NCERT Solutions for Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes – Free PDF Download

Chapter 6 (the first chapter of Chemistry Part II) covers Haloalkanes and Haloarenes — classification and IUPAC nomenclature, nature of the C–X bond, methods of preparation, physical and chemical properties, substitution mechanisms (S_N1 and S_N2) and their stereochemistry, reactions of haloarenes, and the uses/environmental effects of polyhalogen compounds. This chapter was Chapter 10 under the old (pre-2023) numbering, and shifted to Chapter 6 only because four chapters were deleted earlier in Part I — its own 22 exercises (6.1–6.22) are unchanged and none were dropped; only the deep preparation/structural chemistry of polyhalogen compounds (DDT, freons, iodoform) was trimmed to “uses and environmental effects” at the syllabus level. Each answer below is checked for chemical accuracy before publishing.

NCERT Exercise Solutions

6.1 Classify the following halides as alkyl, allylic, benzylic or vinylic: (i) CH₃CH=CHCH₂Cl (ii) C₆H₅CH₂Br (iii) CH₂=CHCl (iv) CH₃CH₂CH₂Br.
Ans: (i) Allylic (Cl is on an sp³ carbon directly adjacent to a C=C double bond). (ii) Benzylic (Br is on an sp³ carbon directly attached to the benzene ring). (iii) Vinylic (Cl is directly on an sp², double-bonded carbon). (iv) Alkyl (1°) (Br is on a simple sp³ carbon, not adjacent to any ring or double bond).

6.2 Write the IUPAC names and classify (1°/2°/3°) the following: (i) (CH₃)₂CHCH₂Br (ii) CH₃CH₂CH(Cl)CH₂CH₃ (iii) (CH₃)₃CBr.
Ans: (i) 1-Bromo-2-methylpropane, 1°. (ii) 3-Chloropentane, 2°. (iii) 2-Bromo-2-methylpropane (tert-butyl bromide), 3°.

6.3 Give the IUPAC name and one method of preparation for a compound commonly known as tert-butyl chloride.
Ans: IUPAC name: 2-Chloro-2-methylpropane. It is readily prepared by treating tert-butyl alcohol with concentrated HCl at room temperature (Lucas reagent, ZnCl₂+conc. HCl): (CH₃)₃COH+HCl→(CH₃)₃CCl+H₂O, proceeding rapidly via an S_N1 mechanism because the intermediate tertiary carbocation is comparatively stable.

6.4 Arrange CH₃Cl, CH₂Cl₂, CHCl₃ and CCl₄ in order of decreasing dipole moment, with reason.
Ans: CH₃Cl>CH₂Cl₂>CHCl₃>CCl₄ (measured values ≈1.87D, 1.60D, 1.04D, 0D). As the number of C–Cl bonds increases, their individual bond dipoles increasingly oppose and cancel each other by symmetry, so the net molecular dipole moment falls steadily; in perfectly symmetric, tetrahedral CCl₄ the four C–Cl dipoles cancel completely, giving a net dipole moment of zero.

6.5 Which C₅H₁₀ alkane gives only a single monochloro product on free-radical chlorination? Explain.
Ans: Neopentane, (CH₃)₄C. All 12 of its hydrogen atoms are chemically equivalent (attached to four identical methyl groups around a central quaternary carbon), so replacing any one of them by chlorine gives exactly the same product, (CH₃)₃CCH₂Cl, regardless of which hydrogen reacts.

6.6 Draw the structures of all the isomeric alkyl bromides with molecular formula C₄H₉Br and name them by the IUPAC system. Which of these is (a) a 1° halide (b) a 2° halide (c) a 3° halide?
Ans: There are 4 isomers: 1-bromobutane CH₃CH₂CH₂CH₂Br (1°), 2-bromobutane CH₃CH₂CH(Br)CH₃ (2°), 1-bromo-2-methylpropane (CH₃)₂CHCH₂Br (1°), and 2-bromo-2-methylpropane (CH₃)₃CBr (3°). (a) 1°: 1-bromobutane and 1-bromo-2-methylpropane. (b) 2°: 2-bromobutane. (c) 3°: 2-bromo-2-methylpropane.

