Haloalkanes and Haloarenes • Topic 1 of 3

Nomenclature & Preparation

Replace one or more hydrogen atoms of a hydrocarbon by a halogen atom (F, Cl, Br, I) and you get a halogen derivative. When the halogen is bonded to an sp3 carbon of an open chain or an aliphatic ring, the compound is a haloalkane (alkyl halide); when it is bonded directly to an sp2 carbon of an aromatic ring, it is a haloarene (aryl halide). This single structural difference controls almost everything in the chapter, so classify the carbon first.

Classification

By the number of halogen atoms: mono- (one X), di- (two X) and poly-halogen compounds. By the hybridisation and position of the C bearing X:

  • Alkyl halides — X on sp3 carbon; further called 1° (primary), 2° (secondary) or 3° (tertiary) depending on how many carbons are attached to the C–X carbon. CH3CH2Cl is 1°, (CH3)2CHCl is 2°, (CH3)3CCl is 3°.
  • Allylic halides — X on an sp3 carbon next to a C=C (e.g. CH2=CH–CH2Cl, allyl chloride).
  • Benzylic halides — X on an sp3 carbon attached to a benzene ring (C6H5CH2Cl, benzyl chloride).
  • Vinylic halides — X on an sp2 carbon of C=C (CH2=CHCl, vinyl chloride).
  • Aryl halides — X on an sp2 carbon of an aromatic ring (C6H5Cl, chlorobenzene).

Vinylic and aryl halides resist nucleophilic substitution; allylic and benzylic halides are unusually reactive because the intermediate cation is resonance-stabilised. So this classification is not decoration — it predicts reactivity.

IUPAC nomenclature

Treat halogen as a substituent prefix (fluoro, chloro, bromo, iodo) on the parent alkane. Number the chain to give the lowest locant set; when there is a choice, the first point of difference decides. List substituents alphabetically. Thus CH3CH2CH2Br is 1-bromopropane and (CH3)2CHCl is 2-chloropropane. Dihalides on the same carbon are gem-dihalides (1,1-) and on adjacent carbons are vic-dihalides (1,2-). For arenes the halogen and its position are named directly: 1-chloro-2-methylbenzene (o-chlorotoluene).

Nature of the C–X bond

Halogens are more electronegative than carbon, so the C–X bond is polar: carbon carries a partial positive charge (δ+) and the halogen δ−. The electrophilic carbon is the site nucleophiles attack. Down the group the C–X bond lengthens and weakens (C–F > C–Cl > C–Br > C–I in bond enthalpy), which is why iodides are the most reactive and fluorides the least in substitution.

Methods of preparation

From alcohols: R–OH reacts with HX (reactivity HI > HBr > HCl; HCl needs anhydrous ZnCl2 for 1°/2°), or better with PCl3, PCl5 or SOCl2. Thionyl chloride (the Darzens procedure) is preferred because the by-products SO2 and HCl are gases and the product is pure.

From hydrocarbons by halogenation: free-radical halogenation of alkanes gives a mixture; allylic/benzylic positions can be selectively brominated. Addition of HX to alkenes follows Markovnikov’s rule (H to the carbon with more H’s), but with peroxides HBr adds anti-Markovnikov (peroxide / Kharasch effect, free-radical). Addition of X2 to alkenes gives vic-dihalides.

Halide exchange: the Finkelstein reaction converts R–Cl/R–Br to R–I using NaI in dry acetone (NaCl/NaBr precipitate, driving the equilibrium). The Swarts reaction makes alkyl fluorides by heating R–X with metallic fluorides such as AgF, Hg2F2, CoF2 or SbF3.

Haloarenes from diazonium salts (Sandmeyer / Balz–Schiemann): arylamines diazotised to ArN2+X− give chloro/bromo arenes with Cu2Cl2/Cu2Br2 (Sandmeyer) and aryl fluorides via the fluoroborate (Balz–Schiemann); aryl iodides form directly with KI.

Physical properties

Haloalkanes are colourless when pure. Boiling points rise with molar mass (more electrons → stronger van der Waals forces) and fall with chain branching. For the same alkyl group, b.p. order is RI > RBr > RCl > RF. They are heavier than water (density rises down the group) and essentially insoluble in water because they cannot form strong H-bonds with it.

Haloalkane classification and preparation routesClassification by C-X carbon1° / 2° / 3°alkyl (sp3)allyl / benzylsp3, resonancevinyl (sp2)unreactivearyl (sp2)unreactiveKey preparation routesR-OH + SOCl2→R-Cl + SO2 + HClR-Cl + NaI (dry acetone)→R-I + NaCl ↓ (Finkelstein)R-X + AgF→R-F + AgX (Swarts)CH2=CHCH3 + HBr/peroxide→CH3CH2CH2Br (anti-Markovnikov)ArN2+Cl- + Cu2Cl2→Ar-Cl + N2 (Sandmeyer)
Solved Examples
1
Worked Example
Give the IUPAC name and classify (1°/2°/3°): (CH3)3C–CH2Br.
Solution
  1. The longest chain containing the C–Br carbon has 4 carbons (a butane skeleton): the bromomethyl carbon plus the three-carbon framework gives 1-bromo-2,2-dimethylpropane.
  2. Number from the Br end so Br gets locant 1 and the two methyls sit on C2.
  3. The carbon bearing Br is attached to only one other carbon, so it is a primary (1°) halide (the neighbouring quaternary carbon does not change this classification).

