Welcome to your study notes on Haloalkanes. This topic explores how polar organic molecules undergo substitution reactions, how we can measure their reactivity experimentally, and the vital role haloalkane chemistry plays in environmental global policy.

What you'll learn:

• How haloalkanes react via nucleophilic substitution, using curly arrows to show electron movement.
• How to experimentally investigate and explain the trends in the rates of hydrolysis of different carbon–halogen bonds.
• How chlorofluorocarbons (CFCs) and other radicals catalytically destroy the Earth's protective ozone layer.

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1. Structure and Bond Polarity

Haloalkanes are saturated organic compounds containing one or more halogen atoms (fluorine, chlorine, bromine, or iodine) covalently bonded to a carbon chain.

The reactivity of haloalkanes is primarily driven by the polarity of the carbon–halogen bond ($C-X$). Halogen atoms are highly electronegative compared to carbon.

Because the halogen atom ($X$) has a higher electronegativity than carbon, it pulls the electron density in the $C-X$ $\sigma$-bond toward itself. This creates a permanent dipole across the bond:

• The carbon atom becomes electron-deficient, carrying a partial positive charge ($\delta +$).
• The halogen atom carries a partial negative charge ($\delta -$).

The electron-deficient carbon ($C^{\delta +}$) acts as an attractive target for electron-rich species known as nucleophiles.

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2. Nucleophiles and Nucleophilic Substitution

Before looking at reactions, we must precisely define what a nucleophile is.

Common nucleophiles you need to know for A-Level Chemistry include:

• The hydroxide ion, ${OH}^{-}$ (specifically from aqueous alkalis like $NaOH$ or $KOH$).
• Water molecules, $H_{2}O$.
• Ammonia molecules, ${NH}_3$.

All nucleophiles must possess at least one lone pair of electrons on an electronegative atom, which they can donate to form a new dative covalent bond with a carbon atom.

Hydrolysis of Primary Haloalkanes by Aqueous Alkali

When a primary haloalkane is warmed with an aqueous alkali (such as aqueous sodium hydroxide, $NaOH(aq)$), it undergoes a hydrolysis reaction. This is a type of nucleophilic substitution.

During this reaction, the halogen atom is substituted by a hydroxyl ($-OH$) group to form an alcohol and a halide ion.

$$R-X+{OH}^{-}\to R-OH+X^{-}$$

The Reaction Mechanism

To explain how this reaction occurs, we use a reaction mechanism. A mechanism uses curly arrows to map the movement of pairs of electrons.

1. The nucleophile (${OH}^{-}$) approaches the electron-deficient carbon ($C^{\delta +}$) from the opposite side of the halogen atom to minimise charge repulsion.
2. A curly arrow starts from a lone pair on the oxygen of the hydroxide ion and points directly to the $C^{\delta +}$ atom, indicating the formation of a new dative $C-O$ covalent bond.
3. Simultaneously, the polar carbon-halogen bond breaks heterolytically. A curly arrow starts from the middle of the $C-X$ bond and points directly to the halogen atom ($X$), showing both electrons in the bond moving onto the halogen to form a stable halide ion leaving group ($X^{-}$).

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3. Investigating the Rates of Hydrolysis (PAG 7)

We can compare how quickly different carbon-halogen bonds break by carrying out a classic practical investigation (part of your Practical Endorsement, PAG 7).

The Experimental Setup

To compare the hydrolysis rates of 1-chlorobutane, 1-bromobutane, and 1-iodobutane, we set up three separate test tubes in a warm water bath ($50{}^{\circ }C$). Each tube contains:

• The specific haloalkane.
• Aqueous silver nitrate (${AgNO}_3(aq)$) — this acts as the source of water (the nucleophile) and silver ions (${Ag}^{+}$) to detect the released halide ions.
• Ethanol — which acts as a co-solvent.

Because haloalkanes are insoluble in water, they would normally form two immiscible layers, preventing a reaction. Ethanol dissolves both the organic haloalkanes and the inorganic aqueous silver nitrate, allowing them to mix and react in a single homogeneous phase.

