Cambridge AS & A Level9701

Halogen compounds

Chemistry 9701 Chapter Notes

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Halogen compounds
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In this chapter

  1. 1Introduction to Halogenoalkanes
  2. 2Nucleophilic Substitution Reactions
  3. 3SN1 and SN2 Mechanisms
  4. 4Reactivity Trends in Halogenoalkanes
  5. 5Elimination Reactions
  6. 6Uses and Environmental Concerns

1. Introduction to Halogenoalkanes

Halogenoalkanes, also known as haloalkanes, are organic compounds derived from alkanes by replacing one or more hydrogen atoms with halogen atoms (F, Cl, Br, I). Their general formula for a monohalogenoalkane is CnH2n+1X, where X is a halogen. The bond between carbon and the halogen (the C-X bond) is polar because halogens are more electronegative than carbon. This creates a partial positive charge (δ+) on the carbon atom and a partial negative charge (δ-) on the halogen atom. This polar nature is the key to their reactivity, making the electron-deficient carbon atom a target for attack by species rich in electrons.

General Formula: CnH2n+1X (where X = F, Cl, Br, or I)

Key term

Halogenoalkane: An alkane in which one or more hydrogen atoms have been substituted by halogen atoms.

Common pitfall

Forgetting that the classification (primary, secondary, tertiary) depends on the number of carbon atoms attached to the carbon bearing the halogen, not the number of hydrogens.

Worked example 13 marks

Classify the following halogenoalkanes as primary (1°), secondary (2°), or tertiary (3°):(a) 1-chlorobutane,(b) 2-chlorobutane,(c) 2-chloro-2-methylpropane.

  1. 1

    Step 1: Understand the classification. A primary halogenoalkane has the halogen on a carbon bonded to one other carbon atom. A secondary has the halogen on a carbon bonded to two other carbons. A tertiary has the halogen on a carbon bonded to three other carbons.

  2. 2

    Step 2: Draw the structure of 1-chlorobutane (CH3CH2CH2CH2Cl). The Cl atom is on a carbon bonded to only one other carbon atom. Therefore, it is a primary (1°) halogenoalkane.

  3. 3

    Step 3: Draw the structure of 2-chlorobutane (CH3CHClCH2CH3). The Cl atom is on a carbon bonded to two other carbon atoms. Therefore, it is a secondary (2°) halogenoalkane.

  4. 4

    Step 4: Draw the structure of 2-chloro-2-methylpropane ((CH3)3CCl). The Cl atom is on a carbon bonded to three other carbon atoms. Therefore, it is a tertiary (3°) halogenoalkane.

Recap

  • Halogenoalkanes contain a polar carbon-halogen (C-X) bond.
  • The carbon atom in the C-X bond carries a partial positive charge (δ+).
  • The halogen atom in the C-X bond carries a partial negative charge (δ-).
  • They are classified as primary (1°), secondary (2°), or tertiary (3°) based on the carbon atom the halogen is attached to.

Quick check

  1. Draw the skeletal formula for 2-bromopropane and state whether it is primary, secondary or tertiary.2 marks

2. Nucleophilic Substitution Reactions

The δ+ carbon atom in a halogenoalkane is an electron-deficient centre, making it susceptible to attack by nucleophiles. A nucleophile is an electron-pair donor. In a nucleophilic substitution reaction, the nucleophile attacks the δ+ carbon, and the halogen atom is displaced, leaving as a halide ion (X-). The halide ion is known as the leaving group. Common nucleophiles include the hydroxide ion (OH-), the cyanide ion (CN-), and ammonia (NH3). The conditions determine the product: aqueous conditions with gentle warming favour substitution.

General Equation: Nu:⁻ + R-X → R-Nu + X⁻

Hydrolysis (forms an alcohol): R-X + OH⁻(aq) → R-OH + X⁻

Formation of a nitrile: R-X + CN⁻(ethanolic) → R-CN + X⁻

Formation of a primary amine: R-X + 2NH₃(ethanolic) → R-NH₂ + NH₄⁺X⁻

Key term

Nucleophile: A species with a lone pair of electrons that it can donate to an electron-deficient centre (a nucleus lover).

Worked example 13 marks

1-bromopropane is warmed with aqueous sodium hydroxide.(a) Name the type of reaction.(b) Write an equation for the reaction.(c) Name the organic product.

  1. 1

    Step 1 (a): Identify the reactants. A halogenoalkane (1-bromopropane) reacts with a nucleophile (OH-) in aqueous solution. This is a nucleophilic substitution reaction.