Answer: 1-bromo-2,2-dimethylpropane (neopentyl bromide), a primary halide.

2
Worked Example
Why is thionyl chloride (SOCl2) preferred over PCl5 for converting an alcohol to an alkyl chloride?
Solution
  1. SOCl2 reacts: R–OH + SOCl2 → R–Cl + SO2↑ + HCl↑.
  2. Both by-products (SO2 and HCl) are gases that escape, so the alkyl chloride is obtained pure and easily isolated.
  3. With PCl5 the by-product POCl3 is a liquid that contaminates the product and must be separated.

Answer: Because SOCl2 gives gaseous by-products (SO2, HCl) that escape, yielding a pure product, whereas PCl5 leaves liquid POCl3 behind.

3
Worked Example
Predict the major product when propene reacts with HBr (a) in the absence and (b) in the presence of benzoyl peroxide.
Solution
  1. (a) Without peroxide the addition is ionic and follows Markovnikov’s rule: H+ adds to the terminal CH2 (more H’s), Br− to the central carbon, via the more stable 2° carbocation.
  2. So the product is 2-bromopropane, (CH3)2CHBr.
  3. (b) With peroxide HBr adds by a free-radical chain (the more stable 2° radical forms), so Br adds to the terminal carbon — anti-Markovnikov.
  4. Product is 1-bromopropane, CH3CH2CH2Br. (This peroxide effect is seen only with HBr, not HCl or HI.)

Answer: (a) 2-bromopropane (Markovnikov); (b) 1-bromopropane (anti-Markovnikov, peroxide effect).

4
Worked Example
How would you convert 1-chlorobutane into 1-iodobutane? Name the reaction and explain why it works.
Solution
  1. Use the Finkelstein reaction: CH3CH2CH2CH2Cl + NaI →(dry acetone) CH3CH2CH2CH2I + NaCl.
  2. NaI is soluble in acetone but NaCl is not, so NaCl precipitates out.
  3. Removing NaCl shifts the equilibrium to the right (Le Chatelier), driving the conversion to completion.

Answer: Treat 1-chlorobutane with NaI in dry acetone (Finkelstein reaction); NaCl precipitates and pushes the equilibrium toward 1-iodobutane.

5
Worked Example
Arrange CH3Cl, CH3Br and CH3I in increasing order of boiling point and explain.
Solution
  1. All three are similar in shape; the difference is molar mass and the number of electrons (polarisability).
  2. Heavier halogen → more electrons → stronger London dispersion forces → higher boiling point.
  3. So CH3Cl < CH3Br < CH3I.

Answer: CH3Cl < CH3Br < CH3I, because boiling point rises with increasing molar mass and polarisability.

6
Worked Example
Starting from aniline, outline how chlorobenzene is prepared. Why is direct chlorination of benzene not used to introduce a -Cl exactly where needed?
Solution
  1. Diazotise aniline with NaNO2 + HCl at 0–5°C: C6H5NH2 → C6H5N2+Cl− (benzenediazonium chloride).
  2. Treat with cuprous chloride (Cu2Cl2/HCl): C6H5N2+Cl− → C6H5Cl + N2↑ (Sandmeyer reaction).
  3. The diazonium route places Cl exactly where the –NH2 was, giving a clean single product; direct chlorination is fine for chlorobenzene itself but the diazonium method gives positional control for substituted arenes.

Answer: Aniline → benzenediazonium chloride (NaNO2/HCl, 0–5°C) → chlorobenzene (Cu2Cl2, Sandmeyer); the diazonium route fixes the halogen position precisely.

Key Points

  • Alkyl halides have X on sp3 C (1°/2°/3°); allyl/benzyl are resonance-activated; vinyl/aryl have X on sp2 C and are unreactive to substitution.
  • IUPAC: halogen is a prefix (fluoro/chloro/bromo/iodo); number for the lowest locant set and list substituents alphabetically.
  • The C–X bond is polar (C is δ+); bond strength falls C–F > C–Cl > C–Br > C–I, so iodides are most reactive.
  • Preparation: from alcohols (HX, PCl3/PCl5, SOCl2), from alkenes (Markovnikov HX; anti-Markovnikov HBr/peroxide), Finkelstein (NaI/acetone → R–I), Swarts (AgF → R–F).
  • Haloarenes from diazonium salts: Sandmeyer (Cu2X2) for Cl/Br, Balz–Schiemann (via fluoroborate) for F; b.p. order RI > RBr > RCl > RF and all are denser than water.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.Which of the following is a tertiary (3°) alkyl halide?
Explanation: In (CH3)3CCl the carbon bearing Cl is attached to three other carbons, so it is tertiary. The first is 1°, the second 2°, and vinyl chloride is a vinylic halide.
Q2.The Finkelstein reaction (R–Cl + NaI in dry acetone → R–I + NaCl) goes to completion mainly because:
Explanation: NaI dissolves in acetone but NaCl does not; precipitation of NaCl removes product from the equilibrium and shifts it to the right (Le Chatelier).
Q3.Addition of HBr to propene in the presence of benzoyl peroxide gives mainly:
Explanation: Peroxide triggers a free-radical (anti-Markovnikov) addition through the more stable secondary radical, so Br adds to the terminal carbon giving 1-bromopropane.
Q4.The correct order of boiling points is:
Explanation: Boiling point rises with molar mass and polarisability (stronger London forces): CH3Cl < CH3Br < CH3I.
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