Home / Notes / Class 12 / Chemistry / Alkyl halides

Alkyl halides: Class 12 Chemistry

Written by Dr. Ayesha Nadeem (Chemistry)

Classification and naming of alkyl halides, methods of preparation, physical properties, nucleophilic substitution (SN1 and SN2), elimination reactions, Grignard reagents and common uses with environmental concerns.

Key points

Alkyl halides (haloalkanes) have the general formula R-X, where X is F, Cl, Br or I attached to an sp³ carbon. They are classified by the carbon holding the halogen. In a primary halide the carbon is bonded to one other carbon (CH₃CH₂Br). In a secondary halide it is bonded to two (CH₃CHBrCH₃). In a tertiary halide it is bonded to three ((CH₃)₃CBr).

IUPAC names treat the halogen as a prefix on the longest chain: CH₃CH₂Cl is chloroethane, CH₃CHBrCH₃ is 2-bromopropane and (CH₃)₃CCl is 2-chloro-2-methylpropane. Common names use alkyl and halide, such as ethyl chloride and isopropyl bromide.

Preparation of alkyl halides:

  • From alcohols with HX (with ZnCl₂ for HCl), PCl₅, PCl₃ or thionyl chloride. R-OH + SOCl₂ → R-Cl + SO₂ + HCl is very convenient because the by-products are gases.
  • By addition of HX or X₂ to alkenes, as in CH₂=CH₂ + HBr → CH₃CH₂Br.
  • By free radical halogenation of alkanes, which gives a mixture of products and is therefore not a good laboratory method.
  • Iodides by halogen exchange (Finkelstein reaction): R-Cl + NaI in dry acetone → R-I + NaCl.

The C-X bond is polar because halogens are more electronegative than carbon. The carbon carries a partial positive charge, so it is attacked by nucleophiles. Boiling points are higher than those of alkanes with the same carbon number because of dipole forces and greater mass. They rise from fluoride to iodide and fall with branching. Alkyl halides are insoluble in water because they cannot form hydrogen bonds with it, but they dissolve in organic solvents.

Bond strength falls in the order C-F > C-Cl > C-Br > C-I. So reactivity in substitution rises from chlorides to iodides. This is why iodoalkanes react fastest.

Nucleophilic substitution (a nucleophile replaces X):

  • With aqueous KOH it gives an alcohol: R-X + OH⁻ → R-OH + X⁻.
  • With alcoholic KCN it gives a nitrile R-CN, which adds one carbon to the chain.
  • With ammonia it gives an amine and with NaOR it gives an ether (Williamson synthesis).
  • With aqueous or alcoholic AgNO₃ it gives a precipitate of AgX (white AgCl, cream AgBr, yellow AgI), which is used to identify the halogen.

The SN2 mechanism is a one-step reaction. The nucleophile attacks from the side opposite to the leaving group, and the rate depends on both the halide and the nucleophile (second order). It is fastest for primary halides and gives inversion of configuration. The SN1 mechanism has two steps. The halide first ionises slowly to a carbocation, and the nucleophile then attacks it. The rate depends only on the halide (first order). It is fastest for tertiary halides, because their carbocations are most stable, and it gives a mixture of both configurations (racemisation).

Elimination (dehydrohalogenation): heating an alkyl halide with alcoholic KOH removes HX and forms an alkene. CH₃CH₂Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O. Aqueous KOH favours substitution while alcoholic KOH favours elimination. When two alkenes are possible, Saytzeff's rule says the more substituted alkene is the major product.

Grignard reagents are made by reacting an alkyl halide with magnesium in dry ether: R-X + Mg → R-MgX. The carbon bonded to Mg behaves as a carbanion and is a strong nucleophile and base. Water must be absent, since R-MgX + H₂O → R-H + Mg(OH)X. Reaction with carbon dioxide followed by acid gives a carboxylic acid, and reaction with aldehydes and ketones gives alcohols.

Uses: solvents (dichloromethane, chloroform), starting materials in synthesis, anaesthetics in the past (chloroform), refrigerants and fire extinguishers. CFCs (chlorofluorocarbons) damaged the ozone layer, because UV light breaks the C-Cl bond to release chlorine radicals that destroy ozone, so their use is now restricted. Some halogenated pesticides such as DDT persist in the environment and build up in food chains.

Study tip: Make one page with the reagent chart: aqueous KOH, alcoholic KOH, KCN, NH₃, Mg in ether, AgNO₃. Most reaction questions come from this chart. In mechanism answers show the curved arrows and the inversion or carbocation clearly, and state which halide class (primary or tertiary) suits which mechanism.

Definitions and flashcards

Say the meaning first, then open the card to check.

Alkyl halide

A compound R-X in which a halogen atom replaces a hydrogen of an alkane.

Nucleophile

A species with a lone pair that attacks an electron-poor carbon.

Leaving group

The atom or group that departs with the bonding electrons in a substitution.

SN2

Bimolecular substitution in a single step with inversion of configuration.

SN1

Unimolecular substitution through a carbocation intermediate.

Dehydrohalogenation

Removal of HX from an alkyl halide to form an alkene.

