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

9. Alkenes and Alkynes

Dehydration Reaction

Organic Chemistry

9. Alkenes and Alkynes

Dehydration Reaction

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6:51 minutes

Problem 93a

Textbook Question

Although 2-methyl-1,2-propanediol is an unsymmetrical vicinal diol, only one product is obtained when it is dehydrated in an acidic solution.
a. What is this product?

Verified step by step guidance

Step 1: Begin by identifying the structure of 2-methyl-1,2-propanediol. It is a vicinal diol, meaning the hydroxyl (-OH) groups are on adjacent carbon atoms. The molecule has a methyl group attached to the second carbon.

Step 2: Understand the dehydration reaction mechanism in acidic conditions. Dehydration involves the removal of water (H₂O) from the molecule, typically facilitated by an acid catalyst like H₂SO₄ or H₃PO₄.

Step 3: Analyze the protonation step. In acidic conditions, one of the hydroxyl groups is protonated, forming a good leaving group (water). This step is crucial for initiating the reaction.

Step 4: Consider the carbocation intermediate formed after the departure of water. The stability of the carbocation is key to determining the product. In this case, the molecule forms a tertiary carbocation (most stable) due to the methyl group on the second carbon.

Step 5: Examine the final step, where a double bond is formed through elimination of a proton from an adjacent carbon atom. This results in the formation of an alkene. The product is 2-methylpropene, as it is the most stable alkene formed under these conditions.

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Key Concepts

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

Vicinal Diols

Vicinal diols, also known as glycols, are organic compounds that contain two hydroxyl (-OH) groups attached to adjacent carbon atoms. In the case of 2-methyl-1,2-propanediol, the presence of these two -OH groups allows for potential dehydration reactions, where water is eliminated to form alkenes. Understanding the structure and reactivity of vicinal diols is crucial for predicting the products of their dehydration.

Dehydration Reaction

A dehydration reaction involves the removal of water from a compound, often resulting in the formation of a double bond. In acidic conditions, the hydroxyl groups of vicinal diols can be protonated, leading to the formation of a more stable carbocation intermediate. This process is essential for understanding how 2-methyl-1,2-propanediol can yield a single product upon dehydration, despite its unsymmetrical nature.

Carbocation Stability

Carbocation stability is a key factor in determining the outcome of dehydration reactions. Tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects from surrounding alkyl groups. In the case of 2-methyl-1,2-propanediol, the dehydration leads to the formation of a stable tertiary carbocation, which ultimately results in the formation of a single alkene product, illustrating the influence of carbocation stability on reaction pathways.

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Related Practice

Textbook Question

Propose a mechanism for each reaction.

(a)

(b)

Textbook Question

Suggest an alkene that, in two steps, could be converted into each of the following ketones. Each sequence should involve a pinacol rearrangement.

(c)

Textbook Question

The following reaction is called the pinacol rearrangement. The reaction begins with an acid-promoted ionization to give a carbocation. This carbocation undergoes a methyl shift to give a more stable, resonance-stabilized cation. Loss of a proton gives the observed product. Propose a mechanism for the pinacol rearrangement.

Textbook Question

Treatment of the following alcohol was expected to give alkene A. Instead, B was produced as the major product. Suggest a mechanism by which B formed.

Textbook Question

A student was studying terpene synthesis, and she wanted to make the compound shown here. First she converted 3-bromo-6-methylcyclohexene to alcohol A. She heated alcohol A with sulfuric acid and purified one of the components (compound B) from the resulting mixture. Compound B has the correct molecular formula for the desired product.

Textbook Question

What is the product of the following reaction?

Textbook Question

Which of the following alcohols dehydrates the fastest when heated with acid?

Textbook Question

Propose mechanisms for the following reactions. Additional products may be formed, but your mechanism only needs to explain the products shown.

(a)

(Hint: Hydride shift)

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