Nucleophilic Substitution Reactions: Mechanisms and Electron Movement
The General Mechanism of Nucleophilic Substitution
- At its core, a nucleophilic substitution reaction involves an electron-rich species, called the nucleophile (Nu⁻), donating a pair of electrons to an electron-deficient carbon atom in a substrate molecule, R-X.
- This carbon is bonded to a leaving group, X⁻, which gets displaced during the reaction.
The general equation for this reaction is:
Here’s a step-by-step breakdown of what happens:
- Nucleophilic Attack:
- The nucleophile uses its lone pair of electrons to form a new covalent bond with the carbon atom in the substrate.
- Bond Breaking:
- At the same time, the bond between the carbon and the leaving group breaks, transferring the bonding electrons to the leaving group.
- Formation of Products:
- The reaction produces a new molecule, R-Nu, and the leaving group, X⁻.
Key Players in the Reaction
- Nucleophile (Nu⁻):
- An electron-rich species, such as OH⁻, CN⁻, or NH₃, that donates a pair of electrons.
- Substrate (R-X):
- Often a halogenoalkane, where the carbon atom bonded to the halogen is electron-deficient due to the polar nature of the C–X bond.
- Leaving Group (X⁻):
- A group that departs with the bonding electrons.
- Good leaving groups (e.g., halide ions like Cl⁻ or Br⁻) are stable and weakly basic.
Tip
A good leaving group is one that can stabilize the negative charge it acquires after leaving, making the reaction more favorable.
Note
Other details of the nucleophilic substitution will be covered by HL students in the later sections.
Example
Halogen Substitution Reaction
Let’s examine the reaction between bromoethane (CH₃CH₂Br) and hydroxide ions (OH⁻), which produces ethanol (CH₃CH₂OH) and a bromide ion (Br⁻):
Step-by-Step Mechanism
- Nucleophilic Attack: The hydroxide ion (OH⁻) uses its lone pair of electrons to attack the electron-deficient carbon atom in bromoethane.
- Bond Breaking: The bond between the carbon and bromine breaks, transferring the bonding electrons to the bromine atom.
- Formation of Products: The final products are ethanol (CH₃CH₂OH) and a bromide ion (Br⁻).
Factors Affecting Nucleophilic Substitution Reactions
Several factors influence the rate and outcome of nucleophilic substitution reactions:
- Strength of the Nucleophile:
- Stronger nucleophiles (e.g., OH⁻, CN⁻) react faster because they donate electrons more readily.
- Nature of the Leaving Group:
- Good leaving groups (e.g., Br⁻, I⁻) facilitate the reaction.
- Fluoride (F⁻) is a poor leaving group due to its strong bond with carbon.
- Polarity of the Solvent:
- Polar solvents stabilize charged intermediates and transition states, aiding the reaction.
- Type of Substrate:
- Primary halogenoalkanes often undergo a one-step mechanism (SN2), while tertiary halogenoalkanes typically follow a two-step mechanism (SN1).
Note
The distinction between SN1 and SN2 mechanisms will be covered in detail in later sections.
Reflection and Practice
Self review
- Can you identify the nucleophile, substrate, and leaving group in a given reaction? Are you able to correctly draw curly arrows to show electron movement?
- Explain why bromide ions (Br⁻) are better leaving groups than fluoride ions (F⁻).
- Predict the products of the reaction between 1-bromopropane and ammonia (NH₃).
Theory of Knowledge
In Theory of Knowledge, consider this: How do models like curly arrows help chemists understand and predict chemical reactions? What are the limitations of representing reactions this way?
