Elimiation At Sp3 Centers Flashcards

1
Q

What is sigma conjugation?

A

Delocalisation of electrons though sigma bonds

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2
Q

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

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

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

A

Tertiary, secondary, primary

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4
Q

Why is the order for E2 reactions substrates opposite to SN2

A

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)

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5
Q

What conditions favour E2 reactions

A
  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)
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6
Q

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

A

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

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7
Q

Why do elimination reactions favour high temperatures?

A

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

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8
Q

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

A

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.

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9
Q

What is Hammonds postulate?

A

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.

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10
Q

Why are more substituted alkenes more stable?

A

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)

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11
Q

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

A

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

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12
Q

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

A

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

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13
Q

can E1 and E2 reactions form a less substituted alkene?

A

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

E2: yes with a bulky base

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14
Q

Which stereoisomer does E1 reactions prefer and why?

A

E alkene preferred due to less steric hindrance

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15
Q

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

A

specific

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16
Q

What overlap is required for E2 reaction to proceed?

A

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

17
Q

Which conformation is preferred for an E2 reaction and why?

A

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

18
Q

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

A

E2 prefers E alkene due to lower steric hindrance

19
Q

What are the 3 products formed in an E2 reaction?

A

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

20
Q

What is the requirement for E2 elimination of cyclohexanes

A

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

21
Q

why must E2 reactions be antiperiplanar

A

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

22
Q

Why can E1cB reactions occur and what are they?

A

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

23
Q

Explain how the E1cB reaction occurs

A

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

24
Q

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

A

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

25
Q

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

A

concerted reaction

26
Q

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

A

E due to steric hindrance

27
Q

which uses a weaker base: E1 or E2 reactions?

A

E1