ChemistryISC Class 12 · Chapter 6

Haloalkanes and Haloarenes ISC Class 12 Chemistry: Notes

Q: Which mechanisms are in the ISC 2027 syllabus for this chapter?

SN1 and SN2 for haloalkanes, electrophilic substitution (chlorination, nitration, sulphonation) of haloarenes, and nucleophilic substitution of Cl by -OH and -NH₂ in haloarenes.

Q: Is the directive influence of halogen needed for disubstituted rings?

No. The syllabus limits directive influence to monosubstituted compounds only.

Q: Do I need the Hunsdiecker reaction?

Yes. Preparation of haloalkanes from silver salts of fatty acids (Hunsdiecker) is listed alongside Finkelstein and Swarts.

Q: What exactly is needed on chloroform and iodoform?

Their preparation and properties: chloroform from ethanol or acetone and bleaching powder, its oxidation to phosgene, the role of ethanol, the carbylamine reaction; iodoform preparation, the iodoform test, and conversion to ethyne.

Q: Which polyhalogen compounds must I know?

Dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons and DDT: uses and environmental effects, plus the structures of freons and DDT.

Q: How many marks does this chapter carry?

It is part of the Organic Chemistry group, 31 of the 70 theory marks, shared with four other organic chapters.

Updated 15 October 2026 · Fact-checked

Haloalkanes and Haloarenes is Chapter 6 of ISC Class 12 Chemistry. It covers naming and classification, the C-X bond, preparation (Finkelstein, Swarts, Hunsdiecker, Sandmeyer, Gattermann), SN1 and SN2 mechanisms with optical rotation, elimination by Saytzeff's rule, Wurtz, Fittig and Grignard reactions, chloroform, iodoform, freons and DDT.

What Haloalkanes and Haloarenes is about

Haloalkanes (alkyl halides, R-X) have a halogen bonded to an sp³ carbon; haloarenes (aryl halides, Ar-X) have it bonded to an sp² carbon of a benzene ring. This chapter opens the Organic Chemistry group, the largest in the paper (31 of the 70 theory marks), and supplies many conversions and mechanism questions.

The C-X bond is polar (carbon δ+), so haloalkanes are attacked by nucleophiles. Primary halides react by the one-step SN2 mechanism (rate = k[RX][Nu⁻], inversion of configuration), tertiary halides by the two-step SN1 mechanism through a carbocation (rate = k[RX], racemisation). With alcoholic KOH they undergo β-elimination to alkenes, giving mainly the more substituted alkene (Saytzeff's rule). With sodium they couple (Wurtz reaction); with magnesium in dry ether they form Grignard reagents.

You must know preparation from alkanes, alkenes, alcohols (PX₃, PCl₅, SOCl₂), halide exchange (Finkelstein, Swarts) and silver salts of acids (Hunsdiecker), and the products with NaOH, water, NaI, NH₃, amines, KCN and AgCN, KNO₂ and AgNO₂, silver salts and LiAlH₄. Chloroform and iodoform need preparation and properties (phosgene, carbylamine, the iodoform test).

In haloarenes, resonance gives the C-X bond partial double-bond character, so nucleophilic substitution needs drastic conditions (e.g. Dow's process at 623 K and 300 atm). The halogen is deactivating but ortho-para directing in electrophilic substitution (chlorination, nitration, sulphonation, all with mechanism). Add the Wurtz-Fittig and Fittig reactions, reduction to benzene, Grignard formation, and the uses and environmental effects of CH₂Cl₂, CHCl₃, CCl₄, iodoform, freons and DDT.

