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ISC 2027
All chaptersChemistry · Unit 6

Haloalkanes and Haloarenes

10 articles28 formulas57 ways the board asks it
CHEClassification & Preparation

Preparation & Conversions

Haloalkanes are made from alcohols, from alkanes (free-radical halogenation), and from alkenes (addition of HX or halogen); haloarenes come from arenes or via diazonium salts. "Convert X to Y" questions test whether you can pick the right reagent for each transformation.

Alcohol to halide with SOCl2SOCl_2
R-OH+SOCl2→R-Cl+SO2 ⁣↑+HCl ⁣↑R\text{-}OH + SOCl_2 \rightarrow R\text{-}Cl + SO_2\!\uparrow + HCl\!\uparrow
Darzens method; gaseous by-products leave a pure halide
Finkelstein reaction
R-Cl+NaI→dry acetoneR-I+NaCl ⁣↓R\text{-}Cl + NaI \xrightarrow{\text{dry acetone}} R\text{-}I + NaCl\!\downarrow
NaClNaCl is insoluble in acetone and precipitates, driving the equilibrium right
Swarts reaction
R-Cl  (or R-Br)+AgF→R-F+AgClR\text{-}Cl \;(\text{or } R\text{-}Br) + AgF \rightarrow R\text{-}F + AgCl
Hg2F2Hg_2F_2, CoF2CoF_2 or SbF3SbF_3 may replace AgFAgF to make fluoroalkanes
Markovnikov vs peroxide addition
CH3CH=CH2+HBr→CH3CHBrCH3;→peroxideCH3CH2CH2BrCH_3CH{=}CH_2 + HBr \rightarrow CH_3CHBrCH_3 \quad ; \quad \xrightarrow{\text{peroxide}} CH_3CH_2CH_2Br
Markovnikov gives 2∘2^\circ halide; peroxide (anti-Markovnikov) gives 1∘1^\circ halide, only with HBrHBr
  • From alcohols: R-OHR\text{-}OH with HXHX, PCl5PCl_5, PCl3PCl_3 or SOCl2SOCl_2. SOCl2SOCl_2 is best — by-products SO2SO_2 and HClHCl escape as gases, leaving a pure halide.
  • Reactivity of HX with a given alcohol: HI>HBr>HClHI > HBr > HCl; reactivity of alcohols with a given HX: 3∘>2∘>1∘3^\circ > 2^\circ > 1^\circ.
  • Addition of HX to alkenes follows Markovnikov's rule; anti-Markovnikov (peroxide / Kharasch effect) occurs only with HBr.
  • Finkelstein reaction: R-Cl+NaI→dry acetoneR-I+NaCl ⁣↓R\text{-}Cl + NaI \xrightarrow{\text{dry acetone}} R\text{-}I + NaCl\!\downarrow — NaCl precipitates and drives the equilibrium forward.
  • Swarts reaction: R-Cl/Br+AgF  (or Hg2F2)→R-FR\text{-}Cl/Br + AgF \;(\text{or } Hg_2F_2) \rightarrow R\text{-}F, used to make fluoroalkanes.
  • Free-radical halogenation of alkanes gives mixtures; chlorination is less selective than bromination.
  • Aryl halides are made via diazonium salts (Sandmeyer / Gattermann), since direct halogenation of benzene needs a Lewis-acid carrier such as FeCl3FeCl_3.
  • From alkenes, adding X2X_2 (Cl2Cl_2 or Br2Br_2) gives a vicinal dihalide; decolourisation of bromine water is the classic alkene test.
  • Concentrated HClHCl with 1∘1^\circ/2∘2^\circ alcohols needs anhydrous ZnCl2ZnCl_2 (Lucas reagent) as catalyst; 3∘3^\circ alcohols react with conc. HCl alone.
  • Sandmeyer: C6H5N2+Cl−→CuCl/HClC6H5ClC_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl (and CuBr/HBrCuBr/HBr for bromobenzene); Gattermann uses CuCu powder with HXHX.
  • Iodobenzene is made directly from the diazonium salt with KIKI (no copper catalyst needed); fluorobenzene via the Balz–Schiemann route (HBF4HBF_4, then heat).
  • Free-radical halogenation: bromination is more selective for the more substituted carbon (3∘>2∘>1∘3^\circ > 2^\circ > 1^\circ H abstraction) because the bromine radical is less reactive and more selective.
C2H5OH+SOCl2→pyridineC2H5Cl+SO2+HCl\mathrm{C_2H_5OH} + \mathrm{SOCl_2} \xrightarrow{\text{pyridine}} \mathrm{C_2H_5Cl} + \mathrm{SO_2} + \mathrm{HCl}
