Class 12Chemistry · Organic ChemistryFull chapter

Haloalkanes and Haloarenes

The whole chapter in one place — read it, then test yourself. Clear notes, key equations, a practice quiz, and worked NCERT solutions & PYQs.

Classification and Nomenclature of Haloalkanes and Haloarenes

Quick answer Halogen derivatives of hydrocarbons are classified as haloalkanes (halogen on an sp3 carbon) or haloarenes (halogen on an aromatic-ring carbon), and are named systematically using IUPAC rules.

When one or more hydrogen atoms of a hydrocarbon are replaced by halogen atoms (F, Cl, Br, I), the products are called halogen derivatives. If the halogen is attached to an sp3 hybridised carbon, the compound is a haloalkane (alkyl halide, general formula CnH2n+1X). If the halogen is attached directly to an sp2 hybridised carbon of a benzene ring, the compound is a haloarene (aryl halide).

Halogen compounds are classified as mono-, di-, tri- or, generally, poly-halogen compounds based on the number of halogens present. Haloalkanes are further classified by the carbon bearing the halogen: primary (1°) — halogen on a carbon joined to only one other carbon, e.g. CH3CH2Cl; secondary (2°) — joined to two other carbons, e.g. (CH3)2CHCl; tertiary (3°) — joined to three other carbons, e.g. (CH3)3CCl.

Position relative to a double bond or ring gives further names: an allylic halide has X on an sp3 carbon next to a C=C, e.g. CH2=CH–CH2Cl (allyl chloride); a benzylic halide has X on an sp3 carbon attached directly to a ring, e.g. C6H5–CH2Cl (benzyl chloride); a vinylic halide has X on a doubly-bonded sp2 carbon, e.g. CH2=CHCl (vinyl chloride); an aryl halide has X directly on the aromatic ring, e.g. C6H5Cl (chlorobenzene).

IUPAC nomenclature treats halogen as a substituent prefix (fluoro-, chloro-, bromo-, iodo-) on the parent hydrocarbon chain. Numbering of the chain first gives the lowest locant to the principal suffix feature (such as a C=C double bond, "-ene"), and only when there is still a choice does numbering fall back to giving the lowest locants to the substituent prefixes, with the group cited first alphabetically getting the lower number in a tie.

Worked example: Classify and name CH2=CH–CH2–Cl. The Cl sits on an sp3 carbon that is directly bonded to a carbon of the C=C bond, so it is an allylic halide. For IUPAC numbering, the carbon–carbon double bond (the principal suffix "-ene") is given priority over the halo- substituent prefix when the lowest locant is assigned, so the three-carbon chain is numbered starting from the =CH2 end: the double bond falls between C1 and C2, and the Cl-bearing carbon is C3. The IUPAC name is therefore 3-chloroprop-1-ene (common name: allyl chloride).

General formula, haloalkane CₙH₂ₙ₊₁X One halogen X (F, Cl, Br or I) replaces one H of alkane CₙH₂ₙ₊₂.
General formula, monohaloarene ArX (e.g. C₆H₅X) Halogen bonded directly to an sp2 ring carbon.
Remember
  • Haloalkane: X on sp3 carbon (R–X); haloarene: X directly on an aromatic sp2 ring carbon (Ar–X).
  • Haloalkanes are classed as 1°, 2° or 3° by how many carbons are attached to the carbon bearing X.
  • Allylic (X on sp3 C next to C=C), benzylic (X on sp3 C attached to a ring), vinylic (X on sp2 C=C carbon) and aryl (X on ring) halides are distinguished by the environment of the C–X carbon.
  • IUPAC names use halo- prefixes (fluoro, chloro, bromo, iodo); a C=C double bond takes priority over the halo- prefix for the lowest locant, and only then are substituent locants (and, in a tie, alphabetical order) used to decide numbering.
  • Common (trivial) names such as allyl chloride, benzyl chloride and chloroform remain in wide use alongside IUPAC names.

Nature of the C–X Bond and Methods of Preparation

Quick answer The C–X bond is polar and its strength decreases down the halogen group; haloalkanes are prepared from alcohols, from hydrocarbons by free-radical or addition reactions, and by halide-exchange reactions.

Carbon is less electronegative than every halogen, so the C–X bond is polarised with a partial positive charge on carbon and partial negative charge on the halogen. As the halogen atom gets larger down the group (F → I), the C–X bond becomes longer (poorer orbital overlap) and its bond enthalpy decreases: C–F is the shortest and strongest, C–I the longest and weakest of the series.

From alcohols — the –OH group is replaced by halogen. With HX (X = Cl, Br, I) in the presence of a catalyst such as anhydrous ZnCl2 (this mixture with conc. HCl is the Lucas reagent), 3° alcohols react almost instantly at room temperature, 2° alcohols react within a few minutes, and 1° alcohols show no visible reaction at room temperature (they need heating) — a difference used as a qualitative test to distinguish 1°, 2° and 3° alcohols. Cleaner routes avoid this variability: R–OH + SOCl2 → R–Cl + SO2↑ + HCl↑ (thionyl chloride method) is preferred because both by-products are gases, leaving a pure alkyl chloride; PCl5 and PCl3 (or red phosphorus with Br2/I2 generated in situ) are used similarly for chlorides, bromides and iodides.