Weightage: Haloalkanes and Haloarenes is part of the Organic Chemistry group, which carries 31 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Classification, nomenclature and the C-X bond: Classification by the carbon bearing X: primary (1°, CH₃CH₂Cl), secondary (2°, (CH₃)₂CHCl), tertiary (3°, (CH₃)₃CCl); also allylic, benzylic, vinylic and aryl halides; mono-, di- (gem- and vic-) and trihalo derivatives. General formula of monohaloalkanes $C_{n} H_{2 n + 1} X$ . Naming: common names use alkyl + halide (isopropyl chloride, tert-butyl bromide); IUPAC names use halo + alkane with lowest locants (2-chloropropane, 2-bromo-2-methylpropane, 1,2-dibromoethane; CHCl₃ trichloromethane). C-X bond: carbon is sp³; halogen is more electronegative, so C^δ+ - X^δ−. Bond length rises C-F < C-Cl < C-Br < C-I and bond enthalpy falls in the same order, so reactivity in substitution is RI > RBr > RCl > RF. Physical properties: methyl chloride, methyl bromide, ethyl chloride and methyl fluoride are gases at room temperature; higher members (including bromoethane) are liquids or solids. Boiling points rise with molar mass (RI > RBr > RCl > RF) and fall with branching (less surface area). They are only slightly soluble in water (cannot form strong enough hydrogen bonds to replace those in water) but dissolve in organic solvents; bromides and iodides are denser than water.; Worked example
Give IUPAC names and classify: (a) (CH₃)₃CBr, (b) CH₃CHClCH₂CH₃, (c) CH₂=CHCH₂Cl.
Try it first, then show the solutionSolution
(a) 2-bromo-2-methylpropane: tertiary.
(b) 2-chlorobutane: secondary.
(c) 3-chloroprop-1-ene: primary and allylic.
Answer: As above
Preparation of haloalkanes: From alkanes (free-radical, mixture of products): $C H_{4} + C l_{2} h ν C H_{3} C l + H C l$ . From alkenes and HX: Markovnikov addition, H goes to the carbon already richer in H: propene + HBr → 2-bromopropane (with peroxide, HBr adds anti-Markovnikov to give 1-bromopropane). From alcohols: $C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$ ; $3 C_{2} H_{5} O H + P C l_{3} \to 3 C_{2} H_{5} C l + H_{3} P O_{3}$ (PBr₃ and PI₃ made in situ from red P and Br₂ or I₂); thionyl chloride in pyridine is best because the by-products are gases: $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$ . Halide exchange: Finkelstein reaction (iodides) in dry acetone: $C_{2} H_{5} B r + N a I \to C_{2} H_{5} I + N a B r$ , driven forward because NaBr/NaCl is insoluble in acetone. Swarts reaction (fluorides) with AgF, Hg₂F₂, CoF₂ or SbF₃: $C H_{3} B r + A g F \to C H_{3} F + A g B r$ . Hunsdiecker reaction: the silver salt of a fatty acid refluxed with Br₂ in CCl₄ gives a bromoalkane with one carbon fewer: $C H_{3} C O O A g + B r_{2} \to C H_{3} B r + C O_{2} + A g B r$ .; Worked example
Why is thionyl chloride preferred to PCl₅ for converting ethanol to chloroethane?
Try it first, then show the solutionSolution
With SOCl₂ the by-products SO₂ and HCl are gases and escape.
With PCl₅, liquid POCl₃ forms and must be separated by distillation.
Answer: Gaseous by-products leave pure chloroethane.
Nucleophilic substitution: SN2 and SN1: SN2 (bimolecular): the nucleophile attacks carbon from the side opposite to the leaving group; in one step a five-coordinate transition state forms (Nu…C…X) and X⁻ leaves. Rate = k[RX][Nu⁻], second order. The product has inverted configuration (Walden inversion), like an umbrella turning inside out: an optically active (R) halide gives the (S) product. Favoured by primary halides (least crowded), strong nucleophiles and polar aprotic solvents. Reactivity: CH₃X > 1° > 2° > 3°. SN1 (unimolecular): step 1 (slow) the C-X bond ionises to a planar carbocation; step 2 (fast) the nucleophile attacks either face. Rate = k[RX], first order. An optically active halide gives a racemic mixture (partial racemisation in practice). Favoured by tertiary halides (most stable carbocation, +I effect and hyperconjugation), benzylic and allylic halides, and polar protic solvents. Reactivity: 3° > 2° > 1° > CH₃X. Optical rotation: chiral molecules (four different groups on carbon) rotate plane-polarised light; d (+) clockwise, l (−) anticlockwise; a racemic mixture is optically inactive.; Worked example
Two isomers of C₅H₁₁Cl, one with an unbranched chain and one with a tertiary carbon bearing Cl, are hydrolysed by aqueous KOH. Which follows rate = k[RCl] and which rate = k[RCl][OH⁻]? Name both.
Try it first, then show the solutionSolution
Rate = k[RCl] is SN1, which needs a stable carbocation: the tertiary halide, 2-chloro-2-methylbutane, (CH₃)₂CClCH₂CH₃.
Rate = k[RCl][OH⁻] is SN2, favoured for an unhindered primary halide: 1-chloropentane, CH₃(CH₂)₄Cl.
Answer: SN1: 2-chloro-2-methylbutane; SN2: 1-chloropentane
Reactions of haloalkanes with nucleophiles: Using bromoethane: aqueous NaOH/KOH → ethanol: $C_{2} H_{5} B r + K O H \to C_{2} H_{5} O H + K B r$ . Water (slow): $C_{2} H_{5} B r + H_{2} O \to C_{2} H_{5} O H + H B r$ . NaI in acetone → iodoethane. Ammonia (alcoholic, heated): $C_{2} H_{5} B r + N H_{3} \to C_{2} H_{5} N H_{2} + H B r$ ; excess halide gives secondary, tertiary amines and a quaternary salt. Primary amine: $C_{2} H_{5} B r + C_{2} H_{5} N H_{2} \to (C_{2} H_{5})_{2} N H + H B r$ . Secondary amine: $C_{2} H_{5} B r + (C_{2} H_{5})_{2} N H \to (C_{2} H_{5})_{3} N + H B r$ . KCN (ionic, attacks through C) → nitrile: $C_{2} H_{5} B r + K C N \to C_{2} H_{5} C N + K B r$ (adds one carbon). AgCN (covalent, attacks through N) → isocyanide: $C_{2} H_{5} B r + A g C N \to C_{2} H_{5} N C + A g B r$ . KNO₂ (ionic, through O) → alkyl nitrite: $C_{2} H_{5} B r + K N O_{2} \to C_{2} H_{5} O N O + K B r$ . AgNO₂ (through N) → nitroalkane: $C_{2} H_{5} B r + A g N O_{2} \to C_{2} H_{5} N O_{2} + A g B r$ . Silver salt of a fatty acid → ester: $C_{2} H_{5} B r + C H_{3} C O O A g \to C H_{3} C O O C_{2} H_{5} + A g B r$ . LiAlH₄ (hydride) → alkane: $4 C_{2} H_{5} B r + L i A l H_{4} \to 4 C_{2} H_{6} + L i B r + A l B r_{3}$ . CN⁻ and NO₂⁻ are ambident nucleophiles: the ionic potassium salts give the C- or O-bonded product, the covalent silver salts the N-bonded one.; Worked example
Bromoethane gives different products with KCN and AgCN. Name them and explain.
Try it first, then show the solutionSolution
KCN is ionic: free CN⁻ attacks through the more nucleophilic carbon, giving ethyl cyanide (propanenitrile), C₂H₅CN.
AgCN is covalent (Ag-C bond); only the nitrogen lone pair is free, so attack is through N, giving ethyl isocyanide, C₂H₅NC.
Answer: Propanenitrile with KCN; ethyl isocyanide with AgCN
Elimination, Wurtz reaction and Grignard reagent: β-Elimination (dehydrohalogenation): heating with alcoholic KOH removes H from the β-carbon and X from the α-carbon, giving an alkene. Saytzeff's rule: the more substituted alkene (fewer H on the double-bonded carbons) is the major product. 2-Bromobutane gives but-2-ene (major, about 80%) and but-1-ene (minor). Aqueous KOH favours substitution (OH⁻ acts as nucleophile); alcoholic KOH favours elimination (ethoxide acts as a strong base). Tertiary halides eliminate most readily. Wurtz reaction: with sodium in dry ether, two alkyl groups join: $2 C H_{3} B r + 2 N a \to C_{2} H_{6} + 2 N a B r$ ; used to make symmetrical alkanes with an even number of carbons. Grignard reagent: $C_{2} H_{5} B r + M g \to C_{2} H_{5} M g B r$ in dry ether; R is strongly nucleophilic and basic, so moisture destroys it ( $C_{2} H_{5} M g B r + H_{2} O \to C_{2} H_{6} + M g (O H) B r$ ). Grignard reagents are used to make alcohols, acids and hydrocarbons.; Worked example
What is the major product when 2-bromo-2-methylbutane is heated with alcoholic KOH? Why?
Try it first, then show the solutionSolution
