Cambridge AS & A Level9701

Halogenoalkanes

Chemistry 9701 Chapter Notes

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Halogenoalkanes
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In this chapter

  1. 1Introduction to Halogenoalkanes
  2. 2Nucleophilic Substitution Reactions
  3. 3Mechanisms: Sₙ1 and Sₙ2
  4. 4Elimination Reactions
  5. 5Reactivity Trends in Halogenoalkanes
  6. 6Uses and Environmental Impact of Halogenoalkanes

1. Introduction to Halogenoalkanes

Halogenoalkanes are organic compounds where one or more hydrogen atoms in an alkane have been replaced by a halogen atom (F, Cl, Br, or I). The bond between the carbon and the halogen (C-X) 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. This electron-deficient carbon atom is susceptible to attack by nucleophiles. Halogenoalkanes are classified as primary (1°), secondary (2°), or tertiary (3°) depending on the number of other carbon atoms attached to the carbon bearing the halogen. A primary halogenoalkane has one carbon atom attached, a secondary has two, and a tertiary has three.

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

Primary (1°): RCH₂X

Secondary (2°): R₂CHX

Tertiary (3°): R₃CX

Key term

Polar Bond: A covalent bond in which electrons are shared unequally between two atoms, resulting in one atom being slightly positive (δ+) and the other slightly negative (δ-).

Common pitfall

Students often misclassify halogenoalkanes by counting the total carbons in the chain, instead of counting the number of carbons directly bonded to the C-X carbon.

Worked example 13 marks

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

  1. 1

    Step 1: Draw the structure of each compound.

  2. 2

    (a) 1-chloropropane: CH₃CH₂CH₂Cl

  3. 3

    (b) 2-bromobutane: CH₃CHBrCH₂CH₃

  4. 4

    (c) 2-chloro-2-methylpropane: (CH₃)₃CCl

  5. 5

    Step 2: Identify the carbon atom bonded to the halogen.

  6. 6

    Step 3: Count the number of other carbon atoms directly attached to this carbon.

  7. 7

    (a) In 1-chloropropane, the carbon bonded to Cl is attached to one other carbon atom. Therefore, it is a primary (1°) halogenoalkane.

  8. 8

    (b) In 2-bromobutane, the carbon bonded to Br is attached to two other carbon atoms. Therefore, it is a secondary (2°) halogenoalkane.

  9. 9

    (c) In 2-chloro-2-methylpropane, the carbon bonded to Cl is attached 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 has a partial positive charge (δ+).
  • They are classified as primary (1°), secondary (2°), or tertiary (3°).
  • This classification is crucial for determining the reaction mechanism (Sₙ1 or Sₙ2).

Quick check

  1. Draw the structure of 2-iodopentane and classify it.2 marks
  2. Why is the carbon-chlorine bond in chloroethane polar?1 mark

2. Nucleophilic Substitution Reactions

Due to the polar C-X bond, the electron-deficient carbon atom in a halogenoalkane is attacked by nucleophiles. A nucleophile is an electron-pair donor, often a negative ion or a molecule with a lone pair of electrons. In a nucleophilic substitution reaction, the nucleophile replaces the halogen atom. The halogen atom leaves as a halide ion (X⁻). Key examples include hydrolysis to form an alcohol, reaction with cyanide to form a nitrile, and reaction with ammonia to form a primary amine.

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

Nitrile Formation: R-X + CN⁻(ethanolic) → R-CN + X⁻ (extends the carbon chain by one)

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

Key term

Nucleophile: An electron-rich species (an ion or molecule) that has a lone pair of electrons which it can donate to an electron-deficient centre (like the δ+ carbon in a halogenoalkane).

Examiner insight

Examiners look for clear understanding of which part of the nucleophile attacks and what the leaving group is. For example, in the reaction with CN⁻, it is the carbon atom of the cyanide ion that forms the new bond.

Common pitfall

Forgetting that two moles of ammonia are required in the overall equation for amine formation: one acts as the nucleophile and the second acts as a base to remove a proton from the intermediate.

