Elimination Reactions

Question 1

Give the requirements of an E1 reaction

Answer 1

Secondary or tertiary as you need stable carbocation
Need a weak base as it does not partake in RDS is formation of carbocation by departure of leaving group
Need a very good leaving group as departure is RDS
Carried out in a polar protic solvent
High temperature

Question 2

What type of alkene is formed in E1?

Answer 2

E alkenes are favoured as they are lower in energy than Z alkenes as substituents can get further apart from each other and there is less steric hinderance

Question 3

What is the rate-determining step of an E1 reaction

Answer 3

Unimolecular where the LG departs and a carbocation is formed

Question 4

Draw the general mechanism of E1 reaction

Answer 4

Refer to notes

Question 5

Why is a more substituted alkene more stable and forms faster?

Answer 5

Because the pi bond is stabilised when the empty pi star orbital can interact with filled orbitals of parallel C-C and C-H. The more of either bonds the more stable the alkene. Removal of proton will also lead to a multi-substituted carbocation which is lower in energy in E1

Question 6

Give the general mechanism of an E2 reaction

Answer 6

Refer to notes

Question 7

What is the rate-determining step of an E2 reaction?

Answer 7

Bimolecular that must have a high concentration of a base - the base attacks the proton leading to deprotonation and the leaving group leaves

Question 8

What are the requirements for an E2 reaction?

Answer 8

A high concentration of strong base is needed however if a bulky base is used then the least substituted alkene (Hoffman product) is produced
LG is good to moderate needed as attack and deprotonation and departure of LG is RDS
Need antiperiplanar orientation of leaving group and proton
Polar aprotic solvents e.g. DMSO, acetone which do not solvate (hydrogen bond with) the strong base and allow it to attack

Question 9

What is the stereochemistry of E2 reaction products like?

Answer 9

E alkenes are favoured and product results from which proton is anti-periplanar to the LG. Reaction is stereo specific and depends on the diastereoisomer of the starting material

Question 10

What is the new pi bond made from in E2 reactions?

Answer 10

Formed by the overlap of the C-H sigma bond with the C-X sigma* antibonding orbital

Question 11

What happens if you alter the base where it is more hindered in E2?

Answer 11

It alters the mechanism and the base will attack the less hindered H giving more regioselectivity for less hindered alkenes

Question 12

What is the RDS of the E1cb mechanism

Answer 12

Departure of LG

Question 13

Give the reaction process and reaction coordinate of an E1cb mechanism

Answer 13

Acidic proton is removed next to carbonyl group which is quite acidic which leaves an anion with the LG which is stable as charge can be delocalised onto the carbonyl group. LG departs

Question 14

Give the reaction mechanism of an E1cb reaction

Answer 14

Refer to notes

Question 15

What determines the regioselectivity?

Answer 15

Location of double bonds depend on location of acidic proton and the LG

Question 16

What is the stereoselectivity of an E1cb reaction?

Answer 16

May be stereoselective and E alkene is favoured as it is less sterically hindered

Question 17

What factors favour elimination over substitution?

Answer 17

Hard nucleophiles as the C-H sigma* bond is higher in energy as the atoms are less electronegative
Increasing branching and strongly basic hindered nucleophiles

Question 18

Why can methyl halides not eliminate?

Answer 18

As there is no appropriately placed proton

Question 19

Give the requirements of an E1cb reaction

Answer 19

Moderate leaving group
Strong base and heat
Anion stabilising electron withdrawing group
Trans (E alkene) products dominate
Polar protic solvent needed e.g. alcohol and/or water

Question 20

What is the stereochemistry of E1 reactions like?

Answer 20

It is stereoselective and mainly forms E alkene due to less steric hinderance as bulky groups are on opposite sides, reducing steric strain, but some Z alkene as well, as the carbocation allows for free rotation before proton removal

Question 21

What happens if the leaving group is too good in an E1cb reaction?

Answer 21

E2 may occur before the conjugate base can be formed