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Chapter 6 · Class 12 Chemistry
Haloalkanes and Haloarenes
1 exercises9 questions solved
Exercise 6.1Preparation, Properties and Reactions of Halogen Compounds
Q1
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl, vinyl or aryl halides: (i) (CH₃)₂CHCH(Cl)CH₃ (ii) CH₂=CHCH₂Cl (iii) ClCH₂C≡CH (iv) (C₂H₅)₂CHBr (v) C₆H₅CH(CH₃)Cl
Solution
(i) (CH₃)₂CHCH(Cl)CH₃:
IUPAC: 2-chloro-3-methylbutane
Type: Alkyl halide (secondary)
(ii) CH₂=CHCH₂Cl:
IUPAC: 3-chloroprop-1-ene (or allyl chloride)
Type: Allylic halide (Cl on carbon adjacent to C=C)
(iii) ClCH₂C≡CH:
IUPAC: 3-chloroprop-1-yne
Type: Allylic-type (propargylic); classified as alkyl halide
(iv) (C₂H₅)₂CHBr:
IUPAC: 3-bromopentane
Type: Alkyl halide (secondary)
(v) C₆H₅CH(CH₃)Cl:
IUPAC: (1-chloroethyl)benzene
Type: Benzylic halide (secondary benzylic)
Q2
Which one of the following is the most reactive towards an SN2 reaction? (i) CH₃Br (ii) CH₃I
Solution
CH₃I (iodomethane) is more reactive towards SN2.
In SN2 reactions, reactivity depends on the leaving group ability.
Leaving group ability: I⁻ > Br⁻ > Cl⁻ > F⁻
(I⁻ is the best leaving group — C–I bond is longest and weakest)
However, both CH₃Br and CH₃I are methyl halides (primary), so steric effects are the same.
The better leaving group (I⁻ in CH₃I) makes it more reactive in SN2.
Answer: CH₃I is more reactive towards SN2.
Q3
Predict the order of reactivity of the following compounds in SN1 reactions: (CH₃)₃CCl, (CH₃)₂CHCl, CH₃CH₂Cl, CH₃Cl.
Solution
SN1 reactivity depends on the stability of the carbocation intermediate:
More stable carbocation → faster SN1 reaction
Carbocation stability: Tertiary > Secondary > Primary > Methyl
(CH₃)₃CCl → tertiary carbocation (most stable)
(CH₃)₂CHCl → secondary carbocation
CH₃CH₂Cl → primary carbocation
CH₃Cl → methyl carbocation (least stable)
Order of SN1 reactivity:
(CH₃)₃CCl > (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl
Q4
Why is it that though aryl and vinyl halides are less reactive than alkyl halides towards nucleophilic substitution reactions but are more reactive towards electrophilic aromatic substitution reactions?
Solution
Why less reactive toward nucleophilic substitution (SN reactions):
1. The C–X bond has partial double bond character (resonance with π system/ring)
2. The carbon attached to X has sp² hybridisation → shorter, stronger C–X bond
3. The carbon is less electrophilic due to electron donation from π system/ring
4. Backside attack (SN2) is hindered by the planar, rigid structure
5. Carbocations on sp² carbons (for SN1) are destabilised (no alkyl substituents providing hyperconjugation)
Why more reactive toward electrophilic aromatic substitution (EAS):
1. Halogens are electron-withdrawing by induction but electron-donating by resonance
2. The lone pairs on X donate electron density into the ring → ring becomes electron-rich
3. This activates the ring toward electrophiles
4. Halogens are ortho/para directors in EAS reactions
5. So even though halogens deactivate the ring slightly overall, they activate it more than they would if only induction were considered
Q5
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Solution
C₅H₁₀ with degree of unsaturation = (2×5+2–10)/2 = 1 → one degree of unsaturation.
Since it does NOT react with Cl₂ in dark (no electrophilic addition), it is NOT an alkene.
Does NOT react with Cl₂ in dark rules out alkenes and alkynes (which react via addition).
Reacts with Cl₂ in bright sunlight → free radical substitution.
Gives only ONE monochloro product → all H atoms are equivalent.
