NCERT Solutions for Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes
These Class 12 Chemistry Chapter 6 solutions cover Haloalkanes and Haloarenes from the NCERT textbook (Reprint 2026–27). You get every Intext Question and every numbered Exercise reproduced verbatim and solved step by step — IUPAC nomenclature, structures, SN1 and SN2 mechanisms, dehydrohalogenation, dipole moments, conversions and the chemistry of haloarenes — written in exam-ready style for the CBSE board examination.
Chapter 6, Haloalkanes and Haloarenes, deals with organic compounds in which one or more hydrogen atoms of a hydrocarbon are replaced by halogen atoms. In haloalkanes (alkyl halides, R–X) the halogen is bonded to an sp3 carbon, whereas in haloarenes (aryl halides) it is bonded to an sp2 carbon of an aromatic ring. The chapter covers their classification, IUPAC nomenclature, the polar nature of the C–X bond, methods of preparation (from alcohols, hydrocarbons, halogen exchange, Sandmeyer’s reaction), physical properties, and chemical reactions — especially nucleophilic substitution (SN1 and SN2), elimination (β-elimination, Saytzeff rule), and reactions with metals (Grignard, Wurtz, Wurtz–Fittig). Stereochemistry — chirality, enantiomers, inversion and racemisation — is used as a tool to understand the mechanisms. Finally it discusses important polyhalogen compounds (CH2Cl2, CHCl3, CHI3, CCl4, freons, DDT) and their environmental effects.
Key Concepts & Reactions
C–X bond: polar because halogen is more electronegative than carbon; carbon bears δ+ and halogen δ–. Bond length increases C–F < C–Cl < C–Br < C–I; bond enthalpy decreases in the same order.
Boiling point order: for the same alkyl group, RI > RBr > RCl > RF; b.p. decreases with branching; haloalkanes are heavier than water (Br, I, polychloro) and only slightly soluble in water.
SN2: single step, second-order kinetics, backside attack → inversion of configuration; reactivity 1° > 2° > 3° (steric).
SN1: two steps via carbocation, first-order kinetics, polar protic solvents → racemisation; reactivity 3° > 2° > 1° (carbocation stability). Allylic/benzylic halides are highly reactive (resonance-stabilised cation).
Elimination (β-elimination): alcoholic KOH removes H from β-carbon and X from α-carbon to give an alkene; the more substituted alkene predominates (Saytzeff/Zaitsev rule).
Haloarenes are far less reactive towards nucleophilic substitution because of resonance (partial double-bond character of C–X), sp2 hybridisation (shorter, stronger bond), instability of the phenyl cation, and nucleophile–ring repulsion. Electron-withdrawing groups (–NO2) at o– and p-positions activate them.
Finkelstein reaction: R–Cl/Br + NaI →[dry acetone] R–I + NaCl/NaBr↓
6.1 Write structures of the following compounds:
(i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert-Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec-butyl-2-methylbenzene
ANSWER(i) 2-Chloro-3-methylpentane: CH3–CHCl–CH(CH3)–CH2–CH3 (Cl on C-2, –CH3 on C-3).(ii) 1-Chloro-4-ethylcyclohexane: a cyclohexane ring with –Cl on C-1 and –C2H5 on C-4 (para positions of the ring).(iii) 4-tert-Butyl-3-iodoheptane: CH3CH2–CHI–CH[C(CH3)3]–CH2CH2CH3 (heptane chain; –I on C-3, –C(CH3)3 on C-4).(iv) 1,4-Dibromobut-2-ene: BrCH2–CH=CH–CH2Br.(v) 1-Bromo-4-sec-butyl-2-methylbenzene: a benzene ring with –Br on C-1, –CH3 on C-2 and –CH(CH3)CH2CH3 (sec-butyl) on C-4.
6.2 Why is sulphuric acid not used during the reaction of alcohols with KI?
ANSWERH2SO4 cannot be used with KI for converting an alcohol to an alkyl iodide because sulphuric acid converts KI into the corresponding acid HI, and being a strong oxidising agent it then oxidises the liberated HI to I2. As a result HI (the actual reagent) is consumed and the yield of alkyl iodide falls drastically.2 KI + H2SO4 → 2 HI + K2SO4; 2 HI + H2SO4 → I2 + SO2 + 2 H2O. Hence non-oxidising H3PO4 is preferred.
