NCERT Solutions for Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes (English Medium)
🧬 Get NCERT Solutions for Class 12 Chemistry Chapter 6 – Haloalkanes and Haloarenes with accurate answers and step-by-step explanations in a simple, exam-friendly format. This chapter covers classification & nomenclature, preparation methods, physical properties, SN1 & SN2 reactions, nucleophilic substitution, elimination reactions, aryl halides (haloarenes), and important topics like reactivity order, uses and environmental effects. Great for CBSE boards and NEET/JEE revision. ✅
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Q.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) p-ClC6H4CH2CH(CH3)2
(xi) m-ClCH2C6H4CH2C(CH3)3
(xii) o-Br–C6H4CH(CH3)CH2CH3
(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) p-ClC6H4CH2CH(CH3)2
(xi) m-ClCH2C6H4CH2C(CH3)3
(xii) o-Br–C6H4CH(CH3)CH2CH3
Answer ✅ (IUPAC Name + Classification)
(i) (CH3)2CHCH(Cl)CH3
👉 2-chloro-3-methylbutane
✅ Alkyl halide, 2° (secondary)
👉 2-chloro-3-methylbutane
✅ Alkyl halide, 2° (secondary)
(ii) CH3CH2CH(CH3)CH(C2H5)Cl
👉 3-chloro-4-methylhexane
✅ Alkyl halide, 2° (secondary)
👉 3-chloro-4-methylhexane
✅ Alkyl halide, 2° (secondary)
(iii) CH3CH2C(CH3)2CH2I
👉 1-iodo-2,2-dimethylbutane
✅ Alkyl halide, 1° (primary)
👉 1-iodo-2,2-dimethylbutane
✅ Alkyl halide, 1° (primary)
(iv) (CH3)3CCH2CH(Br)C6H5
👉 1-bromo-1-phenyl-3,3-dimethylbutane
✅ Alkyl halide, 2° (secondary)
👉 1-bromo-1-phenyl-3,3-dimethylbutane
✅ Alkyl halide, 2° (secondary)
(v) CH3CH(CH3)CH(Br)CH3
👉 2-bromo-3-methylbutane
✅ Alkyl halide, 2° (secondary)
👉 2-bromo-3-methylbutane
✅ Alkyl halide, 2° (secondary)
(vi) CH3C(C2H5)2CH2Br
👉 1-bromo-2-ethyl-2-methylbutane
✅ Alkyl halide, 1° (primary)
👉 1-bromo-2-ethyl-2-methylbutane
✅ Alkyl halide, 1° (primary)
(vii) CH3C(Cl)(C2H5)CH2CH3
👉 3-chloro-3-methylpentane
✅ Alkyl halide, 3° (tertiary)
👉 3-chloro-3-methylpentane
✅ Alkyl halide, 3° (tertiary)
(viii) CH3CH=C(Cl)CH2CH(CH3)2
👉 3-chloro-5-methylhex-2-ene
✅ Vinyl (vinylic) halide
👉 3-chloro-5-methylhex-2-ene
✅ Vinyl (vinylic) halide
(ix) CH3CH=CHC(Br)(CH3)2
👉 4-bromo-4-methylpent-2-ene
✅ Allyl (allylic) halide, 3° (tertiary allylic)
👉 4-bromo-4-methylpent-2-ene
✅ Allyl (allylic) halide, 3° (tertiary allylic)
(x) p-ClC6H4CH2CH(CH3)2
👉 1-chloro-4-(2-methylpropyl)benzene
✅ Aryl halide (para substituted)
👉 1-chloro-4-(2-methylpropyl)benzene
✅ Aryl halide (para substituted)
(xi) m-ClCH2C6H4CH2C(CH3)3
👉 1-(chloromethyl)-3-(2,2-dimethylpropyl)benzene
✅ Benzyl halide, 1° (primary benzyl)
👉 1-(chloromethyl)-3-(2,2-dimethylpropyl)benzene
✅ Benzyl halide, 1° (primary benzyl)
(xii) o-Br–C6H4CH(CH3)CH2CH3
👉 1-bromo-2-(butan-2-yl)benzene
✅ Aryl halide (ortho substituted)
👉 1-bromo-2-(butan-2-yl)benzene
✅ Aryl halide (ortho substituted)
Explanation (Step by Step) 👉
1) IUPAC naming quick rules
✅ Longest carbon chain choose करो (या aromatic में benzene ring main).✅ Numbering ऐसे करो कि substituents को smallest number मिले.
