The Influence of the Leaving Group on the Rate of Nucleophilic Substitution Reactions
Understanding the Role of the Leaving Group
What is a Leaving Group?
Leaving group
A leaving group is an atom or group of atoms that detaches during a substitution reaction.
In nucleophilic substitution reactions, the nucleophile donates an electron pair to the electron-deficient carbon atom of the halogenoalkane, while the leaving group departs with the bonding electrons.
$$ \text{CH}_3\text{CH}_2\text{Cl} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Cl}^- $$
Here, the hydroxide ion ($ \text{OH}^- $) replaces the chlorine atom, and the chloride ion ($ \text{Cl}^- $) becomes the leaving group.
What Makes a Good Leaving Group?
- The effectiveness of a leaving group depends on its ability to stabilize the negative charge it carries after leaving.
- A good leaving group typically:
- Stabilizes the negative charge:
- When the group departs, it takes the bonding electrons, becoming an anion.
- The more stable this anion, the easier it is for the group to leave.
- Forms a weak bond with carbon:
- Weaker bonds break more easily, facilitating faster reactions.
- Stabilizes the negative charge:
- Halide ions (e.g., $ \text{F}^- $, $ \text{Cl}^- $, $ \text{Br}^- $, $ \text{I}^- $) are common leaving groups in halogenoalkanes.
- Their leaving ability increases down Group 17 of the periodic table due to decreasing bond strength and increasing anion stability.
When evaluating leaving groups, prioritize both bond strength and the stability of the anion formed after leaving.
The Rate of Substitution and the Leaving Group
Bond Strength: The Carbon-Halogen Bond
- The rate of nucleophilic substitution reactions is closely tied to the strength of the carbon-halogen bond.
- Bond strength is measured by bond enthalpy, which represents the energy required to break the bond.
The table below shows the bond enthalpies for carbon-halogen bonds:
| Bond | Bond enthalpy ($\text{kJ mol}^{-1}$) |
|---|---|
| C-F | 492 |
| C-Cl | 324 |
| C-Br | 285 |
| C-I | 228 |
- From the table, you can see that the C–F bond is the strongest, while the C–I bond is the weakest.
- This means that fluoroalkanes are highly resistant to nucleophilic substitution, while iodoalkanes react much more quickly.
Stability of the Leaving Group
- In addition to bond strength, the stability of the halide ion formed as the leaving group plays a critical role.
- Larger halide ions, such as $ \text{I}^- $, are better at dispersing their negative charge due to their size, making them more stable.
- This increased stability enhances the likelihood of the halide ion leaving, speeding up the reaction.
Predicting Relative Rates of Substitution
Let’s compare the rates of substitution for halogenoalkanes containing different halogens:
- Fluoroalkanes ($ \text{C–F} $):
- The C–F bond is extremely strong, and the fluoride ion ($ \text{F}^- $) is a poor leaving group due to its small size and high charge density.
- As a result, fluoroalkanes are virtually inert in nucleophilic substitution reactions.
- Chloroalkanes ($ \text{C–Cl} $):
- The C–Cl bond is weaker than the C–F bond, and the chloride ion ($ \text{Cl}^- $) is a more stable leaving group.
- Chloroalkanes react at a moderate rate.
- Bromoalkanes ($ \text{C–Br} $):
- The C–Br bond is weaker than the C–Cl bond, and the bromide ion ($ \text{Br}^- $) is an even better leaving group.
- Bromoalkanes react faster than chloroalkanes.
- Iodoalkanes ($ \text{C–I} $):
- The C–I bond is the weakest, and the iodide ion ($ \text{I}^- $) is the most stable leaving group.
- Iodoalkanes react the fastest in nucleophilic substitution reactions.
Comparing Reaction Rates
Rank the following halogenoalkanes in order of increasing reaction rate with aqueous sodium hydroxide: $ \text{CH}_3\text{CH}_2\text{Cl} $, $ \text{CH}_3\text{CH}_2\text{Br} $, $ \text{CH}_3\text{CH}_2\text{I} $.
Solution
- The rate of reaction increases as the leaving group improves:
$$
\text{CH}_3\text{CH}_2\text{Cl}< \text{CH}_3\text{CH}_2\text{Br} < \text{CH}_3\text{CH}_2\text{I}
$$ - This is because the bond strength decreases and the stability of the leaving group increases from $ \text{Cl}^- $ to $ \text{Br}^- $ to $ \text{I}^- $.
- Many students mistakenly believe that the most electronegative halogen (fluorine) always makes the best leaving group.
- However, bond strength and the stability of the leaving group are far more significant factors than electronegativity.
- Why is the iodide ion a better leaving group than the fluoride ion?
- How would the rate of substitution change if the nucleophile was weaker, such as water instead of hydroxide?
