AQA A-Level Organic Chemistry 3.3.3 Halogenoalkanes Reactivity and Ozone Depletion
3.3.3

Reactivity and Ozone Depletion

Understand trends in C-X bond reactivity, the silver nitrate test, and free radical ozone depletion by CFCs.

The reactivity of halogenoalkanes is determined by competing factors: the polar nature of the carbon-halogen bond and the strength of the covalent bond. We use relative rates of hydrolysis to demonstrate these properties experimentally, and we study the environmental impact of chlorine radicals on the ozone layer.

🔑 Key Principle

The reactivity of halogenoalkanes is governed by bond enthalpy, not bond polarity. Although the C-F bond is the most polar due to fluorine's high electronegativity, it is the least reactive because the C-F bond is extremely strong (highest bond enthalpy). Conversely, the C-I bond is the weakest (lowest bond enthalpy) and hydrolyses the fastest.

Bond Enthalpy vs Bond Polarity

Two opposing factors affect the rate of reaction of a halogenoalkane:

  1. Bond Polarity: Electronegativity decreases down Group 7. Therefore, the C-F bond is the most polar (\(\text{C}^{\delta+}\text{-F}^{\delta-}\)) and the carbon atom is the most electron-deficient, which should make it easiest for a nucleophile to attack.
  2. Bond Enthalpy: Bond enthalpy decreases down Group 7. As the halogen atoms get larger, the overlap of orbitals becomes less effective, making the C-X covalent bond weaker down the group.

Experimental evidence shows that the rate of hydrolysis increases down Group 7. This proves that bond enthalpy is the dominant factor in determining halogenoalkane reactivity.

Key Definitions

Bond Enthalpy

The amount of energy required to break one mole of a specific covalent bond in the gaseous state. Lower bond enthalpy indicates a weaker, more easily broken bond.

Ozone Depletion

The catalytic destruction of ozone (\(\text{O}_3\)) molecules in the stratosphere, primarily by chlorine free radicals derived from chlorofluorocarbons (CFCs).

Testing Reactivity: Silver Nitrate Hydrolysis

To measure the rate of hydrolysis of halogenoalkanes, they are reacted with water (acting as a nucleophile) in the presence of silver nitrate (\(\text{AgNO}_3\)) and an ethanol solvent (to allow the water and halogenoalkane to mix):

\[ \text{R-X(l)} + \text{H}_2\text{O(l)} \rightarrow \text{R-OH(aq)} + \text{H}^+\text{(aq)} + \text{X}^-\text{(aq)} \]

The halide ions (\(\text{X}^-\)) released react immediately with the silver ions (\(\text{Ag}^+\)) to form insoluble silver halide precipitates:

\[ \text{Ag}^+\text{(aq)} + \text{X}^-\text{(aq)} \rightarrow \text{AgX(s)} \]

By timing how long it takes for each precipitate to appear, we determine the relative rate of reaction:

Halogenoalkane Bond Enthalpy / kJ mol⁻¹ Precipitate Identity Precipitate Colour Rate of Hydrolysis
Fluoroalkane 467 AgF (soluble) No precipitate Extremely slow / no reaction
Chloroalkane 338 AgCl White precipitate Slow
Bromoalkane 276 AgBr Cream precipitate Medium
Iodoalkane 238 AgI Yellow precipitate Fastest
📝 AQA Examiner Tip

When asked to explain the trend in rate of hydrolysis, you must attribute it to decreasing bond enthalpy down the group. Do not write about electronegativity or polarity, as this would predict the opposite trend. You must explicitly state that the C-I bond is weaker/has a lower bond enthalpy than the C-Cl bond, meaning less energy is required to break it.

Ozone Depletion by CFCs

Chlorofluorocarbons (CFCs) were widely used in aerosols, air conditioners, and refrigerators because they are non-toxic, non-flammable, and extremely stable in the troposphere. However, their stability allows them to diffuse into the stratosphere, where they are exposed to high-energy ultraviolet (UV) radiation.

The Free Radical Mechanism

Under UV light, homolytic fission of the C-Cl bond occurs (because the C-Cl bond is weaker than the C-F bond), generating chlorine free radicals (\(\text{Cl}\bullet\)):

1. Initiation:

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

2. Propagation:

The chlorine free radical is highly reactive and attacks ozone (\(\text{O}_3\)), destroying the ozone molecule and forming a chlorine monoxide radical (\(\text{ClO}\bullet\)). The chlorine monoxide radical then reacts with another ozone molecule (or oxygen atom), regenerating the chlorine free radical:

\[ \text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2 \]

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

3. Overall Equation:

Because the chlorine radical is used in the first propagation step and regenerated in the second, it acts as a catalyst. A single chlorine radical can destroy thousands of ozone molecules before terminating:

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

The Catalytic Ozone Cycle

The diagram below shows how the chlorine radical is continuously regenerated in the stratosphere, driving the decomposition of ozone:

Ozone Depletion Catalytic Cycle Cl• Chlorine Radical ClO• Monoxide Radical Attack: + O₃ Releases: + O₂ Regeneration: + O₃ Releases: + 2 O₂ Stratospheric Catalytic Loop

Alternative Compounds

To protect the ozone layer, international agreements (like the Montreal Protocol) banned the use of CFCs. Chemists developed alternatives:

Worked Examples

✏️ Worked Example 1
Describe and explain the relative rates of hydrolysis of 1-chlorobutane and 1-iodobutane when heated with aqueous silver nitrate in ethanol solvent.

Answer:

Observation: 1-iodobutane will form a yellow precipitate of silver iodide much faster than 1-chlorobutane forms a white precipitate of silver chloride.

Explanation: Reactivity is determined by the carbon-halogen bond enthalpy. The C-I bond is longer and weaker (bond enthalpy of 238 kJ mol⁻¹) than the C-Cl bond (bond enthalpy of 338 kJ mol⁻¹). Consequently, less energy is required to break the C-I bond, allowing it to hydrolyse at a significantly faster rate.

✏️ Worked Example 2
Write equations to show how trichlorofluoromethane (\(\text{CFCl}_3\)) decomposes in the stratosphere and acts as a catalyst in ozone destruction.

Answer:

Initiation: UV light breaks the weakest C-Cl bond to form chlorine free radicals:

\[ \text{CFCl}_3 \xrightarrow{uv} \text{CFCl}_2\bullet + \text{Cl}\bullet \]

Propagation Step 1: The chlorine radical destroys ozone:

\[ \text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2 \]

Propagation Step 2: The chlorine radical is regenerated:

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

The chlorine radical is regenerated at the end of propagation, proving it acts as a catalyst.

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