Module 5 Practice Set: Nucleophilic Addition Reactions with Alkenes

HX Addition Sample Question

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

Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• HX addition results in the Markovnikov addition of whatever the halogen is in the HX acid on the reaction arrow. In this reaction, the halogen is bromine.
• All HX addition reactions go through carbocation intermediates, so we need to examine the carbons adjacent to the Markovnikov carbon to determine if a carbocation rearrangement will occur. In this reaction, the carbons adjacent to the Markovnikov carbon are not more substituted and do not provide any additional resonance stabilization so a rearrangement will not occur.
• Therefore, we will add a bromine to the Markovnikov carbon and remove the pi bond. The reaction goes through a carbocation intermediate, so the bromine can bond both above and below the plane of the molecule which results in two products being generated.

Mechanism Steps Explained:

1. In step 1, the pi bond forms a new C-H sigma bond to the hydrogen in HBr which adds it to the anti-Markovnikov carbon. At the same time, the electrons in the H-Br sigma bond both go to bromine. This step results in a carbocation because the Markovnikov carbon loses 1 electron from the pi bond that moved. The hydrogen adds to the anti-Markovnikov carbon because it is more energetically favorable for the carbocation to form on the Markovnikov carbon because it is the more substituted carbon.
2. The Br- generated in step 1 forms a bond to the carbocation by donating 1 electron from a lone pair. The bromine can bond above and below the plane of the molecule, so two products are shown.

 

Acid Catalyzed Hydration Sample Question

For each of the following acid catalyzed hydration reactions, predict the major product(s). Then, provide a detailed electron-pushing mechanism to illustrate the reaction process.

Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• Acid catalyzed hydration results in the Markovnikov addition of an alcohol (OH) when the solvent is H2O. Note that the H2O solvent is in excess, so it will react away all of the H2SO4 added in solution which makes H3O+ the acid in this reaction.
• All acid catalyzed hydration reactions go through a carbocation intermediate, so we need to examine the carbons adjacent to the Markovnikov carbon to determine if a carbocation rearrangement will occur. In this reaction, the carbons adjacent to the Markovnikov carbon are not more substituted and do not provide any additional resonance stabilization so a rearrangement will not occur.
• Therefore, we will add an OH group to the Markovnikov carbon and remove the pi bond. The reaction goes through a carbocation intermediate, so the OH can bond both above and below the plane of the molecule which results in two products being generated.

Mechanism Steps Explained:

1. In step 1, the pi bond forms a new C-H sigma bond to a hydrogen in H3O+ which adds it to the anti-Markovnikov carbon. At the same time, the electrons in the H-O sigma bond both go to oxygen. This step results in a carbocation because the Markovnikov carbon loses 1 electron from the pi bond that moved. The hydrogen adds to the anti-Markovnikov carbon because it is more energetically favorable for the carbocation to form on the Markovnikov carbon because it is the more substituted carbon.
2. The H2O conjugate base generated in step 1 forms a bond to the carbocation by donating 1 electron from a lone pair on the oxygen. The H2O can bond above and below the plane of the molecule, so two products are shown.
3. Once the H2O bonds to the carbocation, it has a formal positive charge. Therefore, it is deprotonated by another H2O so that the OH2+ group is turned into a formally neutral OH group. This forms the final product. Note that a second H2O can be used in this step because it is the solvent, so it is in excess.

 

Alkene Isomerization Sample Question

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

Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• Alkene isomerization will create a more substituted alkene by creating a carbocation, and reforming the pi bond between more substituted carbons. In this example, the alkene has zero substitutions on one side, and one substitution on the other. There is a secondary carbon adjacent to the Markovnikov carbon, so it could be made more stable by moving the alkene over to involve that carbon.
• All alkene isomerization reactions go through carbocation intermediates, so we need to examine the carbons adjacent to the Markovnikov carbon to determine if a carbocation rearrangement will occur. In this reaction, the carbons adjacent to the Markovnikov carbon in the starting molecule are not more substituted and do not provide any additional resonance stabilization so a rearrangement will not occur.
• Therefore, the new alkene will form between the Markovnikov carbon and the most substituted carbon adjacent to it, which is the secondary carbon to the left. The product has a double bond with 2 substitutions, which makes it more stable than the initial alkene which had 1 substitution.