6.3 Write structures of different dihalogen derivatives of propane.
6.4 Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields
(i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides.
ANSWER(i) Neopentane, (CH3)4C — all 12 hydrogens are equivalent, so replacement of any H gives only one monochloride, 1-chloro-2,2-dimethylpropane.(ii) n-Pentane, CH3CH2CH2CH2CH3 — it has three sets of equivalent hydrogens (C-1, C-2, C-3), giving three isomeric monochlorides: 1-chloropentane, 2-chloropentane and 3-chloropentane.(iii) Isopentane (2-methylbutane), (CH3)2CHCH2CH3 — it has four different types of hydrogen, giving four isomeric monochlorides: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane and 1-chloro-3-methylbutane.
6.5 Draw the structures of major monohalo products in each of the following reactions:
(i) cyclohexanol + SOCl2 (ii) cyclohexene + HBr (iii) but-2-ene + HCl (iv) propan-1-ol + PBr3
ANSWER(i) Cyclohexanol with SOCl2 → chlorocyclohexane (C6H11Cl) + SO2 + HCl.(ii) Cyclohexene with HBr (Markovnikov addition) → bromocyclohexane (C6H11Br).(iii) But-2-ene (CH3CH=CHCH3) with HCl → 2-chlorobutane, CH3CH2CHClCH3 (both carbons of the double bond are equivalent).(iv) Propan-1-ol with PBr3 → 1-bromopropane, CH3CH2CH2Br.
6.6 Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
ANSWER(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform. Boiling point rises with increasing molecular mass (and number of halogen atoms), which strengthens van der Waals forces.(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane. Among the two C3H7Cl isomers, the branched isopropyl chloride has a smaller surface area and weaker van der Waals forces, so a lower b.p. than n-propyl chloride; 1-chlorobutane has the highest molecular mass.
6.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
(i) CH3CH2CH2CH2Br or CH3CH2CH(Br)CH3 (ii) CH3CH2CH(Br)CH3 or CH3CH2C(CH3)2Br (iii) CH3CH2CH(Br)CH2CH3 or CH3CH(Br)CH2CH2CH3
ANSWER(i) CH3CH2CH2CH2Br (n-butyl bromide) reacts faster. Being a primary halide there is no steric hindrance to the backside attack of the nucleophile, whereas the second compound is secondary.(ii) CH3CH2CH(Br)CH3 (secondary) reacts faster than the tertiary halide, because SN2 rate falls with increasing steric crowding around the reaction centre (1° > 2° > 3°).(iii) CH3CH(Br)CH2CH2CH3 (2-bromopentane) reacts faster. In the other compound (3-bromopentane) two ethyl groups flank the carbon bearing Br, giving greater steric hindrance and a slower SN2 reaction.
6.8 In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction?
(i) CH3CH2CH(Br)CH3 or CH3CH2C(CH3)2Br (ii) CH3CH2CH(Cl)CH3 or CH3CH2CH2Cl
ANSWER(i) The tertiary halide, CH3CH2C(CH3)2Br, reacts faster. SN1 proceeds through a carbocation, and a 3° carbocation is more stable than a 2° one.(ii) The secondary halide, CH3CH2CH(Cl)CH3, reacts faster, because a 2° carbocation is more stable than the 1° carbocation from n-propyl chloride.
6.9 Identify A, B, C, D, E, R and R′ in the following:
CH3–Br →[Mg, dry ether] A →[D2O] B; R–Br →[Na, dry ether] R–R (= C) where C = CH3CH2CH2CH2CH2CH3; CH3CH2Br →[Na, dry ether] D and a halide E.
ANSWERFrom CH3Br with Mg in dry ether, A = CH3MgBr (methylmagnesium bromide). On treatment with D2O (a source of D+), the Grignard reagent gives B = CH3D (mono-deuteromethane).For the Wurtz reaction giving C = n-hexane (CH3(CH2)4CH3), the starting halide must be R–Br = CH3CH2CH2Br, i.e. R = R′ = n-propyl (C3H7–), since two identical alkyl groups couple.Bromoethane (CH3CH2Br) with Na in dry ether (Wurtz) gives D = n-butane, CH3CH2CH2CH3; E = NaBr is the by-product.(The NCERT scheme uses Grignard formation/hydrolysis and Wurtz coupling; the species above are the products obtained from those steps.)