C₅H₁₀ with one degree of unsaturation and all equivalent H atoms:
This is cyclopentane (C₅H₁₀), where all 10 H atoms are equivalent.
Cyclopentane:
• Does not react with Cl₂ in dark
• Reacts in sunlight (free radical substitution): C₅H₁₀ + Cl₂ → C₅H₉Cl + HCl
• Gives single product since all H equivalent
Hydrocarbon = Cyclopentane
Q6
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
Solution
(i) Ethanol → But-1-yne:
Step 1: C₂H₅OH + HBr → C₂H₅Br (bromoethane)
Step 2: C₂H₅Br + NaNH₂ → CH₂=CH₂ + NaBr (elimination) → acetylene (too many steps)
Better: Ethanol → acetaldehyde → vinyl bromide → but-1-yne (via Grignard)
Or: 2C₂H₅OH → (via acetylene) → CH≡CH → CH≡C⁻Na⁺ + C₂H₅Br → CH≡C–CH₂CH₃ (but-1-yne)
(ii) Ethane → Bromoethene (vinyl bromide):
CH₃CH₃ → (Br₂/hν) → CH₃CH₂Br → (alcoholic KOH, elimination) → CH₂=CH₂ → (Br₂) → CH₂BrCH₂Br → (2 KOH/alc, elimination) → CH≡CH → (HBr, anti-Markovnikov) → CH₂=CHBr
(iii) Propene → 1-nitropropane:
CH₂=CHCH₃ → (HBr/peroxide) → CH₂BrCH₂CH₃ (anti-Markovnikov) → (AgNO₂) → CH₂(NO₂)CH₂CH₃
(iv) Toluene → Benzyl alcohol:
C₆H₅CH₃ → (Cl₂/hν) → C₆H₅CH₂Cl → (NaOH/H₂O) → C₆H₅CH₂OH
Q7
Explain why the following are not recommended for use as solvents: (i) benzene (ii) tetrachloromethane.
Solution
(i) Benzene is not recommended as a solvent because:
• It is a known carcinogen (causes leukemia/cancer with prolonged exposure)
• It is readily absorbed through skin and lungs
• Benzene is listed as a Group 1 carcinogen (IARC)
• It is flammable and volatile
• Less toxic alternatives (toluene, hexane) are available
(ii) Tetrachloromethane (CCl₄) is not recommended because:
• It is hepatotoxic (damages the liver) and nephrotoxic (damages kidneys)
• Prolonged exposure causes liver damage and cirrhosis
• It is suspected carcinogen
• It is an ozone-depleting substance (ODS)
• Has high vapour pressure → significant inhalation risk
• It was used as a dry-cleaning agent and fire extinguisher in the past but banned due to health hazards
Q8
Arrange the following compounds in increasing order of their boiling points: CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄
Solution
Boiling points of halogenated methanes depend mainly on van der Waals forces (London dispersion forces), which increase with molecular mass and surface area.
Molecular masses:
CH₃Cl = 50.5, CH₂Cl₂ = 85, CHCl₃ = 119.5, CCl₄ = 154
Actual boiling points:
CH₃Cl: –24°C
CH₂Cl₂: 40°C
CHCl₃: 61°C
CCl₄: 77°C
Increasing order of boiling points:
CH₃Cl < CH₂Cl₂ < CHCl₃ < CCl₄
The order follows increasing molecular mass and increasing van der Waals interactions.
Q9
Which of the compounds in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH₃Br or CH₃I (ii) (CH₃)₃CCl or CH₃Cl
Solution
(i) CH₃Br vs CH₃I:
Both are methyl halides (same steric environment)
SN2 reactivity depends on leaving group ability:
I⁻ is a better leaving group than Br⁻ (weaker C–I bond, larger, more polarisable I⁻)
Answer: CH₃I reacts faster in SN2
(ii) (CH₃)₃CCl vs CH₃Cl:
SN2 reactivity strongly depends on steric hindrance at the electrophilic carbon.
(CH₃)₃CCl is a tertiary alkyl halide — backside attack is severely hindered by 3 methyl groups.
CH₃Cl is a methyl halide — no steric hindrance.
Answer: CH₃Cl reacts faster in SN2
(CH₃)₃CCl undergoes SN1, not SN2, under normal conditions.
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