Domanda del libro di testo
Which reacts faster in an SN1 reaction?
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Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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TOC.tableOfContents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding31
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds: Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
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Bruice 8th EditionCh. 9 - Substitution and Elimination Reactions of Alkyl HalidesProblema 42
Bruice 8th EditionCh. 9 - Substitution and Elimination Reactions of Alkyl HalidesProblema 42Capitolo 9, Problema 42
Explain why only a substitution product and no elimination product is obtained when the following compound reacts with sodium methoxide:
Guida verificata passo dopo passo1
Step 1: Analyze the structure of the compound. The molecule is a cyclohexane ring with two ethyl groups attached in a trans configuration and an iodine atom attached in an axial position. Sodium methoxide (CH3ONa) is a strong nucleophile and base.
Step 2: Consider the reaction mechanism. Sodium methoxide can act as a nucleophile in a substitution reaction or as a base in an elimination reaction. The axial iodine atom is a good leaving group, making the molecule prone to nucleophilic substitution.
Step 3: Evaluate steric hindrance for elimination. For elimination to occur, a β-hydrogen must be anti-periplanar to the leaving group (iodine). In this case, the β-hydrogens on the cyclohexane ring are equatorial, and none are anti-periplanar to the axial iodine. This geometric constraint prevents elimination.
Step 4: Favor substitution. Since elimination is sterically hindered, substitution becomes the favored pathway. Sodium methoxide attacks the carbon bonded to iodine, replacing the iodine with a methoxy group (CH3O−). This reaction proceeds via an SN2 mechanism due to the strong nucleophile and the primary nature of the carbon attached to iodine.
Step 5: Conclude why elimination does not occur. The lack of anti-periplanar β-hydrogens and the favorable conditions for SN2 substitution explain why only the substitution product is obtained in this reaction.
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Nucleophilic Substitution Reactions
Nucleophilic substitution reactions involve the replacement of a leaving group in a molecule by a nucleophile. In this case, sodium methoxide (CH3ONa) acts as a nucleophile, attacking the carbon atom bonded to the iodine (the leaving group). The reaction typically follows either an SN1 or SN2 mechanism, depending on the structure of the substrate and the conditions.
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01:47
Nucleophiles and Electrophiles can react in Substitution Reactions.
Elimination Reactions
Elimination reactions involve the removal of a small molecule (like water or hydrogen halide) from a larger molecule, resulting in the formation of a double bond. The E2 mechanism is a common type of elimination that requires a strong base and typically occurs in a concerted manner. However, in this scenario, the steric hindrance and the structure of the substrate favor substitution over elimination.
Video consigliato:
00:40Recognizing Elimination Reactions.
Steric Hindrance
Steric hindrance refers to the prevention of reactions due to the spatial arrangement of atoms within a molecule. In this case, the bulky substituents on the cyclohexane ring create significant steric hindrance, making it difficult for the base to access the hydrogen atoms necessary for elimination. This steric effect favors the nucleophilic substitution pathway, leading to the formation of a substitution product instead of an elimination product.
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02:53Understanding steric effects.
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How does the ratio of substitution product to elimination product formed from the reaction of propyl bromide with CH3O− in methanol change if the nucleophile is changed to CH3S−?
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