Textbook Question
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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Table of contents
- 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
Chapter 7, Problem 75a
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.
Verified step by step guidance1
Step 1: Analyze the reaction conditions. The presence of KOH in alcohol suggests that the reaction can proceed via two possible mechanisms: elimination (E2) and substitution (SN2). The solvent (alcohol) and strong base (KOH) favor elimination, but substitution is also possible depending on the structure of the substrate.
Step 2: For the elimination product (CH2=CH-CH3), propose an E2 mechanism. In this mechanism, the strong base (OH⁻) abstracts a proton from the β-carbon (the carbon adjacent to the carbon bonded to the bromine atom). This leads to the formation of a double bond between the α-carbon (the carbon bonded to bromine) and the β-carbon, while bromine leaves as a bromide ion (Br⁻).
Step 3: For the substitution product (CH3-CH(OH)-CH3), propose an SN2 mechanism. In this mechanism, the hydroxide ion (OH⁻) directly attacks the α-carbon, displacing the bromine atom in a single concerted step. This results in the substitution of the bromine atom with a hydroxyl group (-OH).
Step 4: Consider the stereochemistry and regioselectivity of the reaction. The elimination product forms via the most stable alkene (Zaitsev's rule), favoring the more substituted double bond. The substitution product forms via direct nucleophilic attack, which does not involve rearrangement.
Step 5: Summarize the reaction outcomes. The reaction yields two products: the elimination product (CH2=CH-CH3) via the E2 mechanism and the substitution product (CH3-CH(OH)-CH3) via the SN2 mechanism. The relative proportions of these products depend on the reaction conditions, with elimination generally favored in the presence of a strong base and alcohol solvent.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Kinetic Isotope Effect
The kinetic isotope effect refers to the change in reaction rate that occurs when one atom in a molecule is replaced by one of its isotopes. In this case, the C―D bond is stronger than the C―H bond, leading to slower reaction rates when breaking C―D bonds. This concept is crucial for understanding how isotopes can influence reaction mechanisms and product formation.
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Understanding the hydrogen isotopes.
Elimination vs. Substitution Reactions
In organic chemistry, elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond, while substitution reactions involve replacing one atom or group with another. The reaction shown produces both an elimination product (alkene) and a substitution product (alcohol), highlighting the competing pathways that can occur under the influence of a strong base like KOH.
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Reaction Mechanisms
A reaction mechanism is a step-by-step description of how reactants transform into products, detailing the bond-breaking and bond-forming processes involved. Understanding the mechanism for the reaction of bromopropane with KOH is essential for predicting the formation of the elimination and substitution products, as it provides insight into the intermediates and transition states that occur during the reaction.
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Related Practice
Textbook Question
One of the following dichloronorbornanes undergoes elimination much faster than the other. Determine which one reacts faster, and explain the large difference in rates.
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Textbook Question
When the following compound is treated with sodium methoxide in methanol, two elimination products are possible. Explain why the deuterated product predominates by about a 7:1 ratio (refer to Problem 7-75).
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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.
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Textbook Question
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)
b. When the following deuterated compound reacts under the same conditions, the rate of formation of the substitution product is unchanged, while the rate of formation of the elimination product is slowed by a factor of 7.
Explain why the elimination rate is slower, but the substitution rate is unchanged.
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