From hydrocarbons — alkanes undergo free-radical halogenation with X2 under heat or UV light, but this gives a mixture of products (poor for synthesis unless the molecule has only one type of hydrogen); alkenes add HX across the double bond following Markovnikov's rule (H adds to the carbon already bearing more hydrogens, X goes to the more substituted carbon, via the more stable carbocation), except that HBr specifically can add with the opposite (anti-Markovnikov) regiochemistry in the presence of peroxides (the peroxide/Kharasch effect), because the mechanism switches to a free-radical chain.

Halide exchange — the Finkelstein reaction converts R–Cl or R–Br to R–I by heating with NaI in dry acetone; NaCl/NaBr precipitates (insoluble in acetone) and drives the reaction forward. The Swarts reaction converts R–Cl/R–Br to R–F using AgF, Hg2F2, CoF2 or SbF3.

Worked example: Convert ethanol to bromoethane by two routes, and say which avoids carbocation rearrangement. Route 1: CH3CH2OH + HBr (conc., with conc. H2SO4 or NaBr/H2SO4, heat) → CH3CH2Br + H2O. Route 2: 3CH3CH2OH + PBr3 (from red P + Br2) → 3CH3CH2Br + H3PO3. Route 2 (via PX3) does not pass through a free carbocation, so it is the preferred method whenever the substrate is prone to rearrangement (e.g. secondary/allylic alcohols); the HX/acid route can rearrange such substrates.

Alcohol + HX R–OH + HX --ZnCl₂--> R–X + H₂O Reactivity of HX: HI > HBr > HCl; basis of the Lucas test (1° slow/none at RT, 2° faster, 3° instant).
Thionyl chloride method R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑ Best method for pure alkyl chlorides — gaseous by-products escape.
Phosphorus pentachloride R–OH + PCl₅ → R–Cl + POCl₃ + HCl
Markovnikov addition Alkene + HX → major product with X on more substituted C Via the more stable (more substituted) carbocation intermediate.
Peroxide (anti-Markovnikov) effect Alkene + HBr --peroxide--> X on less substituted C Free-radical chain mechanism; specific to HBr only.
Finkelstein reaction R–Cl/R–Br + NaI --dry acetone--> R–I + NaCl↓/NaBr↓
Swarts reaction R–Cl/R–Br + AgF (or SbF₃, CoF₂) → R–F
Remember
  • C–X bond is polar; bond length increases and bond enthalpy decreases down the group: C–F strongest/shortest, C–I weakest/longest.
  • Alcohols are converted to halides by HX/ZnCl2 (Lucas test: 3° instant, 2° within minutes, 1° needs heating), or more cleanly by SOCl2, PCl5 or PCl3.
  • SOCl2 is the best method for alkyl chlorides because both by-products (SO2, HCl) are gases.
  • Alkenes + HX follow Markovnikov addition; HBr with peroxides gives anti-Markovnikov addition (peroxide effect).
  • Finkelstein reaction (NaI/dry acetone) makes alkyl iodides; Swarts reaction (AgF/SbF3 etc.) makes alkyl fluorides.

Physical Properties and Nucleophilic Substitution Reactions

Quick answer Haloalkane boiling point, density and solubility depend on halogen size and number; nucleophilic substitution proceeds by the bimolecular SN2 or unimolecular SN1 pathway, each with characteristic kinetics and stereochemistry.

Physical properties: for the same alkyl group, boiling point rises with the size (polarisability) of the halogen: R–I > R–Br > R–Cl > R–F. Among isomers, boiling point falls with branching (n- > iso- > tert-, since a more compact, spherical shape has less surface area for van der Waals contact). Density increases with the mass and number of halogens present — di- and tri-halomethanes are denser than water, and alkyl iodides are denser still. Haloalkanes and haloarenes are only sparingly soluble in water: although the halogen can form a weak hydrogen bond with water, the energy released is not enough to overcome the stronger hydrogen bonds that must be broken within the water itself, so these compounds dissolve much more readily in organic solvents.

Nucleophilic substitution is the characteristic reaction of haloalkanes: a nucleophile (Nu⁻ or Nu:) replaces the halide leaving group, Nu⁻ + R–X → R–Nu + X⁻. Two limiting mechanisms operate.

In the SN2 (substitution, nucleophilic, bimolecular) pathway, the incoming nucleophile attacks the carbon from the side opposite the leaving group in a single concerted step, passing through a five-coordinate transition state; the leaving group departs as the new bond forms. This backside attack flips the carbon's configuration like an umbrella turning inside out — Walden inversion — so SN2 is stereospecific. Because the transition state is crowded, reactivity falls sharply as substitution increases around the carbon: CH3X > 1° > 2° > 3° (3° halides essentially do not react by SN2). The rate law is second order overall: Rate = k[R–X][Nu⁻].

