Welcome back, everyone identify the missing products. The starting material is isopropyl benzene and it is treated with bromine in the presence of light to produce product A, which is then treated with TBI oxide to produce B and B is treated with hydrogen iodide in the presence of heat to produce product C. We are going to identify products A B and C. Let's start with identifying product A and to do that, we need to consider the fact that we're adding broom in the presence of a light. So that would correspond to free radical mechanism, right? So we are going to form radicals and let's, let's remember that Bromine prefers the most substituted position to replace hydrogen. We noticed that we have a tertiary position within our kiel group. We are not going to consider any S P two hybridized carbon atoms in this mechanism, right. So we are finding our tertiary position and we are going to replace hydrogen with bromine. That's how we get product A. So we can state that for product A, we're getting our benzene ring and for the isopropyl substituent, we're simply replacing hydrogen with bromine at the tertiary position. So we get bromine at that carbon and that would correspond to product A. So we are going to label product A because we have got our partial answer and our product A is treated with TB oxide. Let's remember that it's a strong and bulky base ster hindered, right? Because it's a strong historically inner base. We are generally interested in the Hoffman product during the elimination step, right? We are dealing with an E two reaction. But if we carefully consider our a reactant in this case, a right, it essentially has two beta hydrogens to eliminate which are equivalent to each other. So we only have one possibility to produce an alkene. The two metal groups are equivalent to each other. So we can consider any of them. That means we can consider our mechanism for the reaction to carefully predict the product. And if we re product and as well as expand one of the implicit hydrogen atoms, we basically understand that TB to side or specifically the negative charge on oxygen will attack hydrogen, we will form our double bond followed by the loss of the leaving group. And this will give us an keen. So let's draw the product, we will have our six member ring. And now be careful to look at the carbon atoms that we have, we are going to draw two carbon atoms in the parent as well as the methyl group and the double bond formed, right. So we got in that will be product B. So let's label it as our final product. And now let's consider the formation of C in which we're using hydrogen iodide in this reaction. Because we're using an as the starting material. If we redraw it, we have a basic electro pic addition, right? Because we have an alkene. And whenever we use a molecule in the form of H X, where X is a halogen in the presence of heat, we are going to have an electro pic addition mechanism in which first of all, we have our protonation step. Let's remember that we have to follow the Markov's rule which states that we once formed the more stable carbo calum or basically add hydrogen to the less substituted carbon, right. So this gives us a stable tertiary resonant stabilized carbo ion which is then attacked by the iodide Ron formed to form the final product. So we are ready to draw the structure of C that would be our benzene ring. And now we are going to simply add iodine to the carbon with a positive charge. And that's how we get product. C let's label it as our final answer. So we have obtained the structures of A B and C. In the first reaction, we use the free radical mechanism in which bromine replaced the hydrogen at the most substituted position. Bromine is really selective, right. So the major products will always be obtained at the most substituted position for product B, we had an elimination E two using a bulky base. Generally, we form a Hoffman pro product. But in this case, we only have one type of beta hydrogens. And when we get our L K B, it is treated with hyen iodide using an electro addition mechanism in which we are considering the more cognitive and getting our final product. Thank you for watching and I hope to see you 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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6:34 minutes

Problem 85b

Textbook Question

Identify A–J:

Verified step by step guidance

Step 1: Analyze the first reaction. The starting compound is isopropylbenzene (cumene). It reacts with bromine (Br₂) under UV light (hν), which indicates a free radical halogenation reaction. The bromine will selectively replace a hydrogen atom at the benzylic position (the carbon adjacent to the benzene ring) due to the stability of the benzylic radical.

Step 2: The product of the first reaction (A) is a benzylic bromide. In the second step, this benzylic bromide reacts with tert-butoxide (tert-BuO⁻), a strong base. This is an elimination reaction (E2 mechanism), where the bromine is eliminated along with a proton from the adjacent carbon, forming a double bond. The product (B) is an alkene at the benzylic position.

Step 3: The third reaction involves the addition of hydrogen iodide (HI) to the alkene (B) under heat (Δ). This is an electrophilic addition reaction. HI adds across the double bond, following Markovnikov's rule, where the iodine attaches to the benzylic carbon and the hydrogen attaches to the less substituted carbon.

Step 4: The final product (C) is a benzylic iodide. This compound is formed as a result of the addition of HI to the alkene.

Step 5: To summarize, the sequence of reactions involves free radical halogenation, elimination, and electrophilic addition. Each step transforms the functional group at the benzylic position, leading to the final product.

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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 radicals, which are species with unpaired electrons. These reactions often require initiation steps, such as exposure to light (hv) or heat, to generate radicals from stable molecules. In this context, bromine (Br2) is used to generate a radical intermediate (A) through homolytic cleavage.

Nucleophilic Substitution

Nucleophilic substitution is a fundamental reaction in organic chemistry where a nucleophile replaces a leaving group in a molecule. In the reaction sequence, the tert-butoxide ion (tert-BuO-) acts as a nucleophile, attacking the radical intermediate (A) to form compound B. This step is crucial for the transformation of radicals into more stable products.

Elimination Reactions

Elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond or a ring structure. In this case, the reaction of compound B with hydrogen iodide (HI) under heat (Δ) leads to the elimination of a leaving group and the formation of compound C. Understanding elimination mechanisms is essential for predicting product formation in organic synthesis.

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