Questions: Nucleophiles and Electrophiles: Definitions and Reactivity
3 questions to test your understanding
Score: 0 / 3
Question 1 Multiple Choice
In a protic polar solvent (e.g., water or methanol), which halide ion is the strongest nucleophile?
AF⁻
BCl⁻
CBr⁻
DI⁻
In protic solvents, nucleophilicity follows the reverse of basicity for halides: I⁻ > Br⁻ > Cl⁻ > F⁻. Although F⁻ is the strongest base (most tightly held electrons), it is also the most heavily solvated — protic solvents form strong hydrogen bonds around it, shielding its electrons and slowing attack on electrophiles. Iodide, being large and polarizable with loosely held electrons, is weakly solvated and attacks electrophiles readily. In aprotic solvents the order can reverse, illustrating that nucleophilicity is not the same as basicity.
Question 2 True / False
A stronger base is typically a stronger nucleophile.
TTrue
FFalse
Answer: False
Basicity (affinity for a proton, measured by pKa) and nucleophilicity (rate of attack on a carbon electrophile) are related but distinct properties that can be decoupled by solvent and steric effects. In protic solvents, large, polarizable species like I⁻ are strong nucleophiles despite being weak bases. Bulky bases like tert-butoxide are very basic but poor nucleophiles because steric hindrance blocks approach to carbon. The two properties diverge whenever size, polarizability, or solvation is the dominant factor.
Question 3 Short Answer
Why can an alkene act as a nucleophile even though it carries no formal negative charge and no lone pairs?
Think about your answer, then reveal below.
Model answer: The π-bond of an alkene consists of electron density above and below the plane of the double bond. These electrons are loosely held and accessible to electron-deficient species (electrophiles), making the alkene nucleophilic without requiring a lone pair or negative charge.
Nucleophilicity requires electron-rich sites capable of donating electrons to an electrophile — it does not require a formal charge or a lone pair specifically. The π-electrons in a C=C double bond occupy a diffuse region of space that is easily polarized toward an approaching electrophile. This is why electrophilic addition to alkenes begins with electrophile attack on the π-system: the alkene donates its π-electrons to the electrophile in the first mechanistic step. Recognizing π-bonds as nucleophilic sites is essential for understanding electrophilic addition reactions.