Chapter 10: Haloalkanes and Haloarenes Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 10 – Haloalkanes and Haloarenes (Q10.1 to Q10.25)

Question 10.1 Write the structures of the following compounds:

(i) 2-Chloro-3-methylpentane: CH₃–CH(Cl)–CH(CH₃)–CH₂–CH₃ (ii) 1-Chloro-4-ethylcyclohexane: Cyclohexane ring with Cl at position 1 and ethyl at position 4. (iii) 4-tert-butyl-3-iodoheptane: CH₃–CH₂–C(I)H–CH(C(CH₃)₃)–CH₂–CH₂–CH₃ (iv) 1,4-Dibromobut-2-ene: Br–CH₂–CH=CH–CH₂–Br (v) 1-Bromo-4-sec-butyl-2-methylbenzene: Benzene ring with Br at position 1, sec-butyl at position 4, and methyl at position 2.

Question 10.2 Why is sulphuric acid not used during the reaction of alcohols with KI?

Sulphuric acid oxidizes HI (formed from KI and H₂SO₄) to iodine (I₂), reducing the yield of alkyl iodide. Phosphoric acid (H₃PO₄), a non-oxidizing acid, is used instead.

Question 10.3 Write structures of different dihalogen derivatives of propane:

1,1-Dihalopropane: CH₃–CH₂–C(X)₂H 1,2-Dihalopropane: CH₃–CH(X)–CH₂X 2,2-Dihalopropane: CH₃–C(X)₂–CH₃ 2,3-Dihalopropane: CH₂X–CH(X)–CH₃

Question 10.4 Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that yields:

(i) A single monochloride: Neopentane (ii) Three isomeric monochlorides: n-Pentane (iii) Four isomeric monochlorides: Isopentane (2-methylbutane)

Question 10.5 Draw the structures of major monohalo products:

Example: Addition of HBr to propene yields 2-bromopropane. (Refer specific reactions for textbook cases.)

Question 10.6 Arrange each set of compounds in increasing order of boiling points:

(i) CH₃Cl < CH₃Br < CH₂Br₂ < CHBr₃ (ii) Isopropyl chloride < 1-chloropropane < 1-chlorobutane

Question 10.7 Which alkyl halide would react faster by SN2 mechanism?

Primary alkyl halides react faster than secondary and tertiary alkyl halides due to less steric hindrance.

Question 10.8 In the following pairs of halogen compounds, which undergoes faster SN1 reaction?

Tertiary halides undergo SN1 faster because of the formation of stable tertiary carbocations.

Question 10.9 Identify A, B, C, D, E, R and R₁ in the sequence:

Typical sequence: A = alkene, B = halogenated product, C = alcohol, D = haloalkane, E = amine, R and R₁ = alkyl groups.

Question 10.10 A hydrocarbon C₅H₁₀ does not react with chlorine in the dark but gives a single monochloro product in bright sunlight. Identify the hydrocarbon.

Answer: Cyclopentane.

Question 10.11 Write the conversions for the following:

(i) Ethanol → But-1-yne:

• CH₃CH₂OH + PBr₃ → CH₃CH₂Br (Bromoethane)
• CH₃CH₂Br + alc. KOH → CH₂=CH₂ (Ethene)
• CH₂=CH₂ + NaNH₂ → HC≡CH (Ethyne)
• HC≡CH + CH₃CH₂Br → CH₃CH₂C≡CH (But-1-yne)

(ii) Ethane → Bromoethene:

• C₂H₆ + Cl₂ (hv) → C₂H₅Cl (Chloroethane)
• C₂H₅Cl + alc. KOH → CH₂=CH₂ (Ethene)
• Ethene + Br₂ → CH₂=CHBr (Bromoethene)

(iii) Propene → 1-Nitropropane:

• CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br
• CH₃CH₂CH₂Br + AgNO₂ → CH₃CH₂CH₂NO₂

(iv) Toluene → Benzyl alcohol:

• C₆H₅CH₃ + Cl₂ (hv) → C₆H₅CH₂Cl
• C₆H₅CH₂Cl + aq. NaOH → C₆H₅CH₂OH

(v) Propene → Propyne:

• CH₃CH=CH₂ hydrogenated to propane.
• Propane + Cl₂ (hv) → CH₃CH₂CH₂Cl
• CH₃CH₂CH₂Cl + alc. KOH → HC≡C–CH₃

(vi) Ethanol → Ethyl fluoride:

• CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl
• CH₃CH₂Cl + AgF → CH₃CH₂F

(vii) Bromomethane → Propanone:

• CH₃Br + Mg → CH₃MgBr
• CH₃MgBr + CH₃COCl → CH₃COCH₃ (Propanone)

(viii) But-1-ene → But-2-ene:

• But-1-ene isomerises to But-2-ene under catalytic conditions (Al₂O₃, 600 K).

(ix) 1-Chlorobutane → n-Octane:

• 2CH₃CH₂CH₂CH₂Cl + 2Na → CH₃(CH₂)₆CH₃ + 2NaCl (Wurtz reaction)

(x) Benzene → Biphenyl:

• 2C₆H₆ + Cl₂/Fe → C₆H₅Cl
• 2C₆H₅Cl + 2Na → C₆H₅–C₆H₅ (Biphenyl)

Question 10.12 Explain the following:

(i) Chlorobenzene has a lower dipole moment than cyclohexyl chloride:

In chlorobenzene, resonance between the lone pair on chlorine and the benzene ring induces partial double bond character in the C–Cl bond. This shortens and strengthens the bond, reducing the dipole moment compared to cyclohexyl chloride.