NCERT Exercises (6.1–6.22) — Solutions
6.1 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
ANSWER(i) 2-Chloro-3-methylbutane — secondary alkyl (2°) halide.(ii) 3-Chloro-4-methylhexane — secondary alkyl (2°) halide.(iii) 1-Iodo-2,2-dimethylbutane — primary alkyl (1°) halide.(iv) (2-Bromo-3,3-dimethylbutyl)benzene [1-bromo-1-phenyl-3,3-dimethylbutane] — secondary benzylic (2°) halide (X on the carbon attached to the ring).(v) 2-Bromo-3-methylbutane — secondary alkyl (2°) halide.(vi) 1-Bromo-2,2-diethylbutane (2-bromomethyl… commonly named 1-bromo-2-ethyl-2-methylbutane) — primary alkyl (1°) halide.(vii) 3-Chloro-3-methylpentane — tertiary alkyl (3°) halide.(viii) 2-Chloro-5-methylhex-2-ene [3-chloro… ] — vinyl halide (Cl on an sp2 carbon of C=C).(ix) 4-Bromo-4-methylpent-2-ene [2-bromo-2-methyl… ] — allyl halide, tertiary (3°), X on sp3 carbon next to C=C.(x) 1-Chloro-4-(2-methylpropyl)benzene (p-chloroisobutylbenzene) — aryl halide (Cl bonded to the ring).(xi) 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene (m-position) — primary benzyl (1°) halide (Cl on the –CH2– attached to the ring).(xii) 1-Bromo-2-(1-methylpropyl)benzene (o-sec-butylbromobenzene) — aryl halide.
6.2 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
6.3 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
ANSWER(i) CH3CHClCH(CH3)CH2CH3 (Cl on C-2, CH3 on C-3 of pentane).(ii) 1-Bromo-4-chlorobenzene — benzene ring with –Br and –Cl at para (1,4) positions.(iii) Cyclohexane ring with –Cl on C-1 and –C2H5 on C-4.(iv) ICH2–CH(2-ClC6H4)–CH2CH2CH2CH2CH2CH3 (octane chain; –I on C-1, an o-chlorophenyl group on C-2).(v) CH3CHBrCH2CH3.(vi) CH3CH2–CHI–CH[C(CH3)3]–CH2CH2CH3 (heptane; –I on C-3, tert-butyl on C-4).(vii) Benzene ring: –Br on C-1, –CH3 on C-2, –CH(CH3)CH2CH3 (sec-butyl) on C-4.(viii) BrCH2CH=CHCH2Br.
6.4 Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4
ANSWERCH2Cl2 (dichloromethane) has the highest dipole moment (μ ≈ 1.60 D).In CCl4 the four C–Cl bond dipoles are symmetrically arranged (regular tetrahedron) and cancel completely, so μ = 0. In CHCl3 the three C–Cl dipoles partly oppose the resultant of the C–H bond, giving a smaller net moment (≈ 1.08 D). In CH2Cl2 the two C–Cl dipoles add favourably and are not cancelled, giving the largest resultant.
6.5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
ANSWERSince C5H10 does not add chlorine in the dark, it has no C=C double bond — so it is a cycloalkane, not an alkene. The formula C5H10 (one degree of unsaturation) fits cyclopentane.In sunlight it undergoes free-radical substitution to give a single monochloride, which is possible only if all the hydrogen atoms are equivalent. Cyclopentane has ten equivalent hydrogens, so it gives only chlorocyclopentane (C5H9Cl). Hence the hydrocarbon is cyclopentane.
6.6 Write the isomers of the compound having formula C4H9Br.
6.7 Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.
