SN1, SN2, E1, E2

Question 1

What is a protic solvent and an polar aprotic solvent

Answer 1

A protic solvent is a solvent with H bonded to a strongly electronegative atom such as F, O, N or S. Such solvents undergo hydrogen bonding, and have strong dipoles.

Polar aprotic solvents are polar solvents that are not protic.

Question 2

When would you consider SN2 SN1 E1 E2 reactions?

Answer 2

We can consider these reactions if there is a nucleophile and electrophile reacting and a good leaving group.

Question 3

What does SN2 favour?

Answer 3

SN2 favours:
1. Less substituted > More substituted. Tertiary is infeasible.
2. More reactive nucleophile > less reactive
Charged > neutral
More basic > Less basic
Weak nuc — neutral = unfeasible, eg. H2O, Ethanol.
3. Polar aprotic solvent > Polar protic solvent. [SN2 & E2]

Question 4

What does SN1 favour?

Answer 4

SN1 favours:
1. A more stable carbocation intermediate formed > a less stable carbocation intermediate. Reaction is infeasible with a methyl carbocation.
2. More polar > Less polar solvent
Protic > polar aprotic

Note: Strength of base no longer matters as the spontaneous leaving is the slow step.

Question 5

What are the requirements and factors to prefer E2?

Answer 5

E2:
Summarised (no explanation):
Requirements:
1. The leaving group and abstracted H are periplanar.
2. Nucleophile is a strong base

Factors which prefer E2:
1. Heat -> More preferred. Disfavoured at room temperature. Favoured over substitution when heated. [Applicable to all elimination]
2. More substituted alkene product -> more preferred. Monosub or less → SN2 more favoured if possible.
3. Protic solvents -> Less preferred [SN2 & E2]
Polar aprotic solvents are preferred instead.

Question 6

What is the requirement for and the factors that prefer E1?

Answer 6

E1
Summarised (no explanation):
Requirement:
1. The p orbital in the C⁺ formed can rotate to be aligned to the C-H bond of the abstracted H.

Factors:
1. Heat -> More preferred. Disfavoured at room temperature. Favoured over substitution when heated. [Applicable to all elimination]
2. A more stable carbocation intermediate formed > a less stable carbocation intermediate. Reaction is infeasible with a methyl carbocation.
3. More polar > Less polar solvent
Protic > polar aprotic

Note: Strength of base no longer matters as the spontaneous leaving is the slow step.
Note: Anti-periplanar is no longer needed as a carbocation is formed.n

Question 7

Explain the factor of substitution for SN2

Answer 7

Less substituted electrophilic carbon -> More preferred. A more substituted carbon with more bulky groups causes more steric hinderance, preventing the nucleophile from reaching the carbon to react. The reaction is considered infeasible with a tri-substituted (tertiary) carbon.

Question 8

State the leaving group factor for all SN2, SN1, E2, E1

Answer 8

For them to be feasible, the electrophile must have a good leaving group.
A good leaving group forms a stable anion, or is neutral. These tend to be:
- Conjugate bases of strong acids: Stable, eg. Cl-, Br- (not F-)
- NOT strongly basic anion: eg. OH-, O2-, CH3O
- Neutral molecules: Stable, Eg. H2O from a —⁺OH2 protonated hydroxyl group
- Weaker bond strength leads to better leabing group (C-I vs C-Cl)

Question 9

Explain the effect of using a bulky base (Eg. NaOt-Bu) in Sɴ2 and E2.

Answer 9

Sɴ2 is disfavoured at primary position and not possible in secondary and tertiary positions as the base is sterically hindered from reaching the electrophilic carbon.

E2 is not disfavoured.

Elimination prefers to form = with primary carbons as H abstraction is strongly favoured at unhindered primary H.

Question 10

Explain the solvent factor for SN2 and E2

Answer 10

Protic solvents -> Less preferred. The protic solvent solvates the nucleophile through strong hydrogen bond interactions, surrounding it with solvent molecules. This introduces an energy barrier which reduces the reactivity of the nucleophile as it now has to overcome the hydrogen bonding of the solvation shell to reach and react with the electrophile.
- Polar aprotic solvents are preferred instead, as they only solvate with ion-dipole inteactions, leaving anions (nucleophiles) free to react.

Note: remember H bonding is from H bonded to strongly e- (F O N S) to lone pair on strongly e- F O N S.

Question 11

Explain the solvent effects factor for SN1 and E1

Answer 11

More polar -> More preferred. A polar solvent stabilises the carbocation intermediate by solvating the ion. Solvent molecules orient so that their electron-rich regions face the electron-deficient carbocation, releasing potential energy. The higher the solvent polarity, the stronger the ion–dipole interaction and the greater the stabilisation.
- Protic > polar aprotic. Protic solvents form stronger hydrogen bonds with the leaving group anion, releasing more potential energy and more effectively stabilising the leaving group and allowing it to leave easier.

Question 12

Explain the carbocation intermediate stability factor for SN1 and E1

Answer 12

More stable carbocation formed -> More preferred. The more stable carbocation intermediate leads to a lower energy transition state to the intermediate. This leads to a lower activation energy, increasing the rate of effective collisions, increasing the rate of the reaction and preferring the reaction. Reaction is considered infeasible with a methyl carbocation.

Question 13

Explain the factor of heat in E1 and E2 reactions.

Answer 13

Heat -> More preferred. Disfavoured at room temperature. More favoured than substitution when heated.

At room temperature, eliminations are more disfavoured than substitution as elimination tends to have a higher activation energy than substitution.

