What will be the major product obtained from the reaction of Br 2 with 1-butene if the reaction is carried out in c. ethyl alcohol? d. methyl alcohol?

All right. Hello everyone. So for this question, let's go ahead and draw the major product formed when five methyl hex one E reacts with CL two with part one prop alcohol and part two ethyl alcohol. So first, let's go ahead and draw out our starting material. We have five methyl hex one E which is a six carbon pairing chain with a double bond in between carbons, one and two and a methyl group on position five. Now, the reactions being described here are very similar. Our starting material isn't all keen, we're reacting with one equivalent of C two, which is a halogen and both parts one and two have an alcohol. In addition to the other two reagents recall that this combination of regions is going to result in the formation of a Hallow ether where we have a halogen and an oxy group on the same parrot chain. Now, to be a little bit more precise, we can say that it's described as a viscal Hallow ether because the halogen and the ether functional groups are on adjacent carbons, namely each of the two carbons of the starting all keen receives either the halogen or the ether functional group. So they end up on adjacent carpets recall that the mechanism of this formation or of this reaction is characterized by the formation of a bridged ion intermediate. One of the chlorine atoms from co two is going to install itself on to the carbons of the starting keen to create that bridged ion intermediate. Afterwards, we have an SN two type reaction with the al coxie group of the alcohol. Now to be a little bit more precise, even still, this reaction follows Markov Niko radios selectivity with respect to the alcohol. So the oxy group of the ether is going to add on to the more substituted carbon of the starting double wand. So a few more things to keep in mind here are that in addition to Markov Niko radio selectivity, this is an example of anti addition. So the halogen and the axie group are going to be on opposite sides. And one more bit of information here is that this reaction creates a pair of anatomy. So each of the two reactions described produce one pair of Anan tumors. This is because after the formation of the bridged ion, the alcohol can attack on either side of the ring itself. Now, the ring refers to the bridged ion intermediate which once again could have formed on either side of the starting double bond. All right. So now let's start by drawing out the products for part one. So first, let me start off by drawing out the carbon chain. So that's 23456 carbons and a methyl group. Now because we have to consider more Cogni cov radios selectivity chlorine is going to go on carbon one of the parent chain and the prop boxy group from prop alcohol is going to go on carbon two of the parent chain. Notice here that carbon one is not actually a stereo center. So there is no need to specify stereo chemistry on the lorry. However, because we do end up with a pair of anant omer, I'm going to draw my second stereo isomer where the Propox group is depicted on a dash instead of a witch. All right. So here I paused the recording a little bit just for the sake of moving the product somewhat and I almost forgot my methyl group. So here it is. So now let's go ahead and start with part two here, which is the same reaction. But with ethyl alcohol instead. Now the good news is that the products are going to look very similar to the ones that we drew for part one. The only difference is that instead of having a proxy group on carbon two, we're going to have an epoxy group instead. So we have a six carbon chain with some methyl group on carbon five, you've got chlorine on position one, which has no stereo chemistry since it's not a chiral center. And on carbon two we have an Foxy group instead. Now the first an anime I drew here with the oxy group on a wedge. So my second in an tir is going to have the epoxy group on a dash instead and there you have it. So here are the final answers for parts one and two. So with that being said, thank you so very much for watching and I hope you found this helpful.

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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

Halohydrin

Organic Chemistry

10. Addition Reactions

Halohydrin

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6:39 minutes

Problem 25c,d

Textbook Question

What will be the major product obtained from the reaction of Br₂ with 1-butene if the reaction is carried out in
c. ethyl alcohol?
d. methyl alcohol?

Verified step by step guidance

Identify the type of reaction: The reaction of Br2 with 1-butene in the presence of alcohol (ethyl alcohol or methyl alcohol) is an electrophilic addition reaction. The alcohol acts as a nucleophile in this case, leading to the formation of a bromohydrin.

Understand the mechanism: The reaction proceeds via the formation of a bromonium ion intermediate. The double bond in 1-butene reacts with Br2, forming a three-membered bromonium ion.

Determine the role of the alcohol: The alcohol (ethyl alcohol or methyl alcohol) acts as a nucleophile and attacks the more substituted carbon of the bromonium ion. This occurs because the more substituted carbon is better able to stabilize the partial positive charge.

Predict the regioselectivity: The nucleophilic attack by the alcohol will follow Markovnikov's rule, meaning the -OH group (from the alcohol) will attach to the more substituted carbon, while the bromine will attach to the less substituted carbon.

Write the product structure: For ethyl alcohol, the product will be 1-bromo-2-ethoxybutane. For methyl alcohol, the product will be 1-bromo-2-methoxybutane. These are the major products formed in each case.

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Key Concepts

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

Electrophilic Addition

Electrophilic addition is a fundamental reaction mechanism in organic chemistry where an electrophile reacts with a nucleophile, typically involving alkenes. In the case of 1-butene reacting with Br2, the double bond acts as a nucleophile, attacking the electrophilic bromine molecule, leading to the formation of a bromonium ion intermediate. This step is crucial for understanding how the major product is formed.

Regioselectivity

Regioselectivity refers to the preference of a chemical reaction to yield one structural isomer over others when multiple products are possible. In the reaction of Br2 with 1-butene, the regioselectivity is influenced by the stability of the carbocation intermediates formed during the reaction. Understanding regioselectivity helps predict the major product when the reaction is conducted in different solvents like ethyl or methyl alcohol.

Solvent Effects

The choice of solvent can significantly influence the outcome of a chemical reaction, particularly in polar protic solvents like ethyl and methyl alcohol. These solvents can stabilize charged intermediates and influence the reaction pathway. In the case of the bromination of 1-butene, the solvent may affect the stability of the bromonium ion and the subsequent nucleophilic attack, leading to different major products depending on the solvent used.

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