Module 10 Practice Set: Radical Reactions, Epoxides, and Intramolecular Reactions

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Radical Halogenation Sample Question:

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

Answer Key:

Predicting the Product:

  • In a radical halogenation reaction with Br2, a bromine will replace the most substituted hydrogen in an alkane starting molecule. Bromine will always react with the most substituted hydrogen in the starting molecule because a bromine radical is selective.
  • The most substituted hydrogen in the starting molecule is the dashed hydrogen on the tertiary carbon with the wedged methyl group. Therefore, a bromine will be substituted in place of that hydrogen in the product.
  • The bromine adds both above and below the plane of the molecule, however it does not create a stereocenter in this reaction so only 1 product needs to be shown.

Mechanism Steps Explained:

  1. Initiation: the sigma bond in the Br2 molecule breaks apart with 2 fish hook arrows showing 1 electron going to each bromine from the sigma bond. This step results in 2 bromine radicals.
  2. Propagation: the starting molecule then reacts with 1 of the bromine radicals created in step 1. The bromine radical reacts with the most substituted hydrogen in the starting molecule and forms an H-Br sigma bond. The other electron in the original C-H sigma bond forms a radical on the carbon the hydrogen was bonded to in the starting molecule. The newly formed radical on the starting molecule then reacts with another Br2. The Br2 breaks apart similar to step 1, however one of the bromine atoms forms a bond with the new radical, resulting in the major product. There is also a bromine radical byproduct from this step.
  3. Termination: there are 2 remaining bromine radicals, 1 from step 1 and 1 from step 2. These radicals react together to form Br2. This step is necessary because a reaction can never end with a radical.

 

Radical Bromine and Radical Thiol Addition Sample Question:

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

Answer Key:

Predicting the Product:

  • Radical bromine addition results in an anti-Markovnikov bromine addition, and a Markovnikov hydrogen addition. The anti-Markovnikov carbon is the less substituted carbon in the double bond. In this reaction, the carbon on the right in the double bond is the anti-Markovnikov carbon.
  • When the bromine bonds to the anti-Markovnikov carbon, it bonds above and below the plane of the molecule. The hydrogen bonds to the Markovnikov carbon above and below the plane of the molecule as well. Neither of these additions create a stereocenter, so only 1 product needs to be shown.

Mechanism Steps Explained:

  1. In the first step, the sigma bond in the middle of the peroxide molecule breaks apart and 1 electron from each side of the sigma bond goes to each oxygen. This results in 2 OEt radicals.
  2. Next, 1 of the OEt radicals created in step 1 reacts with HBr. The OEt radical forms an O-H sigma bond with the hydrogen from HBr using 1 of the electrons in the H-Br sigma bond. The bromine from HBr takes the other electron from the sigma bond and becomes a radical.
  3. The bromine radical created in step 2 then reacts with the pi bond from the starting molecule. The bromine forms a sigma bond to the anti-Markovnikov carbon with 1 of the electrons from the pi bond. The other electron in the pi bond forms a radical on the Markovnikov carbon. The bromine bonds both above and below the plane of the molecule however it does not form a stereocenter in this reaction, so only 1 product needs to be shown.
  4. The radical on the Markovnikov carbons reacts with another HBr molecule and forms a bond to the hydrogen using 1 electron from the H-Br sigma bond. The hydrogen bonds both above and below the plane of the molecule however it does not form a stereocenter in this reaction, so only 1 product needs to be shown. The other electron in the sigma bond goes to the bromine and forms a radical. A byproduct of Br2 is formed by the reaction of 2 bromine radicals.

*Note that two OEt radicals are created in step 1, but only 1 OEt radical is used in the reaction mechanism. This means that for every 1 peroxide molecule, 2 major product molecules are formed. This also means that 2 bromine radicals are formed in step 4, and a reaction can never end with a radical which is why a Br2 byproduct is formed.

 

Radical Alkyne Reduction Sample Question:

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

Answer Key:

Predicting the Product:

  • Radical alkyne reduction reduces an alkyne into a trans (E) alkene. Therefore, the starting alkyne will become a trans (E) alkene in the product.