Mechanism Steps Explained:

1. In step 1, the pi bond attacks a hydrogen in H2SO4 and forms a new C-H sigma bond which adds it to the anti-Markovnikov carbon. At the same time, the electrons in the H-O sigma bond both go to oxygen. This step results in a carbocation because the Markovnikov carbon loses 1 electron from the pi bond that moved. The hydrogen adds to the anti-Markovnikov carbon because it is more energetically favorable for the carbocation to form on the Markovnikov carbon because it is the more substituted carbon.
2. The HSO4- generated in step 1 is too bulky to form a bond to the carbocation, so it eliminates a hydrogen adjacent to the carbocation instead, called a beta-hydrogen. HSO4- will eliminate the most substituted beta-hydrogen because that will form the most substituted alkene product. Therefore, the HSO4- eliminates a hydrogen bonded to the secondary carbon to the left of the carbocation. Note that a wedged or dashed hydrogen could be eliminated. This forms the final product.

 

X2 Addition and Halohydrin Formation Sample Question

For each of the following X2 addition reactions, predict the major product(s). Then, provide a detailed electron-pushing mechanism to illustrate the reaction process.

 Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• X2 addition results in the anti-addition of halogens on the Markovnikov and anti-Markovnikov carbons. Anti-addition means that the two halogens will be in opposite planes (one wedged and one dashed). The halogens (X2) in this reaction are bromine.
• Therefore, we will add a bromine to both the Markovnikov and anti-Markovnikov carbons bonded in opposite planes. Two products are generated because the 3-membered ring intermediate can form above and below the plane.

Mechanism Steps Explained:

1. In step 1, the pi bond attacks one of the bromine atoms in Br2 and forms a bond to that bromine. At the same time the bromine that is attacked by the pi bond attacks the molecule at the more substituted carbon in the alkene. The other bromine in Br2 gains both electrons from the Br-Br sigma bond. This forms a 3-membered ring with a formally positive bromine. The ring forms both above and below the plane, so both outcomes are shown.
2. The formally positive bromine delocalizes electron density away from the two carbons in the 3-membered ring, which causes the carbons to become formally positive. The bond to the more substituted carbon in the ring becomes longer and weaker, so the Br- created in step 1 bonds to that carbon and breaks the ring. The Br- has to bond via backside attack, so if the ring is wedged it bonds as a dash and vice versa. Both the wedged and dashed rings can be broken in this manner, so two products are shown.

 

Oxymercuration Sample Question

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

 Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• Oxymercuration results in a Markovnikov alcohol (OH) addition when the solvent in step 1 is H2O. Two products are generated because the 3-membered ring intermediate can form above and below the plane.

Mechanism Steps Explained:

1. In step 1, the pi bond attacks the mercury atom in Hg(OAc)2 and forms a bond to that mercury. At the same time, the mercury that is attacked by the pi bond attacks the molecule at the more substituted carbon in the alkene. One of the OAc groups bonded to the mercury leaves and both electrons from the O-Hg sigma bond go to oxygen. This forms a 3-membered ring with a formally positive mercury. The ring forms both above and below the plane, so both outcomes are shown.
2. The formally positive mercury delocalizes electron density away from the two carbons in the 3-membered ring, which causes the carbons to become formally positive. The bond to the more substituted carbon in the ring becomes longer and weaker, so the H2O solvent bonds to that carbon and breaks the ring. The H2O has to bond via backside attack, so if the ring is wedged it bonds as a dash and vice versa. Both the wedged and dashed rings can be broken in this manner, so two products are shown.
3. The H2O becomes formally positive once it forms a bond to break the ring, so it is deprotonated by another H2O. This turns the formally positive OH2+ group into a formally neutral OH group. Another H2O can be used in this step because H2O is the solvent, so it is in excess. Note that the OAc- generated in step 1 could also be shown deprotonating the OH2+ group in this step.
4. The HgOAc group left over from the 3-membered ring intermediate is removed by the NaBH4 via reduction. One of the hydrogens in the NaBH4 replaces the HgOAc which is why it looks like it disappears. Hydrogens are implicit, so there is no need to show it. No electron pushing arrows are needed for this step.

