Hey, everyone and welcome back, identify the missing products. The starting material is prop one in to L benzene and it is treated with high cham bromide in the presence of peroxide to form product A and product A is treated with sodium cyanide to form product B. Our goal is to identify products A and B. Now, the first reaction we're using Hyro in the presence of peroxide, which represents the anti marconi of Rio chemistry, right? We have addition of high cham to our double bonded carbon atoms and essentially based on anti Marconi of radio chemistry due to the fact that we have peroxide, right? Not just H pr we're going to violate the Marconi of and we're going to add the hydrogen to the more substituted carbon and bromine is the less substituted carbon. In essentially, we can illustrate that reaction by redrawing the starting material, right? So we're going to through it and then we are going to state that for this anti cognitive addition, we're using hydrogen bromide in the presence of peroxide. It can be any peroxide, right? As long as we have peroxide, we're violating the more cognitive and we're adding hyen to our more substituted double bonded carbon atom. So basically, we are going to break our pie bond. And instead of that pie bond, we will have hydrogen and the more substituted carbon. And then we are going to bond bromine to the less substituted carbon. So that's how we get our product. A. Now, the next reaction is the treatment of product A with sodium cyanide. And let's remember that sodium cyanide is an ionic compound. So in a solution, it dissociates into sodium ions which are ion Spectators and cyanide ions. So we can essentially think of this reaction as addition of cyanide, which has a negative charge. So it's a strong muco and we have the primary substrate because we have a primary substrate and a strong nu we're going to observe an S N two reaction in which the electrophysics center is attacked followed by the loss of the leaving group. And this is how we are forming our final product as seen in the mechanism. We are just going to replace bromine with C and right. So that would be our night trial forms. That would be our product B. And now we are ready to conclude, we are going to erase the arrow so that we can clearly see what a product A is. We're going to label products A and to be, we have labeled them in two green boxes and we can just review it to produce A. We used the anti Marconi of addition, because we had peroxide and alkene reacted with hydrogen bromide. So we added hyen, the less substituted carbon and bromine. My apologies, we added hyen, the more substituted carbon and bromine, the less substituted carbon. When we got product A, we had an essence two reaction in which we replaced bromine with Cyano group to get N trial. We have labeled our products and thank you for watching. I hope to see in the next video.

Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

11. Radical Reactions

Radical Reaction

Organic Chemistry

11. Radical Reactions

Radical Reaction

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4:27 minutes

Problem 85c

Textbook Question

Identify A–J:

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Step 1: Analyze the first reaction. The starting compound is an alkene with a phenyl group and an isopropenyl group. The reaction involves HBr in the presence of peroxide. This indicates an anti-Markovnikov addition of HBr due to the radical mechanism initiated by the peroxide.

Step 2: In the anti-Markovnikov addition, the bromine atom will add to the less substituted carbon of the double bond, while the hydrogen atom will add to the more substituted carbon. This results in the formation of compound A, which is a brominated alkane.

Step 3: Analyze the second reaction. Compound A reacts with NaCN. This is a nucleophilic substitution reaction (SN2 mechanism), where the bromine atom is replaced by the cyanide group (CN).

Step 4: The cyanide ion (CN⁻) acts as a nucleophile and attacks the carbon bonded to the bromine atom, displacing the bromine and forming compound B, which contains a nitrile group.

Step 5: Summarize the transformations. The first reaction converts the alkene into a brominated alkane (A) via anti-Markovnikov addition. The second reaction replaces the bromine in A with a nitrile group, forming compound B.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Radical Reactions

Radical reactions involve the formation and reaction of free radicals, which are highly reactive species with unpaired electrons. In the presence of peroxides, the addition of HBr to alkenes can lead to the formation of bromine radicals, which then add to the double bond, resulting in a product that can be further transformed. Understanding the mechanism of radical reactions is crucial for predicting the products of the reaction shown.

Markovnikov's Rule

Markovnikov's Rule states that in the addition of HX (where X is a halogen) to an alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms already attached. This principle helps predict the regioselectivity of the reaction, guiding the formation of product A in the diagram. Recognizing this rule is essential for determining the structure of the intermediate or final products in reactions involving alkenes.

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction mechanism in organic chemistry where a nucleophile replaces a leaving group in a molecule. In the context of the reaction from A to B, NaCN acts as a nucleophile that attacks the electrophilic carbon in product A, leading to the formation of product B. Understanding this mechanism is vital for predicting the outcome of the reaction and the nature of the final product.

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