Domanda del libro di testo
Write a mechanism that explains the formation of the following product. In your mechanism, explain the cause of the rearrangement, and explain the failure to form the Zaitsev product.
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Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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TOC.tableOfContents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
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Wade 9th EditionCh. 7 - Structure and Synthesis of Alkenes; EliminationProblema 74
Wade 9th EditionCh. 7 - Structure and Synthesis of Alkenes; EliminationProblema 74Capitolo 7, Problema 74
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.
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Step 1: Protonation of one of the hydroxyl groups occurs in the presence of sulfuric acid (H2SO4). This converts the hydroxyl group into a better leaving group, forming water.
Step 2: The water molecule leaves, generating a carbocation at the carbon where the hydroxyl group was initially attached. This carbocation is relatively unstable.
Step 3: A methyl group undergoes a 1,2-shift (methyl migration) to stabilize the carbocation. This results in the formation of a resonance-stabilized tertiary carbocation.
Step 4: A deprotonation step occurs, where a proton is removed from the carbocation intermediate. This leads to the formation of a ketone group (C=O) at the migrated carbon.
Step 5: The final product, pinacolone (H3CC(=O)C(CH3)3), is formed as a result of the rearrangement and stabilization of the molecule.
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Carbocation Stability
Carbocations are positively charged species that are highly reactive. Their stability is influenced by the degree of substitution; tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects. In the pinacol rearrangement, the formation of a more stable carbocation through a methyl shift is crucial for the reaction's progression.
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05:58
Determining Carbocation Stability
Methyl Shift
A methyl shift is a type of rearrangement where a methyl group moves from one carbon to an adjacent carbon, resulting in the formation of a more stable carbocation. This process is essential in the pinacol rearrangement, as it allows the reaction to proceed through a more stable intermediate, ultimately leading to the formation of the product, pinacolone.
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03:43Alkyl Shift
Acid-Catalyzed Rearrangement
The pinacol rearrangement is an acid-catalyzed reaction, meaning that the presence of an acid (like H2SO4) facilitates the ionization of the starting material, pinacol. This ionization generates a carbocation, which is a key intermediate in the reaction mechanism. The acid not only promotes the formation of the carbocation but also helps in the subsequent steps leading to the final product.
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03:09Acid Catalyzed
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Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review Problem 4-57)
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Explain why the elimination rate is slower, but the substitution rate is unchanged.
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Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review PROBLEM 4-57)
a. Propose a mechanism to explain each product in the following reaction.
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