β-hydrogens are available on C1 and C3, giving 2-methylbut-1-ene or 2-methylbut-2-ene.
Saytzeff: the more substituted alkene, (CH₃)₂C=CHCH₃ (trisubstituted), is more stable and forms mainly.
Answer: 2-methylbut-2-ene (major)
Chloroform and iodoform: Chloroform (CHCl₃) preparation: ethanol or acetone with bleaching powder (supplies Cl₂ and Ca(OH)₂). From ethanol: oxidation ( $C_{2} H_{5} O H + C l_{2} \to C H_{3} C H O + 2 H C l$ ), chlorination ( $C H_{3} C H O + 3 C l_{2} \to C C l_{3} C H O + 3 H C l$ ), hydrolysis of chloral ( $2 C C l_{3} C H O + C a (O H)_{2} \to 2 C H C l_{3} + (H C O O)_{2} C a$ ). Properties: colourless, sweet-smelling, heavy liquid; formerly an anaesthetic. In air and light it is oxidised to poisonous phosgene: $2 C H C l_{3} + O_{2} \to 2 C O C l_{2} + 2 H C l$ . It is stored in dark, closed bottles filled to the brim, with 1% ethanol added, which converts any phosgene to harmless diethyl carbonate: $C O C l_{2} + 2 C_{2} H_{5} O H \to (C_{2} H_{5})_{2} C O_{3} + 2 H C l$ . Carbylamine reaction (test for primary amines; offensive smell): $C H C l_{3} + C_{6} H_{5} N H_{2} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$ . With nitric acid → chloropicrin (tear gas): $C H C l_{3} + H N O_{3} \to C C l_{3} N O_{2} + H_{2} O$ . With silver powder on heating → ethyne: $2 C H C l_{3} + 6 A g \to C_{2} H_{2} + 6 A g C l$ . Iodoform (CHI₃) is a yellow solid with a characteristic smell, formed from ethanol (or any CH₃CO- or CH₃CH(OH)- compound) with I₂ and NaOH: $C_{2} H_{5} O H + 4 I_{2} + 6 N a O H \to C H I_{3} + H C O O N a + 5 N a I + 5 H_{2} O$ ; this is the iodoform test. Heated with silver powder it also gives ethyne: $2 C H I_{3} + 6 A g \to C_{2} H_{2} + 6 A g I$ . It was used as an antiseptic (because it releases free iodine).; Worked example
How is iodoform obtained from ethanol? Write the equation and say what is observed.
Try it first, then show the solutionSolution
Warm ethanol with iodine and sodium hydroxide solution.
$C_{2} H_{5} O H + 4 I_{2} + 6 N a O H \to C H I_{3} + H C O O N a + 5 N a I + 5 H_{2} O$
A yellow precipitate of iodoform with a characteristic smell appears (the iodoform test).
Answer: I₂ and NaOH, warm; yellow CHI₃ precipitates
Haloarenes: preparation and low reactivity to nucleophiles: Preparation: (1) electrophilic substitution of benzene with Cl₂ or Br₂ and a Lewis acid (anhydrous FeCl₃, FeBr₃, AlCl₃) in the dark: $C_{6} H_{6} + C l_{2} F e C l_{3} C_{6} H_{5} C l + H C l$ . (2) Sandmeyer reaction: benzenediazonium chloride with CuCl/HCl (or CuBr/HBr): $C_{6} H_{5} N_{2} C l C u C l / H C l C_{6} H_{5} C l + N_{2}$ . (3) Gattermann reaction: with copper powder and HCl: $C_{6} H_{5} N_{2} C l C u / H C l C_{6} H_{5} C l + N_{2}$ (lower yield than Sandmeyer). Why haloarenes resist nucleophilic substitution: (1) resonance delocalises a lone pair of X into the ring, giving the C-X bond partial double-bond character; (2) the carbon is sp² (more s character, more electronegative), so the bond is shorter and stronger; (3) the phenyl cation is unstable, so SN1 is impossible; (4) the electron-rich ring repels incoming nucleophiles, and backside attack is blocked. Replacement of Cl by OH (Dow's process): $C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$ at 623 K and 300 atm, then acidify. By NH₂: $2 C_{6} H_{5} C l + 2 N H_{3} + C u_{2} O \to 2 C_{6} H_{5} N H_{2} + 2 C u C l + H_{2} O$ at 475 K, 60 atm. Electron-withdrawing –NO₂ groups at ortho/para positions make substitution easy (addition-elimination through a resonance-stabilised carbanion, the Meisenheimer complex): 2,4,6-trinitrochlorobenzene hydrolyses in warm water.; Worked example
Why does p-nitrochlorobenzene react with NaOH at 443 K whereas chlorobenzene needs 623 K and 300 atm?
Try it first, then show the solutionSolution
OH⁻ adds to the carbon bearing Cl, forming a negatively charged intermediate.
The para -NO₂ group delocalises this negative charge onto its oxygen atoms by resonance, stabilising the intermediate and lowering $E_{a}$ .
Chlorobenzene has no such group, so the intermediate is unstable and drastic conditions are needed.
Answer: The -NO₂ group stabilises the intermediate carbanion.
Electrophilic substitution and other reactions of haloarenes: Directive influence: halogen withdraws electrons by −I (deactivating: slower than benzene) but donates by +R into ortho and para positions, so substitution occurs ortho and para; para is major (less steric hindrance). Mechanism (general): (1) generation of electrophile; (2) attack on the ring forming a resonance-stabilised arenium ion (σ-complex); (3) loss of H⁺ restores aromaticity. Chlorination: $C l_{2} + F e C l_{3} \to C l^{+} + F e C l_{4}^{-}$ ; $C_{6} H_{5} C l + C l_{2} \to C_{6} H_{4} C l_{2} + H C l$ (1,2- and 1,4-dichlorobenzene). Nitration: conc. HNO₃ + conc. H₂SO₄ give $N O_{2}^{+}$ ; $C_{6} H_{5} C l + H N O_{3} \to C_{6} H_{4} C l N O_{2} + H_{2} O$ (1-chloro-2- and 1-chloro-4-nitrobenzene). Sulphonation: conc. H₂SO₄ (SO₃ electrophile): $C_{6} H_{5} C l + H_{2} S O_{4} \to C_{6} H_{4} C l S O_{3} H + H_{2} O$ . Reduction to benzene: Ni-Al alloy with NaOH: $C_{6} H_{5} C l + 2 [H] \to C_{6} H_{6} + H C l$ . Wurtz-Fittig: aryl + alkyl halide with Na in dry ether: $C_{6} H_{5} C l + 2 N a + C H_{3} C l \to C_{6} H_{5} C H_{3} + 2 N a C l$ . Fittig: $2 C_{6} H_{5} C l + 2 N a \to C_{6} H_{5} C_{6} H_{5} + 2 N a C l$ (biphenyl). Grignard: $C_{6} H_{5} B r + M g \to C_{6} H_{5} M g B r$ in dry ether or THF.; Worked example
Convert benzene into ethylbenzene using a coupling reaction.
Try it first, then show the solutionSolution
$C_{6} H_{6} + B r_{2} anh y d . F e B r_{3} C_{6} H_{5} B r + H B r$
$C_{6} H_{5} B r + 2 N a + C_{2} H_{5} B r d r y e t h er C_{6} H_{5} C_{2} H_{5} + 2 N a B r$ (Wurtz-Fittig reaction)
Answer: Brominate (FeBr₃), then Wurtz-Fittig with bromoethane and Na in dry ether
Polyhalogen compounds: uses and environmental effects: Dichloromethane (CH₂Cl₂): solvent, paint remover, propellant, metal cleaning; exposure affects the central nervous system (dizziness, numbness), and direct contact burns skin and eyes. Trichloromethane (CHCl₃): solvent for fats and alkaloids, raw material for freon-22; once an anaesthetic, now replaced as it damages liver and kidneys; forms phosgene in air and light. Tetrachloromethane (CCl₄): solvent, raw material for freons; formerly in fire extinguishers; causes liver damage and is an ozone-depleting substance. Iodoform: antiseptic (releases iodine) but replaced because of its smell. Freons (chlorofluorocarbons, e.g. Freon-12, $C C l_{2} F_{2}$ , made by $C C l_{4} + 2 H F \to C C l_{2} F_{2} + 2 H C l$ with SbCl₅): inert, non-toxic, used in refrigerators, air-conditioners and aerosols. In the stratosphere UV light releases Cl atoms that destroy ozone in a chain reaction, thinning the ozone layer. DDT (p,p'-dichlorodiphenyltrichloroethane): made from chlorobenzene and chloral with conc. H₂SO₄: $2 C_{6} H_{5} C l + C C l_{3} C H O \to (C l C_{6} H_{4})_{2} C H C C l_{3} + H_{2} O$ . A powerful insecticide (malaria-carrying mosquitoes), but non-biodegradable and fat-soluble, so it accumulates up food chains (biomagnification), harming birds (thin eggshells); banned in many countries.; Worked example
Draw the structure of DDT and give one reason it was banned.
Try it first, then show the solutionSolution
A central carbon bearing H, a CCl₃ group and two benzene rings, each with Cl at the para position: $(p - C l C_{6} H_{4})_{2} C H - C C l_{3}$ .
It is not biodegradable and is stored in fatty tissue, so its concentration increases up the food chain (biomagnification), harming fish-eating birds.
Answer: $(C l C_{6} H_{4})_{2} C H C C l_{3}$ ; persistent and biomagnified

