Hello and Welcome to Halogenoalkanes!
Welcome to one of the most exciting and reactive families in organic chemistry: the Halogenoalkanes (or alkyl halides). These molecules are essentially alkanes where one or more hydrogen atoms have been replaced by a halogen atom (F, Cl, Br, or I).
Why are they important? They are fundamental building blocks in chemical synthesis, used to create everything from plastics and refrigerants to complex medicines. However, they are also famous (or infamous) for the environmental damage caused by compounds like CFCs.
In this chapter, we will uncover why these compounds are so much more reactive than simple alkanes, focusing on two crucial reaction types: Nucleophilic Substitution and Elimination.
1. Structure, Polarity, and Uses of Halogenoalkanes
1.1 The Crucial C-X Bond
The key to understanding halogenoalkanes is the bond between the carbon atom and the halogen atom (C-X, where X = F, Cl, Br, I).
- Electronegativity: Halogens are significantly more electronegative than carbon (Fluorine > Chlorine > Bromine > Iodine).
- Polarity: This difference in electronegativity means the electron pair in the C-X bond is pulled closer towards the halogen atom.
- Result: The halogen atom gains a small negative charge (\(\delta-\)) and the carbon atom bonded to it gains a small positive charge (\(\delta+\)). This is a polar covalent bond.
Analogy: The Tug-of-War
Imagine the C-X bond electrons are in a tug-of-war. The halogen is much stronger and pulls the electrons mostly onto its side, leaving the carbon atom slightly exposed and ready to be attacked.
The presence of this \(\delta+\) carbon atom makes halogenoalkanes susceptible to attack by nucleophiles (which love positive charges). This is why they are much more reactive than non-polar alkanes!
1.2 Classification of Halogenoalkanes
Like alcohols, halogenoalkanes are classified based on the number of other carbon atoms attached to the carbon atom bearing the halogen (C-X bond).
- Primary (\(1^\circ\)): The C-X carbon is attached to one other carbon chain. (e.g., Bromoethane)
- Secondary (\(2^\circ\)): The C-X carbon is attached to two other carbon chains. (e.g., 2-Bromopropane)
- Tertiary (\(3^\circ\)): The C-X carbon is attached to three other carbon chains. (e.g., 2-Bromo-2-methylpropane)
1.3 Uses and Environmental Impact (CFCs)
Halogenoalkanes are versatile chemicals:
- They are used as solvents (to dissolve substances).
- They form pharmaceuticals (ingredients in drugs).
- They were historically used as refrigerants and aerosol propellants.
Did you know? The use of Chlorofluorocarbons (CFCs) has been heavily restricted. CFCs are non-toxic, non-flammable, and very stable—but this stability is exactly the problem! When they reach the upper atmosphere, UV light breaks them down, releasing free chlorine radicals which catalyse the destruction of the ozone layer. This led to a global ban on most CFCs.
Quick Review: The C-X bond is polar (\(\text{C}^{\delta+}\text{-}\text{X}^{\delta-}\)), making the carbon atom an easy target for electron-rich species.
2. Nucleophilic Substitution Reactions (Substitution)
2.1 What is a Nucleophile?
A nucleophile (meaning 'nucleus lover') is a species (a molecule or an ion) that is electron-rich and uses a lone pair of electrons to attack a partially positive carbon atom (\(\text{C}^{\delta+}\)).
In substitution reactions, the nucleophile attacks the \(\text{C}^{\delta+}\) and displaces the halogen atom (the leaving group).
Key Nucleophiles in the Syllabus:
- Hydroxide ion (\(\text{OH}^-\)): Forms alcohols.
- Cyanide ion (\(\text{CN}^-\)): Forms nitriles, increasing the carbon chain length by one.
- Ammonia (\(\text{NH}_3\)): Forms amines (and then secondary, tertiary, and quaternary salts).
2.2 The Nucleophilic Substitution Mechanism
The overall reaction for substitution is:
\( \text{R-X} + \text{Nu}^- \rightarrow \text{R-Nu} + \text{X}^- \)
The mechanism (how it happens) is an example of Nucleophilic Substitution.
Step-by-Step Mechanism Outline (General \(\text{S}_{\text{N}}2\) type):
- The nucleophile (\(\text{Nu}^-\), shown with a lone pair) uses a curly arrow starting from its lone pair to attack the \(\text{C}^{\delta+}\) carbon atom.
- Simultaneously, the C-X bond breaks. A second curly arrow starts from the C-X bond and moves towards the halogen atom (X), showing that the pair of bonding electrons leaves with the halogen.