ConditionsThionyl chloride with pyridine, warmed
The examiner's favourite 'why is this the best reagent' answer: both by-products, SO2\mathrm{SO_2} and HCl\mathrm{HCl}, are gases that escape, so the alkyl halide is left pure and no separation is needed.
C6H5N2+Cl−→CuCl/HClC6H5Cl+N2\mathrm{C_6H_5N_2^{+}Cl^{-}} \xrightarrow{\mathrm{CuCl}/\mathrm{HCl}} \mathrm{C_6H_5Cl} + \mathrm{N_2}
ConditionsSandmeyer reaction — diazonium salt with cuprous halide
Aryl halides cannot be made by direct substitution on phenol, so the diazonium route is the only way in. With KI\mathrm{KI} no copper salt is needed.
2 CHI3+6 Ag→ΔC2H2+6 AgI2\,\mathrm{CHI_3} + 6\,\mathrm{Ag} \xrightarrow{\Delta} \mathrm{C_2H_2} + 6\,\mathrm{AgI}
ConditionsIodoform heated with silver powder
Asked as a conversion in both 2024 and 2026 — worth knowing as a one-liner.
C6H5Cl+2 NH3→high pressureCu2O, 475 KC6H5NH2+NH4Cl\mathrm{C_6H_5Cl} + 2\,\mathrm{NH_3} \xrightarrow[\text{high pressure}]{\mathrm{Cu_2O},\ 475\ \mathrm{K}} \mathrm{C_6H_5NH_2} + \mathrm{NH_4Cl}
ConditionsAqueous ammonia, Cu2O\mathrm{Cu_2O}, high temperature and pressure
The harsh conditions are the point: a haloarene will not give up its halogen under ordinary nucleophilic conditions.
Where the marks go
  • Applying the peroxide (anti-Markovnikov) effect to HClHCl or HIHI — it works only with HBrHBr.
  • Forgetting "dry acetone" in Finkelstein or "dry ether" elsewhere; moisture ruins these reactions and forfeits marks.
  • Trying to make aryl halides by direct nucleophilic substitution on benzene; you need a Lewis-acid carrier (FeCl3FeCl_3) for electrophilic halogenation, or a diazonium route.
  • Mixing up the two reactivity orders for HX/alcohol: HI is most reactive among acids, while 3∘3^\circ is most reactive among alcohols.
  • Quoting AgFAgF but writing the product as an iodide or chloride; Swarts specifically installs fluorine.
How the board asks it
  • Conversionreagents for each preparation route2 mkAsked 2024 · 2026
    How will you convert (i)(i) propan-22-ol into 22-iodopropane, (ii)(ii) ethanol into chloroethane using SOCl2SOCl_2, and (iii)(iii) but-11-ene into 11-bromobutane?
  • Give reasonsSOCl2SOCl_2 as best reagent; reactivity order of HXHX
    Account for the following: (i)(i) SOCl2SOCl_2 is preferred over PCl5PCl_5 for preparing pure alkyl chlorides from alcohols, and (ii)(ii) among HClHCl, HBrHBr and HIHI, HIHI reacts fastest with a given alcohol.
  • Predict the productnamed reactions: finkelstein, swarts, sandmeyer
    What happens when (give balanced equations): (i)(i) chloroethane is heated with NaINaI in dry acetone, (ii)(ii) chloromethane is treated with AgFAgF, and (iii)(iii) benzene diazonium chloride is warmed with CuCl/HClCuCl/HCl?
  • Conversiondiazonium route to aryl halides2 mkAsked 2026
    Starting from aniline, give the reagents and conditions required to prepare (i)(i) chlorobenzene, (ii)(ii) iodobenzene and (iii)(iii) fluorobenzene.
  • Give reasonsmarkovnikov vs peroxide (anti-markovnikov) addition
    Name the major product formed when propene reacts with HBrHBr (i)(i) in the absence and (ii)(ii) in the presence of organic peroxide, and give reasons for the difference.
  • Conversionpicking reagents from a fixed list2 mk
    Given only alcoholic and aqueous KOH\mathrm{KOH}, Cl2\mathrm{Cl_2}, HBr\mathrm{HBr}, acidified K2Cr2O7\mathrm{K_2Cr_2O_7}, NaOH\mathrm{NaOH} and soda lime, carry out a stated pair of conversions.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.