In the SN1 (unimolecular) pathway, the C–X bond breaks first, in the slow, rate-determining step, to give a planar carbocation; the nucleophile then attacks this intermediate rapidly from either face. Because the rate depends only on how readily R–X ionises, Rate = k[R–X], independent of nucleophile concentration. Carbocation stability governs reactivity, so the order reverses: 3° > 2° > 1° (3° halides favour SN1). Since the flat carbocation can be attacked from both faces, a chiral substrate gives a largely racemised product (a mixture of both configurations, not a single, clean inversion).

Worked example: Solvolysis of tert-butyl bromide, (CH3)3CBr, in aqueous acetone is first order. At [(CH3)3CBr] = 0.050 mol L⁻¹ the initial rate is 2.5 × 10⁻⁵ mol L⁻¹ s⁻¹. Find the rate constant, and the new rate if [(CH3)3CBr] is doubled to 0.100 mol L⁻¹ while [OH⁻] is independently doubled. Since this is SN1, Rate = k[RX], so k = (2.5 × 10⁻⁵)/(0.050) = 5.0 × 10⁻⁴ s⁻¹. At 0.100 mol L⁻¹: new rate = (5.0 × 10⁻⁴)(0.100) = 5.0 × 10⁻⁵ mol L⁻¹ s⁻¹ — the rate exactly doubles when [RX] doubles, and changing [OH⁻] has no effect at all, since [OH⁻] does not appear in the SN1 rate law.

SN2 rate law Rate = k[R–X][Nu⁻] mol L⁻¹ s⁻¹ · Second order overall; concerted, one step, inversion of configuration.
SN1 rate law Rate = k[R–X] mol L⁻¹ s⁻¹ · First order; two steps via a carbocation; product is largely racemised.
SN2 reactivity order CH₃X > 1° > 2° > 3° Set by steric hindrance to backside attack.
SN1 reactivity order 3° > 2° > 1° > CH₃X Set by carbocation stability.
Boiling-point trend b.p.: R–I > R–Br > R–Cl > R–F (same R) Rises with size/polarisability of the halogen.
Remember
  • For the same R group, b.p. order is R–I > R–Br > R–Cl > R–F; density rises with halogen mass/number; haloalkanes are water-insoluble but organic-solvent-soluble.
  • SN2: single step, backside attack, Walden inversion, rate = k[RX][Nu⁻]; fastest for CH3X, slowest (negligible) for 3° halides.
  • SN1: two steps via a planar carbocation, rate = k[RX] only, largely racemised product; fastest for 3° halides (most stable carbocation).
  • Reactivity trends for SN2 and SN1 are exactly opposite because they are governed by sterics (SN2) versus carbocation stability (SN1) respectively.
  • Polar protic solvents favour SN1 (stabilise ions); polar aprotic solvents favour SN2 (leave the nucleophile more reactive).

Elimination Reactions and Reactions with Metals

Quick answer Haloalkanes heated with alcoholic KOH undergo elimination to give alkenes (Saytzeff's rule), and react with reactive metals such as sodium (Wurtz reaction) and magnesium (Grignard reagent) to form new C–C bonds.

When a haloalkane is heated with alcoholic (not aqueous) KOH, instead of substitution it can undergo dehydrohalogenation: a proton is removed from a carbon adjacent to the one bearing the halogen, and the halide ion leaves, forming a carbon–carbon double bond (an alkene). When more than one type of β-hydrogen is available, Saytzeff's rule predicts that the major product is the more highly substituted (more stable) alkene.

Substitution and elimination are always in competition: aqueous KOH (a good nucleophile, weaker base) favours substitution, while alcoholic KOH (a strong, bulky base) favours elimination; 3° halides, which cannot easily undergo SN2, tend to eliminate especially readily with strong bases.

Reactions with metals: in the Wurtz reaction, two molecules of an alkyl halide react with sodium metal in dry ether to couple into a longer, symmetrical alkane: 2R–X + 2Na → R–R + 2NaX. This method only cleanly gives a single product when both halide molecules are identical (an unsymmetrical mixture of two different halides gives a mixture of three coupling products), so it is mainly used to prepare symmetrical alkanes with an even number of carbon atoms.

Alkyl (or aryl) halides react with magnesium turnings in dry ether to form Grignard reagents, R–Mg–X, powerful carbon nucleophiles used throughout organic synthesis. Grignard reagents must be kept completely free of moisture, since even traces of water destroy them: R–MgX + H2O → R–H + Mg(OH)X.