Observing the Rate of Precipitation

As hydrolysis proceeds, halide ions (${Cl}^{-}$, ${Br}^{-}$, or $I^{-}$) are continuously produced. These immediately react with the silver ions (${Ag}^{+}$) in solution to form insoluble silver halide precipitates:

$${Ag}^{+}(aq)+X^{-}(aq)\to AgX(s)$$

By timing how long it takes for each precipitate to appear, we can determine the relative rates of reaction.

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4. Explaining the Rate Trend: Bond Enthalpy vs. Polarity

The experimental results show a clear trend in the rate of hydrolysis:

$$Iodoalkanes\; (Fastest)>Bromoalkanes>Chloroalkanes\gg Fluoroalkanes\; (Unreactive)$$

This trend can be explained by comparing two competing factors: bond polarity and bond enthalpy.

1. Bond Polarity

If bond polarity were the dominant factor, we would expect fluoroalkanes to react fastest. Fluorine is the most electronegative halogen, making the $C-F$ bond the most polar, which gives the carbon atom the largest $\delta +$ charge to attract nucleophiles.

2. Bond Enthalpy

For substitution to occur, the carbon-halogen covalent bond must be broken. The average bond enthalpies down Group 17 are as follows:

Because the atomic radius of the halogen increases down the group, the shared pair of electrons in the $C-X$ bond is further from the halogen nucleus. This results in a weaker electrostatic attraction and a lower bond enthalpy down the group.

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5. Environmental Concerns: Ozone Depletion

Chlorofluorocarbons (CFCs) are synthetic organohalogen compounds containing only chlorine, fluorine, and carbon (e.g., dichlorodifluoromethane, ${CF}_2{Cl}_2$).

Due to their high chemical stability, low toxicity, and volatility, CFCs were widely used throughout the 20th century as refrigerants, aerosol propellants, and air conditioning blowing agents. However, this stability became an environmental disaster.

Radical Production in the Upper Atmosphere

Because CFCs are exceptionally stable in the lower atmosphere (troposphere), they do not break down. Over decades, they diffuse upward into the stratosphere.

Here, they are exposed to high-energy ultraviolet (UV) radiation, which possesses enough energy to break the relatively weak $C-Cl$ bond homolytically (photodissociation), producing highly reactive chlorine radicals (${Cl}^{\bullet }$).

$${CF}_2{Cl}_2\underrightarrow{UV}{CF}_2{Cl}^{\bullet }+{Cl}^{\bullet }$$

Catalytic Breakdown of Ozone

The chlorine radical (${Cl}^{\bullet }$) acts as a homogeneous catalyst in the destruction of the ozone layer ($O_3$), which protects Earth from harmful UVB radiation. This destruction occurs via a two-step propagation cycle:

• Propagation Step 1: The chlorine radical reacts with an ozone molecule, forming a chlorine monoxide radical and oxygen gas.

$${Cl}^{\bullet }+O_3\to {ClO}^{\bullet }+O_2$$

• Propagation Step 2: The chlorine monoxide radical reacts with a monoatomic oxygen atom (present in the stratosphere), regenerating the chlorine radical catalyst and producing more oxygen gas.

$${ClO}^{\bullet }+O\to {Cl}^{\bullet }+O_2$$

By adding these two propagation steps together and cancelling species that appear on both sides (${Cl}^{\bullet }$ and ${ClO}^{\bullet }$), we get the overall equation for the reaction:

$$O_3+O\to 2O_2$$

Because the ${Cl}^{\bullet }$ radical is regenerated at the end of the cycle, it can go on to destroy up to 100,000 ozone molecules before undergoing a termination step.

Other Radicals Destroying Ozone

Chlorine is not the only radical catalyst responsible for ozone depletion. Nitrogen monoxide radicals (${NO}^{\bullet }$), formed naturally by lightning strikes and artificially by aircraft engines operating at high altitudes, undergo an identical catalytic cycle.

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