Worked example 12 marks

1-bromobutane is heated under reflux with an aqueous solution of sodium hydroxide. Write an equation for the reaction and name the organic product.

  1. 1

    Step 1: Identify the reactants and reaction type. 1-bromobutane is a halogenoalkane, and aqueous NaOH provides the OH⁻ nucleophile. This is nucleophilic substitution (hydrolysis).

  2. 2

    Step 2: The nucleophile (OH⁻) will replace the halogen (Br). The Br leaves as a Br⁻ ion.

  3. 3

    Step 3: The organic product is formed by replacing Br with OH. CH₃CH₂CH₂CH₂Br becomes CH₃CH₂CH₂CH₂OH.

  4. 4

    Step 4: Name the organic product. The 4-carbon alcohol is butan-1-ol.

  5. 5

    Step 5: Write the full equation: CH₃CH₂CH₂CH₂Br + NaOH(aq) → CH₃CH₂CH₂CH₂OH + NaBr(aq)

Worked example 23 marks

What is the organic product when chloroethane reacts with excess ethanolic ammonia? Name the product and the type of reaction.

  1. 1

    Step 1: Identify the reactants. Chloroethane (CH₃CH₂Cl) and ammonia (NH₃). This is nucleophilic substitution to form an amine.

  2. 2

    Step 2: The nucleophile is the ammonia molecule, using its lone pair on the nitrogen atom.

  3. 3

    Step 3: The NH₂ group replaces the Cl atom. The product is CH₃CH₂NH₂.

  4. 4

    Step 4: Name the product. This is ethylamine (or aminoethane).

  5. 5

    Step 5: The reaction type is nucleophilic substitution.

Recap

  • Nucleophiles attack the δ+ carbon atom of the C-X bond.
  • The halogen is displaced as a halide ion (X⁻) in a substitution reaction.
  • Reaction with aqueous OH⁻ forms an alcohol.
  • Reaction with CN⁻ forms a nitrile, which is important for increasing the carbon chain length.
  • Reaction with NH₃ forms a primary amine.

Quick check

  1. What nucleophile is needed to convert 1-iodopropane into butanenitrile?1 mark

3. Mechanisms: Sₙ1 and Sₙ2

Nucleophilic substitution can occur via two different mechanisms, Sₙ1 and Sₙ2, depending on the structure of the halogenoalkane.

Sₙ2 (Substitution, Nucleophilic, Bimolecular): This is a one-step mechanism favoured by primary (1°) halogenoalkanes. The nucleophile attacks the δ+ carbon at the same time as the carbon-halogen bond breaks. The rate depends on the concentration of both the halogenoalkane and the nucleophile. The reaction proceeds via a high-energy transition state where the carbon is partially bonded to both the incoming nucleophile and the leaving halogen.

Sₙ1 (Substitution, Nucleophilic, Unimolecular): This is a two-step mechanism favoured by tertiary (3°) halogenoalkanes. The first step is the slow, rate-determining step where the C-X bond breaks heterolytically to form a stable carbocation intermediate. The second step is a fast attack by the nucleophile on the planar carbocation. The rate depends only on the concentration of the halogenoalkane. Secondary halogenoalkanes can react by a mixture of both mechanisms.

Sₙ2 Rate Equation: Rate = k[R-X][Nu⁻]

Sₙ1 Rate Equation: Rate = k[R-X]

Key term

Carbocation: An ion with a positively charged carbon atom, which is an intermediate in reactions like Sₙ1 and some elimination reactions.

Examiner insight

Examiners award marks for correctly drawn mechanisms, which must include all dipoles, charges, and curly arrows. For Sₙ1, explicitly stating that the first step is 'slow' and the second is 'fast' is often required.

Common pitfall

Drawing curly arrows incorrectly. Remember they must start from a source of electrons (a lone pair or a covalent bond) and point to an electron-deficient atom or where a new bond will form.

Worked example 14 marks

Draw the Sₙ2 mechanism for the reaction of hydroxide ions (OH⁻) with bromoethane (CH₃CH₂Br). Include all relevant dipoles, curly arrows, and the structure of the transition state.