6.7 Starting from (a) 1-butanol (b) 1-chlorobutane (c) but-1-ene, how would you obtain 1-iodobutane in each case?
Ans: (a) 1-Butanol+red P/I₂ (generates HI in situ) → 1-iodobutane+H₂O. (b) 1-Chlorobutane+NaI in dry acetone (the Finkelstein reaction) → 1-iodobutane+NaCl (precipitates out, driving the equilibrium forward). (c) But-1-ene+HI, in the presence of peroxides (anti-Markovnikov/Kharasch addition), adds the iodine atom to the terminal (less substituted) carbon, giving 1-iodobutane.

6.8 What is an ambident nucleophile? Explain with an example.
Ans: An ambident nucleophile has two possible sites through which it can attack an electrophile, giving two different products depending on which site reacts. Cyanide ion, CN⁻, is a classic example: it can attack through carbon (giving an alkyl cyanide/nitrile, R–CN) or through nitrogen (giving an alkyl isocyanide, R–NC). With alkyl halides, attack through carbon predominates, since carbon is the site of higher electron density and provides better orbital overlap in the S_N2 transition state, so the nitrile is the major product.

6.9 Arrange the following in order of increasing S_N2 reactivity: CH₃Cl, CH₃CH₂Cl, (CH₃)₂CHCl, (CH₃)₃CCl.
Ans: (CH₃)₃CCl<(CH₃)₂CHCl<CH₃CH₂Cl<CH₃Cl (i.e. reactivity decreases as branching increases). S_N2 proceeds via a backside attack through a crowded pentacoordinate transition state, so bulkier alkyl groups around the carbon bearing the leaving group create steric hindrance that slows the reaction; methyl halides, with no such hindrance, react fastest.

6.10 Predict the alkene formed on dehydrohalogenation of 2-bromopentane with alcoholic KOH, and state the rule used.
Ans: Two alkenes are possible: pent-1-ene and pent-2-ene. By Zaitsev’s (Saytzeff’s) rule, the more substituted, more stable alkene is the major product, so pent-2-ene (a disubstituted alkene) forms preferentially over the terminal, monosubstituted pent-1-ene.

6.11 How will you convert ethanol into (a) ethyl chloride (b) diethyl ether (c) ethene, using suitable reagents?
Ans: (a) Ethanol+SOCl₂ (or conc. HCl/ZnCl₂, or PCl₃/PCl₅) → ethyl chloride+SO₂+HCl (the SOCl₂ route is cleanest, giving only gaseous by-products). (b) Ethanol+ethanol, conc. H₂SO₄ at 413K (Williamson-type acid-catalysed intermolecular dehydration) → diethyl ether+H₂O. (c) Ethanol+conc. H₂SO₄ at 443K (intramolecular dehydration/elimination, excess acid and higher temperature favour elimination over ether formation) → ethene+H₂O.

6.12 Explain the following: (i) Chlorobenzene has a lower dipole moment than cyclohexyl chloride. (ii) Alkyl halides, though polar, are immiscible with water. (iii) Grignard reagents should be prepared under anhydrous conditions.
Ans: (i) In chlorobenzene, one lone pair on chlorine is delocalised into the aromatic ring by resonance, giving the C–Cl bond partial double-bond character; this shortens/strengthens the bond and reduces its polarity compared to the purely single, unconjugated C–Cl bond in cyclohexyl chloride. (ii) Although alkyl halides are polar, they cannot form hydrogen bonds with water; the energy released when a C–X dipole interacts with water is not enough to compensate for the energy needed to break water’s own extensive hydrogen-bonded network, so the two remain largely immiscible. (iii) Grignard reagents (R–MgX) are extremely strong bases/nucleophiles and react instantly and irreversibly with any trace of water (or other proton source) to give the corresponding alkane, R–H+Mg(OH)X, destroying the reagent — so strictly anhydrous ether and glassware are essential.