- This forms the new compound (R-Nu) and the halide ion (\(\text{X}^-\), the leaving group).
Example: Reaction with Hydroxide Ion
When a halogenoalkane reacts with aqueous potassium or sodium hydroxide, an alcohol is formed:
Example: Bromoethane reacting with aqueous hydroxide ions to form ethanol.
\( \text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^- \)
2.3 The Influence of Bond Enthalpy on Reaction Rate
The speed (rate) of a nucleophilic substitution reaction is primarily determined by how easily the C-X bond breaks, which depends on the Carbon-Halogen bond enthalpy (strength).
- Trend: Down Group 17 (F to I), atomic radius increases, and the C-X bond length increases.
- Result: Longer bonds are weaker bonds, meaning they have lower bond enthalpy.
| C-X Bond | Bond Enthalpy (\(\text{kJ mol}^{-1}\)) | Reactivity |
|---|---|---|
| C-F | High (\(\approx 484\)) | Very Slow |
| C-Cl | Medium (\(\approx 338\)) | Slow |
| C-Br | Low (\(\approx 276\)) | Fast |
| C-I | Very Low (\(\approx 238\)) | Very Fast |
The weakest bond (C-I) breaks most easily, leading to the fastest reaction rate. The strongest bond (C-F) is the hardest to break, leading to the slowest rate.
Memory Aid: The Weakest Link
Think of the C-X bond as a chain. The reaction rate depends on the time it takes to break the chain. The C-I bond is the "weakest link," so iodoalkanes react fastest!
Key Takeaway (Substitution): Nucleophilic substitution is driven by the polar \(\text{C}^{\delta+}\text{-}\text{X}^{\delta-}\) bond. The rate is controlled by the strength of the C-X bond (Iodoalkanes react fastest).
3. Elimination Reactions
3.1 The Alternative Path: Elimination
Halogenoalkanes do not just undergo substitution; they can also undergo elimination to form an alkene.
Elimination involves the removal of the halogen atom (X) and a hydrogen atom (H) from an adjacent carbon atom, resulting in the formation of a \(\text{C}=\text{C}\) double bond.
Overall Reaction Example: Elimination of 2-bromopropane forms propene.
\( \text{CH}_3\text{CH}(\text{Br})\text{CH}_3 + \text{Base} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{HBase} + \text{Br}^- \)
3.2 Conditions Matter: Substitution vs. Elimination
The conditions used dictate whether substitution or elimination dominates. We often use potassium hydroxide (\(\text{KOH}\)) or sodium hydroxide (\(\text{NaOH}\)), but the solvent changes the outcome dramatically:
- To favour Substitution: Use aqueous \(\text{KOH}\) (or \(\text{NaOH}\)) under warming. The \(\text{OH}^-\) acts as a nucleophile.
- To favour Elimination: Use ethanolic (dissolved in ethanol) \(\text{KOH}\) (or \(\text{NaOH}\)) under heating. The \(\text{OH}^-\) acts as a base.
3.3 The Dual Role of the Reagent (\(\text{OH}^-\) or \(\text{NH}_3\))
The hydroxide ion (\(\text{OH}^-\)) is capable of acting as both a nucleophile and a base:
- As a Nucleophile: It attacks the \(\text{C}^{\delta+}\) carbon atom, leading to substitution. (Favoured in aqueous solution).
- As a Base: It attacks and removes a proton (\(\text{H}^+\)) from an adjacent carbon atom, leading to elimination. (Favoured in ethanolic solution).
Ethanol is a non-aqueous, non-polar solvent that makes the hydroxide ion less solvated (less surrounded by solvent molecules), making it a stronger base and promoting elimination.
3.4 The Elimination Mechanism Outline
The mechanism involves three bonds changing simultaneously, driven by the attacking base:
- The base (\(\text{OH}^-\)) attacks and removes a hydrogen atom from the carbon atom adjacent to the C-X group. This H atom must be beta-to the C-X carbon (the \(\alpha\)-carbon).
- The electrons from the C-H bond that just broke move to form a new \(\text{C}=\text{C}\) double bond.
- Simultaneously, the C-X bond breaks, and the halide ion (\(\text{X}^-\)) is eliminated as a leaving group.
Common Mistake Alert: Remember that elimination requires the $\text{OH}^-$ to attack an H atom on a carbon next door to the C-X carbon, not the C-X carbon itself!
Key Takeaway (Elimination): Elimination (forming an alkene) requires a strong base, typically provided by heating the halogenoalkane with ethanolic hydroxide. The reagent acts as a base, abstracting a proton.