E1 vs E2

Question 1

How does the increasing the concentration of base affect E1 and E2?

Answer 1

Increasing the concentration of base or changing to a base with greater basic strength increases the rate of E2 but not that of an E1 this is because the base takes part in the rate determining step of an E2 reaction but not that of an E1 reaction. Therefore a relatively high concentration of a strong base reacts more rapidly with a primary halogenoalkane in an E2 reaction that a relatively low concentration of a weaker base. Aldo a relatively high concentration of a strong base is likely to react with a secondary halogenoalkane in an E2 rather than an E1 reaction

Question 2

What halogenoalkanes undergo which type of elimination reaction?

Answer 2

Primary halogenoalkanes only undergo slow E2 reactions, whereas secondary and tertiary halogenoalkanes undergo faster E2 reactions and also E1 reactions

Question 3

How does increasing the solvent polarity affect the rate of E1 and E2?

Answer 3

Slightly decreases the E2 rate (polar solvents solvate the base)
Significant increase in E1 rate (polar solvents solvate the carbocation and halide ion)

Question 4

How does changing from a polar protic to a polar parotid solvent affect the rate of E1 and E2?

Answer 4

Increase in rate of E2 (aprotic solvents do not solvate the base)
Decrease in the rate of E1 (aprotic solvents solvate the carbocation but not the halide ion)

Question 5

How does the leaving group affect the rate of elimination reactions?

Answer 5

Changing from a poor leaving group to a good leaving group in a halogenoalkane increases the rates of both E2 and E1 - this is because the rate determining step of both elimination reactions involve heterolytic cleavage of the C-X bond and the better the leaving group the faster the leakage of the C-X bond