  2. 2

    Step 2 (b): Write the equation. The bromine atom is substituted by the -OH group. The formula for 1-bromopropane is CH3CH2CH2Br. The equation is: CH3CH2CH2Br + NaOH → CH3CH2CH2OH + NaBr.

  3. 3

    Step 3 (c): Name the product. The product CH3CH2CH2OH has a three-carbon chain with an -OH group on the first carbon. The name is propan-1-ol.

Worked example 22 marks

What would be the organic product when chloroethane reacts with excess ammonia in a sealed tube under heat?

  1. 1

    Step 1: Identify the reaction. This is a nucleophilic substitution where ammonia (NH3) acts as the nucleophile.

  2. 2

    Step 2: Write the overall equation. The chlorine atom is replaced by an amino (-NH2) group. Two moles of ammonia are needed: one acts as the nucleophile and the second reacts with the H+ released to form an ammonium salt. The equation is: CH3CH2Cl + 2NH3 → CH3CH2NH2 + NH4Cl.

  3. 3

    Step 3: Name the product. The organic product is CH3CH2NH2, which is named ethylamine.

Recap

  • Nucleophilic substitution involves an electron-pair donor (nucleophile) attacking the δ+ carbon of a halogenoalkane.
  • The halogen atom is displaced as a halide ion (the leaving group).
  • Reaction with aqueous OH⁻ forms an alcohol (hydrolysis).
  • Reaction with KCN in ethanol forms a nitrile, extending the carbon chain by one carbon.
  • Reaction with excess ammonia in ethanol forms a primary amine.

Quick check

  1. What is the organic product of the reaction between iodoethane and potassium cyanide?1 mark

3. SN1 and SN2 Mechanisms

Nucleophilic substitution can occur via two different mechanisms, SN1 and SN2. The mechanism depends on the structure of the halogenoalkane. SN2 (Substitution Nucleophilic Bimolecular): This is a one-step mechanism favoured by primary halogenoalkanes. The nucleophile attacks the δ+ carbon from the opposite side to the halogen (backside attack). This leads to a high-energy transition state where the carbon is weakly bonded to both the nucleophile and the leaving group, before the C-X bond breaks. The rate of reaction depends on the concentration of both the halogenoalkane and the nucleophile: Rate = k[R-X][Nu⁻]. SN1 (Substitution Nucleophilic Unimolecular): This is a two-step mechanism favoured by tertiary halogenoalkanes. The first and slowest (rate-determining) step is the C-X bond breaking to form a stable carbocation intermediate. In the second step, the nucleophile rapidly attacks the planar carbocation. The rate depends only on the concentration of the halogenoalkane: Rate = k[R-X]. Secondary halogenoalkanes can react by a mixture of both mechanisms.

SN2 Rate Law: Rate = k[R-X][Nu⁻]

SN1 Rate Law: Rate = k[R-X]

Key term

Carbocation: An ion with a positively charged carbon atom, which is a key intermediate in the SN1 mechanism.

Examiner insight

Examiners expect clear, correctly drawn curly arrows originating from a lone pair or bond and pointing to the atom being attacked or bond being formed. Marks are often lost for incorrect or ambiguous arrows.

Worked example 14 marks

Draw the mechanism for the reaction of hydroxide ions with bromoethane. Name the mechanism.

  1. 1

    Step 1: Identify the halogenoalkane type. Bromoethane (CH3CH2Br) is a primary halogenoalkane, so it will react via an SN2 mechanism.

  2. 2

    Step 2: Draw the reactants. Show bromoethane with the C-Br bond polarity (Cδ+ and Brδ-). Show the hydroxide nucleophile (OH-) with a lone pair and negative charge.

  3. 3

    Step 3: Draw the curly arrow for the attack. A curly arrow goes from the lone pair on the oxygen of OH- to the δ+ carbon atom.

  4. 4

    Step 4: Draw the second curly arrow. A second curly arrow goes from the C-Br bond to the Br atom, showing the bond breaking and the bromide ion leaving.

  5. 5

    Step 5: Draw the products. The products are ethanol (CH3CH2OH) and a bromide ion (Br-). The mechanism shows both arrows in a single step.

  6. 6

    Step 6: Name the mechanism. This is an SN2 (Substitution Nucleophilic Bimolecular) mechanism.

Worked example 23 marks

Explain why 2-chloro-2-methylpropane undergoes hydrolysis via an SN1 mechanism.