  1. 1

    Step 1: Draw the reactants, bromoethane and the hydroxide ion. Show the dipole on the C-Br bond (Cδ+ and Brδ-).

  2. 2

    Step 2: Draw a curly arrow from the lone pair of electrons on the oxygen of the OH⁻ ion to the δ+ carbon atom of bromoethane. This shows the nucleophilic attack.

  3. 3

    Step 3: Draw a second curly arrow starting from the C-Br bond and pointing to the Br atom. This shows the C-Br bond breaking.

  4. 4

    Step 4: Both arrows should be drawn in a single step. To show the transition state, draw the central carbon with dashed lines to both the incoming OH group and the outgoing Br atom. The structure should have a negative charge, enclosed in square brackets: [HO---CH₂(CH₃)---Br]⁻.

  5. 5

    Step 5: Draw the final products: ethanol (CH₃CH₂OH) and the bromide ion (Br⁻).

Worked example 25 marks

Describe the Sₙ1 mechanism for the hydrolysis of 2-bromo-2-methylpropane, (CH₃)₃CBr. Explain why it proceeds via this mechanism.

  1. 1

    Step 1: State that this is a tertiary halogenoalkane, which favours the Sₙ1 mechanism due to the stability of the tertiary carbocation intermediate formed.

  2. 2

    Step 2 (Mechanism Step 1): Draw the (CH₃)₃CBr molecule. Show the slow, rate-determining step where the C-Br bond breaks heterolytically. Draw a curly arrow from the C-Br bond to the Br atom. This forms the tertiary carbocation (CH₃)₃C⁺ and a bromide ion Br⁻.

  3. 3

    Step 3 (Mechanism Step 2): Draw the planar tertiary carbocation. The nucleophile is water (H₂O). Draw a curly arrow from a lone pair on the oxygen of a water molecule to the positive carbon of the carbocation. This is a fast step.

  4. 4

    Step 4: This forms an intermediate oxonium ion, (CH₃)₃C-OH₂⁺.

  5. 5

    Step 5: A final, fast deprotonation step occurs. Another water molecule acts as a base, removing a proton. Draw a curly arrow from the O-H bond in the oxonium ion to the positive oxygen atom. This forms the final product, 2-methylpropan-2-ol, ((CH₃)₃COH), and a hydronium ion, H₃O⁺.

Recap

  • Primary halogenoalkanes react via the one-step Sₙ2 mechanism.
  • Tertiary halogenoalkanes react via the two-step Sₙ1 mechanism involving a carbocation intermediate.
  • The Sₙ2 rate depends on [R-X] and [Nu⁻]; the Sₙ1 rate depends only on [R-X].
  • Curly arrows must always show the movement of a pair of electrons, from a lone pair or a bond.
  • Tertiary carbocations are more stable than secondary, which are more stable than primary.

Quick check

  1. Which type of halogenoalkane reacts fastest by an Sₙ1 mechanism and why?2 marks
  2. What does 'bimolecular' mean in the context of an Sₙ2 reaction?1 mark

4. Elimination Reactions

Halogenoalkanes can also undergo elimination reactions to form alkenes. This reaction is favoured by using a hot, ethanolic solution of a strong base, such as sodium hydroxide (NaOH) or potassium hydroxide (KOH). In this case, the hydroxide ion (OH⁻) acts as a base, not a nucleophile. It removes a hydrogen atom (a proton) from a carbon atom adjacent to the one bonded to the halogen. Simultaneously, the C-X bond breaks and a C=C double bond is formed. The products are an alkene, a halide ion, and water. This reaction competes with nucleophilic substitution, and the conditions determine the major product: aqueous conditions favour substitution, while hot ethanolic conditions favour elimination.

General Equation: RCH₂CH₂X + NaOH(ethanolic, hot) → RCH=CH₂ + NaX + H₂O

Key term

Elimination Reaction: A reaction in which a small molecule (like HBr or H₂O) is removed from a larger molecule, typically resulting in the formation of a double or triple bond.

Examiner insight

Questions often ask students to state the reagents and conditions needed to favour one reaction type over the other. Be precise: 'hot ethanolic NaOH' is better than just 'NaOH'.