Chemistry in real life: Haloalkanes and Haloarenes

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Storage of chloroform

Suppose a school laboratory finds an old clear-glass bottle of chloroform, half full, left on a sunny window sill.

Why is this dangerous, and how should chloroform be stored?

Try it first, then show the solutionSolution

In light and air chloroform is oxidised: $2 C H C l_{3} + O_{2} \to 2 C O C l_{2} + 2 H C l$ ; phosgene is extremely poisonous.
Store in dark brown bottles filled to the brim (no air), with about 1% ethanol to convert any phosgene into harmless diethyl carbonate.

Answer: Phosgene may have formed; use dark, full bottles with 1% ethanol.

Memory hook: Light + air + chloroform = phosgene; ethanol is the antidote in the bottle.

Ozone depletion by freons

Imagine an old refrigerator being scrapped with its CCl₂F₂ coolant simply vented into the air.

Explain how this coolant can damage the ozone layer.

Try it first, then show the solutionSolution

Freons are unreactive in the lower atmosphere, so they drift up to the stratosphere.
UV light breaks a C-Cl bond releasing Cl•, which reacts: Cl• + O₃ → ClO• + O₂, and ClO• + O → Cl• + O₂, regenerating Cl• so one atom destroys thousands of O₃ molecules.

Answer: Chlorine radicals from freons catalyse ozone destruction.

Memory hook: One chlorine, thousands of ozones.

Biomagnification of DDT

Suppose a lake is sprayed with DDT so that water contains 0.00005 ppm, and plankton, small fish, large fish and fish-eating birds each concentrate it about ten times more than what they eat.

Estimate the concentration in the birds and explain why it rises.

Try it first, then show the solutionSolution

Four steps of tenfold concentration: 0.00005 × 10⁴ = 0.5 ppm (real food chains can concentrate it even more).
DDT is fat-soluble and not broken down or excreted, so it accumulates at each trophic level.

Answer: About 0.5 ppm, 10 000 times the water level.

Memory hook: Fat-soluble and stubborn: DDT climbs the food chain.

Topics in this chapter

Haloalkanes and Haloarenes formula sheet

Alcohol + PCl₅: $C_{2} H_{5} O H + P C l_{5} \to C_{2} H_{5} C l + P O C l_{3} + H C l$; With PCl₃: $3 C_{2} H_{5} O H + P C l_{3} \to 3 C_{2} H_{5} C l + H_{3} P O_{3}$ .
Alcohol + SOCl₂: $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$; Pyridine; gaseous by-products.
Finkelstein reaction: $C_{2} H_{5} B r + N a I \to C_{2} H_{5} I + N a B r$; Dry acetone; NaBr precipitates.
Swarts reaction: $C H_{3} B r + A g F \to C H_{3} F + A g B r$; Also Hg₂F₂, CoF₂, SbF₃.
Hunsdiecker reaction: $C H_{3} C O O A g + B r_{2} \to C H_{3} B r + C O_{2} + A g B r$; Reflux in CCl₄; one carbon fewer.
Nitrile vs isocyanide: $C_{2} H_{5} B r + K C N \to C_{2} H_{5} C N + K B r$; AgCN gives $C_{2} H_{5} N C$ .
Nitrite vs nitroalkane: $C_{2} H_{5} B r + A g N O_{2} \to C_{2} H_{5} N O_{2} + A g B r$; KNO₂ gives ethyl nitrite $C_{2} H_{5} O N O$ .
Ester from silver salt: $C_{2} H_{5} B r + C H_{3} C O O A g \to C H_{3} C O O C_{2} H_{5} + A g B r$; Ethyl ethanoate.
Reduction with LiAlH₄: $4 C_{2} H_{5} B r + L i A l H_{4} \to 4 C_{2} H_{6} + L i B r + A l B r_{3}$; Alkane formed.
Wurtz reaction: $2 C H_{3} B r + 2 N a \to C_{2} H_{6} + 2 N a B r$; Dry ether.
Grignard reagent: $C_{2} H_{5} B r + M g \to C_{2} H_{5} M g B r$; Dry ether; destroyed by water.
Chloroform to phosgene: $2 C H C l_{3} + O_{2} \to 2 C O C l_{2} + 2 H C l$; Light and air; prevented by 1% ethanol.
Phosgene + ethanol: $C O C l_{2} + 2 C_{2} H_{5} O H \to (C_{2} H_{5})_{2} C O_{3} + 2 H C l$; Harmless diethyl carbonate.
Carbylamine reaction: $C H C l_{3} + C_{6} H_{5} N H_{2} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$; Foul smell: test for primary amines.
Iodoform reaction: $C_{2} H_{5} O H + 4 I_{2} + 6 N a O H \to C H I_{3} + H C O O N a + 5 N a I + 5 H_{2} O$; Yellow precipitate.
Iodoform to ethyne: $2 C H I_{3} + 6 A g \to C_{2} H_{2} + 6 A g I$; Heat with silver powder.
Sandmeyer reaction: $C_{6} H_{5} N_{2} C l C u C l / H C l C_{6} H_{5} C l + N_{2}$; Gattermann uses Cu/HCl.
Dow's process: $C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$; 623 K, 300 atm; then acidify.
Wurtz-Fittig reaction: $C_{6} H_{5} C l + 2 N a + C H_{3} C l \to C_{6} H_{5} C H_{3} + 2 N a C l$; Fittig: $2 C_{6} H_{5} C l + 2 N a \to C_{6} H_{5} C_{6} H_{5} + 2 N a C l$ .
DDT synthesis: $2 C_{6} H_{5} C l + C C l_{3} C H O \to (C l C_{6} H_{4})_{2} C H C C l_{3} + H_{2} O$; Conc. H₂SO₄.

