Unit 2: CH02 Syllabus Node

Halogenoalkanes: Nucleophilic Substitution & Elimination

Bond polarity and reactivity, curly arrow precision, nucleophilic substitution with OH-, CN-, and NH3, base elimination, and stratospheric ozone depletion.

1. Carbon-Halogen Bond Polarity and Reactivity

Halogen atoms are more electronegative than carbon, creating a polar bond:

\[ \text{C}^{\delta+} - \text{X}^{\delta-} \]

The electron-deficient carbon atom (\(\text{C}^{\delta+}\)) is vulnerable to attack by nucleophiles (electron-pair donors).

Bond Enthalpy Controls Reactivity

Although the \(\text{C}-\text{F}\) bond is the most polar, reactivity is dictated by bond enthalpy, not bond polarity:

  • \(\text{C}-\text{F}\): \(467\text{ kJ mol}^{-1}\) (extremely strong, virtually unreactive).
  • \(\text{C}-\text{Cl}\): \(338\text{ kJ mol}^{-1}\).
  • \(\text{C}-\text{Br}\): \(276\text{ kJ mol}^{-1}\).
  • \(\text{C}-\text{I}\): \(238\text{ kJ mol}^{-1}\) (weakest bond, broken most easily).

Order of reactivity: \(\text{Iodoalkanes} > \text{Bromoalkanes} > \text{Chloroalkanes} > \text{Fluoroalkanes}\). Iodoalkanes hydrolyse fastest because the \(\text{C}-\text{I}\) bond has the lowest activation energy.

2. Precise Curly Arrow Drawing Rules

Strict Examiner Marking Standards
  • A curly arrow represents the movement of a pair of electrons.
  • An arrow MUST start either directly on a lone pair of electrons (\(:\)) or at the centre of a covalent bond.
  • The arrow head must terminate directly on the destination atom (or between atoms to form a new bond).
  • Floating arrows or arrows originating from minus signs are penalised immediately.
Nucleophilic Substitution of Bromoethane by Hydroxide :OH- H3C - C(H2) delta + Br delta - H3C - CH2 - OH + :Br- Precise Examiner Rule: Curly arrows MUST start directly from a lone pair of electrons or the exact centre of a covalent bond, pointing to the destination atom

3. Nucleophilic Substitution Reactions

A nucleophile replaces the halogen atom in three primary synthetic pathways:

Nucleophile Reaction Conditions Organic Product Stoichiometric Equation
Hydroxide ion (\(:\text{OH}^-\)) Warm aqueous \(\text{NaOH}\) or \(\text{KOH}\), reflux Alcohol \(\text{R}-\text{Br} + \text{OH}^- \rightarrow \text{R}-\text{OH} + \text{Br}^-\)
Cyanide ion (\(:\text{CN}^-\)) Warm ethanolic \(\text{KCN}\), reflux Nitrile (extends carbon chain by +1) \(\text{R}-\text{Br} + \text{CN}^- \rightarrow \text{R}-\text{CN} + \text{Br}^-\)
Ammonia (\(:\text{NH}_3\)) Excess ethanolic \(\text{NH}_3\), heated under pressure Primary Amine \(\text{R}-\text{Br} + 2\text{NH}_3 \rightarrow \text{R}-\text{NH}_2 + \text{NH}_4\text{Br}\)

Why is EXCESS Ammonia Required?

The primary amine product (\(\text{R}-\text{NH}_2\)) still has a lone pair on its nitrogen atom and can act as a competing nucleophile. Using excess ammonia ensures ammonia molecules outcompete the amine, preventing successive substitutions that form secondary and tertiary amines and quaternary ammonium salts.

4. Elimination Reactions to Form Alkenes

When reacted with hot ethanolic potassium hydroxide (\(\text{KOH}\)) in the absence of water, the hydroxide ion acts as a base (proton acceptor) rather than a nucleophile:

\[ \text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \xrightarrow{\text{ethanol, heat}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} + \text{Br}^- \]

Substitution Conditions

  • Aqueous solvent (water).
  • Warm / moderate temperature.
  • Favoured by primary halogenoalkanes.

Elimination Conditions

  • Ethanolic solvent (pure alcohol).
  • High temperature / reflux.
  • Favoured by tertiary halogenoalkanes.

5. Ozone Depletion by Chlorofluorocarbons (CFCs)

Chlorofluorocarbons (CFCs) were widely used as refrigerants and aerosol propellants due to their stability and non-toxicity. In the stratosphere, intense UV radiation breaks the weak \(\text{C}-\text{Cl}\) bond via homolytic fission:

\[ \text{CF}_2\text{Cl}_2 \xrightarrow{UV} \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet \]

Catalytic Ozone Destruction Cycle

The chlorine radical (\(\text{Cl}^\bullet\)) acts as a homogenous catalyst in a two-step ozone destruction cycle:

\[ \text{Step 1: } \text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2 \]

\[ \text{Step 2: } \text{ClO}^\bullet + \text{O}_3 \rightarrow \text{Cl}^\bullet + 2\text{O}_2 \]

\[ \text{Overall: } 2\text{O}_3 \rightarrow 3\text{O}_2 \]

Because the chlorine radical is regenerated in Step 2, a single \(\text{Cl}^\bullet\) can destroy over 100,000 ozone molecules before termination.

6. Practice Questions

Practice Problem (3 Marks)
Outline the mechanism for the reaction of bromoethane with aqueous potassium hydroxide to form ethanol. Show all relevant lone pairs, dipoles, and curly arrows.

Mechanism marks:

  • Curly arrow from lone pair on \(\text{OH}^-\) to \(\text{C}^{\delta+}\) of bromoethane. (1 mark)
  • Correct dipoles \(\text{C}^{\delta+} - \text{Br}^{\delta-}\). (1 mark)
  • Curly arrow from \(\text{C}-\text{Br}\) bond to \(\text{Br}\) atom, forming \(\text{Br}^-\). (1 mark)