ANSWER(i) From 1-butanol: 3 CH3CH2CH2CH2OH + PI3 → 3 CH3CH2CH2CH2I + H3PO3. (PI3 is generated in situ from red P + I2; alternatively heat with NaI/KI in 95% H3PO4.)(ii) From 1-chlorobutane (Finkelstein reaction): CH3CH2CH2CH2Cl + NaI →[dry acetone] CH3CH2CH2CH2I + NaCl↓.(iii) From but-1-ene (anti-Markovnikov addition of HI in presence of peroxide): CH3CH2CH=CH2 + HI →[peroxide] CH3CH2CH2CH2I. (Peroxide effect is shown only with HBr in NCERT; with HI the practical route is via 1-butanol or Finkelstein.)
6.8 What are ambident nucleophiles? Explain with an example.
ANSWERAmbident nucleophiles are nucleophiles that possess two different nucleophilic centres through which they can attack a substrate, so they can give two different products.Example: the cyanide ion is a resonance hybrid [–:C≡N: ↔ :C=N:–] and can link through C or N. With KCN, attack through carbon gives mainly alkyl cyanides (R–CN); with AgCN (covalent), attack through nitrogen gives mainly isocyanides (R–NC). Similarly the nitrite ion [O=N–O]– links through O to give alkyl nitrites (R–O–N=O) or through N to give nitroalkanes (R–NO2).
6.9 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
ANSWER(i) CH3I reacts faster. Iodide is a larger, more polarisable and weaker base, hence a better leaving group than bromide, so it leaves more readily in the rate-determining backside attack.(ii) CH3Cl reacts faster. Methyl chloride is unhindered, whereas tert-butyl chloride is a bulky 3° halide whose three methyl groups hinder the approaching nucleophile, making SN2 very slow.
6.10 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.
ANSWER(i) 1-Bromo-1-methylcyclohexane → 1-methylcyclohexene (only β-H available on the ring carbons) and methylenecyclohexane (from the CH3 β-H). Major: 1-methylcyclohexene (more substituted, Saytzeff product).(ii) 2-Chloro-2-methylbutane → 2-methylbut-2-ene and 2-methylbut-1-ene. Major: 2-methylbut-2-ene (trisubstituted, Saytzeff).(iii) 2,2,3-Trimethyl-3-bromopentane → 2,3,3-trimethylpent-1-ene (… no β-H on the t-Bu side) and 3,4,4-trimethylpent-2-ene. Major: 3,4,4-trimethylpent-2-ene (more substituted double bond, Saytzeff product).
6.11 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.
6.12 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?
ANSWER(i) In chlorobenzene the carbon bearing Cl is sp2 hybridised (more electronegative, more s-character) and the lone pairs on Cl are in conjugation with the ring (–Cl shows +R), so less electron density is pulled towards Cl and the C–Cl bond is partly a double bond. Hence its dipole moment (≈ 1.69 D) is lower than that of cyclohexyl chloride (≈ 2.38 D), where the carbon is sp3 and the C–Cl bond is a pure single bond.(ii) To dissolve an alkyl halide in water, energy is needed to break the strong hydrogen bonds between water molecules. The new attractions between alkyl halide and water molecules are much weaker (no hydrogen bonding), so too little energy is released. Therefore alkyl halides, though polar, are essentially immiscible with water.(iii) Grignard reagents (R–MgX) are extremely reactive and the C–Mg bond is strongly polar (carbanion-like carbon). Even traces of moisture (a proton source) decompose them to the corresponding alkane: RMgX + H2O → RH + Mg(OH)X. Hence they must be prepared and used under strictly anhydrous (dry-ether) conditions.
6.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
ANSWERFreon 12 (CCl2F2): used as a refrigerant in refrigerators and air-conditioners and as an aerosol propellant. (Now restricted because it depletes the ozone layer.)DDT: a powerful insecticide, especially effective against mosquitoes (malaria) and lice (typhus). (Banned in many countries due to toxicity and persistence.)Carbon tetrachloride (CCl4): used as an industrial solvent, a degreasing agent, feedstock for chlorofluorocarbons, and earlier in fire extinguishers (“Pyrene”) and as a cleaning/spot-removing fluid.Iodoform (CHI3): was used as an antiseptic for wounds — its action is due to the slow liberation of free iodine (now largely replaced because of its objectionable smell).