Mechanism Steps Explained:

  1. Na on its own has 1 valence electron, so it is a radical. The reaction starts when Na donates its radical to one of the carbons in the alkyne. This is shown with a fish hook arrow because the electron moving is a radical and not a pair of electrons. At the same time, one of the pi bonds is pushed to the carbon that did not get the radical and it becomes a lone pair. The carbon that gained the lone pair also becomes formally negative. When the triple bond becomes a double bond, the alkene orients itself in a trans (E) conformation because it is more energetically favorable than a cis (Z) conformation.
  2. The formally negative carbon then forms a bond to one of the hydrogens in the NH3, which removes its formal negative charge.
  3. Another Na donates its radical to the same carbon that received the radical in step 1, which turns the remaining radical into a lone pair and makes that carbon formally negative.
  4. The formally negative carbon then forms a bond to a hydrogen in another NH3 molecule, which makes it formally neutral. This forms the final product.

 

Epoxide Synthesis Sample Question:

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

Answer Key:

Predicting the Product:

  • mCPBA creates an epoxide above and below the plane of the starting molecule on the two carbons in the pi bond.
  • The epoxide creates a stereocenter on one of the carbons bonded to the oxygen, so both the wedged and dashed versions need to be shown.

Mechanism Steps Explained:

  1. The reaction of mCPBA with an alkene is a single step mechanism. The epoxide and byproduct form at the same time with the oxygen in the epoxide coming from the oxygen in the OH group in the mCPBA molecule. Both products are shown because the epoxide can form above and below the plane of the molecule and the creation of the epoxide creates a stereocenter in this reaction.

 

Epoxide Opening Sample Question:

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

Answer Key:

Predicting the Product:

  • NaOH is a base, so the bonds in the epoxide will stay equal length and the OH- will open the epoxide at the least substituted carbon. The H2O solvent is the conjugate base of OH- so it has no effect on the reaction conditions.
  • In this example, the primary carbon is the least substituted, so the OH- breaks open the epoxide at that carbon.
  • When the OH- breaks open the epoxide it has to form a bond via backside attack, so it will bond as a dash because the epoxide is wedged. In this example, the OH that opened the epoxide is shown in the plane because it’s on a primary carbon so rotating it from a dash to in the plane would not effect any stereochemistry. However, keeping it as a dash would also be acceptable.

Mechanism Steps Explained:

  1. The OH- base breaks the epoxide open by forming a bond via backside attack. Therefore, it will bond with a dash and the oxygen in the epoxide keeps its wedged bond. The oxygen in the epoxide becomes an O- because it gained both electrons from the sigma bond that broke. Note that the OH that opened the epoxide is shown in the plane because it’s on a primary carbon so rotating it from a dash to in the plane would not effect any stereochemistry. Showing it as a dash or in the plane are both equally valid.
  2. The O- is then be protonated by an H2O to form an OH group. This forms the final product.

 

Course Description:

Key Features:

Radical Halogenation:

  • Practice predicting the products and drawing reaction mechanisms for radical halogenation reactions.
  • 20 questions.

Radical Bromine and Radical Thiol Addition:

  • Practice predicting the products and drawing reaction mechanisms for radical bromine and thiol addition reactions.
  • 20 questions.

Radical Alkyne Reduction:

  • Practice predicting the products and drawing reaction mechanisms for radical alkyne reduction reactions.
  • 20 questions.

Epoxide Synthesis:

  • Practice the formation of epoxides from different starting materials by predicting the products and drawing reaction mechanisms for epoxide synthesis reactions.
  • 20 questions.

Epoxide Opening:

  • Practice predicting the products and drawing reaction mechanisms for epoxide opening reactions.
  • 20 questions.

Detailed Answer Keys:

  • Step-by-step mechanisms provided for each question.
  • Explanations on predicting reaction products and drawing reaction mechanisms.

Enhanced Learning Experience:

  • Consolidate your understanding of radical reactions and epoxide chemistry through structured practice.
  • Aligns with the content of Module 10 instructional videos, providing a seamless extension of your learning journey.

Recommended Prerequisite:

  • Completion of all instructional videos from Module 10 is strongly recommended to ensure you are well-prepared for this advanced practice module.

Who Should Engage with This Practice Module:

This practice module is designed for learners aiming to deepen their knowledge in organic chemistry, particularly in radical and epoxide reactions. Whether you’re preparing for exams, enhancing your problem-solving skills, or solidifying your understanding of these complex topics, this module offers the comprehensive practice and support you need.

Benefits:

  • Focused practice on advanced organic chemistry topics, with 20 questions per topic.
  • Detailed answer keys with step-by-step mechanisms and explanations to guide your learning.
  • Reinforce your understanding of radical and epoxide chemistry through extensive practice and application.

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