Worked example: A primary alkyl halide of formula C4H9Br (A) is heated with alcoholic KOH to give alkene (B); (B) is treated with HBr to give (C), an isomer of (A); and (A) treated with sodium metal gives octane (D), C8H18, different from the n-octane obtained by treating n-butyl bromide with sodium. Identify (A). Since the Wurtz product from n-butyl bromide would be n-octane, and (D) is a different octane isomer, (A) must be isobutyl bromide, (CH3)2CH–CH2–Br (a primary halide). Dehydrohalogenation removes a β-H from the tertiary carbon to give (B) = (CH3)2C=CH2 (2-methylpropene). Markovnikov addition of HBr to (B) puts Br on the more substituted carbon, giving (C) = (CH3)2CBr–CH3 (tert-butyl bromide), which is indeed a C4H9Br isomer of (A). The Wurtz coupling of two molecules of (A) gives (D) = (CH3)2CH–CH2–CH2–CH(CH3)2, i.e. 2,5-dimethylhexane, which is indeed different from n-octane.

Dehydrohalogenation R–CH₂–CHX–R′ + alc. KOH --Δ--> alkene + KX + H₂O
Saytzeff's rule major alkene = the more substituted (more stable) alkene Governs product ratio when elimination can occur in more than one direction.
Wurtz reaction 2R–X + 2Na --dry ether--> R–R + 2NaX
Grignard reagent formation R–X + Mg --dry ether--> R–MgX
Remember
  • Alcoholic KOH promotes elimination (dehydrohalogenation) of haloalkanes to alkenes; aqueous KOH favours substitution.
  • Saytzeff's rule: the more substituted (more stable) alkene is the major elimination product when a choice of β-hydrogens exists.
  • Wurtz reaction (2R–X + 2Na, dry ether → R–R) couples two alkyl halides; clean only for a single, symmetrical alkyl halide.
  • Grignard reagents (R–MgX, from R–X + Mg in dry ether) are versatile carbanion-equivalents but are destroyed instantly by water, alcohols or CO2/acids.
  • 3° halides tend to favour elimination over substitution with strong/bulky bases because SN2 attack at the crowded carbon is very slow.

Preparation and Reactivity of Haloarenes

Quick answer Haloarenes are made by direct halogenation of arenes or from diazonium salts, and their C–X bond is far less reactive towards nucleophilic substitution than that of haloalkanes because of ring resonance.

Preparation: chlorobenzene and bromobenzene are made by direct electrophilic halogenation of benzene, C6H6 + X2 in the presence of a Lewis-acid catalyst such as FeCl3 or FeBr3, giving mainly the ortho and para products. Direct fluorination is too violent to control, and direct iodination is reversible (the HI formed reduces the product back to benzene), so iodoarenes are instead made via a diazonium-salt route.

Aryl diazonium salts, formed from primary aromatic amines with NaNO2/HCl at 0–5 °C, are versatile intermediates for making every haloarene under mild conditions: the Sandmeyer reaction uses Cu2Cl2/Cu2Br2 (cuprous halides) to give ArCl or ArBr; the closely related Gattermann reaction uses copper powder with the corresponding halogen acid; the Balz–Schiemann reaction heats the diazonium tetrafluoroborate salt to give ArF; and treating the diazonium salt directly with KI gives ArI, with no catalyst needed.

Low reactivity towards nucleophilic substitution: unlike haloalkanes, haloarenes resist nucleophilic substitution under ordinary conditions, for several reinforcing reasons. First, a lone pair on the halogen conjugates (delocalises) into the aromatic π-system by resonance, giving the C–X bond partial double-bond character; this makes the bond shorter and stronger than a normal single C–X bond, so it resists being broken. Second, the ring carbon is sp2 hybridised, which holds the bonding electron pair of C–X more tightly than an sp3 carbon would (sp2 carbon has greater effective electronegativity), reinforcing the bond strength. Third, an SN1 pathway is blocked because the resulting phenyl cation would not be resonance-stabilised (its empty orbital lies in the plane of the ring, perpendicular to the π-system) and is therefore very unstable. Fourth, the electron-rich aromatic ring tends to repel an approaching electron-rich nucleophile. Only under forcing conditions — high temperature and pressure, or with strong electron-withdrawing groups (like –NO2) positioned ortho/para to the halogen — does nucleophilic substitution proceed, e.g. the industrial Dow process, C6H5Cl + NaOH at 623 K/300 atm → C6H5OH + NaCl.

Electrophilic substitution still occurs readily on the ring itself (halogenation, nitration, sulfonation, Friedel–Crafts). Here halogen shows its dual character: it is an ortho, para-director because the same lone-pair resonance donation that strengthens the C–X bond also selectively stabilises the arenium-ion (Wheland) intermediate formed when the electrophile attacks the ortho or para position (not meta). Yet, taken over the whole ring, the strong electron-withdrawing inductive (−I) effect of the electronegative halogen outweighs this resonance donation, so the ring as a whole is deactivated — haloarenes undergo electrophilic substitution more slowly than benzene itself, even though the new substituent still goes mainly to the ortho/para positions.