✅ Halogen prefix: fluoro-, chloro-, bromo-, iodo-
✅ Double bond हो तो: -ene और double bond को priority numbering.
2) Classification trick (super easy)
✅ Alkyl halide: X attached to sp3 carbon (normal chain).• 1°: X वाला carbon केवल 1 carbon से जुड़ा
• 2°: X वाला carbon 2 carbons से जुड़ा
• 3°: X वाला carbon 3 carbons से जुड़ा
✅ Vinyl (vinylic) halide: X directly C=C carbon पर हो (जैसे viii).
✅ Allyl (allylic) halide: X double bond के next carbon पर हो (जैसे ix).
✅ Benzyl halide: X –CH2 benzene ring के साथ attached हो (जैसे xi).
✅ Aryl halide: X directly benzene ring पर हो (जैसे x, xii).
Did You Know? 🤔📌
✅ Vinyl halides (vinylic Cl) आमतौर पर SN1/SN2 में कम reactive होते हैं क्योंकि C–X bond मजबूत होता है और carbon sp2 होता है।✅ Allylic और benzylic halides ज्यादा reactive होते हैं क्योंकि carbocation/transition state resonance से stabilize हो जाता है।
✅ Aryl halides (Cl/Br directly benzene पर) normal SN1/SN2 आसानी से नहीं करते, due to resonance + partial double bond character in C–X.
Q.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
(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
Answer ✅ (IUPAC Names)
(i) CH3CH(Cl)CH(Br)CH3
👉 2-bromo-3-chlorobutane
👉 2-bromo-3-chlorobutane
(ii) CHF2CBrClF
👉 2-bromo-2-chloro-1,1,2-trifluoroethane
👉 2-bromo-2-chloro-1,1,2-trifluoroethane
(iii) ClCH2C≡CCH2Br
👉 1-bromo-4-chlorobut-2-yne
👉 1-bromo-4-chlorobut-2-yne
(iv) (CCl3)3CCl
👉 1-chloro-1,1,2,2,3,3,3-heptachloropropane
👉 1-chloro-1,1,2,2,3,3,3-heptachloropropane
(v) CH3C(p-ClC6H4)2CH(Br)CH3
👉 2-bromo-3,3-bis(4-chlorophenyl)butane
👉 2-bromo-3,3-bis(4-chlorophenyl)butane
(vi) (CH3)3CCH=CClC6H4I-p
👉 1-chloro-2-(4-iodophenyl)-3,3-dimethylbut-1-ene
👉 1-chloro-2-(4-iodophenyl)-3,3-dimethylbut-1-ene
Explanation (Step by Step) 👉
(i) CH3CH(Cl)CH(Br)CH3
👉 Main chain = butane
👉 Substituents at 2 and 3 (tie from both ends)
✅ Alphabetical priority: bromo before chloro → bromo gets smaller locant
✅ Name: 2-bromo-3-chlorobutane
👉 Main chain = butane
👉 Substituents at 2 and 3 (tie from both ends)
✅ Alphabetical priority: bromo before chloro → bromo gets smaller locant
✅ Name: 2-bromo-3-chlorobutane
(ii) CHF2CBrClF
👉 Main chain = ethane (2 carbons)
👉 C-1 has two F → 1,1-difluoro
👉 C-2 has Br, Cl, F → 2-bromo, 2-chloro, 2-fluoro
✅ Total F = 3 → trifluoro
✅ Name: 2-bromo-2-chloro-1,1,2-trifluoroethane
👉 Main chain = ethane (2 carbons)
👉 C-1 has two F → 1,1-difluoro
👉 C-2 has Br, Cl, F → 2-bromo, 2-chloro, 2-fluoro
✅ Total F = 3 → trifluoro
✅ Name: 2-bromo-2-chloro-1,1,2-trifluoroethane
(iii) ClCH2C≡CCH2Br
👉 Main chain = 4 carbons with triple bond → but-2-yne
👉 Triple bond position priority: 2 from either side (tie)
👉 Now choose numbering that gives substituents lowest: from Br side → 1-bromo, 4-chloro
✅ Name: 1-bromo-4-chlorobut-2-yne
👉 Main chain = 4 carbons with triple bond → but-2-yne
👉 Triple bond position priority: 2 from either side (tie)
👉 Now choose numbering that gives substituents lowest: from Br side → 1-bromo, 4-chloro
✅ Name: 1-bromo-4-chlorobut-2-yne
(iv) (CCl3)3CCl
👉 Central carbon attached to Cl and three –CCl3 groups
