Chapter 4: Problem 10
In a protic solvent, which of the following halogens would be the best nucleophile? A \(\mathrm{Br}^{-}\) B \(\mathrm{Cl}^{-}\) C \(\mathrm{F}^{-}\) D \(I^{-}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
nucleophilicity
- High negative charge density
- Low electronegativity
- Small size (in cases of similar atoms, though this changes in protic solvents)
solvation effects
Protic solvents like water, alcohols, and carboxylic acids can form hydrogen bonds with anions. This greatly stabilizes them by surrounding them with a shell of solvent molecules. Bigger ions are less strongly solvated due to their larger radius and lower charge density, making them better nucleophiles in such solvents. Smaller ions like \(\text{F}^{-}\) are more heavily solvated, reducing their nucleophilicity because the solvent shell makes it harder for them to attack electrophiles.
halide ions
- \(\text{F}^{-}\): Very small and highly electronegative. Its nucleophilicity is low in protic solvents because it's heavily solvated.
- \(\text{Cl}^{-}\): Larger than fluoride but still quite small. Less strongly solvated compared to \(\text{F}^{-}\), making it a better nucleophile in protic solvents.
- \(\text{Br}^{-}\): Larger and even less solvated, thus an even better nucleophile in protic solvents.
- \(\text{I}^{-}\): The largest halide ion. It's the least solvated and, therefore, the best nucleophile in protic solvents.
protic solvents
In these contexts, the solvent can stabilize the nucleophile to differing extents:
- Smaller ions, such as \(\text{F}^{-}\), are heavily solvated by hydrogen bonding, which reduces their nucleophilicity.
- Larger ions, like \(\text{I}^{-}\), are less strongly solvated, thus maintaining a higher nucleophilicity.