Worked example: Chlorobenzene undergoes nitration (HNO3/H2SO4) more slowly than benzene, giving mainly a mixture of ortho- and para-nitrochlorobenzene with only a trace of the meta isomer. This is explained exactly as above: the −I effect of Cl slows the overall reaction (deactivation), while resonance donation from the Cl lone pair stabilises the intermediate carbocation specifically for ortho/para attack (o,p-direction), and the bulkier para product usually predominates over the sterically hindered ortho product.

Direct halogenation of benzene C₆H₆ + X₂ --FeX₃--> C₆H₅X + HX X = Cl or Br only; electrophilic aromatic substitution.
Sandmeyer reaction ArN₂⁺Cl⁻ --Cu₂Cl₂/HCl--> ArCl + N₂
Gattermann reaction ArN₂⁺X⁻ --Cu powder/HX--> ArX + N₂
Balz–Schiemann reaction ArN₂⁺BF₄⁻ --Δ--> ArF + N₂ + BF₃
Dow process C₆H₅Cl + NaOH --623 K, 300 atm--> C₆H₅OH + NaCl Shows haloarenes CAN be substituted, but only under forcing conditions.
Remember
  • Chlorobenzene/bromobenzene: direct electrophilic halogenation of benzene with X2/FeX3; direct F2 and I2 routes are impractical.
  • Diazonium-salt routes (Sandmeyer with Cu2X2, Gattermann with Cu/HX, Balz–Schiemann for ArF, KI for ArI) give all haloarenes under mild conditions.
  • C–X in haloarenes is shorter/stronger than in haloalkanes due to resonance (lone-pair conjugation into the ring), so nucleophilic substitution is very sluggish.
  • An aryl cation (needed for SN1) is not resonance-stabilised, so haloarenes cannot follow the SN1 pathway either; substitution needs forcing conditions (e.g. the Dow process).
  • Halogens are electrophilic-substitution deactivators (net −I effect) yet remain ortho/para-directors (resonance stabilisation of the o/p arenium intermediate).

Polyhalogen Compounds and Their Environmental Effects

Quick answer A small set of polyhalogen compounds — dichloromethane, chloroform, carbon tetrachloride, iodoform, freons and DDT — have important industrial and household uses, but several also pose serious environmental hazards.

Compounds bearing more than one halogen atom find everyday and industrial use, though a few require careful handling.

Dichloromethane (CH2Cl2) is a widely used industrial solvent and paint remover. Trichloromethane (chloroform, CHCl3) is an excellent solvent and was historically used as a surgical anaesthetic; however, it is slowly oxidised by air and light into the extremely poisonous gas phosgene (COCl2). For this reason chloroform is stored in dark-coloured bottles that are completely filled (to exclude air) and to which about 1% ethanol is added — the ethanol converts any phosgene formed into harmless diethyl carbonate. Triiodomethane (iodoform, CHI3) was once used as an antiseptic on wounds, its action really coming from the free iodine it slowly liberates rather than from the compound itself.

Tetrachloromethane (carbon tetrachloride, CCl4) is a non-flammable solvent that was formerly used in fire extinguishers (known by the trade name pyrene) and dry cleaning; its use has now been curtailed because of its toxicity and its role, like other chlorinated compounds, in ozone depletion.

Freons (chlorofluorocarbons, CFCs, e.g. Freon-12, CCl2F2) are inert, non-toxic, non-flammable, easily liquefiable gases historically used as refrigerants and aerosol propellants. Their very inertness lets them survive long enough to drift up into the stratosphere, where intense UV radiation photolyses the C–Cl bond to release chlorine free radicals: CF2Cl2 + UV → •CF2Cl + Cl•. Each Cl• then destroys ozone catalytically: Cl• + O3 → ClO• + O2, followed by ClO• + O → Cl• + O2, regenerating the chlorine radical so that a single Cl atom can destroy many thousands of ozone molecules in a chain reaction — this is the chief cause of stratospheric ozone-layer depletion.

DDT (para,para′-dichlorodiphenyltrichloroethane, C14H9Cl5) was, for decades, the most widely used synthetic insecticide because it is cheap and highly effective. Unfortunately it is chemically very stable and non-biodegradable: it persists in soil and water and accumulates in the fatty tissue of animals, becoming increasingly concentrated (biomagnified) as it passes up the food chain. Because of this environmental persistence and toxicity to wildlife, DDT is now banned or heavily restricted in many countries.