✅ Common textbook IUPAC used: 1-chloro-1,1,2,2,3,3,3-heptachloropropane
👉 Central carbon attached to Cl and three –CCl3 groups
✅ Common textbook IUPAC used: 1-chloro-1,1,2,2,3,3,3-heptachloropropane
(v) CH3C(p-ClC6H4)2CH(Br)CH3
👉 Main chain = butane
👉 Numbering from right gives bromo at 2 (better than 3)
👉 Two (4-chlorophenyl) groups at carbon 3 → 3,3-bis(4-chlorophenyl)
✅ Name: 2-bromo-3,3-bis(4-chlorophenyl)butane
👉 Main chain = butane
👉 Numbering from right gives bromo at 2 (better than 3)
👉 Two (4-chlorophenyl) groups at carbon 3 → 3,3-bis(4-chlorophenyl)
✅ Name: 2-bromo-3,3-bis(4-chlorophenyl)butane
(vi) (CH3)3CCH=CClC6H4I-p
👉 Main chain includes double bond → but-1-ene skeleton
👉 Cl on C-1 → 1-chloro
👉 tert-butyl side gives 3,3-dimethyl
👉 p-iodophenyl attached on C-2 → 2-(4-iodophenyl)
✅ Name: 1-chloro-2-(4-iodophenyl)-3,3-dimethylbut-1-ene
👉 Main chain includes double bond → but-1-ene skeleton
👉 Cl on C-1 → 1-chloro
👉 tert-butyl side gives 3,3-dimethyl
👉 p-iodophenyl attached on C-2 → 2-(4-iodophenyl)
✅ Name: 1-chloro-2-(4-iodophenyl)-3,3-dimethylbut-1-ene
Did You Know? 🤔📌
✅ जब दोनों तरफ numbering करने पर locants same आते हैं (जैसे 2 और 3), तब alphabetical order से decide होता है कि किस substituent को छोटा number मिलेगा।✅ Alkynes में multiple bond position को substituents से पहले priority मिलती है while numbering.
✅ Benzene में para (p-) का मतलब 1,4-positions होता है, इसलिए 4-iodophenyl और 4-chlorophenyl forms IUPAC में बहुत common हैं।
Q.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
(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
Answer ✅ (Structures)
(i) 2-Chloro-3-methylpentane
👉 CH3–CH(Cl)–CH(CH3)–CH2–CH3
👉 CH3–CH(Cl)–CH(CH3)–CH2–CH3
(ii) p-Bromochlorobenzene (1-bromo-4-chlorobenzene)
👉 p-Br–C6H4–Cl
Br and Cl are at para (1,4) positions on benzene ring
👉 p-Br–C6H4–Cl
Br and Cl are at para (1,4) positions on benzene ring
(iii) 1-Chloro-4-ethylcyclohexane
👉 Cyclohexane ring with:
✅ C-1 : –Cl
✅ C-4 : –CH2CH3 (ethyl)
(1,4-substitution on cyclohexane)
👉 Cyclohexane ring with:
✅ C-1 : –Cl
✅ C-4 : –CH2CH3 (ethyl)
(1,4-substitution on cyclohexane)
(iv) 2-(2-Chlorophenyl)-1-iodooctane
👉 I–CH2–CH(–C6H4Cl-o)–CH2–CH2–CH2–CH2–CH2–CH3
Phenyl ring has ortho-Cl relative to attachment point
👉 I–CH2–CH(–C6H4Cl-o)–CH2–CH2–CH2–CH2–CH2–CH3
Phenyl ring has ortho-Cl relative to attachment point
(v) 2-Bromobutane
👉 CH3–CH(Br)–CH2–CH3
👉 CH3–CH(Br)–CH2–CH3
(vi) 4-tert-Butyl-3-iodoheptane
👉 CH3–CH2–CH(I)–CH[C(CH3)3]–CH2–CH2–CH3
👉 CH3–CH2–CH(I)–CH[C(CH3)3]–CH2–CH2–CH3
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
👉 Br–C6H3(CH3)(sec-butyl) with positions:
✅ 1 : –Br
✅ 2 : –CH3 (methyl)
✅ 4 : –CH(CH3)CH2CH3 (sec-butyl)
Line form: 1-Br, 2-CH3, 4-CH(CH3)CH2CH3 on benzene
👉 Br–C6H3(CH3)(sec-butyl) with positions:
✅ 1 : –Br
✅ 2 : –CH3 (methyl)
✅ 4 : –CH(CH3)CH2CH3 (sec-butyl)
Line form: 1-Br, 2-CH3, 4-CH(CH3)CH2CH3 on benzene
(viii) 1,4-Dibromobut-2-ene
👉 Br–CH2–CH=CH–CH2–Br
👉 Br–CH2–CH=CH–CH2–Br
Explanation (Step by Step) 👉
✅ Structure लिखने का आसान तरीका:1) Main carbon chain / ring पहचानो (pentane, benzene, cyclohexane, octane, heptane, but-2-ene).