  1. 1

    Step 1: Identify the halogenoalkane. 2-chloro-2-methylpropane is a tertiary halogenoalkane.

  2. 2

    Step 2: Explain the first step of SN1. Tertiary halogenoalkanes react via SN1 because the first step is the formation of a carbocation intermediate: (CH3)3C-Cl → (CH3)3C+ + Cl-.

  3. 3

    Step 3: Explain the stability of the intermediate. The tertiary carbocation ((CH3)3C+) is stabilised by the positive inductive effect of the three methyl groups, which donate electron density towards the positive charge. This stability makes its formation feasible.

  4. 4

    Step 4: Contrast with SN2. An SN2 attack is not possible due to steric hindrance. The three bulky methyl groups physically block the nucleophile from attacking the central carbon atom.

Recap

  • Primary halogenoalkanes favour the one-step SN2 mechanism.
  • Tertiary halogenoalkanes favour the two-step SN1 mechanism involving a carbocation intermediate.
  • The rate of an SN2 reaction depends on [R-X] and [Nu⁻].
  • The rate of an SN1 reaction depends only on [R-X].
  • Tertiary carbocations are more stable than secondary, which are more stable than primary.
  • Steric hindrance prevents SN2 reactions in tertiary halogenoalkanes.

Quick check

  1. Which species is formed in the rate-determining step of an SN1 reaction?1 mark

5. Elimination Reactions

In addition to substitution, halogenoalkanes can undergo elimination reactions to form alkenes. This reaction involves the removal of a hydrogen halide molecule (e.g., HBr). The conditions are crucial and different from substitution: elimination is favoured by using a hot, ethanolic solution of a strong base, such as sodium hydroxide or potassium hydroxide. In this reaction, the hydroxide ion acts as a base, not a nucleophile. It removes a proton (H+) from a carbon atom adjacent to the one bonded to the halogen. Simultaneously, the C-X bond breaks and the electrons form a C=C double bond. Substitution and elimination are competing reactions; aqueous, warm conditions favour substitution, while hot, ethanolic conditions favour elimination.

General Equation: CH₃CH₂X + NaOH(ethanolic) --(heat)--> CH₂=CH₂ + NaX + H₂O

Key term

Elimination Reaction: A reaction in which a small molecule is removed from a larger molecule, typically resulting in the formation of a double or triple bond.

Examiner insight

Examiners frequently test the competition between substitution and elimination. Be precise with the conditions: 'aqueous' and 'warm' for substitution vs 'ethanolic' and 'hot' for elimination.

Worked example 13 marks

2-bromopropane is heated with a solution of potassium hydroxide in ethanol.(a) Name the type of reaction.(b) Write an equation for the reaction.(c) Name the organic product.

  1. 1

    Step 1 (a): Identify the conditions. A halogenoalkane is heated with ethanolic KOH. These are the conditions for an elimination reaction.

  2. 2

    Step 2 (b): Write the equation. A molecule of HBr is eliminated. The OH⁻ removes an H⁺ from either C1 or C3, and the Br⁻ leaves from C2. A C=C double bond forms. CH₃CHBrCH₃ + KOH(ethanolic) → CH₃CH=CH₂ + KBr + H₂O.

  3. 3

    Step 3 (c): Name the product. The organic product is CH₃CH=CH₂, which is propene.

Recap

  • Elimination reactions of halogenoalkanes produce alkenes.
  • The reaction involves removing a hydrogen halide (HX) molecule.
  • The required conditions are a hot, ethanolic solution of a strong base (e.g., NaOH or KOH).
  • Under these conditions, the hydroxide ion acts as a base, removing a proton.
  • Elimination competes with substitution; conditions determine the major pathway.

Quick check

  1. What two reagents are needed to convert 1-chloropropane to propene?2 marks

6. Uses and Environmental Concerns

Halogenoalkanes, particularly chlorofluorocarbons (CFCs) and hydrochlorofluorocarbons (HCFCs), were widely used as refrigerants, aerosol propellants, and solvents. Their useful properties include low reactivity (inertness), low flammability, and suitable volatility. However, the inertness of CFCs allows them to persist in the atmosphere and diffuse up to the stratosphere. Here, high-energy UV radiation breaks the relatively weak C-Cl bond (homolytic fission) to generate highly reactive chlorine free radicals (Cl•). These radicals act as catalysts in the breakdown of the ozone layer, which protects Earth from harmful UV radiation. A single chlorine radical can destroy many thousands of ozone molecules. Due to this environmental damage, the Montreal Protocol (1987) phased out the use of CFCs. They have been replaced by alternatives like hydrofluorocarbons (HFCs) and hydrocarbons, which do not contain chlorine and therefore do not deplete ozone.