Haloalkanes and Haloarenes: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

SN2 mechanism with inversion

'Explain the SN2 mechanism' or 'why the product has inverted configuration'.

Draw

Left: HO⁻ approaching a tetrahedral carbon from the side opposite Br, with three groups (H, H, CH₃) shown.
Middle: transition state in square brackets with dotted HO…C…Br bonds and the three groups flattened in a plane; mark it with ‡ and δ− on HO and Br.
Right: HO-C with the three groups now pointing the other way, plus Br⁻.

Label

nucleophile
backside attack
transition state ‡
leaving group
inversion

Watch out: Write 'rate = k[RX][OH⁻]' under the diagram.

SN1 mechanism with racemisation

Explaining the reaction of tert-butyl bromide or racemisation of an optically active halide.

Draw

Step 1 (slow): R₃C-Br ionising to a planar carbocation R₃C⁺ + Br⁻.
Step 2 (fast): OH⁻ attacking the carbocation from the top face and from the bottom face.
Two products shown as mirror images (50:50).

Label

slow step
planar carbocation (sp²)
attack from either face
racemic mixture

Watch out: Write 'rate = k[RX]' under step 1.

Resonance in chlorobenzene and electrophilic substitution

Explaining low reactivity of haloarenes or ortho-para direction.

Draw

Five resonance structures: the neutral structure, then three with C=Cl⁺ and a negative charge at ortho, para and the other ortho position, and the last Kekulé form.
Then the mechanism: E⁺ attacking the para position, the arenium ion, and loss of H⁺.

Label

C=Cl⁺ partial double bond
δ− at ortho and para
arenium ion
−H⁺

Watch out: One line of text: −I deactivates, +R directs ortho-para.

How the board asks Haloalkanes and Haloarenes

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Conversions	Two- or three-step changes: benzene → biphenyl, iodoform → ethyne, ethanol → iodoethane, chlorobenzene → phenol.	Write each step with reagent, catalyst and conditions (dry ether, FeCl₃, heat).
Named reactions	Finkelstein, Swarts, Hunsdiecker, Wurtz, Wurtz-Fittig, Fittig, Sandmeyer, Gattermann, carbylamine.	One balanced equation with conditions for each.
Mechanisms	SN1 vs SN2 with stereochemistry; electrophilic substitution in chlorobenzene; nucleophilic substitution in haloarenes.	Show curved arrows, intermediate or transition state, rate law and stereochemical result.
Identify A and B	Isomers of C₄H₉Br from rate laws; products of KCN vs AgCN; elimination products.	Link rate law to mechanism, mechanism to type of halide.
Give reasons	Haloarenes less reactive; ethanol added to chloroform; halogen o/p-directing yet deactivating; Grignard needs dry ether.	Use resonance, hybridisation, carbocation stability, phosgene.
Uses and environmental effects	CH₂Cl₂, CHCl₃, CCl₄, iodoform, freons, DDT.	Give one use and one harmful effect each.

Tricks to remember Haloalkanes and Haloarenes

Short memory aids. Say each one aloud once, then use it on a question the same day.

Silver gives N
AgCN → isocyanide (N-bonded); AgNO₂ → nitroalkane (N-bonded). K salts give C- or O-bonded products.
Use it when: Ambident nucleophile questions.
SN1 = 1 molecule in the slow step = 3°
SN1: tertiary, carbocation, racemic. SN2: primary, one step, inversion.
Use it when: Mechanism identification.
Aqueous substitutes, alcoholic eliminates
aq KOH → alcohol; alc KOH → alkene.
Use it when: Predicting products.
Saytzeff: the rich get richer
H is removed from the β-carbon with fewer H; the more substituted alkene wins.
Use it when: Elimination products.
Fittig doubles, Wurtz-Fittig mixes
Fittig: two aryl groups (biphenyl). Wurtz: two alkyl. Wurtz-Fittig: aryl + alkyl (toluene).
Use it when: Coupling reactions.

Haloalkanes and Haloarenes: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Reactions of chloroform · 1 mark · Very short answerBoard level

Fill in the blanks: Chloroform warmed with aniline and alcoholic KOH gives ______, which has an offensive smell; chloroform heated with concentrated nitric acid gives the tear gas ______.

Try it first, then show the solutionSolution

$C H C l_{3} + C_{6} H_{5} N H_{2} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$ (carbylamine reaction); $C H C l_{3} + H N O_{3} \to C C l_{3} N O_{2} + H_{2} O$ .

Answer: phenyl isocyanide; chloropicrin

Example 2: Halogen as a deactivating group · 1 mark · Assertion-reasonBoard level

Assertion: Chlorobenzene undergoes nitration more slowly than benzene.
Reason: The electron-withdrawing −I effect of chlorine outweighs its +R effect, so the ring is less electron-rich than in benzene.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Try it first, then show the solutionSolution

Both statements are true: −I lowers the electron density of the whole ring (deactivation), while +R only decides where the electrophile attacks (ortho and para). The reason explains the assertion.

