Module 9 Practice Set: Substitution and Elimination Reactions

Leaving Groups and Nucleophiles Sample Questions:

Leaving Groups:

In each of the following molecules, identify any leaving groups that are present:

 

 

 

 

Answer Key:

Bromine is a halogen, and all halogens with the exception of fluorine are good leaving groups because they are weak bases once they leave. Weak bases are stable, so bromine leaving is energetically favorable. The bromine in this molecule is also bonded to an sp3 carbon, therefore it can act as a leaving group.

Halogen leaving groups: Cl, Br, I

 

Nucleophiles:

Compare the following pairs of molecules and determine which molecule is the better nucleophile. Explain your answer using different qualities such as formal charge, partial charge, electronegativity, etc.

Answer Key:

Br- is a better nucleophile than I-. This is because bromine is more electronegative than iodine, leading to a larger partial negative charge and higher electron density on the bromine atom. Higher electron density enhances the nucleophilicity of Br-, making it more reactive towards electrophiles compared to I-.

 

Identifying Reaction Conditions Sample Question:

For each given reaction, determine whether the reaction conditions are acidic, basic, or neutral. Indicate your answer clearly. Based on the identified reaction conditions, state which type(s) of reactions could occur under those conditions. The possible reaction pathways to consider are SN1, SN2, E1, or E2. Provide a brief explanation for each choice.

Answer Key:

HCl is an HX acid, so this arrow represents acidic conditions.

Potential reactions: SN1, acid/base

HCl is an acid which rules out SN2 or E2, and it has a small/nucleophilic conjugate base, so it will perform an SN1 reaction. The other reaction HCl can perform is an acid/base reaction where it can protonate a neutral or mildly basic group such as R-OH, R-NH2, etc. because it is a strong acid.

 

SN1 Reaction Sample Question:

For each of the following SN1 reactions, predict the major product. Then, provide a detailed electron-pushing mechanism to illustrate the SN1 reaction process.

Answer Key:

Predicting the Product:

• The starting molecule does not have a leaving group, however it does have an OH group that can be turned into a leaving group via protonation.
• The reaction arrow has HBr, which is an HX acid. That tells us the HBr will turn the OH into an OH2+ leaving group.
• We know this reaction proceeds through a carbocation intermediate because it is an SN1 reaction. Therefore, we need to examine the carbons adjacent to the leaving group to determine if a carbocation rearrangement will occur. In this case, the adjacent carbons are not more substituted and would not provide any resonance stabilization so a carbocation rearrangement will not occur.
• We now know the OH2+ leaving group will be switched with the conjugate base of the HBr acid. Therefore, the final product will have a Br substituted in place of the OH. There is no stereocenter created in the final product, so only one version needs to be drawn.

Mechanism Steps Explained:

1. In the first step, the OH is protonated by HBr to create OH2+. This is done because OH is not a leaving group, but OH2+ is. Note that the Br- conjugate base is not shown in the products of this step, but it will be used later on in the mechanism.
2. Next, OH2+ leaves on its own which forms a carbocation intermediate. The leaving group will always leave on its own in an SN1 reaction to ensure the formation of a carbocation intermediate.
3. The Br- conjugate base generated in step 1 then reacts with the carbocation intermediate. This forms the final product. Note that the Br- bonds with the carbocation above and below the plane, however only one product is shown because the addition of the Br- does not create a stereocenter.

 

E1 Reaction Sample Question:

For each of the following E1 reactions, predict the major product. Then, provide a detailed electron-pushing mechanism to illustrate the E1 reaction process.

Answer Key:

 

Predicting the Product:

• The starting molecule does not have a leaving group, however it does have an OH group that can be turned into a leaving group via protonation.
• The reaction arrow has H2SO4 which is a strong acid. That tells us the H2SO4 will turn the OH into an OH2+ leaving group.
• We know this reaction proceeds through a carbocation intermediate because it is an E1 reaction. Therefore, we need to examine the carbons adjacent to the leaving group to determine if a carbocation rearrangement will occur. In this case, the adjacent carbons are not more substituted and do not offer any resonance stabilization for the carbocation, so no rearrangement will occur.
• H2SO4 has a large/non-nucleophilic conjugate base, so it cannot attack the carbocation intermediate. Instead, it will eliminate the most substituted beta-hydrogen.
• This elimination leads to the formation of an alkene as the final product.

Mechanism Steps Explained:

1. In the first step, the OH is protonated by H2SO4 to create OH2+. This is done because OH is not a leaving group, but OH2+ is. Note that the HSO4- conjugate base is not shown in the products of this step, but it will be used later on in the mechanism.
2. Next, OH2+ leaves on its own which forms a carbocation intermediate. The leaving group will always leave on its own in E1 reactions to ensure the formation of a carbocation intermediate.
3. The carbocation is then reacted with the HSO4- conjugate base, but the carbocation is not attacked because of how big the HSO4- base is. Instead, a beta-hydrogen from the most substituted carbon adjacent to the carbocation is eliminated to form a pi bond. This forms the final product.