Chloroform CHCl₃ Solvent, historical anaesthetic; oxidises to poisonous phosgene (COCl₂) in air + light.
Phosgene COCl₂ Highly toxic gas formed by aerial oxidation of chloroform; neutralised by added ethanol.
Carbon tetrachloride CCl₄ Non-flammable solvent, former fire-extinguisher fluid; toxic, ozone-depleting.
Freon-12 CCl₂F₂ Refrigerant/propellant CFC; source of ozone-destroying Cl radicals in the stratosphere.
DDT C₁₄H₉Cl₅ Persistent insecticide; bioaccumulates and biomagnifies up the food chain.
Remember
  • Chloroform (CHCl3) oxidises in air/light to poisonous phosgene; stored in dark, filled bottles with ~1% ethanol to neutralise any phosgene formed.
  • Carbon tetrachloride (CCl4) was used as a solvent and fire-extinguisher fluid but is now restricted due to toxicity and ozone impact.
  • Iodoform's (CHI3) antiseptic action comes from the iodine it liberates, not the compound itself.
  • Freons/CFCs (e.g. CCl2F2) are inert refrigerant/propellant gases that release Cl radicals in the stratosphere, catalytically destroying ozone.
  • DDT is a persistent, non-biodegradable insecticide that bioaccumulates and biomagnifies up the food chain — now banned/restricted in many countries.

Key equations

Every formula in this chapter, in one place — screenshot it before your exam.

CₙH₂ₙ₊₁X
General formula, haloalkane
ArX (e.g. C₆H₅X)
General formula, monohaloarene
R–OH + HX --ZnCl₂--> R–X + H₂O
Alcohol + HX
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
Thionyl chloride method
R–OH + PCl₅ → R–Cl + POCl₃ + HCl
Phosphorus pentachloride
Alkene + HX → major product with X on more substituted C
Markovnikov addition
Alkene + HBr --peroxide--> X on less substituted C
Peroxide (anti-Markovnikov) effect
R–Cl/R–Br + NaI --dry acetone--> R–I + NaCl↓/NaBr↓
Finkelstein reaction
R–Cl/R–Br + AgF (or SbF₃, CoF₂) → R–F
Swarts reaction
Rate = k[R–X][Nu⁻]
SN2 rate lawmol L⁻¹ s⁻¹
Rate = k[R–X]
SN1 rate lawmol L⁻¹ s⁻¹
CH₃X > 1° > 2° > 3°
SN2 reactivity order
3° > 2° > 1° > CH₃X
SN1 reactivity order
b.p.: R–I > R–Br > R–Cl > R–F (same R)
Boiling-point trend
R–CH₂–CHX–R′ + alc. KOH --Δ--> alkene + KX + H₂O
Dehydrohalogenation
major alkene = the more substituted (more stable) alkene
Saytzeff's rule
2R–X + 2Na --dry ether--> R–R + 2NaX
Wurtz reaction
R–X + Mg --dry ether--> R–MgX
Grignard reagent formation
C₆H₆ + X₂ --FeX₃--> C₆H₅X + HX
Direct halogenation of benzene
ArN₂⁺Cl⁻ --Cu₂Cl₂/HCl--> ArCl + N₂
Sandmeyer reaction
ArN₂⁺X⁻ --Cu powder/HX--> ArX + N₂
Gattermann reaction
ArN₂⁺BF₄⁻ --Δ--> ArF + N₂ + BF₃
Balz–Schiemann reaction
C₆H₅Cl + NaOH --623 K, 300 atm--> C₆H₅OH + NaCl
Dow process
CHCl₃
Chloroform
COCl₂
Phosgene
CCl₄
Carbon tetrachloride
CCl₂F₂
Freon-12
C₁₄H₉Cl₅
DDT

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Classification easy

Which of the following is a haloarene?

Q2 Nomenclature easy

What is the IUPAC name of CH3-CH2-CHCl-CH3?

Q3 Nature of C-X bond easy

Arrange CH3F, CH3Cl, CH3Br and CH3I in decreasing order of C-X bond enthalpy.

Q4 Preparation of haloalkanes medium

Which reagent converts an alcohol to the corresponding alkyl chloride most cleanly, since both by-products escape as gases?

Q5 Preparation of haloalkanes medium

The Finkelstein reaction is used to prepare:

Q6 SN2 mechanism medium

Which alkyl halide reacts fastest by the SN2 mechanism?

Q7 Stereochemistry of substitution medium

In an SN2 reaction on a chiral alkyl halide, the configuration at the reacting carbon:

Q8 SN1 kinetics medium

Hydrolysis of tert-butyl bromide with aqueous NaOH shows a rate that is unaffected by [NaOH]. This indicates the mechanism is:

Q9 SN1 kinetics (numerical) hard

Solvolysis of (CH3)3CBr is first order with k = 5.0x10^-4 s^-1 at a given temperature. If the initial concentration of (CH3)3CBr is 0.040 mol/L, what is the initial rate of the reaction?