Answer: (a)

Example 3: Two conversions · 2 marks · Short answerBoard level

How will you bring about: (a) chloroform to ethyne; (b) bromoethane to ethyl isocyanide? [2]

Try it first, then show the solutionSolution

(a) $2 C H C l_{3} + 6 A g Δ C_{2} H_{2} + 6 A g C l$
(b) $C_{2} H_{5} B r + A g C N \to C_{2} H_{5} N C + A g B r$

Answer: Heat with Ag powder; treat with AgCN

Example 4: Reactivity order in SN2 · 2 marks · Short answerHarder board level

Arrange in increasing order of SN2 reactivity and explain: 2-bromo-2-methylpropane, 1-bromobutane, 2-bromobutane. [2]

Try it first, then show the solutionSolution

SN2 needs backside attack; bulky groups around the carbon hinder the nucleophile.
Order: 2-bromo-2-methylpropane (3°) < 2-bromobutane (2°) < 1-bromobutane (1°).

Answer: 3° < 2° < 1°

Example 5: Why haloarenes resist nucleophiles · 3 marks · Short answerHarder board level

Give three reasons why chlorobenzene is much less reactive than chloroethane towards nucleophilic substitution. [3]

Try it first, then show the solutionSolution

Resonance gives the C-Cl bond partial double-bond character, making it harder to break.
The carbon is sp² hybridised: the bond is shorter and stronger than the sp³ C-Cl bond.
The phenyl cation is unstable (no SN1), and the π-electron cloud repels the approaching nucleophile (no backside SN2 attack).

Answer: Resonance, sp² carbon, unstable phenyl cation and repulsion

Example 6: Electrophilic substitution in chlorobenzene · 5 marks · Long answerHarder board level

(a) Explain with resonance structures why the chlorine atom in chlorobenzene is ortho-para directing although it deactivates the ring. [3]
(b) Write the mechanism for the nitration of chlorobenzene and name the major product. [2]

Try it first, then show the solutionSolution

(a) −I effect of Cl withdraws electron density from the whole ring: deactivating.
+R effect: lone pairs of Cl delocalise into the ring, building negative charge at ortho and para positions (resonance structures with C=Cl⁺).
So electrophiles attack ortho and para, where density is relatively higher; para is major (less hindered).
(b) $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ ; NO₂⁺ attacks the para carbon forming an arenium ion; HSO₄⁻ removes H⁺.
Major product 1-chloro-4-nitrobenzene; minor 1-chloro-2-nitrobenzene.

Answer: o/p-directing by +R, deactivating by −I; 1-chloro-4-nitrobenzene

Example 7: Optical activity in substitution · 3 marks · Short answerHOTS / competency

An optically active sample of (R)-2-bromobutane is hydrolysed under two conditions: (a) with concentrated NaOH in acetone, (b) in aqueous ethanol without added base. Predict the stereochemistry of the 2-butanol formed in each case and explain. [3]

Try it first, then show the solutionSolution

(a) A strong nucleophile at high concentration in an aprotic solvent favours SN2: backside attack gives (S)-butan-2-ol (inversion).
(b) Weak nucleophile and polar protic solvent favour SN1: a planar carbocation is attacked from both faces.
So (b) gives a (largely) racemic mixture, optically inactive.

Answer: (a) inverted (S) product (b) racemic mixture

Example 8: Identify the halide · 3 marks · Short answerHOTS / competency

A compound C₃H₇Br (A) gives propene with alcoholic KOH. Propene with HBr gives B, an isomer of A. A gives propan-1-ol with aqueous KOH. Identify A and B and write the equations. [3]

Try it first, then show the solutionSolution

A gives propan-1-ol, so A = 1-bromopropane, CH₃CH₂CH₂Br.
Elimination: $C H_{3} C H_{2} C H_{2} B r + K O H \to C_{3} H_{6} + K B r + H_{2} O$ (alcoholic).
Markovnikov addition of HBr to propene gives B = 2-bromopropane, CH₃CHBrCH₃: $C_{3} H_{6} + H B r \to C_{3} H_{7} B r$ .

Answer: A: 1-bromopropane; B: 2-bromopropane

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Making a Grignard reagent

Suppose a student tries to prepare ethylmagnesium bromide by adding bromoethane to magnesium turnings in ether, but the flask had been rinsed with water and not dried. No useful Grignard reagent forms and a gas is evolved.

(i) · 1 mark
Write the equation for forming the Grignard reagent.
Answer
Answer: $C_{2} H_{5} B r + M g \to C_{2} H_{5} M g B r$
In dry ether.
(ii) · 1 mark
Which gas is evolved, and why?
Answer
Answer: Ethane.
$C_{2} H_{5} M g B r + H_{2} O \to C_{2} H_{6} + M g (O H) B r$ : the strongly basic ethyl group takes a proton from water.
(iii) · 1 mark
Why is ether used as the solvent?
Answer
Answer: It is dry and aprotic and its oxygen lone pairs coordinate to Mg, stabilising the reagent.
Ether has no acidic H to destroy the reagent.
(iv) · 2 marks
Name the product when bromoethane is instead heated with sodium in dry ether, and the reaction.
Answer
Answer: Butane; Wurtz reaction.
$2 C_{2} H_{5} B r + 2 N a \to C_{4} H_{10} + 2 N a B r$

Two butyl bromides

Imagine a chemist measuring the rate of hydrolysis of 1-bromobutane and 2-bromo-2-methylpropane in aqueous ethanol with dilute NaOH. Doubling [OH⁻] doubles the rate for one compound but has no effect on the other.