As ΔG = ΔH - T ΔS, and ΔS is positive, a higher temperature leads to a more negative ΔG, making the reaction more spontaneous and preferred.

The effect of ΔG outweights the effect of Ea at higher temperatures.

Question 14

State the criteria of nucleophile reactivity for SN2.

Note that this criteria applies to nucleophile strength in general.

Answer 14

A more reactive nucleophile -> More preferred.
- Negatively charged»_space;» Neutral. A nucleophile with a negatively charged atom has a highly reactive region of high electron density, which a neutral nucleophile lacks.
- Lower down in group -> Stronger nucleophile. Lower down the group, the valence electrons experience significantly higher shielding effect due to an extra electron shell, and there is higher nuclear charge due to more protons. The net effect is a weaker electronstatic attraction holding the electrons to the atom. Thus, the attraction needs less force to overcome and the electrons are more easily donated.
- (The group trend is actually a part of this criteria) Less electronegative atoms hold electrons more loosely and thus there is a lower energy barrier to donate them, leading to greater nucleophilicity.
- SN2 is considered infeasible with nucleophiles that are weak nucleophiles (neutral), such as ethanol and water.

Question 15

State and explain the requirements for E2.

Answer 15

The leaving group and abstracted hydrogen must be periplanar. This means both bonds that are broken in the reaction have to be aligned. This alignment is needed for the pi bond formation.

A strong base is required as the C-H bond is not weak, thus a strong base is required to abstract the H.

Question 16

Explain the mechanism, stereoselectivity and regioselectivity and rate equation of SN2.

Answer 16

Mechanism:
- One nucleophilic attack step
- Is a backside attack, where nucleophile attacks from behind the bulky leaving group.

Stereoselectivity: Stereochemistry is inverted at the attacked carbon. Imagine an umbrella flipping when the substituting nucleophile comes in from behind.
Regioselectivity: Prefers the site where SN2 is more preferred

Rate equation: First order with respect to both reactants. Necessarily so as the mechanism is only one step.

Question 17

Explain the mechanism and rate equation of E2.

Answer 17

Mechanism:
1. Concerted - Base abstracts hydrogen, and at the same time the leaving group begins to leave and the double bond begins to form.

Rate equation: One step and involves two reactants, second order overall, first order wrt both reactants

Question 18

Explain the mechanism, stereoselectivity and regioselectivity and rate equation of SN1.

Answer 18

Mechanism:
1. Leaving group leaves spontaneously to form carbocation (SLOW)
2. Nucleophile attacks C+ (FAST)

Stereoselectivity: Stereochemistry is lost as the planar carbocation intermediate allows nucleophilic attack from both sides.
- The final mixture is a mixture of both enantiomers but may not be racemic. An enantiomer may be more preferred.

Regioselectivity:
- Prefers the site where SN1 is more preferred
- Rearrangements possible

Rate reaction: Slow step involves only the electrophile.

Question 19

Explain the mechanism and rate equation of E1.

Answer 19

R: Electrophile, nucleophile
Mechanism:
1. The leaving group spontaneously leaves, forming a carbocation (SLOW)
2. Acid-base: The strong base (nucleophile) abstracts the adjacent H and the C=C forms. (FAST)

Rate equation is first-order with respect to the electrophile.

Question 20

State and explain the requirement for E1.

Answer 20

The p orbital in the C⁺ formed must be able to rotate to be aligned to the C-H bond of the abstracted H. This alignment is needed for the pi bond formation.

Question 21

State two overarching principles in deciding E1 and E2 regioselectivity and stereoselectivity.

Answer 21

Overarching principles:
1. The more stable alkene is formed. This is as the intermediates are similar to the product, thus a more stable product leads to more stable intermediates and a lower activation energy.
2. The H which is easier to abstract is favoured. It leads to a lower activation energy in the H abstraction step.

Question 22

State E1 and E2 regioselectivity and stereoselectivity.

Answer 22

Regioselectivity
Rearrangements of the carbocation intermediate in E1 can occur.

More stable alkene:
1. A more substituted alkene is preferred
2. The alkene with less EWG is preferred

Easier H to abstract:
1. Elimination at an = to form a ≡ is preferred as the C-H bond is weaker in that case.

Stereoselectivity
The stereochemistry where bulky groups are opposite is preferred if not geometrically restricted.

Question 23

Explain why a more substituted alkene product favours E2.

Answer 23

A more stable product is formed, leading to a more stable transition state and a lower Ea.

If the alkene is monosubstituted or less, SN2 will be more favoured than E2 if possible.

Question 24

Explain the steps for deciding what pathways occur.

Answer 24

1. Rule out impossible pathways
   SN2 not possible on tertiary substrates or neutral weak nucleophile
   E2 not possible with weak base
   SN1, E1 not possible on primary substrate
   Bulky base: SN2 disfavoured on primary and impossible on greater
2. Apply the rule that SN2 E2»_space;> SN1 E1
3. Use the various factors to determine which is major and which is minor, if any. In order of importance:
   Heat: Elimination > substitution, Rtp/cool is opposite
   Monosub or less: E2 disfavoured, SN2 more favoured if possible.

Note: Solvent effects are so minor they dont affect which is preferred, just the rate of rxn.

Question 25

If Sɴ2 or E2 are preferred, will they be more significant than Sɴ1 and E1 even if they are also preferred?

Answer 25

E1 and Sɴ1 both have a slow rate determining step of leaving group leaving spontaneously. This is much slower than the direct nucleophilic attack in Sɴ2 and E2 if they are preferred. This takes precedence over all other preferring effects.