Class 12 Chemistry

Chapter 6 Solutions — Haloalkanes and Haloarenes

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Solutions

Overview

Step-by-step NCERT solutions for Haloalkanes and Haloarenes (Chapter 6, CBSE Class 12 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Haloalkanes and Haloarenes textbook chapter.

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All 22 questions in Haloalkanes and Haloarenes are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

1.
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
- (i) (CH₃)₂CHCH(Cl)CH₃
- (ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
- (iii) CH₃CH₂C(CH₃)₂CH₂I
- (iv) (CH₃)₃CCH₂CH(Br)C₆H₅
- (v) CH₃CH(CH₃)CH(Br)CH₃
- (vi) CH₃C(C₂H₅)₂CH₂Br
- (vii) CH₃C(Cl)(C₂H₅)CH₂CH₃
- (viii) CH₃CH=C(Cl)CH₂CH(CH₃)₂
- (ix) CH₃CH=CHC(Br)(CH₃)₂ (x)…
2.
Give the IUPAC names of the following compounds:
- (i) CH₃CH(Cl)CH(Br)CH₃
- (ii) CHF₂CBrClF
- (iii) ClCH₂C≡CCH₂Br
- (iv) (CCl₃)₃CCl
- (v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃
- (vi) (CH₃)₃CCH=CClC₆H₄I-p
3.
Write the structures of the following organic halogen compounds.
- (i) 2-Chloro-3-methylpentane
- (ii) p-Bromochlorobenzene
- (iii) 1-Chloro-4-ethylcyclohexane
- (iv) 2-(2-Chlorophenyl)-1-iodooctane
- (v) 2-Bromobutane
- (vi) 4-tert-Butyl-3-iodoheptane
- (vii) 1-Bromo-4-sec-butyl-2-methylbenzene
- (viii) 1,4-Dibromobut-2-ene
4.
Which one of the following has the highest dipole moment?
- (i) CH₂Cl₂
- (ii) CHCl₃
- (iii) CCl₄
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Write the isomers of the compound having formula C₄H₉Br.
7.
Write the equations for the preparation of 1-iodobutane from
- (i) 1-butanol
- (ii) 1-chlorobutane
- (iii) but-1-ene.
What are ambident nucleophiles? Explain with an example.
9.
Which compound in each of the following pairs will react faster in SN₂ reaction with -OH?
- (i) CH₃Br or CH₃I
- (ii) (CH₃)₃CCl or CH₃Cl
10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
- (i) 1-Bromo-1-methylcyclohexane
- (ii) 2-Chloro-2-methylbutane
- (iii) 2,2,3-Trimethyl-3-bromopentane.
11.
How will you bring about the following conversions?
- (i) Ethanol to but-1-yne
- (ii) Ethane to bromoethene
- (iii) Propene to 1-nitropropane
- (iv) Toluene to benzyl alcohol
- (v) Propene to propyne
- (vi) Ethanol to ethyl fluoride
- (vii) Bromomethane to propanone
- (viii) But-1-ene to but-2-ene
- (ix) 1-Chlorobutane to n-octane
- (x) Benzene to biphenyl
12.
Explain why
- (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
- (ii) alkyl halides, though polar, are immiscible with water?
- (iii) Grignard reagents should be prepared under anhydrous conditions?
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
14.
Write the structure of the major organic product in each of the following reactions:
- (i) CH₃CH₂CH₂Cl + NaI (acetone)
- (ii) (CH₃)₃CBr + KOH
- (iii) CH₃CH(Br)CH₂CH₃ + NaOH
- (iv) CH₃CH₂Br + KCN
- (v) C₆H₅ONa + C₂H₅Cl
- (vi) CH₃CH₂CH₂OH + SOCl₂
- (vii) CH₃CH₂CH=CH₂ + HBr
- (viii) CH₃CH=C(CH₃)₂ + HBr
Write the mechanism of the following reaction: n-BuBr + KCN -> n-BuCN
16.
Arrange the compounds of each set in order of reactivity towards SN₂ displacement:
- (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
- (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane
- (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
19.
How the following conversions can be carried out?
- (i) Propene to propan-1-ol
- (ii) Ethanol to but-1-yne
- (iii) 1-Bromopropane to 2-bromopropane
- (iv) Toluene to benzyl alcohol
- (v) Benzene to 4-bromonitrobenzene
- (vi) Benzyl alcohol to 2-phenylethanoic acid
- (vii) Ethanol to propanenitrile
- (viii) Aniline to chlorobenzene
- (ix) 2-Chlorobutane to 3,4-dimethylhexane
- (x) 2-Methyl-1-propene to…
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
21.
Primary alkyl halide C₄H₉Br
- (a) reacted with alcoholic KOH to give compound (b). Compound
- (b) is reacted with HBr to give
- (c) which is an isomer of (a). When
- (a) is reacted with sodium metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of
- (a) and write the equations for all the reactions.
22.
What happens when
- (i) n-butyl chloride is treated with alcoholic KOH,
- (ii) bromobenzene is treated with Mg in the presence of dry ether,
- (iii) chlorobenzene is subjected to hydrolysis,
- (iv) ethyl chloride is treated with aqueous KOH,
- (v) methyl bromide is treated with sodium in the presence of dry ether,
- (vi) methyl chloride is treated with KCN?