2) Numbering नाम के हिसाब से करो (जहाँ halogen/alkyl positions दिए हैं).
3) दिए गए position पर Cl/Br/I और alkyl group attach कर दो.
✅ Benzene में o-, m-, p- का मतलब याद रखो:
👉 o = 1,2 | m = 1,3 | p = 1,4
Did You Know? 🤔📌
✅ sec-butyl group का structure: –CH(CH3)CH2CH3 (attachment middle carbon से होता है)।✅ tert-butyl group का structure: –C(CH3)3 (attachment central carbon से होता है)।
✅ p-Bromochlorobenzene का IUPAC name अक्सर 1-bromo-4-chlorobenzene भी लिखा जाता है (same compound).
Q.4: Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
Answer ✅
✅
CH2Cl2
(dichloromethane) has the highest dipole moment.Order: CH2Cl2 > CHCl3 > CCl4
Explanation (Step by Step) 👉
1) Dipole moment depends on shape + bond dipoles
Dipole moment (μ) is the net result of all bond dipoles.• If bond dipoles cancel due to symmetry → μ becomes zero or very small.
• If they do not cancel → μ is larger.
2) CCl4 has μ = 0
CCl4 is perfectly tetrahedral and highly symmetric.✅ All four C–Cl bond dipoles cancel completely.
👉 Therefore, dipole moment = 0.
3) CHCl3 has a dipole moment (but smaller)
CHCl3 is tetrahedral but not fully symmetric (3 Cl + 1 H).✅ Dipoles do not cancel completely, so μ is non-zero.
However, the three C–Cl bond dipoles partially cancel each other, so the net μ is moderate.
4) CH2Cl2 has the highest μ
CH2Cl2 has 2 Cl and 2 H arranged tetrahedrally.✅ This arrangement produces a stronger net separation of charge, so dipoles cancel less than in CHCl3.
👉 Therefore, CH2Cl2 has the highest dipole moment.
Did You Know? 🤔📌
✅ Symmetry reduces dipole moment.That’s why many highly symmetric molecules like CCl4, CO2, and BF3 have zero dipole moment even though their bonds are polar.
Q.5: A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound
C5H9Cl in bright sunlight. Identify the hydrocarbon.
Answer ✅
✅ The hydrocarbon is
cyclopentane
(C5H10).
Explanation (Step by Step) 👉
1) C5H10 can be an alkene or a cycloalkane
For the general formula CnH2n, the possibilities are:• Alkene (one double bond)
• Cycloalkane (one ring)
2) “No reaction with Cl2 in the dark” rules out alkene
✅ Alkenes generally react with Cl2 even in the dark via addition across C=C.✅ Since there is no reaction in the dark, the compound is not an alkene.
👉 Therefore, it must be a cycloalkane.
3) A single monochloro product means all H atoms are equivalent
In bright sunlight, chlorination occurs by free-radical substitution (not addition).✅ Getting only one monochloro product means all hydrogen atoms are equivalent, so replacing any H gives the same product.
4) Cyclopentane fits perfectly
✅ Cyclopentane is highly symmetrical (all ring –CH2 groups are equivalent).✅ So chlorination gives only one product: monochlorocyclopentane (C5H9Cl).
👉 Therefore, the hydrocarbon is cyclopentane.
Did You Know? 🤔📌
✅ Chlorination in sunlight (hν) is a classic free-radical reaction:👉 Cl2 → 2Cl· (radicals)
✅ Alkenes react with Cl2 without sunlight because they undergo addition, but alkanes/cycloalkanes usually need sunlight for substitution.