Q10 Reactivity of haloarenes hard

Chlorobenzene is far less reactive than ethyl chloride in nucleophilic substitution mainly because:

Q11 Electrophilic substitution in haloarenes hard

Although Cl is an ortho, para-directing group in electrophilic aromatic substitution on chlorobenzene, it still deactivates the ring overall. This is because:

Q12 Environmental effects hard

Chlorofluorocarbons (Freons) deplete the stratospheric ozone layer chiefly because:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Write the structures of the different dihalogen derivatives of propane.Isomerism of haloalkanes

Propane is CH3-CH2-CH3. Replacing two hydrogen atoms with the same halogen X gives four constitutionally different dihalopropanes:

  • 1,1-Dihalopropane: CH3-CH2-CHX2
  • 1,2-Dihalopropane: CH3-CHX-CH2X
  • 1,3-Dihalopropane: CH2X-CH2-CH2X
  • 2,2-Dihalopropane: CH3-CX2-CH3

(X may be F, Cl, Br or I in each case; in addition, 1,2-dihalopropane exists as a pair of enantiomers because C-2 becomes a stereocentre.)

2 Free-radical bromination (Br2/hv) of 2-methylbutane, (CH3)2CH-CH2-CH3, can give several isomeric monobromo products. Draw them and identify the major product, with reasoning.Free-radical halogenation and radical stability

2-Methylbutane has four distinct sets of hydrogens, so bromination at each site gives a different monobromo product:

  • Substitution at either of the two equivalent methyls attached to C-2 gives 1-bromo-2-methylbutane, BrCH2-CH(CH3)-CH2-CH3 (primary product).
  • Substitution at C-2 (the tertiary C-H) gives 2-bromo-2-methylbutane, CH3-CBr(CH3)-CH2-CH3 (tertiary product).
  • Substitution at C-3 gives 3-bromo-2-methylbutane, CH3-CH(CH3)-CHBr-CH3 (secondary product).
  • Substitution at C-4 gives 1-bromo-3-methylbutane, CH3-CH(CH3)-CH2-CH2Br (primary product).

The major product is 2-bromo-2-methylbutane. Free-radical halogenation proceeds through a carbon radical intermediate, and radical stability follows 3° > 2° > 1°. Even though there is only one tertiary hydrogen (compared with six equivalent primary hydrogens on the two methyls), abstraction of that tertiary hydrogen is so much faster (bromination is highly selective, unlike chlorination) that the tertiary bromide dominates the product mixture.

3 Which one of n-butyl bromide (CH3CH2CH2CH2Br) and sec-butyl bromide (CH3CH2CHBrCH3) would you expect to react faster by the SN2 mechanism, and why?SN2 reactivity and steric hindrance

n-Butyl bromide reacts faster. It is a primary halide, with the C-Br carbon bearing two hydrogens and only one alkyl group, so the nucleophile has an unhindered path for backside attack in the SN2 transition state. sec-Butyl bromide is a secondary halide; the C-Br carbon carries two alkyl groups (CH3 and CH2CH3) which crowd the site of attack, raising steric hindrance and slowing the reaction considerably.

This matches the general SN2 reactivity order CH3X > 1° > 2° > 3°, so n-butyl bromide (1°) reacts faster than sec-butyl bromide (2°).

4 Write the IUPAC names of the following: (i) CH3CH(Cl)CH(Br)CH3 (ii) CHF2CBrClFIUPAC nomenclature of polyhalogen compounds

(i) The compound is a butane chain with Cl on one internal carbon and Br on the other. Numbering from either end gives the locant set {2,3} for the two substituents; when locant sets tie, the substituent that is cited first alphabetically (bromo before chloro) is given the lower number. Numbering so Br is at C-2 and Cl is at C-3 gives the name 2-bromo-3-chlorobutane.

(ii) The molecule is a two-carbon (ethane) chain: one carbon bears H, F, F and the other bears F, Cl, Br. Comparing the two possible locant sets shows numbering from the carbon bearing Br, Cl, F as C-1 gives the lower set {1,1,1,2,2}, compared with {1,1,2,2,2} the other way. This carbon then carries bromo, chloro and fluoro substituents at position 1, and the other carbon carries two fluoro substituents at position 2. The IUPAC name is 1-bromo-1-chloro-1,2,2-trifluoroethane.

5 What are ambident nucleophiles? Explain with a suitable example.Nature of nucleophiles

An ambident nucleophile is a nucleophile that has two different atoms bearing lone pairs/negative charge (usually in resonance with each other), so it can attack an electrophilic carbon through either atom, giving two different products depending on which site bonds to carbon.

Example — cyanide ion, CN⁻: the negative charge is delocalised over both carbon and nitrogen. If the nucleophile attacks through carbon, the product is an alkyl cyanide (nitrile), R-C≡N; if it attacks through nitrogen, the product is an alkyl isocyanide, R-N≡C. A second classic example is the nitrite ion, NO2⁻: attack through oxygen gives an alkyl nitrite (R-O-N=O), while attack through nitrogen gives a nitroalkane (R-NO2).

6 A primary alkyl halide of formula C4H9Br (A), on treatment with alcoholic KOH, gives compound (B). (B) reacts with HBr to give (C), which is an isomer of (A). When (A) is treated with sodium metal, it gives compound (D), C8H18, which is different from the compound formed when n-butyl bromide is treated with sodium. Give the structure of (A) and write equations for all the steps involved.Identification via elimination, addition and Wurtz reaction

Since Wurtz coupling of n-butyl bromide would give n-octane, and (D) is a different C8H18 isomer, (A) cannot be n-butyl bromide; it must be another primary C4H9Br isomer, namely isobutyl bromide (1-bromo-2-methylpropane), (CH3)2CH-CH2-Br.