Sample

Worked examples

3 questions from Haloalkanes and Haloarenes, solved step by step.

Exercises

Which one of the following has the highest dipole moment? (i) CH₂Cl₂ (ii) CHCl₃ (iii) CCl₄

CCl₄ (tetrachloromethane): All four C-Cl bonds are identical and arranged tetrahedrally. The bond dipoles cancel completely by symmetry → net dipole moment = 0 D.

CH₂Cl₂ (dichloromethane): The two C-Cl bond dipoles add up (resultant along the Cl-C-Cl bisector), while the two C-H dipoles are much smaller. The resultant is large: dipole moment ≈ 1.60 D.

CHCl₃ (chloroform): The three C-Cl bond dipoles largely oppose one another (they point toward three corners of the tetrahedron), so their resultant is partly cancelled and reinforced only weakly by the single C-H dipole. Net dipole moment ≈ 1.04 D — smaller than in CH₂Cl₂.

Therefore CH₂Cl₂ has the highest dipole moment: the two large C-Cl dipoles reinforce, giving the largest vector sum of the three molecules.

Answer: CH₂Cl₂ (dichloromethane) has the highest dipole moment (≈1.60 D). Order: CH₂Cl₂ (1.60 D) > CHCl₃ (1.04 D) > CCl₄ (0 D)

Exercises

A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.

Degree of unsaturation for C₅H₁₀ = (2×5 + 2 - 10)/2 = 1.

One degree of unsaturation means one ring or one double bond.

Does not react with Cl₂ in the dark → NOT an alkene (alkenes react with Cl₂ via electrophilic addition in dark).

So the compound is a cycloalkane (ring = 1 degree of unsaturation, no reaction with Cl₂ in dark).

Gives a SINGLE monochloro product in sunlight → free-radical chlorination → all H atoms in the molecule are equivalent.

Among C₅H₁₀ cycloalkanes: cyclopentane (C₅H₁₀) has all 10 hydrogen atoms equivalent (all are CH₂ hydrogens on the ring, in identical environments).

Chlorination of cyclopentane gives only one product: chlorocyclopentane (C₅H₉Cl). ✓

Methylcyclobutane (C₅H₁₀) would give multiple products (non-equivalent H atoms), so it is ruled out.

Answer: Cyclopentane (C₅H₁₀). All 10 hydrogen atoms are equivalent; free-radical chlorination in sunlight gives a single product — chlorocyclopentane (C₅H₉Cl).

Exercises

Write the isomers of the compound having formula C₄H₉Br.

C₄H₉Br has no degree of unsaturation. The C₄H₉ group can be arranged in four ways:

1. 1-Bromobutane (n-butyl bromide): CH₃CH₂CH₂CH₂Br — primary halide.

2. 2-Bromobutane (sec-butyl bromide): CH₃CH(Br)CH₂CH₃ — secondary halide.

3. 1-Bromo-2-methylpropane (isobutyl bromide): (CH₃)₂CHCH₂Br — primary halide.

4. 2-Bromo-2-methylpropane (tert-butyl bromide): (CH₃)₃CBr — tertiary halide.

These are all four structural isomers of C₄H₉Br.

Answer: Four isomers: (1) CH₃CH₂CH₂CH₂Br (1-bromobutane, 1°); (2) CH₃CH(Br)CH₂CH₃ (2-bromobutane, 2°); (3) (CH₃)₂CHCH₂Br (1-bromo-2-methylpropane, 1°); (4) (CH₃)₃CBr (2-bromo-2-methylpropane, 3°)

All 22 questions in Haloalkanes and Haloarenes are solved step by step in the Solutions PDF above.

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Chapter 6 Solutions — Haloalkanes and Haloarenes

Overview

What these solutions cover

Exercises

Worked examples

Which one of the following has the highest dipole moment? (i) CH₂Cl₂ (ii) CHCl₃ (iii) CCl₄

A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.

Write the isomers of the compound having formula C₄H₉Br.

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