Predict the major products of the following reactions. (c) 2-methylpent-2-ene + BH 3 ⋅THF (d) the product from part (c) + H 2 O 2 /OH −

Hi everyone and welcome back in today's video, we are looking at the following problem which states for each of the following reactions determine the major products. In the first reaction we have a reaction between two metal hex two in and boring and tetra hydro Furin. And then the second reaction, we have the product from the first part reacting with hydrogen peroxide and hydroxide. Let's start by drawing the original structure our substrate as two metal hacks to in. So because it's hacks, we are going to draw a six member chain At position number two we have a metro substitute print and also we have a double bond because we have two in so we have our starting elkin. Let's draw it a bit more clearly. There's a double bond a disposition and basically the reaction is boring, boring and tetra hydro urine. That's why you're in here is a solvent that stabilizes boring because it forms a dime. Er and we don't want that. We want to make sure that the equilibrium favors the presence of boring. And recall that this essentially as boring onto the double bond. In addition. Don't forget that for this reaction we have and see more cosmic of addition. Now because this is the ante Markovic of addition. We will be adding hydrogen to a more substituted position. We have to lKOh substations over here. So we are going to add hygienic disposition and this is a less substituted position. So beach to the remaining part of the molecule will go here. So let's throw that out, hydrogen could be drawn separately from the remaining part of the molecule which is beached, sea hydrogen is more electro negative. That's not usually the case, but it's the case here and born is less electro negative. So it's partially positive. And therefore as a result, hydrogen would be attracted by this carbon atom and born would form a bond with this carbon atom. We have an addition in which we will not have a double bond anymore. That's how addition works. Were breaking the double bond because we have an addition of two different atoms along the double bond. So we may simply stated that this position will have hydrogen and in this position we will have born our first product would still have that hexane chain. But at that position we are adding hydrogen. We will not be showing it. It's just implicit and at position number three, we are adding BH two. Well done. That's our first product. And now for the second reaction, we're taking that product. As the problem suggests, it says the product from the first part. So let's redraw the product that we got and we are using hydrogen peroxide in hydroxide. Now notice that if you look at these two steps all together, this is the hydro operation oxidation reaction. Recall that when you're using boring and the tri interfere and followed by hydrogen peroxide and hydroxide. You are forming the encima cosmic of alcohol in which a wage goes to a less substantive position. So in this stuff, hydroxide will essentially be replacing B. H. Two. We are replacing beach to group with hydroxide. No need to worry about material chemistry. That's fine in general. That's retention of configuration. So if there was actually a dash or which shown here, we would make sure that the original configuration would be retained. We're not worried about that because there's no requirement for us to indicate stereo chemistry. So we're simply replacing Beach too with a wage minus. Let's draw the final alcohol At position # three. We are adding an alcohol group. Now this is our final product for the second reaction. We may now label our final answers. Our final answers are essentially the use to products and as you can see, the correct answer for this multiple choice question is a If you have any questions, don't hesitate to leave them down below in the comments section, Thank you for watching this video.

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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

10. Addition Reactions

Hydroboration

Organic Chemistry

10. Addition Reactions

Hydroboration

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5:51 minutes

Problem 10c,d

Textbook Question

Predict the major products of the following reactions.
(c) 2-methylpent-2-ene + BH₃⋅THF
(d) the product from part (c) + H₂O₂/OH⁻

Verified step by step guidance

Step 1: Recognize that the reaction in part (c) involves hydroboration of 2-methylpent-2-ene with BH₃·THF. Hydroboration is an anti-Markovnikov addition of hydrogen (H) and boron (BH₂) across the double bond. The boron atom will attach to the less substituted carbon of the double bond, while the hydrogen will attach to the more substituted carbon.

Step 2: Draw the structure of 2-methylpent-2-ene and identify the double bond. Determine which carbon is less substituted (fewer alkyl groups attached) and which is more substituted. This will guide the placement of the BH₂ group and the H atom.

Step 3: After the hydroboration step, the product will be an organoborane intermediate where the BH₂ group is attached to the less substituted carbon of the double bond. Ensure the stereochemistry is considered, as hydroboration is a syn addition (both H and BH₂ add to the same face of the double bond).

Step 4: In part (d), the organoborane intermediate reacts with H₂O₂ (hydrogen peroxide) in the presence of OH⁻ (a base). This step is an oxidation reaction that replaces the BH₂ group with an OH group, forming an alcohol. The product will retain the anti-Markovnikov regiochemistry established in the hydroboration step.

Step 5: Draw the final product of part (d), ensuring the OH group is attached to the less substituted carbon of the original double bond. Confirm that the stereochemistry is consistent with the syn addition from the hydroboration step.

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Key Concepts

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

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step reaction process used to convert alkenes into alcohols. In the first step, an alkene reacts with borane (BH3) to form an organoborane intermediate, which is then oxidized in the second step by hydrogen peroxide (H2O2) in the presence of a base (OH−). This reaction is notable for its anti-Markovnikov addition, where the hydroxyl group attaches to the less substituted carbon of the alkene.

Regioselectivity

Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others when multiple products are possible. In the context of hydroboration-oxidation, the regioselectivity is influenced by the mechanism of the reaction, leading to the formation of the alcohol at the less substituted carbon atom of the alkene, which is a key feature of this reaction pathway.

Stereochemistry

Stereochemistry is the study of the spatial arrangement of atoms in molecules and how this affects their chemical behavior. In the case of the hydroboration-oxidation reaction, the stereochemistry of the product can be influenced by the formation of the organoborane intermediate, which can lead to the generation of chiral centers. Understanding stereochemistry is crucial for predicting the specific isomers formed in the final product.

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