Common pitfall

Confusing the conditions for substitution and elimination. Remember: aqueous and warm for substitution (OH⁻ as nucleophile), ethanolic and hot for elimination (OH⁻ as base).

Worked example 13 marks

2-bromopropane is heated with ethanolic potassium hydroxide. Identify the organic product and write an equation for the reaction.

  1. 1

    Step 1: Identify the reactants and conditions. A secondary halogenoalkane (2-bromopropane) with a strong base (KOH) in ethanol and heat. These conditions strongly favour elimination.

  2. 2

    Step 2: The OH⁻ ion will act as a base, removing a proton from a carbon adjacent to the C-Br carbon. The adjacent carbons are C1 and C3, which are equivalent.

  3. 3

    Step 3: A proton is removed from C1 (or C3), the C-Br bond breaks, and a double bond forms between C1 and C2. The molecule loses HBr.

  4. 4

    Step 4: The organic product is propene (CH₃CH=CH₂).

  5. 5

    Step 5: Write the full equation: CH₃CHBrCH₃ + KOH(ethanolic) → CH₂=CHCH₃ + KBr + H₂O.

Worked example 24 marks

Heating 2-bromobutane with ethanolic NaOH can produce two different structural isomers of butene. Draw and name both isomers and predict which would be the major product.

  1. 1

    Step 1: The halogen is on C2. Elimination can occur by removing a proton from either adjacent carbon: C1 or C3.

  2. 2

    Step 2: Elimination involving C1: A proton is removed from C1, forming a double bond between C1 and C2. Product is but-1-ene (CH₂=CHCH₂CH₃).

  3. 3

    Step 3: Elimination involving C3: A proton is removed from C3, forming a double bond between C2 and C3. Product is but-2-ene (CH₃CH=CHCH₃). Note that but-2-ene exists as E/Z isomers.

  4. 4

    Step 4: Predict the major product using Zaitsev's (Saytzeff's) rule, which states that the more substituted (more stable) alkene is the major product. But-2-ene is a di-substituted alkene (two H atoms on the C=C are replaced by alkyl groups), while but-1-ene is mono-substituted. Therefore, but-2-ene is the major product.

  5. 5

    Answer: The two isomers are but-1-ene and but-2-ene. But-2-ene is the major product.

Recap

  • Elimination reactions form alkenes from halogenoalkanes.
  • The required conditions are a hot, ethanolic solution of a strong base (e.g., KOH).
  • The hydroxide ion acts as a base, removing a proton.
  • Substitution and elimination are competing reactions.
  • For asymmetric halogenoalkanes, the major product is usually the most substituted alkene (Zaitsev's rule).

Quick check

  1. What are the two roles that a hydroxide ion can play when reacting with a halogenoalkane?2 marks
  2. What conditions favour the elimination reaction of a halogenoalkane over substitution?2 marks

6. Uses and Environmental Impact of Halogenoalkanes

The low reactivity of some halogenoalkanes, particularly those containing fluorine, makes them useful. Their properties, such as low flammability, low toxicity, and volatility, have led to their use as refrigerants, aerosol propellants, solvents, and anaesthetics (e.g., halothane, CF₃CHBrCl). However, a class of halogenoalkanes called chlorofluorocarbons (CFCs) has caused significant environmental damage. CFCs (e.g., CCl₂F₂) are very unreactive and are not broken down in the lower atmosphere (troposphere). They diffuse up to the stratosphere, where they are exposed to high-energy ultraviolet (UV) radiation. This UV radiation is strong enough to break the weaker C-Cl bond via homolytic fission, producing highly reactive chlorine radicals (Cl•). These chlorine radicals then act as catalysts in the decomposition of ozone (O₃) into oxygen (O₂), depleting the ozone layer that protects the Earth from harmful UV radiation. Due to this, CFCs have been largely replaced by alternatives like hydrofluorocarbons (HFCs) and hydrocarbons.