Grignard reagent

An organomagnesium halide R-MgX made in dry ether.

Racemisation

Formation of equal amounts of two mirror-image forms.

Short questions with model answers

These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.

Why are alkyl halides insoluble in water although they are polar? (2 marks)

To dissolve, they would have to break the strong hydrogen bonds between water molecules. The attraction between alkyl halide and water molecules is weaker, so the energy is not repaid and they stay insoluble.

Why is the order of reactivity RI > RBr > RCl in nucleophilic substitution? (2 marks)

The C-I bond is longest and weakest and the iodide ion is the best leaving group. The C-Cl bond is shorter and stronger. So iodides react fastest and chlorides slowest.

Distinguish between SN1 and SN2 reactions. (3 marks)

SN2 is a one-step process with first-order dependence on each reactant, so it is second order overall. It goes with inversion and is favoured by primary halides. SN1 is a two-step process through a carbocation, is first order, gives racemisation and is favoured by tertiary halides.

What happens when chloroethane is heated with aqueous KOH and with alcoholic KOH? (3 marks)

With aqueous KOH substitution occurs and ethanol is formed: C₂H₅Cl + KOH → C₂H₅OH + KCl. With alcoholic KOH elimination occurs and ethene is formed: C₂H₅Cl + KOH → C₂H₄ + KCl + H₂O.

How is a Grignard reagent prepared and why must the apparatus be dry? (3 marks)

It is prepared by reacting an alkyl halide with magnesium turnings in dry ether: R-X + Mg → R-MgX. The reagent is a strong base. Any water would react with it to give an alkane, R-MgX + H₂O → R-H + Mg(OH)X, so the reagent would be destroyed and no useful product would form.

Explain the harm done by chlorofluorocarbons. (5 marks)

CFCs are stable and rise to the stratosphere. UV light breaks the C-Cl bond and releases chlorine radicals. A radical reacts with ozone: Cl· + O₃ → ClO· + O₂, and then ClO· + O → Cl· + O₂. The chlorine radical is regenerated, so one atom destroys many ozone molecules. The thinner ozone layer lets more harmful UV reach the earth, which increases skin cancer and cataracts.

Numericals with worked steps

Calculate the mass of chloromethane formed when 16 g of methane reacts completely with chlorine by substitution to give only CH₃Cl. (C = 12, H = 1, Cl = 35.5)
  1. CH₄ + Cl₂ → CH₃Cl + HCl.
  2. Moles of CH₄ = 16 / 16 = 1 mol.
  3. Ratio 1 : 1, so 1 mol CH₃Cl forms.
  4. Molar mass of CH₃Cl = 12 + 3 + 35.5 = 50.5 g/mol.
  5. Mass = 1 × 50.5 = 50.5 g.

Answer: 50.5 g of CH₃Cl

23 g of ethanol reacts with excess HBr to give bromoethane in 70% yield. Find the mass of bromoethane obtained. (C = 12, H = 1, O = 16, Br = 80)
  1. C₂H₅OH + HBr → C₂H₅Br + H₂O.
  2. Molar mass of ethanol = 46 g/mol; moles = 23 / 46 = 0.5 mol.
  3. Ratio 1 : 1, so theoretical moles of C₂H₅Br = 0.5 mol.
  4. Molar mass of C₂H₅Br = 24 + 5 + 80 = 109 g/mol; theoretical mass = 0.5 × 109 = 54.5 g.
  5. Actual mass = 54.5 × 70 / 100 = 38.15 g.

Answer: 38.15 g of bromoethane

Calculate the percentage of chlorine by mass in chloroform CHCl₃. (C = 12, H = 1, Cl = 35.5)
  1. Molar mass = 12 + 1 + 3 × 35.5 = 12 + 1 + 106.5 = 119.5 g/mol.
  2. Mass of chlorine = 106.5 g.
  3. Percentage = 106.5 / 119.5 × 100 = 89.1%.

Answer: 89.1% chlorine

14 g of ethene reacts completely with HBr. Find the mass of bromoethane formed. (C = 12, H = 1, Br = 80)
  1. CH₂=CH₂ + HBr → CH₃CH₂Br.
  2. Molar mass of ethene = 28 g/mol; moles = 14 / 28 = 0.5 mol.
  3. Ratio 1 : 1, so 0.5 mol of C₂H₅Br forms.
  4. Mass = 0.5 × 109 = 54.5 g.

Answer: 54.5 g of bromoethane

Common mistakes

  • Using aqueous KOH when elimination is wanted. Alcoholic KOH gives alkenes and aqueous KOH gives alcohols.
  • Saying SN1 reactions are fastest for primary halides. SN1 prefers tertiary and SN2 prefers primary.
  • Writing the Grignard reagent as R-Mg instead of R-MgX, or preparing it in the presence of water.
  • Thinking C-F is the most reactive bond because fluorine is the most electronegative. The C-F bond is the strongest.
  • Mixing up the colours of silver halide precipitates: AgCl is white, AgBr cream and AgI yellow.

Practise this chapter

Practice questions and timed test Printable worksheet

Previous: Hydrocarbons Next: Alcohols and phenols