6.14 Write the structure of the major organic product in each of the following reactions:
(i) CH3CH2CH2Cl + NaI →[acetone] (ii) (CH3)3CBr + KOH →[ethanol, heat] (iii) CH3CH(Br)CH2CH3 + NaOH →[water] (iv) CH3CH2Br + KCN (v) C6H5ONa + C2H5Cl (vi) CH3CH2CH2OH + SOCl2 (vii) CH3CH2CH=CH2 + HBr (viii) CH3CH=C(CH3)2 + HBr
ANSWER(i) CH3CH2CH2I (1-iodopropane) + NaCl (Finkelstein reaction).(ii) (CH3)2C=CH2 (2-methylprop-1-ene/isobutylene) — alcoholic KOH causes elimination of a 3° halide.(iii) CH3CH(OH)CH2CH3 (butan-2-ol) — aqueous NaOH gives substitution.(iv) CH3CH2CN (propanenitrile / ethyl cyanide); KCN attacks through carbon.(v) C6H5OC2H5 (ethoxybenzene / phenetole) — Williamson ether synthesis.(vi) CH3CH2CH2Cl (1-chloropropane) + SO2 + HCl.(vii) CH3CH2CHBrCH3 (2-bromobutane) — Markovnikov addition (no peroxide).(viii) (CH3)2CBrCH2CH3 (2-bromo-2-methylbutane) — Markovnikov addition, H to the less-substituted carbon, Br to the more-substituted carbon.
6.15 Write the mechanism of the following reaction: nBuBr + KCN → nBuCN
ANSWERn-Butyl bromide is a primary halide, so it reacts with the cyanide ion by an SN2 (bimolecular) mechanism in a single step. The reaction follows second-order kinetics: rate = k[nBuBr][CN–].Step (single): the nucleophile :C≡N– attacks the δ+ carbon bearing Br from the side opposite to the leaving group. A transition state forms in which carbon is partially bonded to both CN and Br and the three other bonds are planar.CN– + CH3CH2CH2CH2–Br → [NC···C···Br]– (T.S.) → CH3CH2CH2CH2–CN + Br–.As the C–CN bond forms, the C–Br bond breaks simultaneously, and the configuration at carbon is inverted (Walden inversion). The product is n-butyl cyanide (pentanenitrile).
6.16 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.
ANSWERSN2 reactivity decreases with steric crowding (1° > 2° > 3°).(i) 1-Bromopentane (1°) > 2-Bromopentane (2°) > 2-Bromo-2-methylbutane (3°).(ii) 1-Bromo-3-methylbutane (1°) > 2-Bromo-3-methylbutane (2°) > 2-Bromo-2-methylbutane (3°).(iii) All are primary, so order depends on branching near the reaction centre: 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane. The closer and bulkier the branch to C-1, the slower the reaction (neopentyl-type is the slowest).
6.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.
ANSWERC6H5CHClC6H5 (benzhydryl chloride) is hydrolysed more easily. Hydrolysis by aqueous KOH of these benzylic halides proceeds mainly by SN1 through a carbocation.The carbocation from C6H5CHClC6H5 is stabilised by resonance with two phenyl rings, whereas the benzyl cation from C6H5CH2Cl is stabilised by only one. The more stable carbocation forms more easily, so benzhydryl chloride is hydrolysed faster.
6.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
ANSWERThe melting point of a solid depends largely on how well its molecules pack into the crystal lattice. The para-isomer is the most symmetrical, so its molecules fit more closely and neatly into the crystal lattice, giving stronger crystal packing.Because more energy is needed to break this well-packed lattice, p-dichlorobenzene has a higher melting point than the less symmetrical ortho- and meta-isomers.
6.19 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 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide
6.20 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.
ANSWERIn aqueous KOH, KOH is almost completely ionised to give a high concentration of OH– ions. The OH– ion acts as a strong nucleophile and substitutes the halogen, giving an alcohol (nucleophilic substitution).In alcoholic KOH, the alcohol partly converts KOH into alkoxide (RO–); alkoxide and OH– here act mainly as a base that abstracts a β-hydrogen, causing dehydrohalogenation (β-elimination) to give an alkene as the major product.
6.21 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.