 

Acid Catalyzed Hydration vs. Oxymercuration Sample Question

In the following reactions, compare the outcomes of identical molecules with acid catalyzed hydration and oxymercuration reaction conditions. Draw each mechanism, and explain why there is or is not a difference in the major products of each reaction.

Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Comparison:

• Acid catalyzed hydration results in a Markovnikov alcohol addition, and goes through a carbocation intermediate so it is susceptible to a carbocation rearrangement. Oxymercuration also results in a Markovnikov alcohol addition, but it goes through a 3-membered ring intermediate so a carbocation rearrangement is not possible.
• In the starting molecule, the adjacent carbon to the left of the Markovnikov carbon is more substituted than the Markovnikov carbon. Therefore, a 1,2 hydride shift will occur in the acid catalyzed hydration pathway.
• This leads to two different products. Acid catalyzed hydration produces two products with an alcohol on the adjacent carbon to the left of the Markovnikov carbon due to the carbocation rearrangement, whereas oxymercuration produces two products with an alcohol on the Markovnikov carbon.

 

Hydrogenation Sample Question

For each of the following Hydrogenation reactions, predict the major product(s).

Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• Hydrogenation reduces pi bonds by attaching a hydrogen to each carbon in the pi bond via syn addition. The pi bond also goes away as a result of the addition of the hydrogens. In this reaction, there are two pi bonds to react with and the catalyst is Pd/C which can react with double and triple bonds.
• Therefore, the product will be an alkane with no remaining pi bonds. The hydrogens don’t have to be shown because they are implicit, but they are shown in the products to illustrate where and how they added.
• Each pi bond is reacted separately, and each pair of hydrogens are added syn to each other.

 

Ozonolysis Sample Question

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

 Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• To predict the product of an ozonolysis reaction, we first need to split the double bond and break the molecule in two.
• Once the two pieces are formed, an oxygen is added onto each double bond at the site of the breakage.
• The second step on the reaction arrow is (CH3)2S, so nothing happens to the hydrogens coming off the double in the original molecule, and the two pieces formed after the oxygen atoms are added are the final products.

 

Hydroboration Sample Question

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

 Answer Key:

*Stereochemistry is not taken into account in the products of these reactions. The purpose of this worksheet is to consider all possible outcomes for the following reactions.

Predicting the Product:

• Hydroboration of an alkene results in an anti-Markovnikov alcohol (OH) addition and a Markovnikov hydrogen addition. The alcohol and hydrogen are both added in the same plane due to the concerted first step in the mechanism.
• Therefore, we will add an alcohol (OH) to the anti-Markovnikov carbon and a hydrogen to the Markovnikov carbon. Because they are added in the same plane, the methyl substituent on the Markovnikov carbon must be in the opposite plane of the alcohol.
• Two products are shown because the alcohol and hydrogen can be added above and below the plane of the molecule.

Mechanism Steps Explained:

