Page 20 Flashcards
Why does the quality of the leaving group affect the rate of E2 reactions?
A: A better leaving group stabilizes the transition state by leaving more easily, increasing the rate of the reaction.
What is the general order of leaving group reactivity in E2 reactions for halides?
A: \text{R-F} < \text{R-Cl} < \text{R-Br} < \text{R-I} , with \text{R-I} being the best leaving group.
How does bond strength correlate with leaving group ability in halides?
A: Weaker carbon-halogen bonds (e.g., C-I) make better leaving groups than stronger bonds (e.g., C-F).
What is the effect of polar aprotic solvents on E2 reaction rates?
A: Polar aprotic solvents increase the rate of E2 reactions by solvating the base but not the nucleophile.
Why are polar aprotic solvents preferred in E2 reactions?
A: They do not stabilize the nucleophile excessively, allowing the base to remain reactive and promote elimination.
What happens to the E2 reaction rate when a poor leaving group, such as \text{R-F} , is used?
A: The rate of the E2 reaction decreases significantly due to the difficulty of bond cleavage.
How does the leaving group affect the transition state of the E2 reaction?
A: A good leaving group reduces the energy barrier by stabilizing the transition state, facilitating bond breaking.
Which halide leaves most easily during E2 elimination, and why?
A: \text{Iodide (I}^-\text{)} leaves most easily due to its large size and high polarizability, which stabilizes the negative charge.
In what way does a polar aprotic solvent enhance the reactivity of bases in E2 reactions?
A: Polar aprotic solvents reduce solvation of anions, increasing the base’s nucleophilicity and reactivity
Why does the bond to the leaving group break partially in the transition state of an E2 reaction?
A: Partial bond breaking facilitates the simultaneous removal of the proton by the base and formation of the π bond.