ANSWER(a) is a primary C4H9Br whose Wurtz product (d, C8H18) is different from n-octane (the Wurtz product of n-butyl bromide). The only other primary C4H9Br is isobutyl bromide, (a) = (CH3)2CHCH2Br (1-bromo-2-methylpropane).(a) → (b): (CH3)2CHCH2Br →[alc. KOH] (CH3)2C=CH2(b) = 2-methylprop-1-ene.(b) → (c): (CH3)2C=CH2 + HBr (Markovnikov) → (CH3)3CBr (c) = 2-bromo-2-methylpropane (tert-butyl bromide), a (tertiary) isomer of (a) — both are C4H9Br.(a) → (d): 2 (CH3)2CHCH2Br + 2 Na →[dry ether, Wurtz] (CH3)2CHCH2CH2CH(CH3)2(d) = 2,5-dimethylhexane (C8H18), which differs from n-octane.
6.22 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?
Q1. Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?
ANSWERBecause (a) the lone pair on the halogen is in resonance with the ring, giving the C–X bond partial double-bond character (hard to break); (b) the carbon is sp2, more electronegative, holding the C–X electrons tightly so the bond is shorter and stronger; (c) the phenyl cation is unstable, ruling out SN1; and (d) the electron-rich ring repels the incoming nucleophile.
Q2. Why does SN2 reaction of an optically active alkyl halide give a product with inverted configuration?
ANSWERIn SN2 the nucleophile attacks from the side opposite to the leaving group (backside attack). As the new bond forms and the old bond breaks in one step, the three other groups flip over like an umbrella in a strong wind (Walden inversion), so the product has the opposite (inverted) configuration.
Q3. What is the Finkelstein reaction? Why is dry acetone used?
ANSWERIt is the conversion of an alkyl chloride/bromide into an alkyl iodide using NaI: R–Cl + NaI → R–I + NaCl. Dry acetone dissolves NaI but precipitates NaCl/NaBr; removal of the product shifts the equilibrium forward (Le Chatelier’s principle), driving the reaction to completion.
Q4. Why is the boiling point of an iodoalkane higher than that of the corresponding chloroalkane?
ANSWERAn iodine atom is larger and has more electrons than chlorine, so it is more polarisable. This gives stronger van der Waals (London dispersion) forces and a higher molecular mass, both of which raise the boiling point (RI > RBr > RCl > RF).
Q5. State Saytzeff’s rule with an example.
ANSWERSaytzeff’s rule: in a dehydrohalogenation reaction the preferred (major) product is the more substituted, more stable alkene — the one with the greater number of alkyl groups on the doubly bonded carbons. Example: 2-bromobutane with alc. KOH gives but-2-ene as the major product rather than but-1-ene.
Long Answer Type Questions
Q1. Compare SN1 and SN2 mechanisms with respect to kinetics, steps, stereochemistry, effect of substrate and solvent.
ANSWERSN2: one step (no intermediate), second-order kinetics (rate ∝ [substrate][nucleophile]), proceeds via a transition state, gives inversion of configuration, favoured by primary halides (1° > 2° > 3°), strong nucleophiles and polar aprotic solvents. SN1: two steps via a carbocation intermediate, first-order kinetics (rate ∝ [substrate] only), gives racemisation, favoured by tertiary halides (3° > 2° > 1°) and polar protic solvents that stabilise ions; allylic and benzylic halides are especially reactive because the carbocation is resonance-stabilised. For both, leaving-group ability follows R–I > R–Br > R–Cl >> R–F.
Q2. Describe the chemical reactions of haloalkanes (substitution, elimination, reaction with metals) with suitable examples.
ANSWER(1) Nucleophilic substitution: with aq. KOH → alcohol; with NaOR′ → ether; with NaI → alkyl iodide; with KCN → nitrile; with AgCN → isonitrile; with NH3 → amine; with AgNO2 → nitroalkane. (2) Elimination: with alc. KOH a β-H and the halogen are removed to give an alkene (Saytzeff). (3) Reaction with metals: with Mg/dry ether → Grignard reagent (RMgX); with Na/dry ether → Wurtz reaction giving a higher alkane (R–R). These reactions make haloalkanes versatile intermediates in synthesis.
