The hydration of three C 5 H 10 alkene isomers can give 2-methylbutan-2-ol. Draw them.

All right. Hi, everyone. So for this question, it says that the acid catalyzed hydration of three isomers of C nine H 16 and alkene produces 133 tri methyl cyclo hexen 10 draw the isomers. So first, let's go ahead and draw out are resulting alcohol, right? And I'm going to start with drawing the parent chain, which is cyclohexane 10. Now that tells me that I'm going to have a six carbon ring here and a hydroxy group on carbon one. But in addition to that, I have three methyl substituent one is going to be on carbon one and the remaining two are going to be on carbon three. So here, let's go ahead and talk about acid catalyzed hydration first because recall that the acid catalyzed hydration of an alkene is the Marconi competition of a hydroxy group. And what I mean by that right is that of the two carbons in the alkene, the more substituted position is what gets the hydroxy group to form an alcohol. So here, right, first of all, carbon, one of the product here must have been one of the carbons in the al. So from that we have to consider what other carbons are directly connected to carbon one here to see what isomers can form. And what you'll notice here is that I have three carbons directly attaching to the one that bears the hydroxy group. I have this methyl group over here and I have two carbons inside of the rink. Now, it's worth noting that the two carbons inside of the ring are not equivalent to each other because this molecule is not symmetrical. So now rate for each isomer, what you're essentially doing is you're creating the double bond between the carbon that bears the hydroxy group and either one of the three carbons that I've circled in different colors. So the first item where I'm going to draw here is with the double bond in between carbon, one of the ring and the methyl group. So there's one and then my next two isomers are going to be with double bonds connecting to the two carbons inside of the ring. And now he's my third isomer and there you have it. So here are the three possibilities because if you look right at all three isomers, the carbon that bears the hydroxy group in the final product is the more substituted position of all three isomers that we've drawn here. So therefore, Marconi competition holds and that's why these three isomers are going to be your final answer. So with that being said, thank you so 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
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21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
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23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
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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

Acid-Catalyzed Hydration

Organic Chemistry

10. Addition Reactions

Acid-Catalyzed Hydration

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3:19 minutes

Problem 47

Textbook Question

The hydration of three C₅H₁₀ alkene isomers can give 2-methylbutan-2-ol. Draw them.

Verified step by step guidance

Identify the target product, 2-methylbutan-2-ol, and recognize that it is a tertiary alcohol. This means the hydroxyl group (-OH) is attached to a carbon atom that is bonded to three other carbon atoms.

Understand that the hydration of an alkene involves the addition of water (H₂O) across the double bond, following Markovnikov's rule. This rule states that the -OH group will attach to the more substituted carbon of the double bond.

Determine the possible structures of C₅H₁₀ alkenes that can lead to 2-methylbutan-2-ol. These alkenes must have a double bond positioned such that, upon hydration, the -OH group attaches to the tertiary carbon in the final product.

Draw the first isomer: 2-methylbut-2-ene. This structure has the double bond between the second and third carbons, and the methyl group is attached to the second carbon. Hydration of this alkene will directly yield 2-methylbutan-2-ol.

Draw the second and third isomers: 3-methylbut-1-ene and 2-methylbut-1-ene. In these cases, the double bond is positioned differently, but hydration will still lead to the formation of 2-methylbutan-2-ol due to carbocation rearrangement during the reaction mechanism.

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Key Concepts

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

Alkene Isomers

Alkene isomers are compounds that have the same molecular formula but differ in the arrangement of their carbon atoms or the position of their double bonds. In the case of C₅H₁₀, there are several structural isomers, including those with different branching patterns. Understanding these isomers is crucial for predicting the products of reactions such as hydration.

Hydration Reaction

Hydration is a chemical reaction where water is added to an alkene, resulting in the formation of an alcohol. This reaction typically follows Markovnikov's rule, where the hydrogen atom from water adds to the carbon with the most hydrogen substituents, leading to the formation of more stable carbocation intermediates. Recognizing how hydration affects the structure of the resulting alcohol is essential for this question.

2-Methylbutan-2-ol

2-Methylbutan-2-ol is a specific alcohol that can be formed from the hydration of certain C₅H₁₀ alkene isomers. It features a secondary alcohol structure with a methyl group on the second carbon of a butane chain. Identifying the alkene isomers that can yield this particular alcohol through hydration is key to solving the question.

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