6.13 Give the uses of (i) Freon-12 (ii) DDT (iii) Carbon tetrachloride (iv) Iodoform.
Ans: (i) Freon-12 (CCl₂F₂): used as a refrigerant and aerosol-spray propellant (now heavily restricted due to stratospheric ozone depletion). (ii) DDT: a powerful insecticide/pesticide (now banned in most countries owing to its non-biodegradability and bioaccumulation in the food chain). (iii) Carbon tetrachloride: formerly a common industrial solvent and dry-cleaning/fire-extinguisher fluid (largely discontinued, since it can form toxic phosgene gas on exposure to light/heat, and is a known health/ozone hazard). (iv) Iodoform (CHI₃): once used as an antiseptic for wound dressing, its action arising from the small amount of free iodine it liberates rather than from the compound itself (largely superseded by better antiseptics today).

6.14 Which alkyl halide from the following pair reacts faster in an S_N2 reaction: CH₃Br or CH₃I? Justify.
Ans: CH₃I reacts faster. Iodide is a much better leaving group than bromide, since the C–I bond is longer and weaker than the C–Br bond (I being a larger, more polarisable atom that stabilises the developing negative charge on the leaving group better in the S_N2 transition state), following the general leaving-group order I⁻>Br⁻>Cl⁻>F⁻.

6.15 Explain why an alkyl halide, though a polar molecule, is immiscible with water, and predict how its density compares with water as the number of halogen atoms increases.
Ans: As in 6.12(ii), alkyl halides cannot hydrogen-bond with water, so they remain immiscible despite being polar. Regarding density: alkyl monohalides (except fluorides) with up to about four carbons are typically heavier than water because the halogen atom contributes substantial mass, and density increases further with more halogen atoms per molecule and with heavier halogens (I>Br>Cl>F), e.g. CH₂Cl₂, CHCl₃ and CCl₄ are all denser than CH₃Cl and denser than water.

6.16 Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is hydrolysed more rapidly by the S_N1 mechanism, and why?
Ans: C₆H₅CHClC₆H₅ (benzhydryl chloride) hydrolyses faster. Its ionisation generates a carbocation stabilised by resonance delocalisation into two phenyl rings simultaneously (a diarylmethyl/diphenylmethyl cation), which is considerably more stable than the singly-stabilised benzylic cation formed from C₆H₅CH₂Cl; since S_N1 rate depends directly on carbocation stability, the doubly-stabilised cation forms, and reacts, faster.

6.17 p-Dichlorobenzene has a higher melting point than its ortho- and meta-isomers. Explain.
Ans: p-Dichlorobenzene is far more symmetrical than the ortho- and meta-isomers, allowing its molecules to pack much more efficiently and closely into a crystal lattice. This tighter, more regular packing maximises intermolecular (van der Waals) forces between neighbouring molecules, requiring more thermal energy to break the lattice apart, which is why the para-isomer has a distinctly higher melting point than the less symmetric ortho- and meta-isomers.

6.18 Which of the two, aqueous KOH or alcoholic KOH, is preferred for carrying out (i) substitution and (ii) elimination reactions with an alkyl halide? Explain.
Ans: (i) Aqueous KOH favours substitution — the polar protic solvent (water) solvates and stabilises the hydroxide nucleophile and the transition state well, and OH⁻ readily displaces the halide (S_N) to give the corresponding alcohol. (ii) Alcoholic KOH favours elimination — in ethanol, KOH generates the bulkier ethoxide-type basic conditions, and heating with alcoholic KOH promotes the E2 pathway (abstraction of a β-hydrogen and loss of the halide together) to give an alkene, following Zaitsev’s rule for the major product.