(i) · 1 mark
For which compound does doubling [OH⁻] have no effect? Name the mechanism.
Answer
Answer: 2-bromo-2-methylpropane; SN1.
Rate = k[RX]: the slow step is ionisation, which does not involve OH⁻.
(ii) · 1 mark
Write the rate law for the other compound.
Answer
Answer: Rate = k[CH₃(CH₂)₃Br][OH⁻]
SN2, one step involving both.
(iii) · 1 mark
Why does the tertiary halide not react by SN2?
Answer
Answer: Three methyl groups block backside attack (steric hindrance).
It also forms a stable tertiary carbocation, so SN1 is favoured.
(iv) · 2 marks
Which would give an alkene more readily with alcoholic KOH? Name the product.
Answer
Answer: 2-bromo-2-methylpropane; 2-methylpropene.
Tertiary halides eliminate most readily; $(C H_{3})_{3} C B r + K O H \to (C H_{3})_{2} C = C H_{2} + K B r + H_{2} O$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Convert chlorobenzene into phenol. (2 marks)

Step	Marks
$C_{6} H_{5} C l + 2 N a O H \to C_{6} H_{5} O N a + N a C l + H_{2} O$ at 623 K and 300 atm (Dow's process)	1
$C_{6} H_{5} O N a + H C l \to C_{6} H_{5} O H + N a C l$ (acidify)	1

Explain the SN1 mechanism for hydrolysis of tert-butyl bromide, with the rate law. (3 marks)

Step	Marks
Slow ionisation to (CH₃)₃C⁺ and Br⁻ (rate-determining)	1
Fast attack of OH⁻ on the planar carbocation giving (CH₃)₃COH	1
Rate = k[(CH₃)₃CBr], first order	1

Describe the preparation of chloroform from ethanol and bleaching powder with equations, and give its reaction with aniline and KOH. (5 marks)

Step	Marks
$C a O C l_{2} + H_{2} O \to C a (O H)_{2} + C l_{2}$	1
$C_{2} H_{5} O H + C l_{2} \to C H_{3} C H O + 2 H C l$	1
$C H_{3} C H O + 3 C l_{2} \to C C l_{3} C H O + 3 H C l$	1
$2 C C l_{3} C H O + C a (O H)_{2} \to 2 C H C l_{3} + (H C O O)_{2} C a$	1
$C H C l_{3} + C_{6} H_{5} N H_{2} + 3 K O H \to C_{6} H_{5} N C + 3 K C l + 3 H_{2} O$ (offensive smell)	1

Where students lose marks in Haloalkanes and Haloarenes

The slips examiners see most often, and what to write instead.

Mistake: Wrong products in the chloroform-ethanol fill-in
CISCE examiners' comments, 2024: candidates wrote diethyl ether instead of phosgene and ethene instead of ethyl carbonate for the role of ethanol in stored chloroform.
Wrong working
Ethanol prevents formation of diethyl ether and gives ethene.
Right working
Ethanol converts poisonous phosgene (COCl₂) into harmless diethyl carbonate.
Do this: Learn the two equations together.
Usually costs: 1 mark.
Mistake: Skipping steps or conditions in conversions
CISCE examiners' comments, 2024: for benzene to biphenyl, candidates started from chlorobenzene and omitted the anhydrous AlCl₃/FeBr₃ step; many left iodoform to acetylene unattempted, used Na instead of Ag, or omitted heat.
Wrong working
$C_{6} H_{5} C l \to C_{6} H_{5} C_{6} H_{5}$ (benzene step missing); $C H I_{3} + N a \to$
Right working
Halogenate benzene with FeCl₃/FeBr₃ first, then Fittig; heat iodoform with silver powder.
Do this: Start from the compound given; write reagent and condition above every arrow.
Usually costs: 1 mark per step.
Mistake: Calling FeCl₃ a source of nucleophiles
CISCE examiners' comments, 2025: in an assertion-reason item on halogenation of benzene with anhydrous FeCl₃, candidates chose (a) or (d), not seeing that the reason (FeCl₃ prepares a nucleophile) was false.
Wrong working
FeCl₃ makes Cl⁻ which attacks benzene.
Right working
FeCl₃ (a Lewis acid) generates the electrophile Cl⁺; benzene undergoes electrophilic substitution.
Do this: Benzene is electron-rich: it is attacked by electrophiles.
Usually costs: 1 mark.
Mistake: Not linking primary/tertiary halides to SN2/SN1
CISCE examiners' comments, 2025: candidates were unclear which type of reaction (SN1 or SN2) tertiary and primary alkyl halides undergo when identifying isomers of C₄H₉Br from rate data.
Wrong working
A (rate depends on [A] only) is 1-bromobutane.
Right working
Rate = k[RX] → SN1 → tertiary: 2-bromo-2-methylpropane; rate = k[RX][OH⁻] → SN2 → primary: 1-bromobutane.
Do this: Make a two-column table: SN1 / SN2 with rate law, substrate, stereochemistry.
Usually costs: 1-2 marks.