Initiation: CCl₂F₂ --(UV light)--> •CClF₂ + Cl•

Propagation Step 1: Cl• + O₃ → ClO• + O₂

Propagation Step 2: ClO• + O → Cl• + O₂

Overall: O₃ + O → 2O₂

Key term

Free Radical: A highly reactive species containing an unpaired electron, formed by the homolytic fission of a covalent bond.

Fun fact

The ozone layer is expected to recover to 1980 levels over the mid-latitudes by around 2040 and over the Antarctic by around 2066, thanks to the global cooperation of the Montreal Protocol.

Worked example 14 marks

Dichlorodifluoromethane (CCl₂F₂) is a CFC.(a) State one property that made it useful as a refrigerant.(b) Explain, with the aid of equations, how it contributes to the depletion of the ozone layer.

  1. 1

    Step 1 (a): State a useful property. Suitable properties include being non-toxic, non-flammable, volatile, or chemically inert.

  2. 2

    Step 2 (b): Explain the process. In the stratosphere, UV radiation has enough energy to break the C-Cl bond, which is weaker than the C-F bond, in a process called homolytic fission. This produces chlorine free radicals. Equation: CCl₂F₂ --(UV)--> •CClF₂ + Cl•.

  3. 3

    Step 3: Show the catalytic cycle. The chlorine radical then catalyses the breakdown of ozone. It reacts with an ozone molecule, and is then regenerated to react again. Equation 1: Cl• + O₃ → ClO• + O₂. Equation 2: ClO• + O → Cl• + O₂. (Note: O atoms are present in the stratosphere). This shows how one Cl• can destroy many O₃ molecules.

Recap

  • Halogenoalkanes like CFCs were used as refrigerants and propellants due to their inertness and volatility.
  • The inertness of CFCs allows them to reach the stratosphere.
  • In the stratosphere, UV light breaks C-Cl bonds to form chlorine free radicals (Cl•).
  • Chlorine radicals act as catalysts to destroy ozone molecules.
  • The Montreal Protocol led to the phasing out of CFCs to protect the ozone layer.

Quick check

  1. Which bond in CCl₂F₂ is broken by UV radiation in the stratosphere?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define the term 'nucleophile' and give one example of a neutral nucleophilic molecule.2 marks
  2. Draw the structure of 2-bromo-2-methylpropane. State and explain, with reference to its structure, the mechanism by which it reacts with aqueous sodium hydroxide.4 marks
  3. Write a balanced chemical equation for the reaction of 1-chlorobutane with an excess of hot, ethanolic ammonia. Name the organic product and the type of reaction.3 marks
  4. A student hydrolyses 1-chloropropane, 1-bromopropane and 1-iodopropane by warming each with aqueous silver nitrate. (a) Predict the order of the rates of reaction, from fastest to slowest. (b) Explain your answer in terms of chemical bonds. (c) What would be observed in the test tube containing 1-iodopropane?5 marks
  5. Give the reagents and specific conditions needed to convert bromoethane into ethene. Write an equation for this reaction.3 marks
  6. Explain why fluoroalkanes are chemically very unreactive, and state one large-scale use that relies on this property.2 marks
  7. The reaction between 1-bromopropane and potassium cyanide proceeds via an SN2 mechanism. (a) Draw the mechanism for this reaction, showing all relevant dipoles, lone pairs and curly arrows. (b) Explain why this type of reaction is useful in organic synthesis.5 marks
  8. Chlorine radicals (Cl•) formed from CFCs are very effective at destroying ozone. Write two propagation steps to show how a chlorine radical catalyses the breakdown of ozone.2 marks
  9. Compare and contrast the reactions of 2-bromopropane when it is treated with: (i) warm aqueous sodium hydroxide; (ii) hot ethanolic sodium hydroxide. For each reaction, state the type of reaction and name the major organic product.4 marks
  10. An organic compound X has the molecular formula C4H9Br. When X is warmed with aqueous sodium hydroxide, the product is an alcohol Y that cannot be oxidised by acidified potassium dichromate(VI). When X is heated with hot, ethanolic potassium hydroxide, a single alkene Z is formed. Deduce the structures of X, Y and Z, explaining your reasoning.6 marks

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