Step 1 (elimination): (CH3)2CH-CH2-Br + alc. KOH --Δ--> (CH3)2C=CH2 (B, 2-methylpropene) + KBr + H2O

Step 2 (Markovnikov addition of HBr to B): (CH3)2C=CH2 + HBr → (CH3)2CBr-CH3 (C, 2-bromo-2-methylpropane / tert-butyl bromide). (C) is indeed a C4H9Br isomer of (A), as required.

Step 3 (Wurtz reaction of A): 2 (CH3)2CH-CH2-Br + 2Na --dry ether--> (CH3)2CH-CH2-CH2-CH(CH3)2 (D, 2,5-dimethylhexane) + 2NaBr. (D) is a branched C8H18 isomer, different from the n-octane that n-butyl bromide would give, exactly as required by the question.

Previous-year board questions 4

Q1 Out of SN1 and SN2 reactions, which one is stereospecific in nature? Explain with a suitable example. 2023 3 marks

The SN2 reaction is stereospecific. It proceeds through a single concerted step in which the nucleophile attacks the electrophilic carbon from the side directly opposite the leaving group; as the new bond forms and the old one breaks, the configuration at that carbon is completely inverted (Walden inversion). Because every molecule reacting by this pathway is inverted in exactly the same way, a single, pure stereoisomer of starting material gives a single, pure (inverted) stereoisomer of product.

Example: treatment of (S)-2-bromooctane with aqueous NaOH under SN2 conditions gives (R)-octan-2-ol: the spatial arrangement at the reacting carbon is inverted, and since the incoming -OH keeps the same relative CIP priority rank as the departing Br, this spatial inversion also flips the R/S descriptor.

By contrast, SN1 proceeds via a planar carbocation intermediate that can be attacked by the nucleophile from either face with roughly equal probability, so a single enantiomer of starting material gives a largely racemised (near 50:50 mixture of both enantiomers) product; SN1 is therefore not stereospecific.

Q2 Why is chlorobenzene less reactive than methyl chloride towards nucleophilic substitution reactions? Give two reasons. 2022 2 marks

Reason 1 — Resonance (C-Cl bond strengthening): in chlorobenzene, a lone pair on chlorine is delocalised into the aromatic ring by resonance, giving the C-Cl bond partial double-bond character. This makes the bond shorter and considerably stronger than the pure single C-Cl bond in methyl chloride, so it is much harder to break.

Reason 2 — Instability of the phenyl cation: an SN1-type pathway would require the ring carbon to form a planar aryl cation, but this cation cannot be stabilised by resonance with the ring pi-system (its empty orbital is perpendicular to it), making it highly unstable; consequently chlorobenzene cannot ionise the way methyl chloride can. (Additionally, the sp2 ring carbon holds the C-Cl electrons more tightly than an sp3 carbon does.)

Q3 Arrange the following in decreasing order of their reactivity towards SN2 displacement: (I) CH3Cl (II) CH3CH2Cl (III) (CH3)2CHCl (IV) (CH3)3CCl 2023 4 marks

SN2 reactivity depends on how accessible the carbon bearing the leaving group is to backside nucleophilic attack; each additional alkyl group around that carbon adds steric hindrance and slows the reaction. Applying this to the four compounds gives:

I > II > III > IV, i.e. CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl.

CH3Cl (no substituents on the reacting carbon) is most reactive; (CH3)3CCl, with three methyl groups crowding the carbon, is so hindered that it essentially does not react by the SN2 pathway at all (it instead reacts, if at all, via SN1).

Q4 2-Bromopentane, on treatment with alcoholic KOH, gives predominantly which alkene as the major product? Explain using the appropriate rule. 2024 4 marks

2-Bromopentane is CH3-CHBr-CH2-CH2-CH3. Alcoholic KOH promotes E2 elimination, and a beta-hydrogen can be removed from either C-1 (giving pent-1-ene) or C-3 (giving pent-2-ene):

  • Removing a beta-H from C-1: CH2=CH-CH2-CH2-CH3 (pent-1-ene, a monosubstituted/terminal alkene) — minor product.
  • Removing a beta-H from C-3: CH3-CH=CH-CH2-CH3 (pent-2-ene, a disubstituted, more highly substituted alkene) — major product.

By Saytzeff's rule, the more highly substituted (and hence more stable) alkene is the major elimination product, so pent-2-ene (predominantly its more stable E/trans isomer) is formed as the major product, with pent-1-ene as the minor product.

Part of Priodemy for School

Interactive Maths & Science — free with every school on Priodemy EduSuite. Explore more chapters and labs on the Priodemy for School hub.

Ask AI