1. In step 1, the pi bond attacks the boron in BH3 and forms a bond to that boron on the anti-Markovnikov carbon. At the same time, the hydrogen attacks the Markovnikov carbon and forms a new C-H sigma bond. Because this step is concerted, the boron and hydrogen have to add in the same plane. The boron and hydrogen add above and below the plane of the molecule, but only one addition is shown in this mechanism. This step effectively replaces one of the hydrogens on the boron with a bond to the anti-Markovnikov carbon of the starting molecule.
   • Note that the hydrogen forms a bond to the Markovnikov carbon by using the electrons in the B-H sigma bond. This occurs because hydrogen is more electronegative than boron, so hydrogen holds a majority of the electron density in the sigma bond.
2. Step 1 occurs two more times and replaces the other two hydrogens on the boron through an identical mechanistic step. Because it was shown once in step 1, it does not need to be shown two more times and we can write “2x” on the arrow. This step results in the boron now having 3 starting molecules bonded to it.
3. At the same time steps 1 and 2 are occurring, the H2O2 in step 2 on the reaction arrow is reacted with OH-. The OH- base deprotonates H2O2 to which creates H2O and an HOO- nucleophile.
4. The HOO- nucleophile then attacks the boron in the BR3 (R=carbon chain from starting molecule) and forms a bond to that boron. This makes the boron formally negative because it has 4 sigma bonds.
5. Next, a rearrangement occurs where one of the carbon chains detaches its sigma bond from the boron and moves it to the oxygen bonded to boron in the O-O-H chain. At the same time, the OH in the O-O-H chain leaves by taking the electrons in the O-O sigma bond. This step results in one of the carbon chains being attached to an oxygen instead of boron and an OH- byproduct generated from the OH leaving the O-O-H chain.
6. Similar to step 1, steps 4 and 5 occur two more times. This results in an oxygen in between the boron and each carbon chain. Both repeats of steps 4 and 5 also produce OH- byproducts.
7. The OH- byproduct then attacks the boron and forms a sigma bond to it. This makes the boron formally negative because it has 4 sigma bonds.
8. Next, one of the oxygens bonded to a carbon chain takes both electrons from the O-B sigma bond and leaves. This results in a formally negative oxygen bonded to the carbon chain.
9. The formally negative oxygen is protonated by the H2O formed in step 3, which turns the O- into an OH. This forms the final product. Note that there are two products because the hydrogen and boron (which eventually turns into the OH) can add above and below the plane of the molecule in step 1. Therefore, both outcomes are shown.

 

 

Course Description:

Key Features:

• HX Addition:
  • Practice predicting the product and drawing the mechanism for various HX addition reactions with alkenes.
  • 20 questions
• Acid Catalyzed Hydration:
  • Practice predicting the product and drawing the mechanism for various acid catalyzed hydration reactions with alkenes.
  • 20 questions
• Alkene Isomerization
  • Practice predicting the product and drawing the mechanism for various alkene isomerization reactions.
  • 20 questions
• X2 Addition and Halohydrin Formation:
  • Practice predicting the product and drawing the mechanism for various X2 addition and halohydrin formation reactions with alkenes.
  • 20 questions
• Oxymercuration:
  • Practice predicting the product and drawing the mechanism for various oxymercuration reactions with alkenes.
  • 20 questions
• Acid Catalyzed Hydration vs. Oxymercuration:
  • Practice predicting the product and explaining the differences between various acid catalyzed hydration and oxymercuration reactions that start with the same alkene molecule.
  • 20 questions
• Hydrogenation:
  • Practice predicting the product for various hydrogenation reactions with alkenes and alkynes.
  • 20 questions
• Ozonolysis:
  • Practice predicting the product and drawing the mechanism for various ozonolysis reactions with alkenes.
  • 20 questions
• Hydroboration:
  • Practice predicting the product and drawing the mechanism for various hydroboration reactions with alkenes.
  • 20 questions

Detailed Answer Keys:

• Each worksheet includes step-by-step worked solutions.
• Comprehensive explanations clarify key concepts and the reasoning behind each step in the mechanism.
• Focus on electron-pushing mechanisms to reinforce understanding of how electrons move during reactions.

Enhanced Learning Experience:

• Structured practice that covers both fundamental and advanced topics in nucleophilic addition reactions with alkenes.
• Ideal for students preparing for advanced organic chemistry exams or those seeking to strengthen their understanding of reaction mechanisms.

Recommended Prerequisite:

• Completion of Module 1: Formal Charge, Partial Charge, Electron Density, Hybridization, and Resonance Structures.
• Completion of Module 2: Energetics, Newman Projections, and Molecular Orbital Theory instructional videos.
• Completion of Module 3: Acidity and Electron Pushing Arrows.
• Completion of Module 4: Alkenes and Carbocations.
• Completion of Module 5: Nucleophilic Addition Reactions with Alkenes

Who Should Engage:

• Learners aiming to deepen their understanding of nucleophilic addition reactions with alkenes.
• Students seeking to excel in organic chemistry, particularly those interested in mastering reaction mechanisms and product prediction.

Benefits:

• Extensive practice with a wide variety of problems to ensure a thorough understanding of nucleophilic addition reactions.
• Detailed answer keys with explanations to reinforce learning and clarify complex concepts.
• Improved ability to predict molecular behavior and reaction outcomes in organic chemistry.