polarity: The C-X bond is polar, so haloalkane molecules have a dipole moment and attract each other through dipole-dipole forces as well. Alkanes are non-polar and have only weak dispersion forces.
Since more energy is needed to overcome these stronger forces, the boiling point is higher.
For the same alkyl group, the order of boiling points is RI > RBr > RCl > RF, following the increasing size and mass of the halogen.

Q12. What is the Finkelstein reaction and the Swarts reaction? Write the equations.

Answer: Finkelstein reaction: An alkyl chloride or bromide is treated with sodium iodide in dry acetone to give the alkyl iodide.
R-Cl + NaI → R-I + NaCl (in dry acetone)
It works because NaCl and NaBr are insoluble in dry acetone and precipitate out, pulling the equilibrium to the right.
Swarts reaction: An alkyl chloride or bromide is heated with a metallic fluoride such as AgF, SbF₃ or Hg₂F₂ to give the alkyl fluoride.
CH₃-Br + AgF → CH₃-F + AgBr
Both reactions are useful because alkyl iodides and fluorides are difficult to prepare directly from the alcohol.

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