(ii) Alkyl halides are immiscible in water:

Alkyl halides are unable to form hydrogen bonds with water. Their interactions with water are weaker than the hydrogen bonding between water molecules, making them immiscible.

(iii) Grignard reagents must be prepared under anhydrous conditions:

Grignard reagents (RMgX) react instantly with water to form alkanes:

RMgX + H₂O → RH + Mg(OH)X

Hence, anhydrous conditions are essential for their preparation and reactions.

Question 10.13 Give uses of the following compounds:

(a) Iodoform (CHI₃):

• Used earlier as an antiseptic.
• Replaced due to its unpleasant smell and irritation.

(b) Carbon Tetrachloride (CCl₄):

• Solvent for oils, fats, and resins.
• Cleaning agent and was used in fire extinguishers.

(c) Freon-12 (CCl₂F₂):

• Used as a refrigerant in air conditioners and refrigerators.
• Used as an aerosol propellant (now banned due to ozone depletion).

(d) DDT (Dichlorodiphenyltrichloroethane):

• Used as an insecticide to control malaria and typhus.
• Environmental hazard due to bioaccumulation and persistence.

Question 10.14 Give the products formed when:

Reaction	Reaction Equation	Product Formed
1-Bromopropane + alcoholic KOH	CH₃CH₂CH₂Br + alc. KOH → CH₃CH=CH₂ + KBr + H₂O	Propene
Bromobenzene + Mg in dry ether	C₆H₅Br + Mg → C₆H₅MgBr	Phenyl magnesium bromide (Grignard reagent)
Chlorobenzene + Hydrolysis	C₆H₅Cl + 6% NaOH (300°C, 300 atm) → C₆H₅OH + NaCl	Phenol
Ethyl chloride + aqueous KOH	C₂H₅Cl + aq. KOH → C₂H₅OH + KCl	Ethanol
Methyl bromide + Na + dry ether	2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr	Ethane

Question 10.15 Explain:

(i) C–Cl bond length in chlorobenzene is shorter than in CH₃Cl because of partial double bond character due to resonance in chlorobenzene.

(ii) Haloalkanes are more reactive towards nucleophilic substitution than haloarenes because haloarenes have stronger C–X bonds due to resonance and sp² hybridisation.

(iii) Haloarenes are extremely less reactive due to electron-rich aromatic rings that repel nucleophiles and strong C–X bonds from resonance stabilisation.

Question 10.16 Arrange in increasing order of reactivity towards SN2 displacement:

Order: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

Question 10.17 Out of C₆H₅CH₂Cl and C₆H₅CHCl₂, which is more easily hydrolysed?

Answer: C₆H₅CH₂Cl hydrolyses more easily due to stable benzyl carbocation formation.

Question 10.18 Give the products formed when:

Reaction	Reaction Equation	Product Formed
2-Chlorobutane + aq. KOH	CH3CH(Cl)CH2CH3 + KOH → CH3CH(OH)CH2CH3 + KCl	2-Butanol
1-Chlorobutane + alc. KOH	CH3CH2CH2CH2Cl + alc. KOH → CH3CH2CH=CH2 + KCl + H2O	But-1-ene
Benzyl chloride + aq. KOH	C6H5CH2Cl + KOH → C6H5CH2OH + KCl	Benzyl alcohol

Question 10.19 Explain:

(i) Vinyl chloride does not undergo nucleophilic substitution due to strong C–Cl bond and resonance stabilisation.

(ii) KCN gives alkyl cyanides (R–CN) while AgCN gives isocyanides (R–NC) because AgCN forms covalent bonds, leading to nitrogen attack.

Question 10.20 Predict major product formed:

Reaction	Major Product
1-Bromo-2-methylpropane + alc. KOH	2-Methylpropene
1-Bromopropane + aq. NaOH	1-Propanol

Question 10.21 Write mechanism of the reaction:

Reaction: C₂H₅Br + KOH (aq) → C₂H₅OH + KBr

Mechanism (SN2):

• OH^– attacks the carbon atom from the opposite side of Br.
• Transition state is formed.
• Br^– leaves, resulting in the formation of ethanol.

Question 10.22 Give examples of:

• SN2 Reaction: CH₃Br + OH^– → CH₃OH + Br^–
• SN1 Reaction: (CH₃)₃CBr + H₂O → (CH₃)₃COH + HBr

Question 10.23 Which compound will react faster in SN2 reaction?

• (i) CH₃I reacts faster than CH₃Br because I^– is a better leaving group.
• (ii) CH₃Cl reacts faster than (CH₃)₃CCl due to less steric hindrance.

Question 10.24 Write the major product(s) in:

Reaction	Product
CH3CH=CH2 + HBr	2-Bromopropane (Markovnikov addition)
C6H5CH2CH3 + Cl2 (hv)	1-Chloro-1-phenylethane (free radical substitution)

Question 10.25 Draw structures of major monohalo products:

• (i) 1,2-Dichloropropane: CH₃CHClCH₂Cl
• (ii) Chlorobenzene: C₆H₅Cl
• (iii) Benzyl chloride: C₆H₅CH₂Cl