Module 4 Practice Set: Alkenes and Carbocations

E/Z Conformations Example Question:

Label each of the double bonds in the molecules below as E or Z.

Answer Key:

The double bond in this molecule is in an E conformation.

The priority group on carbon 1 is the ethyl group highlighted in orange. We know this because the two atoms directly bonded to carbon 1 are hydrogen and carbon. Carbon has a larger atomic mass than hydrogen, so the ethyl group is the priority group.

The priority group on carbon 2 is the methyl group highlighted in orange. We know this because the two atoms directly bonded to carbon 2 are hydrogen and carbon. Carbon has a larger atomic mass than hydrogen, so the methyl group is the priority group.

The two priority groups are on opposite sides of the double bond, so the double bond is in an E conformation.

*Note that because the double bond only has 2 substitutions, it could also be classified as being in a trans conformation.

 

Alkene Stability Sample Question:

Rank the stability of the following alkenes from 1-4, with 1 being the most stable alkene and 4 being the least stable alkene.

Answer Key:

The alkene with 4 substitutions is the most stable because it has the most substitutions. Therefore, it has the most hyperconjugation which makes its energy the lowest.

The alkene with 3 substitutions is the second most stable because it has the second most substitutions. Therefore, it has the second most hyperconjugation which makes its energy the second lowest.

The alkene with 2 substitutions is the third most stable because it has the third most substitutions. Therefore, it has the third most hyperconjugation which makes its energy the third lowest.

The alkene with 1 substitution is the least stable because it has the lowest number of substitutions. Therefore, it has the least amount of hyperconjugation which makes its energy the highest.

 

Identifying Markovnikov and anti-Markovnikov Carbons Sample Question:

Identify the Markovnikov and anti-Markovnikov carbons in each of the following molecules.

Answer Key:

                                                                               

The Markovnikov and anti-Markovnikov carbons have to be carbons in the pi bond. Therefore, the only two carbons they can be are carbons 2 and 3.

The Markovnikov carbon is carbon 2 because it is the most substituted carbon in the pi bond with 1 substitution.

The anti-Markovnikov carbon is carbon 3 because it is the least substituted carbon in the pi bond with 0 substitutions.

 

Identifying Different Types of Carbocations Sample Question:

Identify if the following carbocations are primary, secondary, or tertiary.

Answer Key:

This carbocation is secondary because there are two substitutions on the formally positive carbon. Those substitutions are highlighted orange in the image above.

 

Ranking Carbocation Stability Sample Question:

Compare the stability of each pair of carbocations provided, and explain why one carbocation is more or less stable than the other.

Answer Key:

The tertiary carbocation is more stable than the secondary carbocation. The tertiary carbocation has one more substitution than the secondary carbocation, so its positive charge is more stabilized by the electron donating effects of the additional substitution. The additional substitution also provide more hyperconjugation, which further stabilizes the tertiary carbocation compared to the secondary carbocation.

 

Carbocation Rearrangement Sample Question:

For each of the carbocations depicted below, label whether it is primary (1°), secondary (2°), or tertiary (3°). Next, determine if a carbocation rearrangement will occur. If a rearrangement will occur, provide the electron pushing mechanism and explain why the rearrangement occurs.

Answer Key:

The initial carbocation is secondary because it has 2 substitutions. The adjacent carbon to the right of the carbocation is primary, so the carbocation will not rearrange to that carbon. The adjacent carbon to the left of the carbocation is tertiary, which would be a more stable carbocation if it had the positive charge. Therefore, the carbocation will undergo a 1,2 hydride shift to move the positive charge to the tertiary carbon.

The rearrangement occurs because the tertiary carbocation is more stable than the secondary carbocation. The tertiary carbocation has more substitutions, so there is more electron donation to the positively charged carbon which provides increased stabilization. There is also more hyperconjugation with more substitutions on the carbocation which further stabilizes the molecule.

The electron pushing arrow is drawn from the C-H sigma bond of the hydrogen shifting. It cannot be drawn from the hydrogen atom because electron pushing arrows always have to start at a pair of electrons.

 

Unsaturation Number Sample Question:

For each of the molecules listed below, determine the unsaturation number.

Answer Key:

The unsaturation number of a molecule can be found 2 different ways. The first way we can find it is by adding up all the rings and pi bonds in the molecule. This molecule has 1 pi bond and 1 ring, so its unsaturation number is 2.

The other way we can find the unsaturation number of a molecule is by using the equation shown below:

 

 

Course Description:

Key Features:

• E/Z Conformations:
  • Practice assigning an E/Z label to different double bonds.
  • 20 questions
• Alkene Stability
  • Practice ranking the relative stability of different alkenes.
  • 20 questions
• Identifying Markovnikov and anti-Markovnikov Carbons
  • Practice identifying Markovnikov and anti-Markovnikov carbons in different alkenes.
  • 20 questions
• Identifying Different Types of Carbocations
  • Practice identifying primary, secondary, and tertiary carbocations.
  • 20 questions
• Ranking Carbocation Stability
  • Practice ranking the relative stability of different carbocations.
  • 20 questions
• Carbocation Rearrangement
  • Practice determining if a carbocation rearrangement will occur, and drawing the electron pushing mechanism for different carbocation rearrangements.
  • 20 questions
• Unsaturation Number
  • Practice finding the unsaturation number of different molecules
  • 20 questions

Detailed Answer Keys:

• Each worksheet includes step-by-step worked solutions.
• Comprehensive explanations that clarify key concepts and problem-solving strategies.

Enhanced Learning Experience:

• Structured practice that covers fundamental and advanced topics in alkene chemistry and carbocation stability.
• Perfect for students preparing for advanced organic chemistry exams.

Recommended Prerequisite:

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

Who Should Engage:

• Learners seeking to deepen their understanding of E/Z conformations, alkene stability, carbocation behavior, and unsaturation numbers.
• Ideal for students aiming to excel in organic chemistry.

Benefits:

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