Elimiation At Sp3 Centers

Question 1

What is sigma conjugation?

Answer 1

Delocalisation of electrons though sigma bonds

Question 2

What are the 3 factors that decide whether it will be a substitution or elimination reaction (say the factors that favour elimination)

Answer 2

1. Basicity (more basic is elimination)
2. Size (larger/bulkier prefers elimination)
3. Higher temperature prefers elimination

Question 3

What is the order of reactivity for E1 and E2 reactions?

Answer 3

Tertiary, secondary, primary

Question 4

Why is the order for E2 reactions substrates opposite to SN2

Answer 4

For E2, tertiary substrates have greater beta carbons, thus the no. of hydrogens increase and the rate of deprotonation increases. The more substituted alkene is the more stable, making the transition state and final product more stable (less energy required thus formed more rapidly)

Question 5

What conditions favour E2 reactions

Answer 5

1. Increasing concentration of base
2. Stronger bases (more acidic more E1 since the protonation of a group can make the leaving group more stable such as H2O)
3. No acid catalyst (those favour E1)

Question 6

Why does the methyl group not go through elimination or SN1 reactions

Answer 6

Carbocation intermediate is too unstable (no sigma donation) for SN1, and no beta carbons to deprotonate for elimination reactions.

Question 7

Why do elimination reactions favour high temperatures?

Answer 7

Gibbs free energy equation, temperature and entropy (entropy does not change for substitution while it increases for elimination reactions)

Question 8

For E2 reactions, is the internal or terminal alkene bond more likely to be formed and why?

Answer 8

The more substituted alkene (internal bond) is the more favoured product because factors that stabilize the product (alkene) will also stabilize the transition state and a more substituted product is lower in energy which means lower activation energy required thus more favoured.

Question 9

What is Hammonds postulate?

Answer 9

If two states, such as a transition state and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will result in only a minor reorganisation of molecular structures.

Question 10

Why are more substituted alkenes more stable?

Answer 10

More substituted means more beta bonds (from neighbouring C-H or C-C) which can overlap favourably with the antibonding pi orbital of the C=C and stabilize it (as it forms a partial MO which is puts the electrons in a lower energy/more stable orbital)

Question 11

What is the role of each step in determining the product of E1 reactions

Answer 11

Step 1: rate determining step (Ea)
Step 2: product determining step (which part is deprotonated)

Question 12

what are the conditions required to form the less substituted alkene for E2 reactions?

Answer 12

changing to a bulky base, the steric hindrance prevents access to certain hydrogens (within a ring for eg.) and thus the base would more easily deprotonate a substituent which forms the less substituted product

Question 13

can E1 and E2 reactions form a less substituted alkene?

Answer 13

E1: no, it always forms the more substituted product (due to the formation of a carbocation)

E2: yes with a bulky base

Question 14

Which stereoisomer does E1 reactions prefer and why?

Answer 14

E alkene preferred due to less steric hindrance

Question 15

What is a reaction that has 100% of one product over another called?

Answer 15

specific

Question 16

What overlap is required for E2 reaction to proceed?

Answer 16

the C-H sigma bond must be in the same plane as the antibonding orbital of the C- leaving group because it allows for the best orbital overlap to form the double bond

Question 17

Which conformation is preferred for an E2 reaction and why?

Answer 17

Antiperiplanar is preferred because there is less steric hindrance and orbital overlap is better (due to less steric hindrance)

Question 18

Which confirmation (E or Z) does E2 prefer and why?

Answer 18

E2 prefers E alkene due to lower steric hindrance

Question 19

What are the 3 products formed in an E2 reaction?

Answer 19

1 terminal alkene
2 internal alkene (E and Z stereochemistry)

Question 20

What is the requirement for E2 elimination of cyclohexanes

Answer 20

both the hydrogen and the leaving group must be axial so that the bonds are antiperiplanar
- equatorial C-H bonds are only antiperiplanar to C-C bonds which means it does not have optimal orbital overlap to C- leaving group bond

Question 21

why must E2 reactions be antiperiplanar

Answer 21

so that the C-H sigma bond and the leaving group pi star orbital (that is to be occupied) have optimal overlap which lowers the energy of the transition state

Question 22

Why can E1cB reactions occur and what are they?

Answer 22

E1cB is when poor leaving groups can become leaving groups due to the presence of a stable anion (due to Elec withdrawing groups) after the hydrogen has been taken by a strong base

Question 23

Explain how the E1cB reaction occurs

Answer 23

A strong base deprotonates the acidic hydrogen resulting in a negative anion.
The re-stabilization due to the negative charge results in the departure of the leaving group and forms the double bond

Question 24

How many steps are in the E1cB reaction and which one is the RDS?

Answer 24

2 steps, second step (departure of leaving group) is the RDS

Question 25

What is the name of a reaction where the leaving group and the double bond is being formed at the same time?

Answer 25

concerted reaction

Question 26

in E1cB reactions, is the E or Z confirmation prefered?

Answer 26

E due to steric hindrance

Question 27

which uses a weaker base: E1 or E2 reactions?

Answer 27

E1