6.19 An alkyl halide C₅H₁₁Br reacts with alcoholic KOH to give only one alkene as the sole elimination product. Identify the alkyl halide and explain.
Ans: The alkyl halide is 2-bromo-2-methylbutane, (CH₃)₂C(Br)CH₂CH₃. Although this tertiary halide has β-hydrogens available on two different sides, symmetry/Zaitsev preference directs elimination overwhelmingly to the single major, most-substituted alkene, 2-methylbut-2-ene, which forms essentially as the sole product under standard E2 conditions with hot alcoholic KOH.

6.20 Identify the major product formed when chlorobenzene is treated with (i) Na/dry ether (Wurtz–Fittig-type conditions with an alkyl halide) and (ii) Mg in dry ether, and explain why chlorobenzene resists nucleophilic substitution under ordinary conditions.
Ans: (i) Chlorobenzene+an alkyl halide+Na in dry ether (Wurtz–Fittig reaction) gives an alkylbenzene, coupling the aryl and alkyl groups. (ii) Chlorobenzene+Mg in dry ether does not readily form a Grignard reagent under ordinary conditions (aryl Grignards require more forcing/specialised conditions than alkyl halides). Chlorobenzene resists ordinary nucleophilic substitution because the C–Cl bond has partial double-bond character (lone-pair delocalisation into the ring, as in 6.12(i)), making it shorter and stronger, and because the aromatic ring’s electron density (further donated into by Cl) repels an incoming nucleophile, both factors making S_N1/S_N2 attack at that carbon very difficult.

6.21 Arrange the following in increasing order of boiling point: bromomethane, bromoform, chloromethane, dibromomethane. Give reasons.
Ans: Chloromethane<bromomethane<dibromomethane<bromoform. Boiling point rises with increasing molecular mass and the number/size of halogen atoms, since heavier, more polarisable halogens increase the strength of van der Waals (London dispersion) forces between molecules; bromine (heavier than chlorine) and more halogen atoms per molecule both raise the boiling point, so bromoform (CHBr₃, three heavy Br atoms) boils highest and chloromethane (one light Cl atom) boils lowest.

6.22 Out of an alkyl halide and an alkyl fluoride, which is more reactive in a nucleophilic substitution reaction, and why is an alkyl fluoride reaction rarely used in the laboratory?
Ans: Heavier alkyl halides (Cl, Br, I) are far more reactive than alkyl fluorides in nucleophilic substitution. The C–F bond is exceptionally strong and short (fluorine’s small size gives excellent orbital overlap with carbon), making F⁻ a very poor leaving group; consequently, alkyl fluorides react far too slowly to be practically useful in typical laboratory S_N1/S_N2 reactions, and are rarely prepared or used this way.

Frequently Asked Questions

What is the difference between the S_N1 and S_N2 mechanisms?
An S_N2 reaction occurs in a single step: the nucleophile attacks the carbon from the side opposite the leaving group, displacing it in one concerted motion, causing an inversion of configuration (Walden inversion) at that carbon; its rate depends on the concentration of both the substrate and the nucleophile, and it is favoured by unhindered (methyl, 1°) substrates. An S_N1 reaction occurs in two steps: the leaving group departs first to form a planar carbocation intermediate, which the nucleophile then attacks from either face, typically giving a racemic mixture (with some net inversion due to ion-pairing effects); its rate depends only on substrate concentration, and it is favoured by substrates that form stable carbocations (3°, benzylic, allylic).

Why are haloarenes far less reactive than haloalkanes towards nucleophilic substitution?
In a haloarene, the halogen’s lone pair is delocalised into the aromatic ring by resonance, giving the C–X bond partial double-bond character (making it shorter and stronger than a normal C–X single bond), and the electron-rich aromatic ring itself repels approaching nucleophiles. Additionally, since the halogen-bearing carbon is sp² hybridised, backside S_N2-type attack is sterically and electronically much harder than at the sp³ carbon of a haloalkane. Together, these factors make haloarenes resistant to nucleophilic substitution under ordinary conditions.