Practice questions on Haloalkanes and Haloarenes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which reagent converts bromoethane into nitroethane?
(a) KNO₂
(b) AgNO₂
(c) HNO₃
(d) NaNO₃
Hint
Silver gives N-bonding.
Answer
(b) AgNO₂
Q2 · 1 mark · MCQBoard level
The Swarts reaction is used to prepare:
(a) alkyl iodides
(b) alkyl fluorides
(c) aryl chlorides
(d) alkanes
Hint
AgF, SbF₃.
Answer
(b)
Q3 · 1 mark · Very short answerBoard level
Fill in the blanks: The reaction of an alkyl halide with an aryl halide and sodium in dry ether is the ______ reaction; two aryl halides with sodium undergo the ______ reaction.
Hint
Coupling reactions.
Answer
Wurtz-Fittig; Fittig
Q4 · 1 mark · Very short answerBoard level
Why must Grignard reagents be prepared in dry conditions?
Hint
Basic R⁻.
Answer
They react with water (any source of H⁺) to form alkanes, destroying the reagent.
Q5 · 2 marks · Short answerHarder board level
Write the Hunsdiecker reaction for silver propanoate and name the product.
Hint
Lose one carbon.
Answer
$C_{2} H_{5} C O O A g + B r_{2} \to C_{2} H_{5} B r + C O_{2} + A g B r$ (reflux in CCl₄); bromoethane.
Q6 · 2 marks · Short answerHarder board level
2-Bromopentane is heated with alcoholic KOH. Name every alkene that can form (count geometrical isomers separately), identify the major product and state the rule used.
Hint
β-hydrogens on C1 and C3; but-2-ene type products show cis-trans isomerism.
Answer
Three alkenes: pent-1-ene, cis-pent-2-ene and trans-pent-2-ene. Pent-2-ene (mainly the more stable trans form) is the major product by Saytzeff's rule: the more substituted alkene predominates.
Q7 · 2 marks · Short answerHarder board level
How will you convert (a) ethanol to iodoethane, (b) chlorobenzene to toluene?
Hint
Two different named reactions.
Answer
(a) $C_{2} H_{5} O H + S O C l_{2} \to C_{2} H_{5} C l + S O_{2} + H C l$ , then Finkelstein: $C_{2} H_{5} C l + N a I \to C_{2} H_{5} I + N a C l$ (dry acetone); (or red P/I₂). (b) Wurtz-Fittig: $C_{6} H_{5} C l + 2 N a + C H_{3} C l \to C_{6} H_{5} C H_{3} + 2 N a C l$ .
Q8 · 3 marks · Short answerHOTS / competency
A tertiary alkyl halide C₅H₁₁Br (A) reacts with alcoholic KOH to give an alkene B which on ozonolysis gives propanone and ethanal. Identify A and B.
Hint
Join the carbonyl fragments with C=C.
Answer
B = (CH₃)₂C=CHCH₃, 2-methylbut-2-ene. A must eliminate to give it as major product: 2-bromo-2-methylbutane, (CH₃)₂CBrCH₂CH₃ (Saytzeff product).
Q9 · 3 marks · Short answerHarder board level
Write the mechanism of the chlorination of chlorobenzene and name the products.
Hint
Generate Cl⁺ first.
Answer
$C l_{2} + F e C l_{3} \to C l^{+} + F e C l_{4}^{-}$ ; Cl⁺ attacks the ortho or para carbon forming an arenium ion stabilised by resonance (including a form with C=Cl⁺); $F e C l_{4}^{-}$ removes H⁺, regenerating FeCl₃ and releasing HCl. Products: 1,4-dichlorobenzene (major) and 1,2-dichlorobenzene.
Q10 · 2 marks · Short answerBoard level
Give one use and one harmful effect each of CCl₄ and freons.
Hint
Ozone.
Answer
CCl₄: solvent / making freons; damages liver and depletes ozone. Freons: refrigerants and aerosol propellants; release Cl radicals that deplete stratospheric ozone.
Q11 · 5 marks · Long answerHarder board level
(a) Distinguish SN1 and SN2 in four ways. [2]
(b) Explain why allyl chloride is hydrolysed faster than n-propyl chloride by SN1. [1]
(c) Write equations for the reaction of bromoethane with (i) excess NH₃, (ii) LiAlH₄. [2]
Hint
Carbocation stability.
Answer
(a) SN1: two steps, rate = k[RX], carbocation intermediate, racemisation, favoured by 3°. SN2: one step, rate = k[RX][Nu⁻], transition state, inversion, favoured by 1°. (b) The allyl carbocation CH₂=CH-CH₂⁺ is stabilised by resonance. (c)(i) $C_{2} H_{5} B r + N H_{3} \to C_{2} H_{5} N H_{2} + H B r$ (ii) $4 C_{2} H_{5} B r + L i A l H_{4} \to 4 C_{2} H_{6} + L i B r + A l B r_{3}$ .
Q12 · 2 marks · Short answerHOTS / competency
2,4,6-Trinitrochlorobenzene is hydrolysed by warm water but chlorobenzene needs 623 K and 300 atm with NaOH. Explain.
Hint
Stabilising the intermediate.
Answer
Nucleophilic attack gives a carbanion intermediate. Three -NO₂ groups at ortho and para positions delocalise its negative charge by resonance, stabilising it and greatly lowering the activation energy. Chlorobenzene has no such stabilisation.
Q13 · 2 marks · Short answerHOTS / competency
Chloroform is a liquid of density 1.48 g cm⁻³. Why does it form the lower layer when shaken with water, and why does it not dissolve?
Hint
Hydrogen bonding.
Answer
It is denser than water, so it forms the lower layer. It cannot form hydrogen bonds strong enough to compensate for breaking those between water molecules, so it is immiscible.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Saytzeff's rule.Show answer
In dehydrohalogenation the more highly substituted alkene is the major product.
2.Finkelstein reaction?Show answer
R-Cl/R-Br + NaI in dry acetone → R-I + NaCl/NaBr.
3.Rate law of SN2?Show answer
Rate = k[RX][Nu⁻]
4.Product of bromoethane + AgCN?Show answer
Ethyl isocyanide, C₂H₅NC.
5.Why is ethanol added to chloroform?Show answer
To convert any phosgene formed into harmless diethyl carbonate.
6.Name a reaction making chlorobenzene from a diazonium salt.Show answer
Sandmeyer (CuCl/HCl) or Gattermann (Cu/HCl).
7.Dow's process product?Show answer
Phenol (via sodium phenoxide).
8.Expand DDT.Show answer
p,p'-Dichlorodiphenyltrichloroethane.

Best order to study it

Learn classification, naming, the C-X bond and physical properties.
Learn all preparations, including Finkelstein, Swarts and Hunsdiecker.
Master SN1 and SN2 with stereochemistry; then the full list of nucleophilic reactions.
Learn elimination (Saytzeff), Wurtz and Grignard; chloroform and iodoform.
Learn haloarene preparation, low reactivity, electrophilic substitution mechanisms and coupling reactions.
Learn uses and environmental effects; practise conversions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name and classify any haloalkane?
Can you write Finkelstein, Swarts and Hunsdiecker equations?
Can you draw SN1 and SN2 mechanisms and state the stereochemistry?
Can you predict products with KCN/AgCN and KNO₂/AgNO₂?
Can you apply Saytzeff's rule?
Can you prepare chloroform and write the phosgene and carbylamine reactions?
Can you explain why haloarenes resist nucleophilic substitution?
Can you write Sandmeyer, Gattermann, Wurtz-Fittig, Fittig and DDT equations?

Haloalkanes and Haloarenes: frequently asked questions

Which mechanisms are in the ISC 2027 syllabus for this chapter?

SN1 and SN2 for haloalkanes, electrophilic substitution (chlorination, nitration, sulphonation) of haloarenes, and nucleophilic substitution of Cl by -OH and -NH₂ in haloarenes.

Is the directive influence of halogen needed for disubstituted rings?

No. The syllabus limits directive influence to monosubstituted compounds only.

Do I need the Hunsdiecker reaction?

Yes. Preparation of haloalkanes from silver salts of fatty acids (Hunsdiecker) is listed alongside Finkelstein and Swarts.

What exactly is needed on chloroform and iodoform?

Their preparation and properties: chloroform from ethanol or acetone and bleaching powder, its oxidation to phosgene, the role of ethanol, the carbylamine reaction; iodoform preparation, the iodoform test, and conversion to ethyne.

Which polyhalogen compounds must I know?

Dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons and DDT: uses and environmental effects, plus the structures of freons and DDT.

How many marks does this chapter carry?

It is part of the Organic Chemistry group, 31 of the 70 theory marks, shared with four other organic chapters.

Based on: CISCE ISC 2027 Chemistry syllabus; CISCE ISC 2027 Chemistry specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Chemistry, 2025; CISCE Analysis of Pupil Performance, ISC Chemistry, 2024