 

SN2 Reaction Sample Question:

For each of the following SN2 reactions, predict the major product. Then, provide a detailed electron-pushing mechanism to illustrate the SN2 reaction process.

Answer Key:

Predicting the Product:

• The starting molecule has a bromine leaving group, and the reaction arrow has a small/non-bulky OH- base. Na is a group 1 cation, so we know that NaOH will split up into Na+ and OH-.
• The bromine is on a primary carbon, so even though OH- will form a bond through backside attack, a change in stereochemistry will not be seen in the product.
• The final product will have an OH in place of the Br, bonded with what appears to be the same stereochemistry.

Mechanism Steps Explained:

1. SN2 reactions are concerted, meaning everything happens in one step. Therefore, the OH- base attacks the carbon the bromine leaving group is on at the same time the bromine leaving group leaves.
2. This concerted step forces the OH- to form a bond to the starting molecule via backside attack, however because the bromine is on a primary carbon, there is no visible change in stereochemistry in the final product.

 

E2 Reaction Sample Question:

For each of the following SN2 reactions, predict the major product. Then, provide a detailed electron-pushing mechanism to illustrate the SN2 reaction process.

Answer Key:

Predicting the Product:

• The starting molecule has a bromine leaving group, and the reaction arrow has a large/bulky OEt- base. Na is a group 1 cation, so we know that NaOEt will split up into Na+ and OEt-. Note that the EtOH solvent is the conjugate acid of the OEt- base so it has no effect on the reaction outcome.
• The bromine is bonded with a wedge to the carbon chain, and because the reaction is E2 we know that the beta-hydrogen that is eliminated has to be bonded with a dash so that it is anti-coplanar to the leaving group. We also want to eliminate the most substituted beta-hydrogen so that the most substituted alkene forms. Therefore, the beta-hydrogen we will eliminate is bonded to the secondary carbon to the right of the leaving group with a dash.
• The final product will have a pi bond between the carbon the leaving group was on and the carbon that had the beta-hydrogen that was eliminated.

Mechanism Steps Explained:

1. E2 reactions are concerted, meaning everything happens in one step. Therefore, the OEt- base eliminates the wedged beta-hydrogen on the secondary carbon to the right of the leaving group at the same time the bromine leaving group leaves.
2. This concerted step forces the OEt- to eliminate a beta-hydrogen that is in the opposite plane (anti-coplanar) relative to the leaving group. This creates an alkene and forms the final product.

 

Substitution vs Elimination Reaction Determination Sample Question:

For each of the following reactions, predict the major product. Then, provide a detailed electron-pushing mechanism to illustrate the reaction process.

Answer Key:

Reaction Determination:

1. First we need to determine the substitution of the carbon the leaving group is bonded to. The OH that will be turned into a leaving group on the starting molecule is on a secondary carbon. This does not rule out any reactions.
2. The reaction conditions are acid because the arrow has HBr which is an HX acid. Acidic conditions tell us the reaction with be SN1 or E1.
3. HBr has a small/nucleophilic conjugate base, which tells us the reaction with be an SN1 reaction.

Predicting the Product:

• The starting molecule does not have a leaving group, however it does have an OH group that can be turned into a leaving group via protonation.
• The reaction arrow has HBr, which is an HX acid. That tells us that the HBr will turn the OH into an OH2+ leaving group.
• We know this reaction proceeds through a carbocation intermediate because it is an SN1 reaction. Therefore, we need to examine the carbons adjacent to the leaving group to determine if a carbocation rearrangement will occur. In this case, the adjacent carbon to the right of the leaving group is more substituted so a rearrangement will occur via a 1,2 methanide shift.
• We now know the OH2+ leaving group will be switched with the conjugate base of the HBr acid. Therefore, the final final product will have a Br substituted on the more substituted carbon carbon adjacent to the OH group in the starting molecule. There is no stereocenter created in the final product, so only one version needs to be shown.

Mechanism Steps Explained:

1. In the first step, the OH is protonated by HBr to create OH2+. This is done because OH is not a leaving group, but OH2+ is. Note that the Br- conjugate base is not shown in the products of this step, but it will be used later on in the mechanism.
2. Next, OH2+ leaves on its own which forms a carbocation intermediate. The leaving group will always leave on its own in SN1 reactions to ensure the formation of a carbocation intermediate.
3. The carbocation then rearranges to the carbon with two methyl groups to the right via a 1,2 hydride shift. This occurs because that shift will change the carbocation from secondary to tertiary.
4. The Br- conjugate base generated in step 1 then reacts with the carbocation intermediate and bonds both above and below the plane of the carbocation. This forms the final product. Note that because no stereocenter was formed, only one version of the product needs to be shown.