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

Propagation Step 1: Cl• (g) + O₃ (g) → ClO• (g) + O₂ (g)

Propagation Step 2: ClO• (g) + O (g) → Cl• (g) + O₂ (g)

Overall Reaction: O₃ (g) + O (g) → 2O₂ (g)

Key term

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

Examiner insight

When explaining ozone depletion, it is vital to state that the chlorine radical is regenerated, explaining why it is a catalyst and can cause so much damage.

Fun fact

The Montreal Protocol, an international treaty signed in 1987 to phase out CFCs, is considered one of the most successful international environmental agreements in history. The ozone layer is now slowly recovering.

Worked example 14 marks

Dichlorodifluoromethane (CCl₂F₂) was once used as a refrigerant but is now banned.(a) Explain why this compound is stable in the troposphere.(b) Write two equations to show how it causes ozone depletion in the stratosphere.

  1. 1

    (a) Stability in Troposphere:

  2. 2

    Step 1: Refer to the bond strengths. CCl₂F₂ contains strong C-Cl and very strong C-F bonds.

  3. 3

    Step 2: Explain that these bonds are too strong to be broken by the conditions in the troposphere (lower atmosphere), making the molecule very inert.

  4. 4

    (b) Ozone Depletion:

  5. 5

    Step 3: Initiation. In the stratosphere, high-energy UV radiation provides enough energy to break the weaker C-Cl bond (not the C-F bond) via homolytic fission, producing a chlorine radical. Equation: CCl₂F₂ → •CClF₂ + Cl•

  6. 6

    Step 4: Propagation. The chlorine radical then catalyses the breakdown of ozone in a two-step cycle. First, it reacts with ozone. Equation: Cl• + O₃ → ClO• + O₂. Second, the ClO• radical reacts with an oxygen atom (also present in the stratosphere) to regenerate the chlorine radical. Equation: ClO• + O → Cl• + O₂. The regenerated Cl• can then destroy another ozone molecule.

Recap

  • The low reactivity of many fluoroalkanes makes them useful as solvents and anaesthetics.
  • CFCs are unreactive in the troposphere but are broken down by UV light in the stratosphere.
  • UV light causes homolytic fission of the C-Cl bond to form chlorine radicals (Cl•).
  • Chlorine radicals act as catalysts to destroy the ozone layer.
  • One chlorine radical can destroy many thousands of ozone molecules.
  • CFCs have been replaced by less harmful alternatives like HFCs.

Quick check

  1. What is the name of the process where a covalent bond breaks to form two radicals?1 mark
  2. Why is the C-Cl bond in a CFC broken by UV light, but not the C-F bond?1 mark

End-of-chapter exercise

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

  1. 1-bromobutane can be converted into butan-1-ol or but-1-ene. For each conversion, state the reagent(s), the essential conditions, and the type of reaction occurring.6 marks
  2. Draw the mechanism for the reaction between 2-chloro-2-methylpropane and a hydroxide ion. Name the mechanism and the organic product.5 marks
  3. Explain why iodoethane undergoes nucleophilic substitution much faster than chloroethane.3 marks
  4. Write a balanced chemical equation for the reaction of 1-chloropropane with excess concentrated ammonia in ethanol. Name the organic product.2 marks
  5. A student suggests that the reaction between bromoethane and potassium cyanide (KCN) produces CH₃CH₂NC. Explain why the major product is actually CH₃CH₂CN.2 marks
  6. Trichloromethane (CHCl₃), also known as chloroform, is a tertiary halogenoalkane. True or False? Justify your answer.2 marks
  7. The breakdown of ozone by chlorine radicals is a chain reaction. (a) What is meant by the term 'radical'? (b) Write the two propagation steps that show how chlorine radicals catalyse the breakdown of ozone.3 marks
  8. 2-bromopentane is heated under reflux with hot, ethanolic sodium hydroxide. Two structural isomers are formed. Draw the structures of both isomers and state which is the major product.3 marks
  9. Compare and contrast the Sₙ1 and Sₙ2 mechanisms of nucleophilic substitution in terms of their reaction steps, the type of halogenoalkane they are favoured by, and their rate equations.6 marks
  10. Halothane (CF₃CHBrCl) is an anaesthetic. Suggest two reasons why compounds containing C-F bonds are often unreactive.2 marks

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