Q3. Why does the presence of a nitro group at ortho and para positions increase the reactivity of haloarenes towards nucleophilic substitution, but not at the meta position?
ANSWERAn –NO2 group is strongly electron-withdrawing. At ortho and para positions it withdraws electron density from the carbon bearing the halogen, making it more open to nucleophilic attack, and the negative charge of the intermediate carbanion (Meisenheimer-type) appears at the carbon bearing the –NO2 group, where it is stabilised by resonance with the nitro group. At the meta position, none of the resonance structures place the negative charge on the carbon bearing –NO2, so the carbanion is not stabilised and no rate enhancement occurs. Hence only o- and p-nitro groups activate haloarenes.
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
Reason: Tertiary carbocations are more stable than primary carbocations.
A-R 2. Assertion: CCl4 has zero dipole moment.
Reason: The four C–Cl bond dipoles are arranged symmetrically and cancel one another.
A-R 3. Assertion: Chlorobenzene has a lower dipole moment than cyclohexyl chloride.
Reason: In chlorobenzene the C–Cl bond has partial double-bond character due to resonance, and the carbon is sp2 hybridised.
A-R 4. Assertion: Grignard reagents must be prepared under anhydrous conditions.
Reason: Grignard reagents react with even traces of water to give alkanes.
A-R 5. Assertion: p-Dichlorobenzene has a higher melting point than its o- and m-isomers.
Reason: The symmetrical para-isomer packs more closely in the crystal lattice.
Answer key: 1-(A), 2-(A), 3-(A), 4-(A), 5-(A).
Common Mistakes & Exam Tips
Common mistakes to avoid
Mixing up the reactivity orders — SN1 is 3° > 2° > 1°, but SN2 is the reverse 1° > 2° > 3°.
Forgetting that aqueous KOH → substitution (alcohol) while alcoholic KOH → elimination (alkene).
Confusing Markovnikov (no peroxide) with anti-Markovnikov / peroxide (only with HBr) addition.
Writing CCl4 as having a high dipole moment — it is actually zero (symmetry); CH2Cl2 is highest among the three.
Saying haloarenes give SN1 — the phenyl cation is unstable, so SN1 is ruled out.
Forgetting Saytzeff’s rule when asked for the major alkene in dehydrohalogenation.
How to score full marks in this chapter
Always show the reagent and condition above the arrow (e.g. “alc. KOH”, “dry ether”, “NaI/acetone”) in conversion questions — marks are given for conditions. For mechanism questions, draw the transition state (SN2) or the carbocation intermediate (SN1) and state the kinetics. Use sub/superscripts correctly and name reactions (Finkelstein, Swarts, Sandmeyer, Wurtz, Wurtz–Fittig) by name. Justify ortho/para directing effect of halogens using resonance and the –I effect.
Frequently Asked Questions
What is Class 12 Chemistry Chapter 6 about?
Chapter 6, Haloalkanes and Haloarenes, covers the classification, IUPAC nomenclature, preparation, physical properties and chemical reactions of alkyl halides (R–X) and aryl halides — including SN1 and SN2 nucleophilic substitution, elimination (Saytzeff rule), reactions with metals (Grignard, Wurtz), the stereochemistry of these reactions, and important polyhalogen compounds such as chloroform, iodoform, CCl4, freons and DDT.
What is the difference between SN1 and SN2 reactions?
SN2 is a single-step, second-order reaction proceeding through a transition state with inversion of configuration, fastest for primary halides. SN1 is a two-step, first-order reaction through a carbocation intermediate with racemisation, fastest for tertiary, allylic and benzylic halides.
Why is chlorobenzene less reactive than chloroethane towards nucleophilic substitution?
In chlorobenzene the C–Cl bond has partial double-bond character due to resonance, the carbon is sp2 (shorter, stronger bond), the phenyl cation is unstable so SN1 is ruled out, and the electron-rich ring repels nucleophiles — making it far less reactive than chloroethane.
Are these Class 12 Chemistry Chapter 6 solutions free?
Yes. All solutions are free and follow the official NCERT Chemistry textbook (Reprint 2026–27), with every intext question and exercise reproduced verbatim and solved step by step.
